SOLUTIONS MANUAL CHAPTER 1 1. The energy contained in a volume dV is
U(ν,T )dV = U (ν ,T )r 2 dr sinθ dθdϕ when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area dA is dE(ν ,T ) = U (ν ,T )dV
dAcos θ 4πr 2
The total energy emitted is dE (ν ,T ) = ∫
cΔt
0
dr∫
π /2 0
2π
dθ ∫ dϕU (ν,T )sin θ cosθ 0
dA 4π
π/2 dA 2π cΔtU (ν ,T ) ∫ dθ sinθ cosθ 0 4π 1 = cΔtdAU (ν ,T ) 4
=
.
By definition of the emissivity, this is equal to EΔtdA . Hence E (ν, T ) =
c U (ν, T ) 4
2. We have
c c 8π hc w(λ ,T ) = U (ν ,T )  dν / dλ = U ( ) 2 = 5
λ λ
λ
1 e
hc/λkT
−1
This density will be maximal when dw(λ ,T ) / dλ = 0 . What we need is A /λ
1 e A 1 d ⎛1 1 ⎞ 1 − (− )) =0 = (−5 5 A / λ 6 5 A / λ 2 A / d λ ⎝ λ e − 1⎠ λ λ e −1 λ e λ −1
Where A = hc / kT . The above implies that with x = A / λ , we must have 5 − x = 5e −x
A solution of this is x = 4.965 so that
hc = 2.898 × 10 −3 m 4.965k
λmaxT =
In example 1.1 we were given an estimate of the sun’s surface temperature as 6000 K. From this we get
λsun max =
28.98 × 10 −4 mK = 4.83 × 10−7 m = 483nm 6 × 10 3 K
3. The relationship is hν = K + W
where K is the electron kinetic energy and W is the work function. Here (6.626 × 10−34 J .s)(3× 10 8 m / s) hν = = = 5.68 × 10 −19 J = 3.55eV −9 λ 350 × 10 m hc
With K = 1.60 eV, we get W = 1.95 eV 4. We use
hc
λ1
−
hc
λ2
= K1 − K2
since W cancels. From ;this we get h= =
1 λ1λ 2 (K − K 2 ) = c λ 2 − λ1 1
(200 × 10 −9 m)(258 × 10−9 m) × (2.3 − 0.9)eV × (1.60 × 10 −19 )J / eV (3 × 10 8 m / s)(58 × 10−9 m)
= 6.64 × 10
−34
J .s
5. The maximum energy loss for the photon occurs in a headon collision, with the photon scattered backwards. Let the incident photon energy be hν , and the backwardscattered photon energy be hν' . Let the energy of the recoiling proton be E. Then its recoil momentum is obtained from E = p 2c 2 + m 2c 4 . The energy conservation equation reads
hν + mc 2 = hν ' + E
and the momentum conservation equation reads hν ' hν =− +p c c
that is
hν = −hν '+ pc
We get E + pc − mc 2 = 2hν from which it follows that p2 c 2 + m2 c 4 = (2hν − pc + mc 2 ) 2
so that 4 h 2ν 2 + 4 hνmc 2 pc = 4 hν + 2mc 2 The energy loss for the photon is the kinetic energy of the proton K = E − mc 2 . Now hν = 100 MeV and mc 2 = 938 MeV, so that pc = 182MeV
and E − mc 2 = K = 17.6 MeV 6. Let hν be the incident photon energy, hν' the final photon energy and p the outgoing electron momentum. Energy conservation reads hν + mc = hν ' + p c + m c 2
2 2
2 4
We write the equation for momentum conservation, assuming that the initial photon moves in the x –direction and the final photon in the ydirection. When multiplied by c it read
i(hν ) = j(hν ') + (ipx c + jpy c) Hence px c = hν ; pyc = − hν '. We use this to rewrite the energy conservation equation as follows: (hν + mc 2 − hν ')2 = m 2c 4 + c 2 ( px2 + py2 ) = m2 c 4 + (hν ) 2 + (hν ') 2 From this we get ⎛ mc 2 ⎞ ⎜ ⎟ hν'= hν ⎝ hν + mc 2 ⎠
We may use this to calculate the kinetic energy of the electron
⎛ mc 2 ⎞⎟ hν ⎜ K = hν − hν ' = hν 1 − 2 = hν ⎝ hν + mc ⎠ hν + mc 2 =
(100keV )2 = 16.4 keV 100keV + 510 keV
Also pc = i(100keV ) + j(−83.6keV)
which gives the direction of the recoiling electron.
7. The photon energy is
(6.63× 10 −34 J.s)(3 × 108 m / s) hν = = = 6.63× 10 −17 J 6 −9 3× 10 × 10 m λ −17 6.63× 10 J = 4.14 × 10 −4 MeV = −19 1.60 × 10 J / eV hc
The momentum conservation for collinear motion (the collision is head on for maximum energy loss), when squared, reads 2
2
⎛ hν ⎞ ⎛ hν ⎞ ⎛ hν '⎞ ⎛ hν ' ⎞ 2 2 ⎝ c ⎠ + p + 2⎝ c ⎠ pηi = ⎝ c ⎠ + p' +2⎝ c ⎠ p' η f Here ηi = ±1, with the upper sign corresponding to the photon and the electron moving in the same/opposite direction, and similarly for η f . When this is multiplied by c2 we get (hν ) 2 + (pc) 2 + 2(hν ) pcηi = (hν ')2 + ( p'c) 2 + 2(hν ') p'cη f The square of the energy conservation equation, with E expressed in terms of momentum and mass reads
(hν ) 2 + (pc) 2 + m 2c 4 + 2Eh ν = (hν ')2 + ( p'c)2 + m2 c 4 + 2E' hν ' After we cancel the mass terms and subtracting, we get
hν(E − η i pc) = hν '(E'− η f p'c) From this can calculate hν' and rewrite the energy conservation law in the form
⎛ E − ηi pc ⎞ E − E'= hν ⎜ − 1⎟ ⎝ E '− p'cη f ⎠ The energy loss is largest if ηi = −1;η f = 1. Assuming that the final electron momentum is not very close to zero, we can write E + pc = 2E and E '− p' c =
(mc 2 )2 so that 2 E'
⎛ 2 E × 2 E' ⎞ E − E '= hν ⎜ ⎝ (mc 2 )2 ⎟⎠ 1 1 It follows that = + 16 hν with everything expressed in MeV. This leads to E' E E’ =(100/1.64)=61 MeV and the energy loss is 39MeV. 8.We have λ’ = 0.035 x 1010 m, to be inserted into
λ'− λ =
6.63 × 10−34 J .s h h (1− cos600 ) = = = 1.23× 10 −12 m me c 2 mec 2 × (0.9 × 10 −30 kg)(3× 10 8 m / s)
Therefore λ = λ’ = (3.501.23) x 1012 m = 2.3 x 1012 m. The energy of the Xray photon is therefore −34
8
(6.63× 10 J .s)(3 × 10 m / s) 5 hν = = = 5.4 × 10 eV −12 −19 λ (2.3× 10 m)(1.6 × 10 J / eV ) hc
9. With the nucleus initially at rest, the recoil momentum of the nucleus must be equal and opposite to that of the emitted photon. We therefore have its magnitude given by p = hν / c , where hν = 6.2 MeV . The recoil energy is
E=
6.2 MeV p2 hν = hν = 1.5 × 10−3 MeV 2 = (6.2 MeV ) 2M 2 Mc 2 × 14 × (940 MeV )
10. The formula λ = 2asinθ / n implies that λ / sinθ ≤ 2a / 3 . Since λ = h/p this leads to p ≥ 3h / 2asin θ , which implies that the kinetic energy obeys
9h2 p2 K= ≥ 2m 8 ma2 sin2 θ Thus the minimum energy for electrons is 9(6.63× 10 −34 J .s)2 K= = 3.35eV 8(0.9 × 10−30 kg)(0.32 × 10−9 m)2 (1.6 × 10−19 J / eV )
For Helium atoms the mass is 4(1.67 × 10−27 kg) / (0.9 × 10 −30 kg) = 7.42 × 103 larger, so that
K= 11. We use K =
K=
33.5eV = 4.5 × 10 −3 eV 7.42 × 103
p2 h2 = with λ = 15 x 109 m to get 2m 2 mλ2
(6.63× 10 −34 J .s)2 = 6.78 × 10−3 eV 2(0.9 × 10−30 kg)(15 × 10−9 m)2 (1.6 × 10−19 J / eV )
For λ = 0.5 nm, the wavelength is 30 times smaller, so that the energy is 900 times larger. Thus K =6.10 eV. 12. For a circular orbit of radius r, the circumference is 2πr. If n wavelengths λ are to fit into the orbit, we must have 2πr = nλ = nh/p. We therefore get the condition pr = nh / 2π = n
which is just the condition that the angular momentum in a circular orbit is an integer in units of . 13. We have a = nλ / 2sinθ . For n = 1, λ= 0.5 x 1010 m and θ= 5o . we get a = 2.87 x 1010 m. For n = 2, we require sinθ2 = 2 sinθ1. Since the angles are very small, θ2 = 2θ1. So that the angle is 10o. 14. The relation F = ma leads to mv 2/r = mωr that is, v = ωr. The angular momentum quantization condition is mvr = n , which leads to mωr2 = n . The total energy is therefore
E=
1 2 1 mv + mω 2 r2 = mω 2 r 2 = n ω 2 2
The analog of the Rydberg formula is
ν( n → n') =
En − En' ω ( n − n') ω = = ( n − n') h h 2π
The frequency of radiation in the classical limit is just the frequency of rotation νcl = ω / 2π which agrees with the quantum frequency when n – n’ = 1. When the selection rule Δn = 1 is satisfied, then the classical and quantum frequencies are the same for all n.
15. With V(r) = V0 (r/a)k , the equation describing circular motion is
v 2 dV 1 ⎛ r⎞ m = = kV0 ⎝ ⎠ dr a r r
k
so that
kV0 ⎛ r ⎞ m ⎝ k⎠
v=
k/2
The angular momentum quantization condition mvr = n reads ⎛ r⎞ ma kV0 ⎝ a⎠ 2
k +2 2
=n
We may use the result of this and the previous equation to calculate k
k k ⎡ n 2 2 ⎤ k +2 1 2 1 1 ⎛ r⎞ ⎛ r⎞ E = mv + V0 ⎝ ⎠ = ( k + 1)V0 ⎝ ⎠ = ( k + 1)V0 ⎢ 2 ⎥ 2 2 2 a a ⎣ ma kV0 ⎦ In the limit of k >>1, we get k
E→
1 (kV ) 2 0
2 k +2
2 ⎡ 2 ⎤ k + 2 2 k k+2 2 ⎢ ma2 ⎥ (n ) → 2 ma2 n ⎣ ⎦
Note that V0 drops out of the result. This makes sense if one looks at a picture of the potential in the limit of large k. For r< a the potential is effectively zero. For r > a it is effectively infinite, simulating a box with infinite walls. The presence of V0 is there to provide something with the dimensions of an energy. In the limit of the infinite box with the quantum condition there is no physical meaning to V0 and the energy scale is provided by 2 / 2ma 2 . 16. The condition L = n implies that
n2 2 E= 2I In a transition from n1 to n2 the Bohr rule implies that the frequency of the radiation is given
ν12 =
2 E1 − E 2 = (n12 − n 22 ) = (n12 − n 22 ) h 2Ih 4π I
Let n1 = n2 + Δn. Then in the limit of large n we have (n12 − n 22 ) → 2n 2 Δn , so that
ν12 →
1 n2 1 L Δn = Δn 2π I 2π I
Classically the radiation frequency is the frequency of rotation which is ω = L/I , i.e. ω L νcl = 2π I We see that this is equal to ν12 when Δn = 1. 17. The energy gap between lowlying levels of rotational spectra is of the order of 2 / I = (1 / 2π ) h / MR 2 , where M is the reduced mass of the two nuclei, and R is their separation. (Equivalently we can take 2 x m(R/2)2 = MR2). Thus
hν =
hc
λ
=
1 h 2π MR 2
This implies that R=
λ 2πMc
=
λ πmc
=
(1.05 × 10 −34 J.s)(10−3 m) = 26nm π (1.67 × 10 −27 kg)(3× 10 8 m / s)
CHAPTER 2 1. We have ∞
∞
ψ (x) = ∫−∞ dkA(k)e ikx = ∫−∞ dk
∞ N N ikx e = dk 2 coskx 2 2 ∫ −∞ k +α k + α2
because only the even part of eikx = coskx + i sinkx contributes to the integral. The integral can be looked up. It yields
ψ ( x) = N so that ψ ( x)  = 2
π −α x  e α N 2π 2
α
2
e −2α x
If we look at A(k)2 we see that this function drops to 1/4 of its peak value at k =± α.. We may therefore estimate the width to be Δk = 2α. The square of the wave function drops to about 1/3 of its value when x =±1/2α. This choice then gives us Δk Δx = 1. Somewhat different choices will give slightly different numbers, but in all cases the product of the widths is independent of α. 2. the definition of the group velocity is vg =
dω 2π dν dν dν = = = −λ 2 dk 2πd (1/ λ ) d (1/ λ ) dλ
The relation between wavelength and frequency may be rewritten in the form
ν 2 − ν 02 =
c2
λ2
so that dν c 2 −λ = = c 1 − (ν 0 / ν ) 2 dλ νλ 2
3. We may use the formula for vg derived above for
ν= to calculate
2π T
ρ
λ−3/2
v g = −λ
2
dν 3 2π T = dλ 2 ρλ
4. For deep gravity waves,
ν = g / 2π λ −1/2 from which we get, in exactly the same way v g =
1 λg . 2 2π
5. With ω = k2/2m, β = /m and with the original width of the packet w(0) = √2α, we have 2 2 w(t ) β 2t 2 t 2 2t2 = 1+ = 1 + = 1 + w(0) 2α 2 2m 2α 2 m 2 w 4 (0)
(a) With t = 1 s, m = 0.9 x 1030 kg and w(0) = 106 m, the calculation yields w(1) = 1.7 x 102 m With w(0) = 1010 m, the calculation yields w(1) = 1.7 x 106 m. These are very large numbers. We can understand them by noting that the characteristic velocity associated with a particle spread over a range Δx is v = /mΔx and here m is very small. (b) For an object with mass 103 kg and w(0)= 102 m, we get 2 2t 2 2(1.05 × 10−34 J .s) 2 t 2 = = 2.2 × 10−54 m2 w 4 (0) (10−3 kg) 2 × (10 −2 m) 4 for t = 1. This is a totally negligible quantity so that w(t) = w(0). 6. For the 13.6 eV electron v /c = 1/137, so we may use the nonrelativistic expression for the kinetic energy. We may therefore use the same formula as in problem 5, that is 2 2 w(t ) β 2t 2 t 2 2t2 = 1+ = 1 + = 1 + w(0) 2α 2 2m 2α 2 m 2 w 4 (0)
We caclulate t for a distance of 104 km = 107 m, with speed (3 x 108m/137) to be 4.6 s. We are given that w(0) = 103 m. In that case −34
2(1.05 × 10 J.s) (4.6s) −2 w(t) = (10 m) 1 + −30 2 −3 4 = 7.5 × 10 m (0.9 × 10 kg) (10 m) −3
2
2
For a 100 MeV electron E = pc to a very good approximation. This means that β = 0 and therefore the packet does not spread.
7. For any massless particle E = pc so that β= 0 and there is no spreading. 8. We have
φ( p) = =
{
∞ 0 ∞ 1 A −μ x −ipx/ ( μ −ik )x dxAe e = dxe + dxe −( μ +ik )x ∫ ∫ ∫ −∞ −∞ 0 2π 2π A ⎧ 1 1 ⎫ A 2μ + ⎨ ⎬= 2 2 2π ⎩ μ − ik μ + ik ⎭ 2π μ + k
}
where k = p/ . 9. We want ∞
∫−∞ dxA2e −2μx  = A 2
{
∞
}
∫−∞ dxe2μx + ∫0 dxe−2μx = A 2 0
1
μ
=1
so that A= μ 10. Done in text. 11. Consider the Schrodinger equation with V(x) complex. We now have
∂ψ (x,t) i ∂ 2ψ (x,t) i = − V (x)ψ (x,t) ∂t 2m ∂x 2 and i ∂ 2ψ *(x,t) i ∂ψ *(x,t) =− + V *(x)ψ (x,t) ∂t 2m ∂x 2 Now
∂ ∂ψ * ∂ψ (ψ * ψ ) = ψ +ψ * ∂t ∂t ∂t 2 i ∂ψ* i i ∂ 2ψ (x,t) i = (− + V * (x) ψ *) ψ + ψ * ( − V (x)ψ (x,t)) 2 2 2m ∂x 2m ∂x i ∂ 2ψ (x,t) i ∂ 2ψ * ( ψ − ψ * ) + (V *−V )ψ *ψ =− 2 2 ∂x 2m ∂x =− Consequently
i ∂ ⎧∂ψ * ∂ψ ⎫ 2ImV (x) ψ −ψ * ⎬ + ψ *ψ ⎨ 2m ∂x ⎩ ∂x ∂x ⎭
2 ∞ ∂ ∞ 2 dx  ψ (x,t)  = ∫−∞ dx(ImV (x))  ψ (x,t) 2 ∂t ∫−∞ We require that the left hand side of this equation is negative. This does not tell us much about ImV(x) except that it cannot be positive everywhere. If it has a fixed sign, it must be negative. 12. The problem just involves simple arithmetic. The class average 〈g〉 = ∑ gn g = 38.5 g
(Δg) = 〈g 〉 − 〈g〉 = ∑ g ng − (38.5) = 1570.81482.3= 88.6 2
2
2
2
2
g
The table below is a result of the numerical calculations for this system (g 
)2/(Δg)2 = λ eλ Ceλ g ng 60 1 5.22 0.0054 0.097 55 2 3.07 0.0463 0.833 50 7 1.49 0.2247 4.04 45 9 0.48 0.621 11.16 40 16 0.025 0.975 17.53 35 13 0.138 0.871 15.66 30 3 0.816 0.442 7.96 25 6 2.058 0.128 2.30 20 2 3.864 0.021 0.38 15 0 6.235 0.002 0.036 10 1 9.70 0.0001 0.002 5 0 12.97 “0” “0” __________________________________________________________
15. We want ∞
1 = 4N 2 ∫−∞ dx so that N =
∞ sin2 kx sin2 t 2 2 = 4N k dt ∫−∞ t 2 = 4πN k x2
1 4πk
16.
We have ⎛α⎞ 〈x 〉 = ⎝ ⎠ π
1/ 2
∫
∞
dxx n e −α x −∞
n
2
Note that this integral vanishes for n an odd integer, because the rest of the integrand is even. For n = 2m, an even integer, we have ⎛α⎞ 〈x 〉 = ⎝ ⎠ π
1/2
2m
⎛α⎞ ⎛ d ⎞ = − ⎝ π ⎠ ⎝ dα ⎠ 1/2
m
∫
⎛α⎞ ⎛ d ⎞ ⎛ π⎞ = − ⎝ π ⎠ ⎝ dα ⎠ ⎝ α ⎠ 1/2
∞ −∞
dxe
−αx 2
m
1/ 2
For n = 1 as well as n = 17 this is zero, while for n = 2, that is, m = 1, this is 1 2π
φ( p) =
17.
∫
1 . 2α
⎛ α ⎞ −α x 2 /2 e ⎝π ⎠ 1/4
∞ −∞
dxe
− ipx /
The integral is easily evaluated by rewriting the exponent in the form −
α 2
x 2 − ix
p
=−
α⎛ 2⎝
2
x+
ip ⎞ p2 − 2 α⎠ 2 α
A shift in the variable x allows us to state the value of the integral as and we end up with 1 ⎛ π ⎞ − p 2 / 2α φ( p) = e π ⎝α⎠ 1/4
2
We have, for n even, i.e. n = 2m,
〈p 2m 〉 =
1 ⎛ π⎞ π ⎝α⎠
1/ 2
∞
2m − p ∫−∞ dpp e m
1 ⎛π⎞ ⎛ d ⎞ ⎛π ⎞ = ⎜− ⎟ ⎜ ⎟ π ⎝ α ⎠ ⎝ dβ ⎠ ⎝ β ⎠ 1/ 2
where at the end we set β =
1
α
2
2
/α
2
=
1/2
. For odd powers the integral vanishes.
18. Specifically for m = 1 we have
We have 1 (Δx)2 = 〈x 2 〉 = 2α
(Δp)2 = 〈p 2 〉 = so that ΔpΔx =
2
α
2
2
. This is, in fact, the smallest value possible for the product of the
dispersions.
22. We have
∫
∞
−∞
=
dxψ *(x)xψ (x) = 1 2π
∫
∞ −∞
1 2π ∞
∫
∞ −∞
dxψ * (x)∫ dpφ (p) −∞
∞
dx ψ * (x)x ∫ dpφ ( p)e ipx/ −∞
∞ ∂ ipx/ ∂φ (p) e = ∫ dpφ * (p)i −∞ ∂p i ∂p
In working this out we have shamelessly interchanged orders of integration. The justification of this is that the wave functions are expected to go to zero at infinity faster than any power of x , and this is also true of the momentum space wave functions, in their dependence on p.
CHAPTER 3. 1. The linear operators are (a), (b), (f) 2.We have
∫
x
−∞
dx' x'ψ (x') = λψ (x)
To solve this, we differentiate both sides with respect to x, and thus get
λ
dψ (x) = xψ (x) dx
A solution of this is obtained by writing dψ / ψ = (1/ λ )xdx from which we can immediately state that
ψ (x) = Ceλx
2
/2
The existence of the integral that defines O6ψ(x) requires that λ < 0. 3, (a)
O2 O6ψ (x) − O6O2ψ (x) x d x dψ (x') = x ∫ dx' x'ψ (x') − ∫ dx' x'2 −∞ −∞ dx dx' x x d = x 2ψ (x) − ∫ dx' (x'2 ψ (x')) + 2∫ dx' x'ψ (x') −∞ −∞ dx' = 2O6ψ (x)
Since this is true for every ψ(x) that vanishes rapidly enough at infinity, we conclude that [O2 , O6] = 2O6 (b) O1O2ψ (x) − O2 O1ψ (x) d 3 ⎛ dψ ⎞ 3 4 dψ − O = O1⎝ x x ψ = x − x ( ) (x ψ ) 2 dx dx ⎠ dx = −3x 3ψ (x) = −3O1ψ (x) so that [O1, O2] = 3O1
4.
We need to calculate
2 a nπx dxx 2 sin 2 ∫ 0 a a
〈x 2 〉 =
With πx/a = u we have
〈x 2 〉 =
2 a3 π a2 π 2 2 duu sin nu = duu2 (1− cos2nu) a π 3 ∫0 π 3 ∫0
The first integral is simple. For the second integral we use the fact that
∫
π
0
⎛ d ⎞ duu 2 cosα u = − ⎝ ⎠ dα
2
∫
π 0
2
⎛ d ⎞ sin απ ducosα u = −⎝ ⎠ dα α
At the end we set α = nπ. A little algebra leads to a2 a2 − 2 2 3 2π n
〈x 2 〉 =
For large n we therefore get Δx = Δp =
πn a
a . Since 〈p 2 〉 = 3
2
n 2π 2 , it follows that a2
, so that ΔpΔx ≈
nπ 3
The product of the uncertainties thus grows as n increases.
π2
2
5. With E n =
2ma
2
n 2 we can calculate
(1.05 × 10 −34 J .s)2 1 = 0.115eV E 2 − E1 = 3 −30 −9 2 2(0.9 × 10 kg)(10 m) (1.6 × 10 −19 J / eV ) We have ΔE =
hc
λ
so that λ =
2π c 2π (2.6 × 10−7 ev.m) = = 1.42 × 10 −5 m ΔE 0.115eV
where we have converted c from J.m units to eV.m units.
6. (a) Here we write 2
2
n =
2ma E
π
2
2
=
2(0.9 × 10
−30
−2
2
kg)(2 × 10 m) (1.5eV )(1.6 × 10 (1.05 × 10 −34 J .s)2 π 2
−19
J / eV )
so that n = 4 x 107 . (b) We have
ΔE =
π2
2
2ma
2nΔn = 2
(1.05 × 10−34 J.s) 2 π 2 2(4 × 107 ) = 1.2 × 10 −26 J −30 −2 2 2(0.9 × 10 kg)(2 × 10 m)
= 7.6 × 10−8 eV 7. The longest wavelength corresponds to the lowest frequency. Since ΔE is proportional to (n + 1)2 – n2 = 2n + 1, the lowest value corresponds to n = 1 (a state with n = 0 does not exist). We therefore have
h
c
λ
π2
2
=3
2ma2 If we assume that we are dealing with electrons of mass m = 0.9 x 1030 kg, then 3 πλ 3π (1.05 × 10 −34 J.s)(4.5 × 10 −7 m) a = = = 4.1× 10−19 m2 −30 8 4(0.9 × 10 kg)(3× 10 m / s) 4mc 10 so that a = 6.4 x 10 m. 2
π 2n 2
2
8. The solutions for a box of width a have energy eigenvalues E n =
with 2ma 2 n = 1,2,3,…The odd integer solutions correspond to solutions even under x → − x , while the even integer solutions correspond to solutions that are odd under reflection. These solutions vanish at x = 0, and it is these solutions that will satisfy the boundary conditions for the “halfwell” under consideration. Thus the energy eigenvalues are given by En above with n even. 9. The general solution is ∞
ψ (x,t) = ∑ Cn un (x)e − iE t / n
n =1
with the Cn defined by a /2
Cn = ∫− a / 2 dxu*n (x)ψ (x,0)
15
= 1.59 × 10
(a) It is clear that the wave function does not remain localized on the l.h.s. of the box at later times, since the special phase relationship that allows for a total interference for x > 0 no longer persists for t ≠ 0. 2 0 dxun (x) .We may work this out by a ∫−q /2 using the solution of the box extending from x = 0 to x = a, since the shift has no physical consequences. We therefore have
(b) With our wave function we have Cn =
a /2
2 a/ 2 2 nπx 2 ⎡ a nπ x ⎤ 2 ⎡ nπ ⎤ Cn = dx sin = − cos = 1− cos ∫ 0 2 ⎦ a a ⎦0 a a a ⎣ nπ nπ ⎣ Therefore P1 = C1 2 =
4
π
2
and P2 = C2 2 =
1
π
2
 (1− (−1)) 2 =
4
π2
10. (a) We use the solution of the above problem to get
Pn = Cn 2 =
4 n π2 2
fn
where fn = 1 for n = odd integer; fn = 0 for n = 4,8,12,…and fn = 4 for n = 2,6,10,… (b) We have ∞
∑P n =1
n
=
4
π
2
1
∑n
2
+
odd
4 1 8 2 = 2 ∑ 2 =1 π n = 2,6,10,,, n π odd n 4
2
∑
Note. There is a typo in the statement of the problem. The sum should be restricted to odd integers. 11. We work this out by making use of an identity. The hint tells us that 5
(sin x)5 = =
⎛ 1 ⎞ ix 1 1 5ix (e − e −ix ) 5 = (e − 5e 3ix + 10e ix − 10e − ix + 5e −3ix − e −5ix ) ⎝ 2i ⎠ 16 2i 1 (sin5x − 5sin 3x + 10sin x) 16
Thus
ψ (x,0) = A (a) It follows that
a 1 (u (x) − 5u3(x) + 10u1 (x)) 2 16 5
ψ (x,t) = A
a 1 − iE t / −iE t / −iE t / u5 (x)e 5 − 5u3 (x)e 3 + 10u1 (x)e 1 ( 2 16
(b) We can calculate A by noting that
∫
a
0
)
dx  ψ (x,0) 2 = 1 . This however is equivalent
to the statement that the sum of the probabilities of finding any energy eigenvalue adds up to 1. Now we have
P5 =
25 a 2 1 a a 100 A ;P3 = A 2 ;P1 = A 2 2 256 2 256 2 256
so that A2 =
256 63a
The probability of finding the state with energy E3 is 25/126. 12. The initial wave function vanishes for x ≤ a and for x ≥ a. In the region in between it πx is proportional to cos , since this is the first nodeless trigonometric function that 2a vanishes at x = ± a. The normalization constant is obtained by requiring that a
1 = N 2 ∫− a dx cos2
πx 2a
⎛ 2a ⎞ π /2 = N 2 ⎝ ⎠ ∫−π / 2 ducos2 u = N 2 a
π
1 . We next expand this in eigenstates of the infinite box potential with a boundaries at x = ± b. We write
so that N =
1 πx ∞ cos = ∑ Cn un (x;b) a 2a n =1 so that b
a
−b
−a
Cn = ∫ dxun (x;b)ψ (x) = ∫ dx un (x;b)
1 πx cos a 2a
In particular, after a little algebra, using cosu cosv=(1/2)[cos(uv)+cos(u+v)], we get
πx πx 1 a dx cos cos = ∫ −a 2b 2a ab
C1 = =
π x(b + a) ⎤ 1 a 1 ⎡ πx(b − a) dx ⎢cos + cos ∫ −a 2ab 2ab ⎥⎦ ab 2⎣
πa 4b ab 2 2 cos 2b π (b − a )
so that
16ab 2 πa 2 2 2 2 cos 2b π (b − a ) 3
2
P1 = C1  =
The calculation of C2 is trivial. The reason is that while ψ(x) is an even function of x, u2(x) is an odd function of x, and the integral over an interval symmetric about x = 0 is zero. Hence P2 will be zero. 13. We first calculate
2 nπ x e ipx/ 1 1 ⎛ a sin = dxeix (nπ /a + p / ) − (n ↔ −n)⎞⎠ a i 4π a ⎝ ∫ 0 a 2π 1 ⎛ e iap / (−1)n − 1 e iap / (−1) n − 1⎞ − ⎜ ⎟ 4π a ⎝ p / − nπ / a p / + nπ / a ⎠ a
φ( p) = ∫0 dx = =
2nπ / a (−1)n cos pa / − 1+ i(−1)n sin pa / 2 2{ 4π a (nπ / a) − (p / ) 1
}
From this we get P( p) = φ (p)  = 2
2n 2π 1− (−1) n cos pa / a3 [(nπ / a)2 − (p / ) 2 ]2
The function P(p) does not go to infinity at p = nπ / a , but if definitely peaks there. If we write p / = nπ / a + ε , then the numerator becomes 1− cosaε ≈ a2ε 2 / 2 and the ⎛ nπ ⎞ = a / 4π . The fact that the denominator becomes (2nπε / a)2 , so that at the peak P⎝ a ⎠ peaking occurs at 2 2 2 p2 π n = 2m 2ma2
suggests agreement with the correspondence principle, since the kinetic energy of the particle is, as the r.h.s. of this equation shows, just the energy of a particle in the infinite box of width a. To confirm this, we need to show that the distribution is strongly peaked for large n. We do this by looking at the numerator, which vanishes when aε = π / 2 , that is, when p / = nπ / a + π / 2a = (n + 1 / 2)π / a . This implies that the width of the
distribution is Δp = π / 2a . Since the xspace wave function is localized to 0 ≤ x ≤ a we only know that Δx = a. The result ΔpΔx ≈ (π / 2) is consistent with the uncertainty principle. 14. We calculate 1/4
⎛ α ⎞ −αx 2 / 2 1 − ipx/ e e ⎝π ⎠ 2π
∞
φ( p) = ∫−∞ dx 1/ 4
1/2
⎛α ⎛ 1 ⎞ = ⎞ ⎝ π ⎠ ⎝ 2π ⎠
∫
∞
2
dxe−α (x − ip/α ) e − p −∞
⎛ 1 ⎞1/ 4 − p 2 / 2α = e ⎝ πα 2 ⎠
2
/2α
2
2
From this we find that the probability the momentum is in the range (p, p + dp) is 1/ 2
⎛ 1 ⎞ − p 2 /α e  φ( p)  dp = ⎝ πα 2 ⎠ 2
2
To get the expectation value of the energy we need to calculate 1 ⎛ 1 ⎞1/ 2 ∞ 2 p2 dpp 2e − p /α 〈 〉= 2 ⎠ ∫−∞ ⎝ 2m 2m πα 1/2
1 ⎛ 1 ⎞ = 2m ⎝ πα 2 ⎠
π 2
(α 2 )3/ 2 =
2
α
2
2m
An estimate on the basis of the uncertainty principle would use the fact that the “width” of the packet is1 / α . From this we estimate Δp ≈ / Δx = α , so that
α (Δp) = 2m 2m 2
E≈
2
The exact agreement is fortuitous, since both the definition of the width and the numerical statement of the uncertainty relation are somewhat elastic.
15. We have
j(x) = = =
⎛ dψ (x) dψ *(x) ψ * (x) − ψ (x)⎞ ⎝ ⎠ dx dx 2im
[(A * e 2im
−ikx
ikx
+ B *e )(ikAe
2
2im
2
ikx
[ik  A  −ik  B  +ikAB *e 2
2
− ikBe 2ikx
− (−ik )  A  −(ik)  B  −(−ik)A * Be =
−ikx
) − c.c)]
− ikA* Be
−2ikx
−2ikx
− ikAB *e
2ikx
]
k [ A 2 −  B 2 ] m
This is a sum of a flux to the right associated with A eikx and a flux to the left associated with Beikx.. 16. Here
du(x) ikx ⎡ ⎤ u(x)e − ikx (iku(x)e ikx + e ) − c.c 2im ⎣ dx ⎦ du(x) k 2 = [(iku2 (x) + u(x) ) − c.c] = u (x) 2im dx m
j(x) =
∂ change sign, and since the ∂x function consists of an odd power of x and/or p, it is an odd function of x. Now the eigenfunctions for a box symmetric about the x axis have a definite parity. So that un (− x) = ±un (x) . This implies that the integrand is antisymmetric under x Æ  x. Since the integral is over an interval symmetric under this exchange, it is zero. (c) Under the reflection x Æ x both x and p = −i
(d) We need to prove that
∫
∞
∞
dx(P ψ (x))* ψ (x) = ∫ dxψ (x)* P ψ (x) −∞ −∞
The left hand side is equal to
∫
∞
∞
dxψ *(− x)ψ (x) = ∫−∞ dy ψ * (y)ψ (− y) −∞
with a change of variables x Æy , and this is equal to the right hand side. The eigenfunctions of P with eigenvalue +1 are functions for which u(x) = u(x), while those with eigenvalue –1 satisfy v(x) = v(x). Now the scalar product is
∫
∞
∞
∞
dxu *(x)v(x) = ∫−∞ dyu *(− x)v(− x) = −∫−∞ dxu *(x)v(x) −∞
so that
∫
∞
−∞
dxu *(x)v(x) = 0
(e) A simple sketch of ψ(x) shows that it is a function symmetric about x = a/2. This means that the integral
∫
a
0
dxψ (x)un (x) will vanish for the un(x) which are odd
under the reflection about this axis. This means that the integral vanishes for n = 2,4,6,…
CHAPTER 4. 1. The solution to the left side of the potential region is ψ (x) = Aeikx + Be−ikx . As shown in Problem 315, this corresponds to a flux
j(x) =
k ( A 2 −  B 2 ) m
The solution on the right side of the potential is ψ (x) = Ce ikx + De −ikxx , and as above, the flux is k  C 2 −  D 2 ) ( m Both fluxes are independent of x. Flux conservation implies that the two are equal, and this leads to the relationship j(x) =
 A 2 +  D 2 = B 2 +  C 2 If we now insert C = S11 A + S12 D B = S21A + S22D
into the above relationship we get *  A 2 +  D 2 = ( S21A + S22D)(S21* A * +S22 D*) + ( S11A + S12D)( S11*A * + S12*D*)
Identifying the coefficients of A2 and D2, and setting the coefficient of AD* equal to zero yields  S21 2 +  S11 2 = 1 2
2
 S22  +  S12  = 1 S12S22* + S11S12* = 0
Consider now the matrix ⎛ S11 tr S =⎜ ⎝ S12
S21 ⎞ ⎟ S22 ⎠
The unitarity of this matrix implies that
S21⎞⎛ S11* ⎟⎜ S22 ⎠⎝ S*21
⎛⎜ S11 ⎝ S12
S12* ⎞ ⎛⎜ 1 0⎞ ⎟ ⎟= S*22 ⎠ ⎝ 0 1⎠
that is,
 S11 2 +  S21 2 = S12 2 +  S22 2 = 1 *
*
S11S12 + S21S22 = 0 These are just the conditions obtained above. They imply that the matrix Str is unitary, and therefore the matrix S is unitary. 2. We have solve the problem of finding R and T for this potential well in the text.We take V0 < 0. We dealt with wave function of the form e ikx + Re −ikx Te
x < −a x>a
ikx
In the notation of Problem 41, we have found that if A = 1 and D = 0, then C = S11 = T and B = S21 = R.. To find the other elements of the S matrix we need to consider the same problem with A = 0 and D = 1. This can be solved explicitly by matching wave functions at the boundaries of the potential hole, but it is possible to take the solution that we have and reflect the “experiment” by the interchange x Æ  x. We then find that S12 = R and S22 = T. We can easily check that 2
2
2
2
2
2
 S11  +  S21  = S12  +  S22  = R  +  T  = 1
Also
* S11S12 + S21S*22 = TR* + RT* = 2Re(TR*)
If we now look at the solutions for T and R in the text we see that the product of T and R* is of the form (i) x (real number), so that its real part is zero. This confirms that the S matrix here is unitary. 3. Consider the wave functions on the left and on the right to have the forms ψ L (x) = Ae ikx + Be− ikx
ψ R (x) = Ce ikx + De −ikx Now, let us make the change k Æ  k and complex conjugate everything. Now the two wave functions read
ψ L (x)'= A *e ikx + B *e − ikx ψ R (x)'= C * e ikx + D * e −ikx Now complex conjugation and the transformation k Æ  k changes the original relations to C* = S11*(− k)A * +S12* (−k)D * *
*
B* = S21 (−k)A * + S22 (− k)D * On the other hand, we are now relating outgoing amplitudes C*, B* to ingoing amplitude A*, D*, so that the relations of problem 1 read C* = S11(k)A * +S12 (k)D * B* = S21 (k)A * + S22 (k)D * * * * This shows that S11(k) = S11 (− k); S22 (k) = S22 (−k); S12 (k) = S21 (−k) . These
result may be written in the matrix form S(k) = S+ (−k) . 4. (a) With the given flux, the wave coming in from x = −∞ , has the form eikx , with unit amplitude. We now write the solutions in the various regions
x
e ikx + Re− ikx
k 2 = 2mE /
−b < x < −a
Aeκx + Be−κx
κ 2 = 2m(V0 − E) /
−a < x < c
Ce ikx + De − ikx
c
Me iqx + Ne −iqx
d
Teikx
2
2
q2 = 2m(E + V1 ) /
2
(b) We now have x <0
u(x) = 0
0
Asinkx
a< x < b
Beκx + Ce −κx
b
e −ikx + Reikx
k 2 = 2mE /
2
κ 2 = 2m(V0 − E ) /
2
The fact that there is total reflection at x = 0 implies that R2 = 1
5. The denominator in (4 ) has the form
D = 2kq cos2qa − i(q2 + k 2 )sin2qa With k = iκ this becomes
D = i(2κqcos2qa − (q − κ )sin2qa) 2
2
The denominator vanishes when tan 2qa =
2tan qa 2qκ = 2 2 1− tan qa q − κ 2
This implies that 2
⎛ q2 − κ 2 ⎞ q2 − κ 2 q2 − κ 2 q 2 + κ 2 tan qa = − ± 1+⎜ ± ⎟ =− 2κq 2κq 2κq ⎝ 2κ q ⎠
This condition is identical with (4 ). The argument why this is so, is the following: When k = iκ the wave functio on the left has the form e −κx + R(iκ )eκx . The function eκx blows up as x → −∞ and the wave function only make sense if this term is overpowered by the other term, that is when R(iκ ) = ∞ . We leave it to the student to check that the numerators are the same at k = iκ. 6. The solution is
u(x) = Aeikx + Beikx = Ceikx + Deikx
xb
The continuity condition at x = b leads to Aeikb + Beikb = Ceikb + Deikb And the derivative condition is (ikAeikb –ikBeikb)  (ikCeikb –ikDeikb)= (λ/a)( Aeikb + Beikb) With the notation Aeikb = α ; Beikb = β; Ceikb = γ; Deikb = δ These equations read
α+β=γ+δ ik(α  β + γ  δ) = (λ/a)(α + β) We can use these equations to write (γ,β) in terms of (α,δ) as follows 2ika λ α+ δ 2ika − λ 2ika − λ 2ika λ β= α+ δ 2ika − λ 2ika − λ
γ=
We can now rewrite these in terms of A,B,C,D and we get for the S matrix 2ika ⎛ ⎜ 2ika − λ S= ⎜ λ 2ikb ⎝ 2ika − λ e
λ
⎞ e −2ikb 2ika − λ ⎟ 2ika ⎟ 2ika − λ ⎠
Unitarity is easily established: 4k a λ + 2 2 =1 2 2 2 4k a + λ 4k a + λ2 ⎛ 2ika ⎞⎛ λ ⎛ λ ⎛ −2ika ⎞ * S11S12 + S12S22* = e −2ikb ⎞ + e −2ikb ⎞ =0 ⎝ 2ika − λ ⎠⎝ −2ika − λ ⎠ ⎝ 2ika − λ ⎠⎝ −2ika − λ ⎠ 2 2
 S11 2 +  S12 2 =
2
The matrix elements become infinite when 2ika =λ. In terms of κ= ik, this condition becomes κ = λ/2a = λ/2a. 7. The exponent in T = eS is
S=
2
=
∫
2
B A
∫
dx 2m(V (x) − E) B
A
dx (2m(
mω 2 2 x 3 ω (x − )) − 2 a 2
where A and B are turning points, that is, the points at which the quantity under the square root sign vanishes. We first simplify the expression by changing to dimensionless variables: x=
The integral becomes
/ mω y; η = a /
/ mω << 1
2∫ dy y 2 − ηy 3 − 1 y2
with η <<1
y1
where now y1 and y2 are the turning points. A sketch of the potential shows that y2 is very large. In that region, the –1 under the square root can be neglected, and to a good approximation y2 = 1/ η . The other turning point occurs for y not particularly large, so that we can neglect the middle term under the square root, and the value of y1 is 1. Thus we need to estimate
∫
1/η
1
2 3 dy y − ηy − 1
The integrand has a maximum at 2y – 3ηy2= 0, that is at y = 2η/3. We estimate the contribution from that point on by neglecting the –1 term in the integrand. We thus get 1/η
2 ⎡ (1− ηy)5/ 2 (1− ηy)3/ 2 ⎤ 8 3 1 ∫2/ 3η dyy 1− ηy = η 2 ⎢⎣ 5 − 3 ⎥⎦ = 135 η 2 2/3η 1/η
To estimate the integral in the region 1 < y < 2/3η is more difficult. In any case, we get a lower limit on S by just keeping the above, so that S > 0.21/η2 The factor eS must be multiplied by a characteristic time for the particle to move back and forth inside the potential with energy ω / 2 which is necessarily of order 1/ω. Thus the estimated time is longer const. 0.2/η 2 than e .
ω
8. The barrier factor is eS where
S=
2
∫
2
b R0
dx
l(l + 1) − 2mE x2
where b is given by the value of x at which the integrand vanishes, that is, with 2mE/ 2 =k2, b = l(l + 1) / k .We have, after some algebra 1
S = 2 l(l + 1) ∫R
0 /b
du 1− u2 u
⎡ 1+ 1− (R / b)2 ⎤ 0 = 2 l(l + 1) ⎢ln − 1− (R0 / b) 2 ⎥ R0 / b ⎢⎣ ⎥⎦ We now introduce the variable f = (R0/b) ≈ kR0 / l for large l. Then
2l
⎡1+ 1− f 2 ⎤ −2l eS eS = ⎢ ⎥ e f ⎣ ⎦
1− f 2
⎛e⎞ ≈⎝ ⎠ 2
−2l
f −2l
for f << 1. This is to be multiplied by the time of traversal inside the box. The important factor is f2l. It tells us that the lifetime is proportional to (kR0)2l so that it grows as a power of l for small k. Equivalently we can say that the probability of decay falls as (kR0)2l. 9. The argument fails because the electron is not localized inside the potential. In fact, for weak binding, the electron wave function extends over a region R = 1/α = 2mE B , which, for weak binding is much larger than a. 10. For a bound state, the solution for x > a must be of the
form u(x) = Ae −αx , where α = 2mEB / . Matching
1 du at x = a u dx
yields −α = f (E B ). If f(E) is a constant, then we immediately know α.. Even if f(E) varies only slightly over the energy range that overlaps small positive E, we can determine the binding energy in terms of the reflection coefficient. For positive energies the wave function u(x) for x > a has the form eikx + R(k)eikx, and matching yields
f (E ) ≈ −α = −ik
e − ika − Re ika 1− Re2ika = −ik e −ika + Re ika 1 + Re2ika
so that R = e −2 ika
k + iα k − iα
We see that R2 = 1. 11. Since the well is symmetric about x = 0, we need only match wave functions at x = b and a. We look at E < 0, so that we introduce and α2 = 2mE/ 2 and q2 = 2m(V0E)/ 2 . We now write down Even solutions: u(x) = coshαx 0
Matching
1 du(x) at x = b and at x = a leads to the equations u(x) dx
α tanhα b = q
Acosqb − Bsinqb Asinqb + B cosqb
−α = q
Acosqa − Bsnqa Asinqa + B cosqa
From the first equation we get B qcosqb − α tanhαbsinqb = A qsinqb + α tanhα bcosqb and from the second B qcosqa + α sinqa = A qsinqa − α cosqa Equating these, crossmultiplying, we get after a little algebra
q 2 sinq(a − b) − α cosq(a − b) = α tanh αb[α sinq(a − b) + qcosq(a − b)] from which it immediately follows that sinq(a − b) αq(tanhα b + 1) = cosq(a − b) q 2 − α 2 tanhα b Odd Solution Here the only difference is that the form for u(x) for 0 < x < b is sinhαx. The result of this is that we get the same expresion as above, with tanhαb replaced by cothαb. 11. (a) The condition that there are at most two bound states is equivalent to stating that there is at most one odd bound state. The relevant figure is Fig. 48, and we ask for the condition that there be no intersection point with the tangent curve that starts up at 3π/2. This means that
λ − y2
=0 y for y ≤ 3π/2. This translates into λ = y2 with y < 3π/2, i.e. λ < 9π2/4. (b) The condition that there be at most three bound states implies that there be at most two even bound states, and the relevant figure is 47. Here the conditon is that y < 2π so that λ < 4π2.
(c) We have y = π so that the second even bound state have zero binding energy. This means that λ = π2. What does this tell us about the first bound state? All we know is that y is a solution of Eq. (454) with λ = π2. Eq.(454) can be rewritten as follows: tan2 y =
1− cos2 y λ − y 2 1− (y 2 / λ ) = = cos2 y (y 2 / λ ) y2
so that the even condition is cos y = y / λ , and in the same way, the odd conditin is sin y = y / λ . Setting λ = π still leaves us with a transcendental equation. All we can say is that the binding energy f the even state will be larger than that of the odd one. 13.(a) As b Æ 0, tanq(ab) Æ tanqa and the r.h.s. reduces to α/q. Thus we get, for the even solution
tanqa = α/q and, for the odd solution, tanqa =  q/α. These are just the single well conditions. (b) This part is more complicated. We introduce notation c = (ab), which will be held fixed. We will also use the notation z = αb. We will also use the subscript “1” for the even solutions, and “2” for the odd solutions. For b large, e z − e − z 1− e −2z −2z z −z = −2z ≈ 1− 2e e +e 1+e coth z ≈ 1 = 2e −2z tanh z =
The eigenvalue condition for the even solution now reads q1α 1 (1+ 1− 2e −2z1 ) 2q1α 1 q12 + α12 −2z1 tan q1c = 2 ≈ (1− 2 e ) q1 − α12 (1− 2e −2z1 ) q12 − α12 q1 − α12 The condition for the odd solution is obtained by just changing the sign of the e2z term, so that tan q2c =
q2α 2 (1+ 1 + 2e −2z2 ) 2q2α 2 q22 + α 22 −2z2 ≈ (1+ e ) q22α 22 (1 + 2e −2z2 ) q22 − α 22 q22 − α 22
In both cases q2 + α2 = 2mV0/ 2 is fixed. The two eigenvalue conditions only differ in the e2z terms, and the difference in the eigenvalues is therefore proportional to e2z , where z here is some mean value between α1 b and α2b. This can be worked out in more detail, but this becomes an exercise in Taylor expansions with no new physical insights. 14. We write 〈x
∞ dV (x) dV (x) 〉 = ∫ dx ψ (x)x ψ (x) −∞ dx dx ∞ dψ ⎤ ⎡d xV − ψ 2V ⎥ = ∫ dx ⎢ (ψ 2 xV )− 2ψ −∞ dx ⎦ ⎣ dx
The first term vanishes because ψ goes to zero rapidly. We next rewrite ∞
−2∫ dx −∞
2 ∞ d2 dψ dψ xVψ = −2 ∫ dx x(E + 2 )ψ −∞ dx dx 2m dx 2 2 ∞ ∞ dψ 2 d ⎛ dψ ⎞ = −E ∫ dxx − dxx −∞ dx ⎝ dx ⎠ dx 2m ∫−∞
Now ∞ ∞ dψ 2 d ∫−∞ dxx dx = ∫−∞ dx dx (x ψ 2 )− ∫−∞ dxψ 2 ∞
The first term vanishes, and the second term is unity. We do the same with the second term, in which only the second integral ⎛ dψ ⎞ ∫−∞ dx ⎝ dx ⎠ ∞
2
remains. Putting all this together we get 2
2 ∞ ∞ dV p2 ⎛ dψ ⎞ 2 〈x 〉 + 〈V 〉 = dx + E ∫−∞ dxψ = 〈 〉+ E dx 2m ∫−∞ ⎝ dx ⎠ 2m
so that
p2 1 dV 〈x 〉=〈 〉 2 dx 2m
CHAPTER 5.
1. We are given
∫
∞
∞
dx(AΨ(x)) *Ψ(x) = ∫−∞ dxΨ(x) * AΨ(x) −∞
Now let Ψ(x) = φ (x) + λψ (x) , where λ is an arbitrary complex number. Substitution into the above equation yields, on the l.h.s.
∫
∞
−∞
dx(Aφ (x) + λAψ (x)) *(φ (x) + λψ (x))
= ∫−∞ dx[(Aφ ) * φ + λ (A φ )* ψ + λ * (Aψ )* φ +  λ 2 (A ψ )* ψ ] ∞
On the r.h.s. we get
∫
∞
−∞
dx(φ (x) + λψ (x)) *(Aφ (x) + λA ψ (x))
= ∫−∞ dx[φ * A φ + λ * ψ * Aφ + λφ * Aψ +  λ 2 ψ * A ψ ] ∞
Because of the hermiticity of A, the first and fourth terms on each side are equal. For the rest, sine λ is an arbitrary complex number, the coefficients of λ and λ* are independent , and we may therefore identify these on the two sides of the equation. If we consider λ, for example, we get
∫
∞
∞
dx(Aφ (x)) *ψ (x) = ∫−∞ dx φ (x) * A ψ (x) −∞
the desired result. 2. We have A+ = A and B+ = B , therefore (A + B)+ = (A + B). Let us call (A + B) = X. We have shown that X is hermitian. Consider now (X +)n = X+ X+ X+ …X+ = X X X …X = (X)n which was to be proved. 3. We have ∞
〈A 2 〉 = ∫−∞ dx ψ * (x)A 2ψ (x) Now define Aψ(x) = φ(x). Then the above relation can be rewritten as
∞
∞
〈A 2 〉 = ∫−∞ dx ψ (x)A φ (x) = ∫−∞ dx (Aψ (x))* φ (x) ∞
= ∫ dx(A ψ (x))* Aψ (x) ≥ 0 −∞
∞
4. Let U = eiH =
n +
∞ ∞ iH (−i) (H ) (−i) (H ) + . Then U = = = e − iH , and thus ∑ ∑ ∑ n! n! n! n= 0 n= 0 n =0 n
n
n
n
n
the hermitian conjugate of eiH is eiH provided H = H+.. 5. We need to show that iH −iH
e e
∞
i n n ∞ (−i)m m =∑ H ∑ H =1 n =0 n! m = o m!
Let us pick a particular coefficient in the series, say k = m + n and calculate its coefficient. We get, with m= k – n, the coefficient of Hk is 1 k i n (−i) k −n k! ∑ n! (k − n)! = k! ∑ n!(k − n)!i n (−i) k −n n= 0 n =0 1 = (i − i)k = 0 k! k
Thus in the product only the m = n = 0 term remains, and this is equal to unity. ∞
6. We write I(λ , λ *) = ∫ dx (φ (x) + λψ (x))* (φ (x) + λψ (x)) ≥ 0. The left hand side, in −∞ abbreviated notation can be written as I(λ , λ *) = ∫  φ 2 + λ * ∫ ψ * φ + λ ∫ φ * ψ + λλ * ∫  ψ 2
Since λ and λ* are independent, he minimum value of this occurs when
∂I = ψ * φ + λ ∫  ψ 2 = 0 ∂λ * ∫ ∂I = φ *ψ + λ * ∫  ψ 2 = 0 ∂λ ∫ When these values of λ and λ* are inserted in the expression for I(λ,λ*) we get * 2 I(λ min ,λ min ) = ∫  φ  −
∫ φ *ψ ∫ ψ *φ ≥0  ψ  ∫ 2
from which we get the Schwartz inequality. 7. We have UU+ = 1 and VV+ = 1. Now (UV)+ = V+U+ so that
(UV)(UV)+ = UVV+U+ = UU+ = 1 8. Let Uψ(x) = λψ(x), so that λ is an eigenvalue of U. Since U is unitary, U+U = 1. Now
∫
∞
−∞
∞
dx (Uψ (x))*Uψ (x) = ∫ dxψ *(x)U +Uψ (x) = −∞
∞
= ∫−∞ dxψ * (x)ψ (x) = 1 On the other hand, using the eigenvalue equation, the integral may be written in the form
∫
∞
∞
dx (Uψ (x))*Uψ (x) = λ * λ ∫−∞ dxψ *(x)ψ (x) = λ 2 −∞
It follows that λ2 = 1, or equivalently λ = eia , with a real. 9. We write
∫
∞
−∞
∞
∞
−∞
−∞
dxφ (x) * φ (x) = ∫ dx (Uψ (x))*Uψ (x) = ∫ dxψ *(x)U +Uψ (x) = ∞
= ∫−∞ dxψ * (x)ψ (x) = 1 10. We write, in abbreviated notation
∫ v v =∫ * a b
(Uua )*Uub = ∫ ua*U +Uub = ∫ u*a ub = δ ab
11. (a) We are given A+ = A and B+ = B. We now calculate
(i [A,B])+ = (iAB – iBA)+ = i (AB)+  (i)(BA)+ = i (B+A+) + i(A+B+) = iBA + iAB = i[A,B] (b) [AB,C] = ABCCAB = ABC – ACB + ACB – CAB = A(BC – CB) – (AC – CA)B = A [B,C] – [A,C]B (c) The Jacobi identity written out in detail is [A,[B,C]] + [B,[C,A]] + [C,[A,B]] =
A(BC – CB) – (BC – CB)A + B(CA – AC) – (CA  AC)B + C(AB – BA) – (AB – BA)C = ABC – ACB – BCA + CBA + BCA – BAC – CAB + ACB + CAB – CBA – ABC + BAC It is easy to see that the sum is zero. 12. We have
eA B eA = (1 + A + A2/2! + A3/3! + A4/4! +…)B (1  A + A2/2!  A3/3! + A4/4! …) Let us now take the term independent of A: it is B. The terms of first order in A are AB – BA = [A,B]. The terms of second order in A are A2B/2! – ABA + BA2/2! = (1/2!)(A2B – 2ABA + BA2) = (1/2!)(A(AB – BA) – (AB – BA)A) = (1/2!){A[A,B][A,B]A} = (1/2!)[A,[A,B]] The terms of third order in A are A3B/3! – A2BA/2! + ABA2/2! – BA3. One can again rearrange these and show that this term is (1/3!)[A,[A,[A,B]]]. There is actually a neater way to do this. Consider λA − λA F(λ ) = e Be
Then
dF(λ ) = e λA ABe−λA − e λA BAe−λA = e λA [A,B]e −λA dλ Differentiating again we get d 2 F(λ ) = e λA [A,[A,B]]e −λA dλ 2 and so on. We now use the Taylor expansion to calculate F(1) = eA B eA. 1 1 F''(0) + F '''(0) + .., 3! 2! 1 1 = B+ [A,B] + [A,[A,B]] + [A,[A,[A,B]]] + ... 2! 3!
F(1) = F (0) + F'(0) +
13. Consider the eigenvalue equation Hu = λu. Applying H to this equation we get
H2 u = λ 2u ; H3 u = λ3u and H4u = λ4u . We are given that H4 = 1, which means that H4 applied to any function yields 1. In particular this means that λ4 = 1. The solutions of this are λ = 1, 1, i, and –i. However, H is hermitian, so that the eigenvalues are real. Thus only λ = ± 1 are possible eigenvalues. If H is not hermitian, then all four eigenvalues are acceptable.
14. We have the equations (1) (2) Bu(1) a = b11ua + b12 ua (1) (2) Bu(2) a = b21ua + b22ua
(2) Let us now introduce functions (v (1) a ,v a ) that satisfy the equations (1) (2) (2) Bv(1) a = b1v a ;Bv a = b2 v a . We write, with simplified notation,
v1 = α u1 + β u2 v2 = γ u1 + δ u2 The b1  eigenvalue equation reads b1v1 = B ( α u1 + β u2) = α (b11 u1 + b12u2) + β (b21u1 + b22u2) We write the l.h.s. as b1(α u1 + β u2). We can now take the coefficients of u1 and u2 separately, and get the following equations α (b1 – b11) = βb21 β (b1 – b22) = αb12
The product of the two equations yields a quadratic equation for b1, whose solution is
b11 + b22 (b11 − b22 ) ± + b12b21 2 4 2
b1 =
We may choose the + sign for the b1 eigenvalue. An examination of the equation involving v2 leads to an identical equation, and we associate the – sign with the b2 eigenvalue. Once we know the eigenvalues, we can find the ratios α/β and γ/δ. These suffice, since the normalization condition implies that α2 + β2 = 1 and γ2 + δ2 = 1 15. The equations of motion for the expectation values are
d i i p2 i p 〈x〉 = 〈[H ,x]〉 = 〈[ , x]〉 = 〈 p[ p, x]〉 = 〈 〉 2m dt m m d i i 1 〈p〉 = 〈[H, p]〉 = − 〈[p, mω12 x 2 + ω 2 x]〉 = −mω12 〈x〉 − ω 2 dt 2 16. We may combine the above equations to get
d2 ω2 2 2 〈x〉 = −ω1 〈x〉 − dt m The solution of this equation is obtained by introducing the variable X = 〈x〉 +
ω2 mω12
The equation for X reads d2X/dt2 =  ω12 X, whose solution is X = Acosω1 t + Bsinω1 t
This gives us
〈x〉t = −
ω2 + Acosω1 t + B sinω1 t mω12
At t = 0 〈x〉0 = −
ω2 +A mω12
〈p〉 0 = m
d 〈x〉 = mBω1 dt t = 0
We can therefore write A and B in terms of the initial values of < x > and < p >,
〈x〉t = −
⎛ ω2 ω2 ⎞ 〈p〉 0 sinω1 t 2 + ⎜ 〈x〉0 + 2 ⎟ cosω1 t + mω1 ⎝ mω1 ⎠ mω1
17. We calculate as above, but we can equally well use Eq. (553) and (557), to get
d 1 〈x〉 = 〈 p〉 dt m d ∂V (x,t) 〈p〉 = −〈 〉 = eE 0cosωt dt ∂x Finally
∂H d 〈H 〉 = 〈 〉 = eE0ω sin ωt〈x〉 ∂t dt 18. We can solve the second of the above equations to get
〈p〉 t =
eE 0
ω
sin ωt + 〈p〉 t =0
This may be inserted into the first equation, and the result is 〈x〉t = −
〈 p〉t = 0 t eE0 + 〈x〉t = 0 2 (cosω t − 1) + m mω
CHAPTER 6 19. (a) We have Aa> = aa> It follows that = a = a if the eigenstate of A corresponding to the eigenvalue a is normalized to unity. The complex conjugate of this equation is * = = a* If A+ = A, then it follows that a = a*, so that a is real. 13. We have
〈ψ  (AB) +  ψ 〉 = 〈(AB)ψ  ψ 〉 = 〈Bψ  A +  ψ 〉 = 〈ψ  B + A +  ψ 〉 This is true for every ψ, so that (AB)+ = B+A+ 2.
TrAB = ∑ 〈n  AB  n〉 = ∑ 〈n  A1B  n〉 n
n
= ∑ ∑ 〈n  A  m〉〈m  B  n〉 = ∑ ∑ 〈m  B  n〉〈n  A  m〉 n
m
n
m
= ∑ 〈m  B1A  m〉 = ∑ 〈m  BA  m〉 =TrBA m
m
3. We start with the definition of n> as  n〉 =
1 (A + )n  0〉 n!
We now take Eq. (647) from the text to see that A  n〉 =
1 n n + n + n−1 + n −1 A(A )  0〉 = (A )  0〉 = (A )  0〉 = (n − 1)! n! n! N
+ + n 4. Let f (A ) = ∑ Cn (A ) . We then use Eq. (647) to obtain n=1
n  n − 1〉
N
N
Af (A + )  0〉 = A ∑ Cn (A + ) n  0〉 = ∑ nCn (A + ) n−1  0〉 n=1
=
n=1
d df (A + ) + n C (A )  0〉 =  0〉 ∑ dA + n =1 n dA + N
5. We use the fact that Eq. (636) leads to
x=
2mω
(A + A + )
mω (A + − A) 2
p= i We can now calculate 〈k  x  n〉 = =
2mω 2mω
〈k  A + A +  n〉 =
(
2mω
(
n〈k  n − 1〉 + k 〈k − 1  n〉
)
)
nδk ,n −1 + n + 1δ k ,n +1
which shows that k = n ± 1. 6. In exactly the same way we show that 〈k  p  n〉 = i
mω mω + 〈k  A − A  n〉 = i ( n + 1δk ,n+1 − nδ k,n −1 ) 2 2
7. Let us now calculate
〈k  px  n〉 = 〈k  p1 x  n〉 = ∑ 〈k  p  q〉〈q  x  n〉 q
We may now use the results of problems 5 and 6. We get for the above i 2
∑(
k δ k −1,q − k + 1δk +1,q )( nδ q,n−1 + n + 1δ q,n+1 )
q
i ( kn δkn − (k + 1)nδ k +1,n−1 + k(n + 1)δ k −1,n +1 − (k + 1)(n + 1)δ k +1,n+1 ) 2 i = (−δ kn − (k + 1)(k + 2)δ k +2,n + n(n + 2)δ k,n +2 ) 2 =
To calculate 〈k  xp  n〉 we may proceed in exactly the same way. It is also possible to abbreviate the calculation by noting that since x and p are hermitian operators, it follows that
〈k  xp  n〉 = 〈n  px  k〉*
so that the desired quantity is obtained from what we obtained before by interchanging k and n and complexconjugating. The latter only changes the overall sign, so that we get
〈k  xp  n〉 = −
i (−δ kn − (n + 1)(n + 2)δ k ,n+ 2 + (k + 1)(k + 2)δ k +2,n ) 2
8.The results of problem 7 immediately lead to 〈k  xp − px  n〉 = i δkn
9. This follows immediately from problems 5 and 6. 10. We again use
x= p= i
2mω
(A + A + )
mω (A + − A) 2
to obtain the operator expression for x2 =
(A + A + )(A + A + ) = (A 2 + 2A + A + (A + )2 + 1) 2mω 2mω mω mω (A + − A)(A + − A) = − (A 2 − 2A + A + (A + ) 2 − 1) p2 = − 2 2
where we have used [A,A+] = 1. The quadratic terms change the values of the eigenvalue integer by 2, so that they do not appear in the desired expressions. We get, very simply 〈n  x 2  n〉 =
(2n + 1) 2mω mω (2n + 1) 〈n  p2  n〉 = 2 14. Given the results of problem 9, and of 10, we have
(Δx)2 =
(2n + 1) 2mω mω (Δp)2 = (2n + 1) 2
and therefore 1 ΔxΔp = (n + ) 2 15. The eigenstate in Aα> = αα> may be written in the form
 α 〉 = f (A + )  0〉 It follows from the result of problem 4 that the eigenvalue equation reads df (A + )  0〉 = αf (A + )  0〉 Af (A )  0〉 = + dA +
The solution of df (x) = α f(x) is f(x) = C eαx so that αA +
 α 〉 = Ce
 0〉
The constant C is determined by the normalization condition <αα> = 1 This means that ∞ 1 (α *) n ⎛ d ⎞ αA + α *A αA + = 〈0  e e  0〉 = 〈0  e  0〉 ∑ 2 ⎝ dA + ⎠ C n! n =0 n
∞ + 2  α 2n  α 2n 〈0  eαA  0〉 = ∑ = e α  n! n =0 n= 0 n! ∞
=∑
Consequently C = e −α 
2
/2
We may now expand the state as follows  α 〉 = ∑  n〉〈n  α 〉 = ∑  n〉〈0  n
n
A n αA + Ce  0〉 n!
αn 1 ⎛ d ⎞ αA + 〈0  e  0〉 = C  n〉 = C∑  n〉 ⎝ dA + ⎠ n! n! n n
The probability that the state α> contains n quanta is
 α 2n ( α 2 )n −α  2 Pn = 〈n  α 〉  = C = e n! n! 2
2
This is known as the Poisson distribution. Finally
〈α  N  α 〉 = 〈α  A + A  α 〉 = α * α = α 2 13. The equations of motion read dx(t) i i p 2 (t) p(t) = [H, x(t)]= [ ,x(t)] = dt 2m m dp(t) i = [mgx(t), p(t)] = − mg dt
This leads to the equation 2 d x(t) = −g dt 2
The general solution is x(t) =
1 2 p(0) gt + t + x(0) 2 m
14. We have, as always
dx p = dt m Also dp i 1 = [ mω 2 x 2 + eξx, p] 2 dt 1 i ⎛1 ⎞ = ⎝ mω 2 x[x, p] + mω 2[x, p]x + eξ[x, p]⎠ 2 2 = −mω 2 x − eξ Differentiating the first equation with respect to t and rearranging leads to eξ eξ d 2x 2 = − ω 2 (x + ) 2 = −ω x − dt m mω 2
The solution of this equation is x+
eξ = Acosω t + B sinωt mω 2 eξ ⎞ p(0) ⎛ = x(0) + sinωt 2 cosωt + ⎝ mω ⎠ mω
We can now calculate the commutator [x(t1),x(t2)], which should vanish when t1 = t2. In this calculation it is only the commutator [p(0), x(0)] that plays a role. We have p(0) p(0) sinωt1,x(0)cos ωt2 + sin ωt2 ] mω mω i ⎛ 1 ⎞ (cosωt1 sinωt2 − sinω t1 cosωt 2 = sinω (t 2 − t1 ) =i ⎝ mω ⎠ mω
[x(t1 ),x(t2 )] = [x(0)cosω t1 +
16. We simplify the algebra by writing
mω = a; 2
2mω
=
1 2a
Then ⎛ π⎞ n!⎝ mω ⎠
1/ 4
n
1 d ⎞ − a2x 2 ⎛ un (x) = v n (x) = ⎝ ax − e 2a dx ⎠
Now with the notation y = ax we get 2 2 1 d −y 2 )e = (y + y)e − y = 2ye− y 2 dy 2 2 1 d v 2 (y) = (y − )(2ye −y ) = (2y 2 − 1 + 2y 2 )e −y 2 dy
v1 (y) = (y −
= (4 y 2 − 1)e −y Next
2
v 3 (y) = (y −
[
1 d 2 ) (4 y 2 − 1)e −y 2 dy
]
= (4y 3 − y − 4y + y(4 y 2 − 1))e − y = (8y 3 − 6y)e − y
2
2
mω x 2
The rest is substitution y =
17. We learned in problem 4 that
df (A + )  0〉 dA +
+ Af (A )  0〉 =
Iteration of this leads to n
+
d f (A ) A f (A )  0〉 =  0〉 n dA + +
n
We use this to get ∞
e f (A )  0〉 = ∑ λA
+
λn
n= 0 n!
∞
A f (A )  0〉 = ∑ n
+
λn ⎛ d ⎞
n
⎝ +⎠ n= 0 n! dA
f (A + )  0〉 = f (A + + λ )  0〉
18. We use the result of problem 16 to write
e λA f (A + )e −λA g(A + )  0〉 = e λA f (A + )g(A + − λ )  0〉 = f (A + + λ )g(A +)  0〉 Since this is true for any state of the form g(A+)0> we have λA + −λA + e f (A )e = f (A + λ )
In the above we used the first formula in the solution to 16, which depended on the fact that [A,A+] = 1. More generally we have the BakerHausdorff form, which we derive as follows: Define λA + −λA F(λ ) = e A e Differentiation w.r.t. λ yields dF(λ ) = e λA AA+e −λA − eλA A + Ae−λA = e λA [A,A + ]e −λA ≡ e λA C1e −λA dλ Iteration leads to
d 2 F(λ ) = e λA [A,[A,A + ]]e −λA ≡ e λA C2e −λA 2 dλ ....... d n F(λ ) λA = e [A,[A,[A,[A,....]]..]e −λA ≡ eλA Cn e −λA n dλ with A appearing n times in Cn. We may now use a Taylor expansion for
σ n d n F(λ ) ∞ σ n λA −λA = ∑ e Cn e F(λ + σ ) = ∑ dλ n n =0 n! n =0 n! ∞
If we now set λ = 0 we get ∞
σn
n =0
n!
F(σ ) = ∑
Cn
which translates into σA + −σA + + e A e = A + σ [A, A ] +
σ2 2!
[A,[A, A + ]] +
σ3 3!
[A,[A,[A, A + ]]] + ...
Note that if [A,A+] = 1 only the first two terms appear, so that σA + −σA + + + e f (A )e = f (A + σ [A,A ]) = f (A + σ )
19. We follow the procedure outlined in the hint. We define F(λ) by e
λ(aA + bA + )
=e
λaA
F(λ )
Differentiation w.r.t λ yields + λaA λA λaA (aA + bA )e F(λ ) = aAe F (λ ) + e
dF(λ ) dλ
−λaA on the left yields The first terms on each side cancel, and multiplication by e
dF (λ ) = e −λaA bA +eλaA F (λ ) = bA + − λab[ A, A + ]F (λ ) dλ When [A,A+] commutes with A. We can now integrate w.r.t. λ and after integration Set λ = 1. We then get F(1) = e bA
+
− ab[A ,A + ]/2
+
= e bA e −ab / 2
so that +
+
e aA + bA = e aAe bA e −ab / 2 20. We can use the procedure of problem 17, but a simpler way is to take the hermitian conjugate of the result. For a real function f and λ real, this reads +
+
e −λA f ( A)e λA = f ( A + λ ) Changing λ to λ yields +
+
e λA f ( A )e −λA = f ( A − λ ) The remaining steps that lead to +
+
e aA + bA = e bA e aA e ab /2 are identical to the ones used in problem 18. 20. For the harmonic oscillator problem we have x=
2mω
+
(A + A )
This means that eikx is of the form given in problem 19 with a = b = ik
/ 2mω
This leads to
e ikx = e ik
/ 2m ω A +
e ik
/2mω A − k 2 / 4mω
e
Since A0> = 0 and <0A+ = 0, we get 〈0  e ikx  0〉 = e −
k 2 / 4mω
21. An alternative calculation, given that u0 ( x ) = (mω / π )1/ 4 e −mωx
⎛ mω ⎞ ⎝π ⎠
1/ 2
∫
∞ −∞
ikx −mωx 2
dxe e
⎛ mω ⎞ =⎝ π ⎠
1/ 2
∫
∞
−∞
The integral is a simple gaussian integral and
dxe
∫
−
mω
∞
−∞
dye
(x −
/2
, is
ik k2 )2 − 2mω 4 mω
−m ωy 2 /
cancels the factor in front. Thus the two results agree.
2
e
=
π which just mω
CHAPTER 7 1. (a) The system under consideration has rotational degrees of freedom, allowing it to rotate about two orthogonal axes perpendicular to the rigid rod connecting the two masses. If we define the z axis as represented by the rod, then the Hamiltonian has the form L2x + L2y L2 − L2z H= = 2I 2I where I is the moment of inertia of the dumbbell. (b) Since there are no rotations about the z axis, the eigenvalue of Lz is zero, so that the eigenvalues of the Hamiltonian are
( + 1) 2I 2
E= with = 0,1,2,3,…
(c) To get the energy spectrum we need an expression for the moment of inertia. We use the fact that I = Mred a
2
where the reduced mass is given by M red =
MC M N 12 × 14 = M nucleon = 6.46 M nucleon MC + M N 26
If we express the separation a in Angstroms, we get I = 6.46 × (1.67 × 10
−27
kg)(10
−10
m / A) aA = 1.08 × 10 2
2
−46
2
aA
2
The energy difference between the ground state and the first excited state is 2 / 2I which leads to the numerical result −34
∆E =
−4
(1.05 × 10 J.s) 1 6.4 × 10 = eV − 46 2 2 × −19 1.08 × 10 aA kg.m a2A (1.6 × 10 J / eV )
2. We use the connection
2
x y z = sinθ cos φ; = sin θ sinφ ; = cosθ to write r r r
3 iφ 3 x + iy e sin θ = − ( ) 8π 8π r
Y1,1 = − Y1,0 =
3 cosθ = 4π
Y1,−1 = (− 1)Y1,1 = *
3 z ( ) 4π r
3 −iφ e sinθ = 8π
3 x − iy ( ) 8π r
Next we have
Y2,2 =
15 2iφ 2 e sin θ = 32π
15 (cos2φ + i sin2φ )sin2 θ 32π
=
15 (cos2 φ − sin2 φ + 2isin φ cosφ )sin2 θ 32π
=
15 x 2 − y 2 + 2ixy 32π r2
Y2,1 = −
15 iφ 15 (x + iy)z e sinθ cos θ = − 8π 8π r2
and
5 5 2z 2 − x 2 − y 2 2 Y2,0 = (3cos θ − 1) = 16π 16π r2 We may use Eq. (746) to obtain the form for Y2,−1 and Y2,−2 .
3. We use L± = Lx ± iLy to calculate Lx =
1 i (L+ + L − ); L y = (L− − L+ ) . We may now 2 2
proceed
1 1 〈l,m1  L+  l,m 2 〉 + 〈l,m1  L −  l,m 2 〉 2 2 i i 〈l,m1  L y  l,m 2 〉 = 〈l,m1  L−  l,m 2 〉 − 〈l,m1  L+  l,m 2 〉 2 2 〈l,m1  L x  l,m 2 〉 =
and on the r.h.s. we insert
〈l,m1  L +  l,m2 〉 = (l − m2 )(l + m2 + 1)δm1 ,m2 +1 〈l,m1  L −  l,m2 〉 = (l + m2 )(l − m2 + 1)δm1 ,m2 −1 4. Again we use Lx =
1 i (L+ + L − ); L y = (L− − L+ ) to work out 2 2
L2x =
1 (L + L− )(L + + L− ) = 4 + 1 = (L2+ + L2− + L2 − L2z + Lz + L2 − L2z − Lz ) 4 1 1 1 1 = L2+ + L2− + L2 − L2z 4 2 2 4
We calculate 〈l,m1  L +  l,m 2 〉 = (l − m 2 )(l + m 2 + 1)〈l,m1  L +  l,m 2 + 1〉 2
= 2 ((l − m 2 )(l + m2 + 1)(l − m 2 − 1)(l + m 2 + 2) δ m1 ,m 2 +2 1/2
and 〈l,m1  L −  l,m 2 〉 = 〈l,m 2  L+  l,m1 〉 * 2
2
which is easily obtained from the preceding result by interchanging m1 and m2. The remaining two terms yield 2
1 2 2 2 〈l,m1  (L − Lz )  l,m 2 〉 = (l(l + 1) − m 2 )δ m1 ,m 2 2 2 The remaining calculation is simple, since 〈l,m1  L y  l,m 2 〉 = 〈l,m1  L − Lz − L x  l,m 2 〉 2
5. The Hamiltonian may be written as
2
2
2
L2 − L2z L2z H= + 2I1 2I3
whose eigenvalues are
1 2 l(l + 1) 2 1 +m − 2I3 2I1 2I1 where –l ≤ m ≤ l. (b) The plot is given on the right. 2
(c) The spectrum in the limit that I1 >> I3 is just E =
2 m , 2I3
with m = 0,1,2,…l. The m = 0 eigenvalue is nondegenerate, while the other ones are doubly degenerate (corresponding to the negative values of m).
−iφ
6. We will use the lowering operator L− = e (−
∂ ∂ + icot θ ) acting on Y44. Since ∂θ ∂φ
we are not interested in the normalization, we will not carry the factor. Y43 ∝ e −iφ (−
∂ ∂ + i cot θ )[e 4 iφ sin4 θ] ∂θ ∂φ
= e 3iφ {−4 sin3 θ cos θ − 4 cot θ sin 4 θ }= −8e 3 iφ sin 3 θ cosθ
Y42 ∝ e −iφ (−
∂ ∂ + i cot θ )[e 3iφ sin 3 θ cosθ ] ∂θ ∂φ
= e 2iφ {−3sin2 θ cos 2 θ + sin 4 θ − 3sin 2 θ cos2 θ}= = e 2iφ {−6sin 2 θ + 7sin4 θ} Y41 ∝ e −iφ (−
∂ ∂ + icot θ )[e 2iφ (−6sin2 θ + 7sin 4 θ ] ∂θ ∂φ
iφ
{
iφ
{24sin θ cosθ − 42sin
}
= e 12sinθ cos θ − 28sin θ cosθ − 2(−6sinθ cos θ + 7sin θ cosθ ) =e
3
3
3
cosθ}
Y40 ∝ e −iφ (−
∂ ∂ + i cot θ )[e iφ (4sinθ − 7sin3 θ )cos θ ] ∂θ ∂φ
= {(−4cos θ + 21sin 2 θ cosθ )cosθ + (4 sin2 θ − 7sin4 θ ) − (4 cos2 θ − 7sin 2 θ cos2 θ } = {−8 + 40sin2 θ − 35sin4 θ}
7. Consider the H given. The angular momentum eigenstates  ,m〉 are eigenstates of the
Hamiltonian, and the eigenvalues are ( + 1) E= + α m 2I 2
with − ≤ m ≤ . Thus for every value of there will be (2 +1) states, no longer degenerate. 8. We calculate
[x,Lx ] = [x, ypz − zpy ] = 0 [y,Lx ] = [y, ypz − zpy ] = z[py , y] = −iz [z,L x ] = [z, ypz − zpy ] = − y[ pz,z] = i y [x,Ly ] = [x,zp x − xpz ] = −z[px ,x] = iz [y,Ly ] = [y,zp x − xpz ] = 0 [z,L y ] = [z,zp x − xpz ] = x[ pz,z] = −i x
The pattern is cyclical (x ,y) i z and so on, so that we expect (and can check) that [x,Lz ] = −iy [y,Lz ] = ix [z,L z ] = 0 9. We again expect a cyclical pattern. Let us start with
[px ,Ly ] = [px ,zpx − xpz ] = −[px , x]pz = ipz and the rest follows automatically.
10. (a) The eigenvalues of Lz are known to be 2,1,0,1,2 in units of . (b) We may write
(3 / 5)Lx − (4 / 5)Ly = n • L where n is a unit vector, since nx + ny = (3 / 5) + (−4 / 5) = 1. However, we may well 2
2
2
2
have chosen the n direction as our selected z direction, and the eigenvalues for this are again 2,1,0,1,2. (c) We may write
2L x − 6Ly + 3Lz = 2 2 + 62 + 32
6 2 L − L + 3Lz 7 x 7 y
= 7n • L Where n is yet another unit vector. By the same argument we can immediately state that the eigenvalues are 7m i.e. 14,7,0,7,14.
11. For our purposes, the only part that is relevant is
xy + yz + zx = sin2 θ sin φ cosφ + (sinφ + cos φ )sinθ cos θ 2 r e 2iφ − e −2iφ e iφ − e − iφ e iφ + e − iφ 1 = sin2 θ + sinθ cosθ + 2i 2i 2 2 Comparison with the table of Spherical Harmonics shows that all of these involve combinations of = 2 functions. We can therefore immediately conclude that the 2 probability of finding = 0 is zero, and the probability of finding 6 iz one, since this value corresponds to = 2 .A look at the table shows that
32π Y2,2 ; e −2iφ sin2 θ = 15
e 2iφ sin 2 θ =
e iφ sin θ cosθ = −
32π Y 15 2,−2
8π Y2,1; e − iφ sinθ cos θ = 15
8π Y 15 2,−1
Thus
e iφ − e −iφ e iφ + e − iφ xy + yz + zx 1 2 e 2iφ − e −2iφ = sin θ + sin θ cos θ + 2 2i 2i 2 2 r =
1 32π 1 32π −i + 1 8π i + 1 8π Y2,2 − Y2,−2 − Y2,1 + Y2,−1 4i 15 15 15 4i 15 2 2
This is not normalized. The sum of the squares of the coefficients is
2π 2π 4π 4π 12π 4π + + + = = , so that for normalization purposes we must multiply 15 15 15 15 15 5 5 . Thus the probability of finding m = 2 is the same as getting m = 2, and it is by 4π P±2 =
5 2π 1 = 4π 15 6
Similarly P1 = P1 , and since all the probabilities have add up to 1, P±1 =
1 3 imφ
12.Since the particles are identical, the wave function e must be unchanged under the rotation φ φ + 2π/N. This means that m(2π/N ) = 2nπ, so that m = nN, with n = 0,±1,±2,±3,… The energy is 2 2 2 2 m N 2 E= = n 2MR 2 2MR 2
The gap between the ground state (n = 0) and the first excited state (n =1) is 2
2
N ∆E = → ∞ as N → ∞ 2MR 2
If the cylinder is nicked, then there is no such symmetry and m = 0,±1,±23,…and 2
∆E =
2MR 2
CHAPTER 8 1. The solutions are of the form ψ n1 n 2 n 3 (x, y,z) = un1 (x)un 2 (y)un 3 (z) where un (x) =
2 nπ x sin ,and so on. The eigenvalues are a a π 2 2 2 ( n + n2 + n3 ) 2ma 2 1 2
E = E n1 + E n 2 + E n 3 =
2
2. (a) The lowest energy state corresponds to the lowest values of the integers {n1,n2,n3}, that is, {1,1,1)Thus π = ×3 2 ma 2 2
E ground
2
2π 2 the energies are In units of 2ma 2 nondegenerate) {1,1,1} 3 {1,1,2},(1,2,1},(2,1,1} 6 (triple degeneracy) {1,2,2},{2,1,2}.{2,2,1} 9 (triple degeneracy) {3,1,1},{1,3,1},{1,1,3} 11 (triple degeneracy) (nondegenrate) {2,2,2} 12 {1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,1,2},{3,2,1} 14 (6fold degenerate) {2,2,3},{2,3,2},{3,2,2}17 (triple degenerate) {1,1,4},{1,4,1},{4,1,1}18 (triple degenerate) {1,3,3},{3,1,3},{3,3,1} 19 (triple degenerate) {1,2,4},{1,4,2},{2,1,4},{2,4,1},{4,1,2},{4,2,1}21 (6fold degenerate)
3. The problem breaks up into three separate, here identical systems. We know that the energy for a onedimensional oscillator takes the values ω(n + 1/ 2) , so that here the energy eigenvalues are
E = ω (n1 + n2 + n3 + 3 / 2) 3 ω . 2 4. The energy eigenvalues in terms of ω with the corresponding integers are The ground state energy correspons to the n values all zero. It is
(0,0,0) (0,0,1) etc (0,1,1) (0,0,2) etc (1,1,1),(0,0,3),(0,1,2) etc (1,1,2),(0,0,4),(0,2,2),(0,1,3) (0,0,5),(0,1,4),(0,2,3)(1,2,2) (1,1,3)
3/2 5/2 7/2 9/2 11/2
degeneracy 1 3 6 10 15
13/2
21
(0,0,6),(0,1,5),(0,2,4),(0,3,3) (1,1,4),(1,2,3),(2,2,2), (0,0,7),(0,1,6),(0,2,5),(0,3,4) (1,1,5),(1,2,4),(1,3,3),(2,2,3) (0,0,8),(0,1,7),(0,2,6),(0,3,5) (0,4,4),(1,1,6),(1,2,5),(1,3,4) (2,2,4),(2,3,3) (0,0,9),(0,1,8),(0,2,7),(0,3,6) (0,4,5)(1,1,7),(1,2,6),(1,3,5) (1,4,4),(2,2,5) (2,3,4),(3,3,3)
15/2
28
17/2
36
19/2
45
21/2
55
5. It follows from the relations x = ρ cosφ ,y = ρ sin φ that
dx = dρ cosφ − ρ sinφdφ ; dy = dρsinφ + ρ cosφdφ Solving this we get
dρ = cosφ dx + sin φdy; ρ dφ = − sinφ dx + cosφ dy so that
∂ ∂ρ ∂ ∂φ ∂ ∂ sinφ ∂ = + = cosφ − ∂x ∂x ∂ρ ∂x ∂φ ∂ρ ρ ∂φ and ∂ ∂ρ ∂ ∂φ ∂ ∂ cos φ ∂ = + = sin φ + ∂y ∂y ∂ρ ∂y ∂φ ∂ρ ρ ∂φ We now need to work out
∂2 ∂2 + = ∂x 2 ∂y 2 (cosφ
∂ sinφ ∂ ∂ sinφ ∂ ∂ cosφ ∂ ∂ cos φ ∂ − )(cosφ − ) + (sin φ + )(sinφ + ) ρ ∂φ ρ ∂φ ρ ∂φ ρ ∂φ ∂ρ ∂ρ ∂ρ ∂ρ
A little algebra leads to the r.h.s. equal to 1 ∂ ∂ 2 + 2 ∂ρ ρ ∂φ 2 2
2
The timeindependent Schrodinger equation now reads
2 ∂2 Ψ(ρ,φ ) 1 ∂2 Ψ(ρ,φ ) + V ( ρ)Ψ(ρ,φ ) = EΨ(ρ,φ ) − + 2 ∂φ 2 2m ∂ρ 2 ρ
The substitution of Ψ(ρ,φ ) = R( ρ)Φ(φ ) leads to two separate ordinary differential equations. The equation for Φ(φ ) , when supplemented by the condition that the solution is unchanged when φ φ + 2π leads to Φ(φ ) =
1 imφ e 2π
m = 0,±1,±2,...
and the radial equation is then d R(ρ ) m 2mE 2mV ( ρ) − 2 R( ρ) + 2 R( ρ) = R(ρ ) 2 dρ ρ 2 2
2
6. The relation between energy difference and wavelength is 2π
c
λ
=
1 1 2 2 m red c α 1− 4 2
so that
λ=
16π me 1+ 3 me cα 2 M
where M is the mass of the second particle, bound to the electron. We need to evaluate this for the three cases: M = mP; M =2mp and M = me. The numbers are
λ (in m) = 1215.0226 × 10 −10 (1 +
me ) M
= 1215.68
for hydrogen
=1215.35
for deuterium
= 2430.45
for positronium
7. The ground state wave function of the electron in tritium (Z = 1) is
ψ 100 (r) =
2 1 4π a0
3/ 2
e −r /a 0
This is to be expanded in a complete set of eigenstates of the Z = 2 hydrogenlike atom, and the probability that an energy measurement will yield the ground state energy of the Z = 2 atom is the square of the scalar product
2 1 ∫ d r 4π a0
3/2
3
e
− r /a 0
2 4π
2 2 a0
3/ 2
e
− 2r /a 0
3
8 2 ∞ 8 2 a0 16 2 = 3 ∫ r 2 dre −3r /a 0 = 3 2!= a0 0 a0 3 27 Thus the probability is P =
512 729
8. The equation reads E −mc 2Zα E 1 (Z α ) − − )ψ (r) = 0 2 2 c c r r2 Compare this with the hydrogenlike atom case 2
2 4
2
−∇ ψ + (− 2
2mE B 2mZe 2 −∇ ψ (r) + − 2 4πε 0 2 2
1 ψ (r) = 0 r
and recall that d 2 d ( + 1) −∇ = − 2 − + dr r dr r2 2
2
We may thus make a translation E 2 − m 2c 4 → −2mc 2 E B 2Z α E 2mZe →− c 4πε 0 2 2
−
( + 1) − Z 2α 2 → ( + 1)
Thus in the expression for the hydrogenlike atom energy eigenvalue 2
2 2
1 mZ e 2mE B = − 2 4πε 0 (nr + + 1) 2 we replace by * , where * ( * +1) = ( + 1) − (Zα ) , that is, 2
1/ 2
2 1 1 2 * = − + + − (Zα ) 2 2
mZe Zα E E −m c We also replace by and 2mE B by − 4πε 0 c c2 2
2
2 4
We thus get 2 2 Zα E = m c 1 + 2 (n r + * +1) 2
−1
2 4
For (Zα) << 1 this leads to 1 2 1 2 2 E − mc = − mc (Zα ) 2 ( nr + * +1) 2 This differs from the nonrelativisric result only through the replacement of by * . 9. We use the fact that Ze 1 mc (Zα ) 〈T 〉nl − 〈 〉 nl = E nl = − 4πε 0 r 2n 2 2
2
2
Since Ze 1 Ze Z Ze 2 mcα mc Z α 〈 〉nl = = 2 = 4πε 0 r 4πε 0 a0 n 4πε 0 n 2 n2 2
2
2
2
we get 1 mc Z α = 〈V (r )〉 nl 2 2n 2 2
〈T 〉nl =
2
2
10. The expectation value of the energy is 2
2 2 2 10 4 3 1 E1 + E2 + − E2 + 〈 E〉 = E2 6 6 6 6
mc 2α 2 16 20 1 mc 2α 2 21 =− =− + 2 36 36 2 2 2 36
Similarly 20 40 2 2 2 16 〈L 〉 = × 0 + × 2 = 36 36 36
Finally 9 1 10 16 〈L z 〉 = × 0 + × 1+ ×0+ × (−1) 36 36 36 36 1 =− 36
2
2
11. We change notation from α to β to avoid confusion with the finestructure constant that appears in the hydrogen atom wave function. The probability is the square of the integral
β ∫ d r π
3/2
3
e
4 Zβ = 1/ 4 π a0
3/2
4 Zβ = 1/ 4 π a0
3/2
2 2
−β r / 2
∫
∞
0
2 Z 4π a0
r 2 dre −β
3/ 2
e − Zr/a 0
r / 2 − Zr/ a 0
2 2
e
d ∞ −β 2 r2 / 2 − Zr/ a 0 e −2 2 ∫0 dre dβ
The integral cannot be done in closed form, but it can be discussed for large and small a0 β . 12. It follows from 〈
d (p • r )〉 = 0 that 〈[ H, p • r ]〉 = 0 dt
Now p2 1 1 ∂V 2 [ pi pi + V (r ), x j p j ] = (−i ) p + ix j = −i − r • ∇V (r ) 2m m m ∂x j
As a consequence 2
〈
p 〉 = 〈r • ∇V (r )〉 m
If 2
V ( r) = −
Ze 4πε 0 r
then 2
Ze 〈r • ∇V ( r )〉 = 〈 〉 4πε 0 r
so that 2
1 Ze 1 〈T 〉 = 〈 〉 = − V ( r )〉 2 4πε 0 r 2 13. The radial equation is
d 2 2 d 2m 1 l(l + 1) 2 2 2 R(r) + 2 E − mω r − R(r) = 0 + dr 2 r dr 2mr 2 2 With a change of variables to ρ =
mω r and with E = λω / 2 this becomes
d 2 2 d R( ρ) + λ − ρ 2 − l(l +2 1) R( ρ) = 0 2 + dρ ρ dρ ρ We can easily check that the large ρ behavior is e The function H(ρ) defined by R(ρ ) = ρ e l
−ρ 2 / 2
−ρ 2 / 2
and the small ρ behavior is ρl.
H (ρ )
obeys the equation
l +1 d 2 H( ρ) dH (ρ) + 2 − ρ + (λ − 3− 2l)H (ρ ) = 0 2 dρ dρ ρ Another change of variables to y = ρ2 yields
d 2 H(y) l + 3 / 2 dH (y) λ − 2l − 3 + −1 + H (y) = 0 dy y dy 2 4y This is the same as Eq. (827), if we make the replacement
2l → 2l + 3 / 2
λ − 1→
λ − 2l − 3 4
This leads to the result that
λ = 4nr + 2l + 3 or, equivalently E = ω(2nr + l + 3 / 2) (b)
While the solution is La ( y ) with a = nr and b = (2l + 3)/4
CHAPTER 9
A+ =
1. With
0 1 0 0 0
0 0 2 0 0
0 0 0 3 0
0 0 0 0 4
0 0 0 0 0
we have
(A + )2 =
0 1 0 0 0
0 0 2 0 0
0 0 0 3 0
0 0 0 0 4
0 0 0 0 0
0 1 0 0 0
0 0 2 0 0
0 0 2 0 0
0 0 0 3 0
0 0 0 0 4
0 0 0 0 0
0 0 2 0 0
0 0 0 6 0
0 0 0 3 0
0 0 0 0 4
0 0 0 = 0 0
0 0 0 0 0
0 0 0 0 0 = 0 0 3.2.1 0 0
0 0 2 0 0
0 0 0 6 0
0 0 0 0 12
0 0 0 0 0
0 0 0 0 0
0 0 0 0 4.3.2
0 0 0 0 0
0 0 0 0 0
It follows that
(A + )3 =
0 1 0 0 0
0 0 0 0 12
The next step is obvious: In the 5 x 5 format, there is only one entry in the bottom leftmost corner, and it is 4.3.2.1. 2. [The reference should be to Eq. (636) instead of Eq. (64) + x= (A + A ) = 2mω 2mω from which it follows that
0 1 0
1 0 2
0
0
0
0
0 2 0 3 0
0 0 3 0 4
0 0 0 4 0
0 0 0 0 0
2 x = 2mω
1
0
2.1
0
0
3
0
3.2
2.1
0
5
0
0
3,2
0
7
0
0
4.3
0
0 0 4.3 0 9
− 1
0
0
0
− 2
0
0
2
0
− 3
0
0
0
0
3. The procedure here is exactly the same. We have mω + mω p= i (A − A) = i 2 2
0 1
3
0
0
4
0 0 0 − 4 0
from which it follows that mω 2 − p = 2
1
0
− 2.1
0
0
3
0
− 3.2
2.1
0
5
0
0
− 3.2
0
7
0
0
− 4.3
0
4. We have
u 1 = A+ u 0 =
0 1 0 0 0
0 0 2 0 0
0 0 0 3 0
0 0 0 0 4
0 0 0 0 0
0 1 1 0 0 = 0 0 0 0 0
0 − 4.3 0 9 0
5. We write
1 u2 = (A + )2 u0 = 2!
0 0 2 0 0
0 0 0 6 0
0 0 0 0 12
0 1 0 0 0 0 0 0 = 1 0 0 0 0 0 0
0 0 0 0 0
Similarly
0 0 1 u3 = (A + )3 u0 = 0 3! 3.2.1 0
0 1 0 0 0 0 0 0 = 0 0 0 1 0 0 0
0 0 0 0 4.3.2
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0 0 0 0 = 0 0 0 0 0 0 1
and
0 0 1 0 u4 = (A + ) 4 u0 = 4! 0 4.3.2.1
0 0 0 0 0
The pattern is clear. un is represented by a column vector with all zeros, except a 1 in the (n+1)th place.
6. (a)
0 0 1 / 2 0 1 0 3/2 0 0 1 1 2 3 〈H 〉 = (1 2 1 0)ω = ω 0 5 /2 0 6 1 2 6 0 0 0 0 7 / 2 0 1 (b) 〈x 2 〉 = (1 2 1 0) 2m ω 6
1
0
0
3
0
0
5
0
0
2 0
2
0 1 6 1 2 2 = (3 + ) 3 0 6 1 2mω 0 7
1 〈x〉 = (1 2 1 0) 2m ω 6
0
1
1
0
0
0
0
1 1 mω 0 2 〈p 〉 = (1 2 1 0) 2 − 2 6 0
1 mω 〈p〉 = (1 2 1 0)i 2 6
2
2
0
0 1 0 1 2 2 = (1+ 2) ω3 1 2m 3 6 0 0
0
3 0
− 2
3
0
0
5
− 6
0
− 1
0
0
− 2
0
2
0
0
0
0 1
3
0 1 − 6 1 2 mω 2 = (3 − ) 2 3 0 6 1 0 7
0 1 2 0 1 =0 − 3 6 1 0 0
(c) We get (∆x) = 2
2 2 5 2 (1− );(∆p) = (3 − ) 2m ω 3 3 2m ω 3
(∆x)(∆p) = 2.23
7. Consider −3 19 / 4e − iπ /3
19 / 4e iπ / 3 u1 u1 u = λ u 2 6 2
Suppose we choose u1=1. The equations then lead to (λ + 3) + 19 / 4e iπ / 3 u2 = 0 19 / 4e
−iπ /3
+ (6 − λ )u2 = 0
(a) Dividing one equation by the other leads to (λ + 3)(λ  6) =  19/4
The roots of this equation are λ = 7/2 and λ = 13/2. The values of u 2 corresponding to the two eigenvalues are 1 −iπ /3 −iπ /3 e ;u2 (13 / 2) = − 19e 19
u2 (−7 / 2) = (b) The normalized eigenvectors are 1 19 ; 20 −e −iπ / 3
iπ /3 1 e 20 19
It is easy to see that these are orthogonal. (c) The matrix that diagonalizes the original matrix is, according to Eq. (955)
U=
1 − 19e iπ /3 1 20 19e − iπ / 3 1
It is easy to check that
0 13 / 2 U + AU = 0 −7 / 2 8. We have, as a result of problem 7, A = UAdiagU
+
From this we get e = Ue A
A diag
e13/ 2 0 + U = U −7/2 U 0 e +
The rest is rather trivial matrix multiplication.
9, The solution of
1 1 1 1
1 1 1 1
1 1 1 1
1 a a b 1 b λ = c 1 c 1 d d
is equivalent to solving a + b + c + d = λa = λb = λc = λd
1 1 1 One solution is clearly a = b = c = d with λ = 4. The eigenvector is 2 1 1 We next observe that if any two (or more) of a , b , c , d are not equal, then λ = 0. These are the only possibilities, so that we have three eigenvalues all equal to zero. The Eigenvectors must satisfy a + b + c + d = 0, and they all must be mutually orthogonal. The following choices will work
1 1 −1 ; 2 0 0
0 1 1 0 11 ; 2 1 2 −1 −1 −1
10.An hermitian matrix A can always be diagonalized by a particular unitary matrix U, such that +
UAU = Adiag Let us now take traces on both sides: TrUAU = TrU UA = TrA while TrAdiag = ∑ an +
+
n
Where the an are the eigenvalues of A.
1 1 1 1 ... 1 1 1 1 ... 11. The product of two N x N matrices of the form M = 1 1 1 1 ... is 1 1 1 1 ... . . . . ... N N N N ... N N N N ... N N N N ... . Thus M2 = N M . This means that the eigenvalues can only be N N N N ... . . . . ... N or zero. Now the sum of the eigenvalues is the trqice of M which is N (see problem 10). Thus there is one eigenvalue N and (N –1) eigenvalues 0. 1 0 1 1 12. We found that the matrix U = i 0 −i has the property that 2 0 2 0 + M 3 = U(Lz / )U . We may now calculate
M1 ≡ U (Lx / )U + = 1 1 1 = i 2 2 0
0 0 2
1 0 1 0 1 −i −i 1 0 1 0 0 0 0 1 0 1 i
0 0 0 1 2 = 0 0 0 0 1 0 0
and +
M 2 ≡ U ( Ly / )U = 1 1 1 i = 2 2 0
0 0 2
1 0 −i 0 1 −i −i i 0 −i 0 0 0 0 i 0 1 i
We can easily check that
0 0 0 0 2 = 0 0 1 0 0 1 0
0 0 1 0 0 0 0 0 00 0 1 [M1, M2 ] = 0 0 0 0 0 1 − 0 0 10 0 0 = 1 0 0 0 1 0 0 1 01 0 0 0 1 0 = − 1 0 0 = iM 3 0 0 0
This was to be expected. The set M1, M2 and M3 give us another representation of angular momentum matrices.
13. We have AB = BA. Now let U be a unitary matrix that diagonalizes A. In our case we have the additional condition that in
UAU + = Adiag
a1 0 0 a2 = 0 0 0 0
0 0 a3 0
0 0 0 a4
all the diagonal elements are different. (We wrote this out for a 4 x 4 matrix) Consider now
U[A,B]U + = UAU +UBU + − UBU +UAU + = 0 This reads as follows (for a 4 x 4 matrix)
a1 0 0 a2 0 0 0 0
0 0 a3 0
0 b11 0 b21 0 b31 a4 b41
b12 b13 b22 b23 b32 b33 b42 b43
b14 b11 b24 b21 = b34 b31 b44 b41
b12 b13 b22 b23 b32 b33 b42 b43
b14 a1 0 b24 0 a2 b34 0 0 b44 0 0
0 0 a3 0
0 0 0 a4
If we look at the (12) matrix elements of the two products, we get, for example a1b12 = a2b12
and since we require that the eigenvalues are all different, we find that b12 = 0. This argument extends to all offdiagonal elements in the products, so that the only matrix elements in UBU= are the diagonal elements bii.
14. If M and M+ commute, so do the hermitian matrices (M + M+) and i(M – M+). Suppose we find the matrix U that diagonalizes (M + M+). Then that same matrix will diagonalize i( M – M+), provided that the eigenvalues of M + M+ are all
different. This then shows that the same matrix U diagonalizes both M and M+ separately. (The problem is not really solved, till we learn how to deal with the situation when the eigenvalues of A in problem 13 are not all different).
CHAPTER 10 1. We need to solve
0 −i u u = ± 2 i 0 v 2 v
1 1 2 i The – eigenstate can be obtained by noting that it must be orthogonal to the + state, and 1 1 this leads to χ − = . 2 −i
For the + eigenvalue we have u = iv , s that the normalized eigenstate is χ + =
2, We note that the matrix has the form
σ z cos α + σ x sinα cosβ + σ y sin α sin β ≡ σ • n n = (sinα cos β ,sinα sin β ,cosα ) This implies that the eigenvalues must be ± 1. We can now solve cosα sin αe iβ
u sin αe − iβ u = ± v − cos α v
For the + eigenvalue we have u cosα + v sinα eIβ = u. We may rewrite this in the form
α
α
2v sin cos e 2 2
−iβ
= 2usin
2
α 2
From this we get
α cos 2 χ+ = α e iβ sin 2 The – eigenstate can be obtained in a similar way, or we may use the requirement of orthogonality, which directly leads to
α −iβ e sin 2 χ− = α − cos 2
α cos 2 The matrix U = α iβ e sin 2
e −iβ sin
α
2 − cos 2
α
has the property that 1 0 sin αe − iβ U = 0 −1 − cosα
cosα + U sin αe iβ as is easily checked. The construction is quite simple.
0 0 3 /2 0 0 1/ 2 0 0 Sz = 0 −1/ 2 0 0 0 0 0 −3 / 2 To construct S+ we use (S+ ) mn = δ m ,n +1 (l − m + 1)(l + m) and get
0 0 S+ = 0 0
3
0
0
2
0
0
0
0
0 0 3 0
We can easily construct S = (S+)+. We can use these to construct 1 Sx = (S+ + S− ) = 2 2
and
0 3
3
0
0
2
0
2
0
0
0
3
0 0 3 0
0 −i 3 0 i i 3 Sy = (S− − S+ ) = 2 2 0 2i 0 0
0 −2i 0 0 −i 3 i 3 0 0
The eigenstates in the above representation are very simple:
1 0 0 0 0 1 0 0 χ 3/ 2 = ; χ 1/ 2 = ; χ −1/2 = ; χ −3/2 = 0 0 1 0 0 0 0 1
5. We first need the eigenstates of (3Sx + 4Sy)/5. The eigenvalues will be ± / 2 since the operator is of the form S • n, where n is a unit vector (3/5,4/5,0).The equation to be solved is 3 4 ( σ x + σ y ) χ± = ± χ± 2 5 5 2
In paricular we want the eigenstate for the –ve eigenvalue, that is, we want to solve
0 3 + 4i 5
3− 4i 5 u = − u v 0 v
This is equivalent to (34i) v = 5u A normalized state is
1 3 − 4i . 50 −5
The required probability is the square of
3 − 4i 1 (2 1) 1 −5 = 5 50
1 (6 − 8i − 5) = 250
1 (1− 8i) 250
This number is 65/250 = 13/50.
6. The normalized eigenspinor of Sy corresponding to the negative eigenvalue was found 1 1 . The answer is thus the square of in problem 1. It is 2 −i
1 (4 65
7)
1 1 1 = (4 − 7i) 2 −i 130
which is 65/130 = 1/2.
7. We make use of σ xσ y = iσ z =  σ yσ x and so on, as well as σ x = 1and so on, to work out 2
(σ x Ax + σ y Ay + σ z Az )(σ x Bx + σ y By + σ z Bz ) = Ax Bx + Ay By + AzBz + iσ z (Ax By − Ay Bx ) + iσ y (AzBx − Ax Bz ) + iσ x (Ay Bz − AzBy ) = A • B + iσ • A × B
8. We may use the material in Eq. (1026,27)., so that at time T, we start with 1 e ψ (T ) = iωT 2 e − iωT
with ω = egB / 4me . This now serves as an initial state for a spin 1/2 particle placed in a magnetic field pointing in the y direction. The equation for ψ is according to Eq. (1023)
i
0 −i dψ (t) ψ (t) = ω dt i 0
a(t) we get da = −ωb; db = ωa . The solutions are in general Thus with ψ (t) = dt dt b(t) a(t) = a(T )cosω (t − T ) − b(T )sinω (t − T ) b(t) = b(T )cosω (t − T ) + a(T )sinω (t − T )
e − iωT e iωT We know that a(T ) = ;b(T ) = 2 2 So that cosωT − e sinω T 1 e ψ (2T ) = iω T − iωT 2 e cosω T + e sin ωT −iωT
iωT
The amplitude that a measurement of Sx yields / 2 is
−iω T cosω T − e iωT sin ωT 1 1 e = (1 1) 2 2 e iωT cos ωT + e − iωT sin ωT
= (cos 2 ωT − isin 2 ωT ) 1 4 4 2 Thus the probability is P = cos ω T + sin ωT = (1 + cos 2ωT ) 2 A + Bz α β equal to A + σ • B = 9. If we set an arbitrary matrix γ δ Bx + iBy
Bx − iB y A − Bz
we see that allowing A, Bx … to be complex we can match all of the α,β,…. (b) If the matrix M = A + σ • B is to be unitary, then we require that (A + σ • B)(A * + σ • B*) =  A 2 +A σ • B* + A * σ • B + B• B* +iσ • B × B* = 1 which can be satisfied if
 A 2 +  Bx 2 +  By 2 +  Bz 2 = 1 ABx * + A * Bx + i(By Bz *− By * Bz ) = 0 ......... If the matrix M is to be hermitian, we must require that A and all the components of B be real. 10. Here we make use of the fact that (σ • a)(σ • a) = a • a ≡ a in the expansion 2
i2 i3 i4 (σ • a) 2 + (σ • a) 3 + (σ • a) 4 + ... 2! 3! 4! 3 1 1 a = 1− a 2 + (a 2 ) 2 + ...+ i σ • aˆ (a − + ...) 2! 3! 4! sina = cosa + iσ • aˆ sina = cosa + iσ • a a
e iσ • a = 1 + i(σ • a) +
11. We begin with the relation 2
2 2 S = σ 1 + σ 2 = (σ 1 + σ 2 + 2σ 1 • σ 2 ) 2 2 4 2
2
from which we obtain σ1 • σ 2 = 2 S( S + 1) − 3 . This = 3 for a singlet and +1 for a triplet state. We now choose eˆ to point in the z direction, so that the first term in S12 is equal to 3σ 1 zσ 2 z . (a) for a singlet state the two spins are always in opposite directions so that the first term is –6 and the second is +3. Thus S12 X singlet = 0 (b) For a triplet the first term is +1 when Sz = 1 and Sz = 1 and –1 when Sz =0. This means that S12 acting on a triplet state in the first case is 31= 2, and in the second case it is –31=  4. Thus (S12 − 2)(S12 + 4)X triplet = 0
12. The potential may be written in the form
V ( r) = V1 ( r ) + V2 ( r) S12 + V3 ( r )[2S( S + 1) − 3] For a singlet state S12 has expectation value zero, so that V(r) = V1(r) – 3V3(r) For the triplet state S12 has a value that depends on the z component of the total spin. What may be relevant for a potential energy is an average, assuming that the two particles have equal probability of being in any one of the three Sz states. In that case the average value of Sz is (2+24)/3= 0 1 (1) (2) (1) (2) ( χ + χ − − χ − χ + ), if one of the 2 electrons is in the “up” state, the other must be in the “down” state. 13. (a) It is clear that for the singlet state, ψ singlet =
(b). Suppose that we denote the eigenstates of Sy by ξ± . These are, as worked out in problem 1,
ξ(1) + =
1 1 (1) 1 1 ;ξ− = 2 i 2 −i
The spinors for particle (1) may be expanded in terms of the ξ± thus:
1 0
1 1 1 1 1 1 (ξ + ξ− ) + = 2 2 i 2 + 2 −i
0 1
i 1 1 1 1 i (ξ − ξ− ) − = 2 2 i 2 + 2 −i
χ+ = = χ− = =
Similarly, for particle (2), we want to expand the spinors in terms of the η± , the eigenstates of Sx
η(2) + =
1 1 1 1 (2) ; η− = 2 1 2 −1
1 0
1 1 1 1 1 1 (η + η − ) + = 2 + 2 2 1 2 −1
0 1
1 1 1 1 1 1 (η − η − ) − = 2 + 2 2 1 2 −1
thus
χ+ = = χ− = =
We now pick out, in the expansion of the singlet wave function the coefficient of ξ+ η+ and take its absolute square.Some simple algebra shows that it is
(1) (2)

1 1 1− i 2 1 = 4 2 2 2 iβ
iβ
9. The state is (cosα 1 χ + + sin α1e 1 χ − )(cos α 2 χ + + sinα 2 e 2 χ − ) . We need to calculate the scalar product of this with the three triplet wave functions of the twoelectgron system. It is easier to calculate the probability that the state is found in a singlet state, and then subtract that from unity. The calculation is simple (1)
(1)
(2)
(2)
1 (1) (2) (1) (2) (1) iβ (1) (2) iβ (2) ( χ + χ − − χ − χ + )  (cosα 1 χ + + sin α1e 1 χ − )(cos α 2 χ + + sin α 2 e 2 χ − )〉 2 1 (cosα1 sin α 2 e iβ2 − sinα 1e iβ1 cos α 2 ) = 2
〈
The absolute square of this is the singlet probability. It is 1 2 2 2 2 Ps = (cos α 1 sin α 2 + cos α 2 sin α 1 + 2sin α1 cos α1 sinα 1 cosα 2 cos(β1 − cos β 2 )) 2 and Pt = 1  Ps
14. We use J = L +S so that J2 = L2 + S2 + 2L.S, from which we get L •S =
1 [J (J + 1) − L(L + 1) − 2] 2 2
since S = 1. Note that we have taken the division by into account. For J = L + 1 this takes on the value L; for J = L, it takes on the value –1, and for J = L – 1 it is –L – 1. We therefore find J = L + 1:
V = V1 + LV2 + L V3
J=L
V = V1 − V2 + V3
J = L −1
2
V = V1 − (L + 1)V2 + (L + 1)2 V3
CHAPTER 11 1. The first order contribution is 2
+ + E = λ 〈n  x  n〉 = λ 〈n  (A + A )(A + A )  n〉 2m ω (1) n
2
To calculate the matrix element 〈n  A 2 + AA + + A + A + (A + ) 2  n〉 we note that
A +  n〉 = n + 1  n + 1〉; 〈n  A = n + 1〈n + 1  so that (1) the first and last terms give zero, and the second and third terms yield (n + 1) + (n – 1)=2n. Thus the first order shift is
En = λ (1)
n mω
The second order calculation is quite complicated. What is involved is the calculation of
En = λ (2)
2
2m ω
2
+ 2
+ 2
〈n  (A + A )  m〉〈m  (A + A )  n〉 ∑ ω(n − m) m ≠n
This is manageable but quite messy. The suggestion is to write 2
1 p 2 2 2 + mω x + λx H= 2m 2 This is just a simple harmonic oscillator with frequency
λ 1 λ2 ω* = ω + 2λ / m = ω + − + ... ωm 2 ω 3 m2 2
Whose spectrum is 1 1 1 1 λ λ E n = ω * (n + ) = ω (n + ) + (n + ) − 3 2 (n + ) + ... 2 2 ωm 2 2ω m 2 2
The extra factor of 1/2 that goes with each n is the zeropoint energy. We are only interested in the change in energy of a given state n> and thus subtract the zeropoint energy to each order of λ. Note that the first order λ calculation is correct.
2. The eigenfunction of the rotator are the spherical harmonics. The first order energy shift for l = 1 states is given by
2π
π
0
0
∆E = 〈1,m  E cosθ 1,m〉 = E ∫ dφ ∫ sinθdθ cos θ  Y1.m 
2
For m = ±1, this becomes π
2π E ∫ sinθ dθ cosθ 0
3E 3 2 sin θ = 8π 4
∫
1
−1
duu(1 − u ) = 0 2
The integral for m = 0 is also zero. This result should have been anticipated. The eigenstates of L2 are also eigenstates of parity. The perturbation cosθ is odd under the reflection r  r and therefore the expectation value of an odd operator will always be zero. Since the perturbation represents the interaction with an electric field, our result states that a symmetric rotator does not have a permanent electric dipole moment. The second order shift is more complicated. What needs to be evaluated is  〈1, m  cosθ  L, M〉  ∑ E1 − E L L ,M (L ≠1)
2
∆E
(2)
=E
2
2
L( L + 1) . The calculation is simplified by the fact that only L = 0 and L = 2 2I terms contribute. This can easily be seen from the table of spherical harmonics. For L =1 we saw that the matrix element vanishes. For the higher values we see that cosθ Y1,±1 ∝ Y2,±1 and cosθ Y1,0 ∝ aY2,0 + bY0,0 . The orthogonality of the spherical harmonics for different values of L takes care of the matter. Note that because of the φ integration, for m = ±1 only the L = 2 ,M = ± 1 term contributes, while for the m = 0 term, there will be contributions from L = 0 and L = 2, M = 0. Some simple integrations lead to with E L =
2
∆E
(2) m =±1
2
2 IE 1 2 IE 1 (2) =− 2 ; ∆E m = 0 = − 2 15 60
3. To lowest order in V0 the shift is given by 2
2 V0 L 2 nπx ∆E = dxx sin ∫ L L L 0 2V L = 20 L π
2
∫
π
0
duusin2 nu =
V0
π
2
∫
π 0
1 duu(1− cos2nu) = V0 2
The result that the energy shift is just the value of the perturbation at the midpoint is perhaps not surprising, given that the square of the eigenfunctions do not, on the average, favor one side of the potential over the other.
λ
0 0 E 0 0 consists of two boxes which can be diagonalized 0 2E σ 0 σ 0 u E λ u separately. The upper left hand box involves solving = η v λ E v The result is that the eigenvalues are η = E ± λ . The corresponding eigenstates are easily 1 1 worked out and are for the two cases. 2 ±1 2E σ a a For the lower left hand box we have to solve = ξ . Here we find that the σ 0 b b E λ 4. The matrix 0 0
eigenvalues are ξ = E ± E + σ . The corresponding eigenstates are 2
2
σ 1 2 2 2 2 N E +σ ) . 2 2 respectively, with 2 = σ + (− E ± N − E ± E + σ 5. The change in potential energy is given by 3e 2 2 1 2 e2 V1 = − r + 3 R − 3 4πε 0 r 8πε0 R =0 elsewhere
r≤ R
Thus R
∆E = ∫ d r ψ nl (r )V1ψ nl (r ) = ∫ r drV1 Rnl (r) 3
*
2
2
0
We may now calculate this for various states. Z n = 1 ∆E10 = 4 a0
3
∫
R 0
2 2 3e 2 1 2 e r 2 dre −2Zr /a 0 − R − r + 3 4πε 0 r 8πε 0 R 3
With a change of variables to x = r/Za0 and with ρ = ZR/a0 this becomes
Ze2 ρ 2 3 x2 1 −2 x x ∆E10 = 4 dx − + + e ∫ 3 x 4πε 0 a0 0 2 ρ 2ρ Since x << 1 we may approximate e results is
−2x
≈ 1− 2x , which simplifies the integrals. What
Ze2 4 2 ∆E10 = ρ + ... 4πε 0 a0 10 A similar calculation yields
1 Ze2 ρ 2 3 x2 1 − x Ze 2 1 2 2 x dx(1− x) − ∆E 20 = + 3 + e ≈ ρ + ... ∫ 0 2 4πε 0 a0 x 2 ρ 2ρ 4πε 0 a0 20 and ∆E 21 =
1 Ze2 ρ 2 3 x 2 1 −x Ze2 1 2 4 x dxx − + 2ρ 2 ρ 3 + x e ≈ 4πε a 1120 ρ + ... 24 4πε 0 a0 ∫0 0 0
6. We need to calculate λ 〈0  x  0〉. One way of proceeding is to use the expression 4
x=
(A + A + ) 2mω
Then 2
+ + + + 〈0  (A + A )(A + A )(A + A )(A + A )  0〉 2m ω
λ 〈0  x 4  0〉 = λ The matrix element is
〈0  (A + A + )(A + A + )(A + A + )(A + A + )  0〉 = +
+
+
+
〈0  A (A + A )(A + A )A  0〉 = 〈1 (A + A + )(A + A + ) 1〉 =
[〈0  +
][
]
2〈2   0〉 + 2  2〉 = 3
Thus the energy shift is ∆E = 3λ
2mω
2
It is easy to see that the same result is obtained from mω 1/4 − mωx 2 /2 ∫−∞ dx(λx ) π e ∞
4
2
7. The first order perturbation shift is
2ε b π x nπx ∆E n = dx sin sin ∫ 0 b b b 2ε π = ∫0 dusinu(sin nu)2
2
π
=
2ε 1 1+ 2 π 4n − 1
8. It follows from [p, x] = −i that −i = 〈a  px − xp  a〉 = = ∑ {〈a  p  n〉〈n  x  a〉 − 〈a  x  n〉〈n  p  a〉} n
Now 〈a  p  n〉 = m〈a 
dx im im  n〉 = 〈a  Hx − xH  n〉 = (E a − E n )〈a  x  n〉 dt
Consequently 〈n  p  a〉 = 〈a  p  n〉* = −
im (E a − E n )〈n  x  a〉
Thus −i = ∑ n
2im (E a − E n )〈a  x  n〉〈n  x  a〉
from which it follows that 2
∑ (E n − E a )  〈a  x  n〉  = 2m n 2
9. For the harmonic oscillator, with a> = 0>, we have
〈n  x  0〉 =
〈n  A +  0〉 = 2mω
δ 2mω n,1
This means that in the sum rule, the left hand side is
2 = ω 2mω 2m
as expected.
10. For the n = 3 Stark effect, we need to consider the following states:
l = 2 : ml = 2,1,0,1,2 l = 1 : ml = 1,0,1 l = 0 : ml = 0 In calculating matrix element of z we have selection rules ∆ l = 1 (parity forbids ∆ l = 0) and, since we are dealing with z, also ∆ ml = 0. Thus the possible matrix elements that enter are 〈2,1 z 1,1〉 = 〈2,−1 z 1,−1〉 ≡ A
〈2,0  z 1,0〉 ≡ B 〈1,0  z  0,0〉 ≡ C The matrix to be diagonalized is
0 A 0 0 0 0 0
A 0 0 0 0 0 0
0 0 0 0 0 0 0 B 0 B 0 C 0 C 0 0 0 0 0 0 0
0 0 0 0 0 0 A
0 0 0 0 0 A 0
The columns and rows are labeled by (2,1),(1,1) (2,0) (1,0),(0,0),(2,1), (1,1). The problem therefore separates into three different matrices. The eigenvalues of the submatrices that couple the (2,1) and (1,1) states, as well as those that couple the (2,1) and (1,1) states are
λ=±A where ∞
A = ∫ dΩY21 cos θY11 ∫ r drR32 (r)rR31(r) *
0
2
0 B 0 The mixing among the ml = 0 states involves the matrix B 0 C 0 C 0 Whose eigenvalues are λ = 0, ± B + C .. Here 2
2
∞
B = ∫ dΩY20* cos θY10 ∫ r 2 dr R32 (r)rR31 (r) 0
∞
C = ∫ dΩY cosθY00 ∫ r drR31 (r)rR30 (r) * 10
2
0
The eigenstates of the A submatrices are those of σ x , that is
1 1 . The eigenstates of 2 ±1
the central 3 x 3 matrix are
C B 1 1 ± B2 + C 2 0 ; B 2 + C 2 2(B 2 + C 2 ) C −B with the first one corresponding to the λ = 0 eigenvalue.
11. For a onedimensional operator (labeled by the x variable) we introduced the raising and lowering operators A+ and A. We were able to write the Hamiltonian in the form 1 + H x = ω (A A + ) 2 We now do the same thing for the harmonic oscillator labeled by the y variable. The raising and lowering operators will be denoted by B+ and B, with 1 + H y = ω (B B + ) 2 The eigenstates of H x + H y are
(A + ) n (B + ) m  m,n〉 =  0,0〉 n! m! where the ground state has the property that A 0,0> = B 0.0> = 0 The perturbation may be written in the form
H1 = 2λ xy =
λ + + (A + A )(B + B ) mω
(a) The first order shift of the ground state is
〈0,0  H1  0,0〉 = 0 since every single of the operators A,…B+ has zero expectation value in the ground state. (b) Consider the two degenerate states 1,0> and 0,1>. The matrix elements of interest to us are <1,0(A+A+)(B + B+)1,0> = <0,1(A+A+)(B + B+)0,1> = 0 <1,0(A+A+)(B + B+)0,1> = <0,1(A+A+)(B + B+)1,0> = <1,0(A+A+)(B + B+)1,0> = 1 Thus in degenerate perturbation theory we must diagonalize the matrix
0 h where h =
h 0
λ . The eigenvalues are ±h , and the degenerate levels are split to mω E = ω (1±
λ ) mω 2
(c) The second order expression is
λ mω −
2
 〈0,0  (A + A + )(B + B + )  k,n〉 2 = ∑ − ω (k + n) k ,n
λ2  〈1,1 k,n〉 2 λ2 = − mω 3 ∑ (k + n) 2mω 3 k ,n
The exact solution to this problem may be found by working with the potential at a classical level. The potential energy is 1 2 2 2 mω (x + y ) + λ xy 2 Let us carry out a rotation in the x – y plane. The kinetic energy does not change since p2 is unchanged under rotations. If we let x = x'cosθ + y'sinθ y = − x'sinθ + y'cosθ
then the potential energy, after a little rearrangement, takes the form 1 1 2 2 2 2 ( m ω − λ sin2θ )x' +( mω + λ sin2θ )y' +2λ cos2θ x' y' 2 2 If we choose cos2θ = 0, so that sin2θ = 1, this reduces to two decoupled harmonic oscillators. The energy is the sum of the two energies. Since 1 1 mω x2 = mω 2 − λ 2 2 1 1 mω y2 = mω 2 + λ 2 2 the total energy for an arbitrary excited state is 1 1 E k,n = ω x (k + ) + ω y (n + ) 2 2 where λ λ 2 ωx = ω (1− 2λ / mω ) = ω − − + ... m ω 2m 2ω 3 2 λ λ 2 1/2 ωy = ω (1 + 2λ / m ω ) = ω + − + ... m ω 2m 2ω 3 2 1/ 2
12. Thespectral line corresponds to the transition (n = 4,l = 3) (n = 3,l = 2). We must therefore examine what happens to these energy levels under the perturbation H1 =
e L•B 2m
eB L. 2m z In the absence of the perturbation the initial state is (2l + 1) = 7fold degenerate, with the Lz level unchanged, and the others moved up and down in intervals of eB/2m. The final state is 5fold degenerate, and the same splitting occurs, with the same intervals. If transitions with zero or ±1 change in Lz/ , the lines are as shown in the figure on the right.
We define the z axis by the direction of B , so that the perturbation is
What will be the effect of a constant electric field parallel to B? The additional perturbation is therefore
H2 = −eE0 • r = −eE 0 z and we are only interested in what this does to the energy level structure. The perturbation acts as in the Stark effect. The effect of H1 is to mix up levels that are degenerate, corresponding to a given ml value with different values of l. For example, the l = 3, ml = 2 and the l = 2, ml = 2 degeneracy (for n = 4)will be split. There will be a further breakdown of degeneracy. 13. The eigenstates of the unperturbed Hamiltonian are eigenstates of σ z . They are
1 0 corresponding to E = E0 and corresponding to E =  E0. 0 1 The first order shifts are given by
α
u 1 = λα β 0
α
u 0 = λβ β 1
(1 0)λ u * (0 1)λ u *
for the two energy levels. The second order shift for the upper state involves summing over intermediate states that differ from the initial state. Thus, for the upper state, the intermediate state is just the lower one, and the energy denominator is E0 – ( E0) = 2E0. Thus the second order shift is
α u 0 α u1 λ 2  u 2 (1 0)u * β 1(0 1) u* β0 = 2E 2E 0 0
λ2
For the lower state we get
α u 1 α u 0 λ2  u 2 (0 1) u* β0(1 0)u * β 1 = − 2E −2E 0 0
λ2
The exact eigenvalues can be obtained from
det This leads to
E0 + α − ε u =0 u* − E0 + β − ε
ε=λ =λ
α +β 2
α +β 2
± (E 0 − λ ± (E 0 − λ
α −β 2
) +λ u 2
2
2
α−β
1 λ 2  u 2 )(1+ + ... 2 2 2 E0
(b) Consider now
E0 u H = v − E0 where we have dropped the α and β terms. The eigenvalues are easy to determine, and they are
ε = ± E 02 + λ2 uv a The eigenstates are written as and they satisfy b E 0 v
u a a = ± E 02 + λ2 uv − E 0 b b
For the upper state we find that the unnormalized eigenstate is
λu 2 2 E 0 + λ uv − E 0 For the lower state it is
− λu 2 2 E 0 + λ uv + E 0 The scalar product − λ  u  +[( E 0 + λ uv ) − E 0 ]= λ u( u * −v ) ≠ 0 2
2
2
2
2
2
which shows that the eigenstates are not orthogonal unless v = u*.
CHAPTER 12. 1. With a potential of the form V(r) =
1 2 2 mω r 2
the perturbation reduces to
1 1 dV (r) ω 2 2 2 H1 = = 2 2 S• L 2 (J − L − S ) 4mc 2m c r dr 2 (ω ) = 2 ( j( j + 1) − l(l + 1) − s(s + 1)) 4mc 2
where l is the orbital angular momentum, s is the spin of the particle in the well (e.g. 1/2 for an electron or a nucleon) and j is the total angular momentum. The possible values of j are l + s, l + s – 1, l + s –2, …l – s. 3 The unperturbed energy spectrum is given by E n r l = ω (2n r + l + ). Each of the 2 levels characterized by l is (2l + 1)fold degenerate, but there is an additional degeneracy, not unlike that appearing in hydrogen. For example nr =2, l = 0. nr =1, l = 2 , nr = 0, l = 4 all have the same energy. A picture of the levels and their spinorbit splitting is given below.
2. The effects that enter into the energy levels corresponding to n = 2, are (I) the basic Coulomb interaction, (ii) relativistic and spinorbit effects, and (iii) the hyperfine structure which we are instructed to ignore. Thus, in the absence of a magnetic field, the levels under the influence of the Coulomb potential consist of 2n2 = 8 degenerate levels. Two of the levels are associated with l = 0 (spin up and spin down) and six
levels with l = 0, corresponding to ml = 1,0,1, spin up and spin down. The latter can be rearranged into states characterized by J2, L2 and Jz. There are two levels characterized by j = l –1/2 = 1/2 and four levels with j = l + 1/2 = 3/2. These energies are split by relativistic effects and spinorbit coupling, as given in Eq. (1216). We ignore reduced mass effects (other than in the original Coulomb energies). We therefore have
1 1 1 3 ∆E = − mec 2α 4 3 − n j + 1/ 2 4n 2 1 2 4 5 = − me c α 64 2
j =1 /2
1 1 = − me c 2α 4 64 2
j = 3/2
(b) The Zeeman splittings for a given j are
∆E B =
e B 2 mj 2me 3
j =1 /2
=
eB 4 mj 2me 3
j = 3/2
1 eB 2 4 −5 −5 me c α ≈ 1.132 × 10 eV , while for B = 2.5T = 14.47 × 10 eV , 128 2m e so under these circumstances the magnetic effects are a factor of 13 larger than the relativistic effects. Under these circumstances one could neglect these and use Eq. (1226).
Numerically
3. The unperturbed Hamiltonian is given by Eq. (1234) and the magnetic field interacts both with the spin of the electron and the spin of the proton. This leads to
H=A
Sz I S• I +b z 2 +a
Here A=
m I•S 4 4 α me c 2 gP e 2 3 MP
a= 2
eB 2m e
b = −gP
eB 2M p
Let us now introduce the total spin F = S + I. It follows that
1 2 3 2 3 2 S• I − 2 = 2 F(F + 1) − 2 4 4 1 for F = 1 = 4 3 =− for F = 0 4 We next need to calculate the matrix elements of aSz + bI z for eigenstates of F2 and Fz . These will be exactly like the spin triplet and spin singlet eigenstates. These are
〈1,1 aSz + bI z 1,1〉 = 〈 χ +ξ+  aSz + bIz  χ +ξ+ 〉 =
1 (a + b) 2
2
1 〈1,0  aSz + bI z 1,0〉 = 〈 χ +ξ− + χ −ξ+  aSz + bI z  χ +ξ− + χ +ξ− 〉 = 0 2 1 〈1,−1  aSz + bIz 1,−1〉 = 〈 χ −ξ−  aSz + bIz  χ −ξ− 〉 = − (a + b) 2 And for the singlet state (F = 0) 2
1 1 〈1,0  aSz + bI z  0,0〉 = 〈 χ +ξ− + χ −ξ+  aSz + bI z  χ +ξ− − χ +ξ −〉 = (a − b) 2 2 2
1 〈0,0  aSz + bI z  0,0〉 = 〈 χ +ξ− − χ −ξ+  aSz + bIz  χ +ξ− − χ +ξ− 〉 = 0 2
Thus the magnetic field introduces mixing between the 1,0> state and the 0.0> state. We must therefore diagonalize the submatrix
(a − b) / 2 (a − b) / 2 − A / 4 0 A /2 A/4 = + (a − b) / 2 −3A / 4 0 − A / 4 (a − b) / 2 −A /2 The second submatrix commutes with the first one. Its eigenvalues are easily determined to be ± A / 4 + (a − b) / 4 so that the overall eigenvalues are 2
2
−A /4 ±
A / 4 + (a − b) / 4 2
2
Thus the spectrum consists of the following states:
F = 1, Fz = 1
E = A / 4 + (a + b) / 2
F =1, Fz = 1
E = A / 4 − (a + b) / 2
F = 1,0; Fz = 0
E = −A / 4 ± (A / 4 + (a − b) / 4 2
2
We can now put in numbers. For B = 104 T, the values, in units of 106 eV are 1.451, 1.439, 0(1010), 2.89 For B = 1 T, the values in units of 106 eV are 57.21,54.32, 54.29 and 7 x 106.
4. According to Eq. (1217) the energies of hydrogenlike states, including relativistic + spinorbit contributions is given by En, j = −
2 2 1 1 m e c (Zα ) 1 1 3 2 4 1 m ec (Zα ) 3 − 2 − n j + 1/ 2 4n 2 (1+ m e / M p ) n 2
The wavelength in a transition between two states is given by
λ=
2π c ∆E
where ∆E is the change in energy in the transition. We now consider the transitions n =3, j = 3/2 n = 1, j = 1/2 and n = 3 , j = 1/2 n = 1, j = 1/2.. The corresponding energy differences (neglecting the reduced mass effect) is 1 1 1 1 2 2 2 4 1 m ec (Zα ) (1− ) + m e c (Z α ) (1− ) 2 9 2 4 27
(3,3/21,1/2)
∆E =
(3,1/21,1/2)
1 1 1 3 2 2 2 4 1 me c (Zα ) (1− ) + m e c (Z α ) (1− ) 2 9 2 4 27
We can write these in the form (3,3/21,1/2)
∆E 0 (1 +
13 2 (Zα ) ) 48
(3,1/21,1/2)
∆E 0 (1 +
1 2 (Zα ) ) 18
where ∆E 0 =
1 2 2 8 m ec (Zα ) 2 9
The corresponding wavelengths are
(3,3/21,1/2)
λ 0 (1−
13 2 −9 (Zα ) ) = 588.995 × 10 m 48
(3,1/21,1/2)
λ 0 (1−
1 2 −9 (Zα ) ) = 589.592 × 10 m 18
We may use the two equations to calculate λ0 and Z. Dividing one equation by the other we get, after a little arithmetic Z = 11.5, which fits with the Z = 11 for Sodium. (Note that if we take for λ0 the average of the two wavelengths, then , using λ 0 = 2π c / ∆E 0 = 9π / 2mc(Zα )2 , we get a seemingly unreasonably small value of Z = 0.4! This is not surprising. The ionization potential for sodium is 5.1 eV instead of Z2(13.6 eV), for reasons that will be discussed in Chapter 14) 2
2 1 p 4. The relativistic correction to the kinetic energy term is − . The energy 2mc 2 2m shift in the ground state is therefore 2
p2 1 1 1 〈0  m ω 2 r 2 ) 2  0〉 ∆E = −  0〉 = − 2 2 〈0  (H − 2mc 2 2m 2mc To calculate < 0  r2  0 > and < 0  r4  0 > we need the ground state wave function. We know that for the onedimensional oscillator it is 1/4
mω − mωx 2 / 2 u0 (x) = e π so that for the three dimensional oscillator it is u0 (r) = u0 (x)u0 (y)u0 (z) =
mω π
3/ 4
e
It follows that 〈0  r  0〉 = ∫ 2
∞ 0
mω 4π r dr π
= 4π =
3/2
2
mω π
3 2m ω
3/ 2
mω
2 − mω r 2 /
re 5/2
∫
∞ 0
=
4 −y 2
dyy e
− mω r 2 / 2
We can also calculate
〈0  r  0〉 = ∫ 4
∞ 0
= 4π
mω 4πr dr π mω π
15 = 4 mω We made use of
∫
∞
0
3/ 2
2
3/ 2
mω
4 − mω r 2 /
re 7/2
∫
∞ 0
=
6 − y2
dyy e
2
n −z
dzz e = Γ(n + 1) = nΓ(n)
and
1 Γ( ) = π 2
Thus ∆E = −
2 2 1 2 4 15 1 3 3 2 3 m + ω − ω ω m ω 4 m ω 2mω 4 2mc 2 2 2
=−
15 (ω ) 2 2 32 mc
6. (a) With J = 1 and S = 1, the possible values of the orbital angular momentum, such that j = L + S, L + S –1…L – S can only be L = 0,1,2. Thus the possible states are 3 3 3 S1 , P1, D1 . The parity of the deuteron is (1)L assuming that the intrinsic parities of the proton and neutron are taken to be +1. Thus the S and D states have positive parity and the P state has opposite parity. Given parity conservation, the only possible 3 admixture can be the D1 state. (b)The interaction with a magnetic field consists of three contributions: the interaction of the spins of the proton and neutron with the magnetic field, and the L.B term, if L is not zero. We write H = −M p • B − M n • B − M L • B
Mp =
eg p e S p S p = (5.5792) 2M 2M
egn e S n Sn = (−3.8206) 2M 2M e ML = L 2M red
where M n =
We take the neutron and proton masses equal (= M ) and the reduced mass of the two3 particle system for equal masses is M/2. For the S1 stgate, the last term does not contribute.
If we choose B to define the z axis, then the energy shift is
−
S pz eB 3 S 〈 S1  gp + gn nz 3 S1 〉 2M
We write
S pz S g p + gn S pz + Snz g p − gn S pz − Snz gp + gn nz = + 2 2 It is easy to check that the last term has zero matrix elements in the triplet states, so Sz 1 that we are left with (g p + gn ) , where Sz is the zcomponent of the total spin.. 2 Hence 〈 S1  H1  S1 〉 = − 3
3
3B g p + gn ms 2M 2
where ms is the magnetic quantum number (ms = 1,0,1) for the total spin. We may therefore write the magnetic moment of the deuteron as e g p + gn e µeff = − S = −(0.8793) S 2M 2 2M The experimental measurements correspond to gd = 0.8574 which suggests a small 3 admixture of the D1 to the deuteron wave function.
1
CHAPTER 13 me −4 = (1− 5.45 × 10 )me 1 + me / M p me −4 = (1− 2.722 × 10 )m e (b) electrondeuteron system mr = 1 + me / M d m (c) For two identical particles of mass m, we have mr = 2
1. (a) electronproton system mr =
2. One way to see that P12 is hermitian, is to note that the eigenvalues ±1 are both real. Another way is to consider
∑ ∫ dx dx ψ
* ij
(x1, x 2 )P12ψ ij (x1, x2 ) =
∑ ∫ dx dx ψ
* ij
(x1, x 2 )ψ ji (x 2 , x1 ) =
∑ ∫ dy dy ψ
* ji
(y 2 ,y1 )ψ ij (y1 ,y 2 ) = ∑ ∫ dy1dy 2 (P12ψ ij (y1 ,y 2 )) *ψ ij (y1 ,y 2 )
1
2
i, j
1
2
i, j
1
j,i
2
j,i
3. If the two electrons are in the same spin state, then the spatial wave function must be antisymmetric. One of the electrons can be in the ground state, corresponding to n = 1, but the other must be in the next lowest energy state, corresponding to n = 2. The wave function will be 1 ψ ground (x1, x 2 ) = (u1(x1)u2(x 2) − u2 (x1)u1 (x 2 )) 2 2π 2 2 2 n ≡ εn 2ma 2 Only two electrons can go into a particular level, so that with N electrons, the lowest N/2 levels must be filled. The energy thus is
4. The energy for the nth level is E n =
N /2
E tot = ∑ 2εn ≈ 2ε 2
n =1
3
1 N εN = 3 2 12
3
If N is odd, then the above is uncertain by a factor of εN which differs from the above by (12/N )ε, a small number if N is very large. 2
5. The problem is one of two electrons interacting with each other. The form of the interaction is a square well potential. The reduction of the twobody problem to a oneparticle system is straightforward. With the notation
2
x1 + x 2 ; P = p1 + p2 , the wave function has the form 2 ψ (x1, x 2 ) = e iPX u(x), where u(x) is a solution of x = x1 − x 2 ; X =
2 d 2 u(x) − + V (x)u(x) = Eu(x) m dx 2 Note that we have taken into account the fact that the reduced mass is m/2. The spatial interchange of the two electrons corresponds to the exchange x x .Let us denote the lowest bound state wave function by u0(x) and the next lowest one by u1(x). We know that the lowest state has even parity, that means, it is even under the above interchange, while the next lowest state is odd under the interchange. Hence, for the two electrons in a spin singlet state, the spatial symmetry must be even, and therefor the state is u0(x), while for the spin triplet states, the spatial wave function is odd, that is, u1(x). 1 1 6. With P = p1 + p2 ; p = ( p1 − p2 ); X = (x1 + x 2 ); x = x1 − x 2 , the Hamiltonian 2 2 becomes 2
H=
2
1 p 1 P 2 2 2 2 + Mω X + + µω x 2M 2 2µ 2
with M = 2m the total mass of the system, and µ = m/2 the reduced mass. The energy spectrum is the sum of the energies of the oscillator describing the motion of the center of mass, and that describing the relative motion. Both are characterized by the same angular frequency ω so that the energy is 1 1 E = ω (N + ) + ω (n + ) = ω (N + n + 1) ≡ ω (ν + 1) 2 2 The degeneracy is given by the number of ways the integer ν can be written as the sum of two nonnegative integers. Thus, for a given ν we can have (N,n) = (ν ,0),(ν − 1,1),(ν − 2,2),...(1,ν − 1).(0,ν ) so that the degeneracy is ν + 1. Note that if we treat the system as two independent harmonic oscillators characterized by the same frequency, then the energy takes the form 1 1 E = ω (n1 + ) + ω ( n2 + ) = ω ( n1 + n 2 + 1) ≡ ω (ν + 1) 2 2 which is the same result, as expected.
3
7. When the electrons are in the same spin state, the spatial twoelectron wave function must be antisymmetric under the interchange of the electrons. Since the two electrons do not interact, the wave function will be a product of the form 1 ( un ( x1 ) uk ( x 2 ) − uk ( x1 )un ( x 2 )) 2
π 2 2 (n + k ) . The lowest state corresponds to n = 1, 2ma 2 k = 2, with n2 + k2 = 5 . The first excited state would normally be the (2,2) state, but this is not antisymmetric, so that we must choose (1,3) for the quantum numbers. 2
2
with energy E = E n + E k =
8. The antisymmetric wave function is of the form
π − µ (x − a) /2 −µ (x + a) / 2 − µ (x + a) /2 −µ (x − a) / 2 e e −e e µ2 π = N 2 e −µ a e − µ (x + x )/ 2 e − µ (x − x )a − e − µ (x − x )a µ N
(
2
2
2
2
1
2
2
2
2
2
2 1
2 2
2
2
2
1
(
2
2
2
2
1
1
2
)
)
Let us introduce the center of mass variable X and the separation x by
x x x1 = X + ; x 2 = X − 2 2 The wave function then becomes
ψ = 2N
π − µ a −µ e e µ2 2 2
2
X2
e
−µ 2 x 2 / 4
sinh µ ax 2
To normalize, we require
∫
∞
−∞
∞
dX ∫ dx  ψ  = 1 2
−∞
Some algebra leads to the result that
N
π 1 1 2 = µ 2 1− e −2 µ
2 2
a
The second factor is present because of the overlap. If we want this to be within 1 part in −2( µa )2 ≈ 1/ 500 , i.e. µa = 1.76, or a = 0.353 a 1000 away from 1, then we require that e 2 2 2 4 mc mc 3 d nm R fi = (Zα ) 2 . π a0 2∆E .
4
9. Since e
ψ= 2
−( µa) 2
1− e
e
−2(µa )
2
−( µX )2
e
−( µx )2 / 4
sinh µ ax 2
the probability density for x is obtained by integrating the square of ψ over all X. This is a simple Gaussian integral, and it leads to
2e −2( µa ) P(x)dx = 2 1− e −2( µa) 2
π 1 −( µx ) /2 e sinh2 (µ 2 ax)dx 2 µ 2
It is obvious that ∞
〈X〉 = ∫ dXXe
−2( µX ) 2
=0
−∞
since the integrand os an odd function of X.
10. If we denote µx by y, then the relevant quantities in the plot are e −y e
−y 2 / 2
2
/2
2
sinh 2y and
2
sinh (y / 2).
11. Suppose that the particles are bosons. Spin is irrelevant, and the wave function for the two particles is symmetric. The changes are minimal. The wave function is
ψ = 2N
π − µ a −µ e e µ2 2 2
2
X2
e
−µ 2 x 2 / 4
cosh µ xa 2
with
N
1 µ2 1 = π 2 1+ e −2 µ
2
a2
and
π − µ 2 x 2 /2 2e −2 µ a P(x) = e cosh2 (µ 2 ax) −2 µ 2 a 2 µ 2 1+e 2 2
The relevant form is now P(y) = e
− y 2 /2
cosh κy which peaks at y = 0 and has extrema at 2
5
− y coshκy + 2κ sinhκy = 0 , that is, when tanhκy = y / 2κ which only happens if 2κ > 1. Presumably, when the two centers are close together, then the peak occurs in between; if they are far apart, there is a slight rise in the middle, but most of the time the particles are around their centers at ± a. 2
12. The calculation is almost unchanged. The energy is given by E = pc =
πc 2 2 2 n1 + n 2 + n3 L
so that in Eq. (1358)
R 2 = n12 + n 22 + n32 = (E F / cπ ) 2 L2 Thus
π EFL N= 3 π c
3
and 1/3
3n E F = πc π
13. The number of triplets of positive integers {n1,n2,n3} such that n1 + n2 + n3 = R = 2
2
2
2
2mE 2 L 2π 2
is equal to the numbers of points that lie on an octant of a sphere of radius R, within a 1 2 thickness of ∆n = 1. We therefore need 4π R dR . To translate this into E we use 8 2 2 2 2RdR = (2mL / π )dE . Hence the degeneracy of states is N(E)dE = 2 ×
1 3 m 2m 4πR(RdR) = L 8 3π 2
E dE
To get the electron density we had to multiply by 2 to take into account that there are two electrons per state.
14. Since the photons are massless, and there are two photon states per energy state, this problem is identical to problem 12. We thus get
6
2
E 2 n +n + n = R = L π c 2 1
2 2
2 3
2
or R = EL / π c . Hence 2
E 1 2 3 N(E)dE = 4πR dE = L 3 3 2 dE 8 cπ
15. The eigenfunctions for a particle in a box of sides L1,L2, L3 are of the form of a product
u(x,y,z) =
8 n πx n πy n πz sin 1 sin 2 sin 3 L1 L2 L3 L1L2 L3
and the energy for a massless partticle, for which E = pc is n n n n n +n E = cπ 12 + 22 + 32 = cπ 1 2 2 + 32 L1 L2 L 3 a L 2
2
2
2
2
2
Note that a << L . thus the lowlying states will have n1 = n2 = 1, with n3 ranging from 1 upwards. At some point the two levels n1 = 2, n2=1 and n1=1 and n2 = 2 will provide a new “platform” upon which n3 = 1,2,3,… are stacked. With a = 1 nm and L = 103 nm, for n1 = n2 = 1 the n3 values can go up to 103 before the new platform starts.
16. For nonrelativistic particles we have
2 n12 + n 22 n32 E= + 2 2m a2 L 17. We have π 3n EF = 2M π 2
2
2/ 3
where M is the nucleon mass, taken to be the same for protons and for neutrons, and 4π 3 r A , the number where n is the number density. Since there are Z protons in a volume 3 0 densities for protons and neutrons are np =
3 Z 3 A− Z ; nn = 3 4π r0 A 4π r03 A
7
Putting in numbers, we get E Fp
Z = 65 A
2/ 3
Z MeV ; E Fn = 65 1− A
2/3
MeV
For A = 208, Z = 82 these numbers become E Fp = 35MeV; E Fn = 47MeV .
1
CHAPTER 14 1. The spinpart of the wave function is the triplet (2) χ (1) + χ+
ms = 1 ms = 0
1 (2) (χ (1) χ (2) + χ (1) − χ+ ) 2 + −
ms = −1
(2) χ (1) − χ−
This implies that the spatial part of the wave function must be antisymetric under the interchange of the coordinates of the two particles. For the lowest energy state, one of the electrons will be in an n = 1, l = 0 state. The other will be in an n = 2, l = 1, or l = 0 state. The possible states are
1 (u (r )u (r ) − u100 (r2)u21m (r1)) 2 100 1 21m 2 1 (u (r )u (r ) − u100 (r2)u200 (r1 )) 2 100 1 200 2
m = 1,0,−1
Thus the total number of states with energy E2 + E1 is 3 x 4 = 12 2. For the triplet state, the first order perturbation energy shifts are given by 1
∆E 21m = ∫ ∫ d 3 r1 d 3 r2 
2
(u100 (r1)u21m (r2 ) − u100 (r2 )u21m (r1 ))2
e2 4πε 0  r1 − r2 
1 e2 2 ∆E 200 = ∫ ∫ d r1 d r2  (u100 (r1 )u200 (r2) − u100 (r2)u200 (r1 )) 4πε  r − r  2 0 1 2 The l = 1 energy shift uses twelectron wave functions that have an orbital angular momentum 1. There is no preferred direction in the problem, so that there cannot be any dependence on the eigenvalue of Lz. Thus all three m values have the same energy. The l = 0 energy shift uses different wave functions, and thus the degeneracy will be split. Instead of a 12fold degeneracy we will have a splitting into 9 + 3 states. The simplification of the energy shift integrals reduces to the simplification of the integrals in the second part of Eq. (1429). The working out of this is messy, and we only work out the l = 1 part. 3
The integrals
3
∫ ∫ d rd r 3
3
1
2
∞
∞
→ ∫ r1 dr1 ∫ r2 dr2 ∫ dΩ1 ∫ dΩ2 and the angular parts only come 0
2
2
0
through the u210 wave function and through the 1/r12 term. We use Eqs. (1426) – (1429) to get, for the direct integral
2
2
∞ ∞ e r12 dr1 ∫ r22 dr2 R10 (r1 )2 R21 (r2 )2 ∫ 0 4πε 0 0 2 1 3 dΩ dΩ ∫ 1∫ 2 4π 4π cosθ 2
2
∑ PL (cosθ12 ) L
rL +1
where θ12 is the angle between r1 and r2. We make use of an addition theorem which reads PL (cosθ12 ) = PL (cosθ1 )PL (cosθ 2 ) (L − m)! m rL PL (cosθ1 )PLm (cosθ 2 )cos mφ 2 L<+1 r> m =1 (L + m)!
+ 2∑
Since the sum is over m = 1,2,3,…the integration over φ2 eliminates the sum, and for all practical purposes we have
∑ PL (cos θ12 ) L
rL +1 r>L +1
The integration over dΩ1 yields 4πδL0 and in our integral we are left with ∫ dΩ 2 (cosθ 2) 2 = 4π / 3. The net effect is to replace the sum by 1 / r> to be inserted into the radial integral. (b) For the exchange integral has the following changes have to be made: In the radial integral, R10(r1 ) R21(r2 ) → R10 (r1 )R21(r1 )R10 (r2 )R21 (r2 ) 2
2
In the angular integral 1 3 3 2 (cosθ 2 ) → cosθ1 cosθ 2 4π 4π (4π )2 In the azimuthal integration again the m ≠ 0 terms disappear, and in the rest there is a product of two integrals of the form
∫ dΩ
3 cosθPL (cosθ ) = 4π
4π δ 3 L1
1 r< inserted into the radial integral. 3 r>2 For the l = 0 case the same procedure will work, leading to
The net effect is that the sum is replaced by
3
2
∞ ∞ e 1 2 2 r1 dr1 ∫ r2 dr2 [R10 (r1 )R20 (r2 )][R10 (r1 )R20 (r2 ) − R10 (r2 )R20(r1 )] ∫ 0 r> 4πε 0 0
The radial integrals are actually quite simple, but there are many terms and the calculation is tedious, without teaching us anything about physics. To estimate which of the(l = 0,l = 0) or the (l = 0, l = 1) antisymmetric combinations has a lower energy we approach the problem physically. In the twoelectron wave function, one of the electrons is in the n = 1, l = 0 state. The other electron is in an n = 2 state. Because of this, the wave function is pushed out somewhat. There is nevertheless some probability that the electron can get close to the nucleus. This probability is larger for the l = 0 state than for the l = 1 state. We thus expect that the state in which both electrons have zero orbital angular momentum is the lowerlying state.
3. In the ground state of orthohelium, both electroNs have zero orbital angular momentum. Thus the only contributions to the magnetic moment come from the electron spin. An electron interacts with the magnetic field according to H=−
ge ge ge s1 • B − s2 • B = − S• B 2m e 2m e 2m e
The value of g is 2, and thus coefficient of B takes on the values −
e m , where 2m e 1
m1 = 1,0,1. 4. We assume that ψ is properly normalized, and is of the form
 ψ 〉 = ψ 0 〉 + ε  χ 〉 The normalization condition implies that
〈ψ  ψ 〉 = 1 = 〈ψ 0  ψ 0 〉 + ε * 〈 χ  ψ 0 〉 + ε 〈ψ 0  χ 〉 + εε *〈 χ  χ 〉 so that
ε * 〈χ  ψ 0 〉 + ε 〈ψ 0  χ 〉 + εε *〈 χ  χ 〉 = 0 Now 〈ψ  H  ψ 〉 = 〈ψ 0 + εχ  H  ψ 0 + εχ 〉 = E 0 + ε * E 0 〈 χ  ψ 0 〉 + ε E 0 〈ψ 0  χ 〉+  ε  〈 χ  H  χ 〉 2
= E 0 +  ε 2 〈 χ  H − E 0  χ 〉
4
where use has been made of the normalization condition.Thus the expectation value of H differs from the exact value by terms of order ε2. 5. We need to calculate
E(α ) =
∫
∞ 0
−αr 2 d2 2 d 1 2 2 4π r 2 dre −α r − + mω r e 2+ r dr 2 2m dr
∫
With a little algebra, and using
0
∞
0
2
4πr 2 dre −2α r n −y
dyy e = n!, we end up with
3mω α + 2m 2α 2 2
E(α ) =
∫
∞
2
This takes its minimum value when dE(α ) / dα = 0 . This is easily worked out, and leads 2 to α = 3mω / . When this is substituted into E(α) we get
E min =
3ω
The true ground state energy is bound to lie below this value. The true value is
3 ω so 2
that our result is pretty good.
4. The Schrodinger equation for a bound state in an attractive potential, with l = 0 reads
r 2 d2 2 d − ψ (r)− V0  f ( )ψ (r) = − E Bψ (r) 2+ r dr r0 2m dr With the notation x = r/r0 , u0(x) = x ψ(x), λ = 2m  V0  r0 / ; 2
2
α 2 = 2mEB r02 / 2 this
becomes 2
d u0 (x) 2 − α u0 (x) + λf (x)u0 (x) = 0 dx 2 Consider, now an arbitrary function w(x) which satisfies w(0)= 0 (like u0(0)) , and define
η[w] =
∫
∞ 0
dw(x) 2 dx + α 2 w 2 (x) dx
∫
∞ 0
dxf (x)w 2 (x)
5
We are asked to prove that if η = λ + δλ and w (x) = u0(x) + δ u(x) , then as δ u(x) 0, δλ 0. We work to first order in δu(x) only. Then the right hand side of the above equation, written in abbreviated form becomes
∫ (u '
2
0
+α u0 ) + 2∫ (u0 'δu' +α u0δu) 2
2
∫
f (u0 + 2u0δu)
2
2
∫ (u ' +α u ) − 2 ∫ (u 'δu' +α = ∫ fu ∫ fu 2
2 2 0
0
2
0
2 0
=
u0δu) ∫ (u0 ' 2 +α 2 u02 )
2 0
∫ fu
2 0
In the above, the first term is just η [u0], and it is easy to show that this is just λ. The same form appears in the second term. For the first factor in the second term we use d
∫ dx (u 'δu')= ∫ dx dx (u ' δu)− ∫ dxu '' δu 0
0
0
The first term on the right vanishes because the eigenfunction vanishes at infinity and because δu(0) = 0. Thus the second term in the equation for η [w] becomes 2
∫ fu
2 0
∫ δu[−u '' +α 0
2
u0 − λfu0 ]
Thus η → λ as δu 0. 5. We want to minimize 〈ψ  H  ψ 〉 = ∑ ai H ij a j subject to the condition that *
i, j
〈ψ  ψ 〉 = ∑ a a = 1 . The method of Lagrange multipliers instructs us to minimize * i i
i
F (ai , ai ) = ∑ ai H ij a j − λ ∑ ai ai *
*
*
ij
i
The condition is that ∂F / ∂ai = 0 . The condition implies that *
∑H
ij
a j = λ ai
j
Similarly ∂F / ∂ai = 0 implies that
∑a H * i
= λa j
*
ij
i
Thus the minimization condition yields solutions of an eigenvalue equation for H.
6
6. Consider the expectation value of H evaluated with the normalized trial wave function 1/ 2
2 2 β ψ (x) = e − β x / 2 π
Then an evaluation of the expectation value of H yields, after some algebra, ∞ 2 d 2 E(β ) = ∫ dxψ *(x) − + V (x)ψ (x) 2 −∞ 2m dx
=
β π
β β + 2m π 2
=
2 2 β 4 2 − β 2x 2 dx ∫−∞ 2m (β − β x )e + π ∞
2
∫
∞ −∞
dxV (x)e
∫
∞
dxV (x)e −∞
− β 2x 2
−β 2 x 2
The question is: can we find a value of β such that this is negative. If so, then the true value of the ground state energy will necessarily be more negative. We are given the fact that the potential is attractive, that is, V(x) is never positive. We write V(x) =  V(x) and ask whether we can find a value of β such that
β π
∫
∞ −∞
dx  V (x)  e
−β 2 x 2
2 β 2 > 2m
For any given V(x) we can always find a square “barrier” that is contained in the positive form of V(x). If the height of that barrier is V0 and it extends from –a to +a , for example, then the left side of the above equation is always larger than L(β ) =
a β −β V0 ∫ dxe − a π
2 2
x
Our question becomes: Can we find a β such that 4m 2 2 L(β ) > β It is clear that for small β such that β2a2 << 1, the left hand side is approximated by β 4mV0 2a . This is linear in β so that we can always find a β small enough so that the π 2 left hand side is larger than the right hand side. 7. The data indicates a resonance corresponding to a wavelength of 20.61 nm. This corresponds to an energy of
7
2π (1.054 × 10 −34 J.s)(3 × 108 m / s) h = = 60.17eV λ (20.61× 10 −9 m)(1.602 × 10−19 J / eV ) c
above the ground state. The ground state has energy – 78.98 eV, while the ground state of He+ has a binding energy of a hydrogenlike atom with Z = 2, that is, 54.42 eV. This means that the ionization energy of He is (78.9854.42)eV = 24.55 eV above the ground state. Thus when the (2s)(2p) state decays into He+ and an electron, the electron has an energy of (60.17 – 24.55)eV = 35.62 eV. This translates into 6 v = 2E / m = 3.54 × 10 m / s . The first excited state of the He+ ion lies 54.42(11/4)=40.82 eV above the ground state of He+ , and this is above the (2s)(2p) state. 8. To calculate the minimum of
E(α1 ,α 2 ,...) =
〈ψ (α1 ,α 2 ,...)  H  ψ (α1,α 2 ,...)〉 〈ψ (α 1,α 2 ,...)  ψ (α 1,α 2 ,...)〉
we set ∂E / ∂αi = 0, i = 1,2,3.... This implies that
〈
∂ψ ∂ψ ∂ψ ∂ψ  ψ 〉 + 〈ψ  〉  H  ψ 〉 + 〈ψ  H  〉 〈ψ  H  ψ 〉 〈 ∂α i ∂α i ∂α i ∂α i − =0 〈ψ  ψ 〉 2 〈ψ  ψ 〉
This is equivalent to
〈
∂ψ ∂ψ  H  ψ 〉 + 〈ψ  H  〉= ∂α i ∂α i
∂ψ ∂ψ E(α1 ,α 2 ,..) 〈  ψ 〉 + 〈ψ  〉 ∂α i ∂α i Let us now assume that H depends on some parameter λ.. To calculate the minimum we must choose our parameters α i to depend on λi. We may rewrite the starting equation by emphasizing the dependence of everything on λ, as follows
E(λ )〈ψ (λ )  ψ (λ )〉 = 〈ψ (λ )  H  ψ (λ )〉 Let us now differentiate with respect to λ , noting that We get
∂ ∂α ∂ =∑ i ∂λ i ∂λ ∂α i
8
dE(λ ) ∂α ∂ψ ∂ψ (λ ) 〈ψ (λ )  ψ (λ )〉 + E(λ )∑ i 〈  ψ (λ )〉 + 〈ψ (λ )  〉 = dλ ∂α i i ∂λ ∂α i = 〈ψ (λ ) 
∂H ∂α ∂ψ ∂ψ (λ )  ψ (λ )〉 + ∑ i 〈  H  ψ (λ )〉 + 〈ψ (λ )  H  〉 ∂λ ∂α i i ∂λ ∂α i
Since we have shown that
〈
∂ψ ∂ψ  H  ψ 〉 + 〈ψ  H  〉= ∂α i ∂α i
∂ψ ∂ψ E(α1 ,α 2 ,..) 〈  ψ 〉 + 〈ψ  〉 ∂α i ∂α i we obtain the result that dE(λ ) ∂H 〈ψ (λ )  ψ (λ )〉 = 〈ψ (λ )  ψ (λ )〉 dλ ∂λ With normalized trial wave functions we end up with dE(λ ) ∂H = 〈ψ (α1 ,α 2 ,..)   ψ (α1 ,α 2 ,..)〉 dλ ∂λ A comment: The Pauli theorem in Supplement 8A has the same form, but it deals with exact eigenvalues and exact wave functions. Here we find that the same form applies to approximate values of the eigenvalue and eigenfunctions, provided that these are chosen to depend on parameters {α} which minimize the expectation value of the Hamiltonian (which does not depend on these parameters).
9. With the trial wave function 1/ 2
2 2 β ψ (x) = e − β x / 2 π
we can calculate
9
β E(β ) = π =
β π
−∞
dxe
∞
∫−∞ dxe
− β 2 x 2 /2
−β 2 x 2
2 d 2 22 + λx 4 e −β x /2 − 2 2m dx
2 2 4 2 4 (β − β x ) + λ x 2m
3λ β + 4 2m 4β 2
=
∫
∞
2
1/3
We minimize this by setting ∂E / ∂β = 0 , which leads to β = 2
6m λ . When this is 2
inserted into the expression for E, we get 2/3
2/ 3
2 1/ 3 2 (4 λ )1/ 3 6 + 3 12/ 3 = 1.083 λ1/ 3 E min = 46 2m 4 2m This is quite close to the exact value, for which the coefficient is 1.060 10. With the Hamiltonian 2
p 4 H= + λx 2m we first choose (1/2m) as the parameter in the FeynmanHellmann theorem. This leads to 〈0  p  0〉 = 2
∂E min 4 1/3 = 0.890( mλ ) ∂(1 / m)
If we choose λ as the parameter, then 2 ∂E min 〈0  x  0〉 = = 0.353 ∂λ 2mλ
2/3
4
11. We start from
〈ψ  H  ψ 〉 = E0 ≤ 〈ψ  ψ 〉
∑a H a ∑a a * i
ij
j
ij
* i i
i
We now choose for the trial vector one in which all the entries are zero, except that at the kth position there is 1, so that ai = δ ik . This leads to
10
E 0 ≤ H kk
(k is not summed over)
We may choose k = 1,2,3,…Thus the lowest eigenvalue is always smaller than the the smallest of the diagonal elements. 12. With the system’s center of mass at rest, the twobody problem reduces to a onebody problem, whose Hamiltonian is 2
H=
1 p 2 2 + µω r 2µ 2
where µ is the reduced mass, whose value is m/2. (a) The two particles are in an l = 0 state which means that the ground state wave function only depends on r, which is symmetric under the interchange of the two particles (Recall that r = r1 − r2  ). Thus the electrons must be in a spinsinglet state, and the ground state wave function is
ψ (r ) = u0 (r)X singlet where 3/4
u0 (r) = u0 (x)u0 (y)u0 (z) =
µω − µωr2 / 2 e π
(We use u0(x) from Eq. (655)). (b) To proceed with this we actually have to know something about the solutions of the simple harmonic oscillator in three dimensions. The solution of this was required by Problem 13 in Chapter 8. We recall that the solutions are very similar to the hydrogen atom problem. There are two quantum numbers, nr and l. Here l = 0, so that the first excited singlet state must correspond to nr = 1. In the spin triplet state, the spinwave function is symmetric, so that the spatial wave function must be antisymmetric. This is not possible with l = 0! To actually obtain the wave function for the first excited singlet state, we look at the equation for H(ρ), with H(ρ) of the form a + bρ2. Since 2
d H 1 dH − ρ) + 4H = 0 2 + 2( ρ dρ dρ We get H (ρ)= 12ρ2/3 and the solution is 2 2 − ρ 2 /2 u1 (r) = N (1− ρ )e 3
11
1/2
µω where ρ = r . The normalization constant is obtained from the requirement that N
2
∫
∞
0
2
r dr(1−
2µω 2 2 −µωr2 / r ) e =1 3
so that 6 µω N = π
3/2
2
(c) The energy shift to lowest order is ∞ 2µω 2 2 −µ ωr 2 / δ (r) 2 2 2 ∆E = ∫ r dr C 2 N (1− r ) e = CN 0 r 3
13. The energy is given by
J (J + 1) 1 2 2 E = Mred ω (R − R0 ) + 2Mred R 2 2 2
If we treat the vibrational potential classically, then the lowest state of energy is characterized by R = R0. The vibrational motion changes the separation of the nuclei in the molecule. The new equilibrium point is given by R1 , which is determined by the solution of
J(J + 1) 2 ∂E 2 ∂R R = 0 = Mred ω (R1 − R0 ) − M red R13 1 Let R1 = R0 + ∆. Then to first order in ∆, J(J + 1) ∆= 2 2 3 Mred ω R0 2
If we now insert the new value of R1 into the energy equation, we find that only the rotational energy is changed (since the vibrational part is proportional to ∆2). The rotational energy is now E rot =
J(J + 1)2 2Mred R02 (1 + 2∆ / R0 )
J(J + 1) 2 = 2 − (J(J + 1)) 3 2M red R0 M red ω 2 R06 2
4
12
The sign of the second term is negative. The sign is dictated by the fact that the rotation stretches the molecule and effectively increases its moment of inertia. 14. In the transition J = 1 J = 0 we have
2π c 2 (2 − 0) = 2M red R λ 2
∆E = so that R2 =
λ 1 λ 1 1 1 = = + 2π c M red 2π c Mnucleon 12 16
= (1.127 × 10−10 m)2 The internuclear separation is therefore 0.113 nm, and the momentu of inertia is M red R = 1.45 × 10 2
− 46
kg.m
2
15. (a) The two nuclei are identical. Since the twoelectron state is a spatially symmetric spin 0 state, we can ignore the electrons in discussing the lowest energy states of the molecule. In the ground state, the two protons will be in the symmetric L = 0 state, so that they must be in a spinantisymmetric S = 0 state. For the spinsymmetric S = 1 state, the spatial wave function must be antisymmetric, so that the lowest energy state will have L = 1.
(b) The lowest energy state that lies above the ground state of L = 0, and is also a spin S = 0 state must have L = 2. Thus the change in energy in the transition is
6 2 2π c 2 ∆E = 2 (2(2 + 1) − 0) = 2 = M pR M pR λs We have used the fact that the reduced mass of the twoproton system is Mp/2. For the S = 1 system, the state above the lowest L = 1 state is the L = 3 state, and here
∆E =
10 2 2π c 2 ( 3(3 + 1) − 1(1+ 1) ) = 2 2 = M pR M pR λt
The singlet and triplet wavelengths are easily calculated once we know R. Note that these are not exactly the same, but can be looked up.
1
CHAPTER 15 1. With the perturbing potential given, we get C(1s → 2 p) =
∞ eE 0 iωt − γt 〈φ 210  z  φ100 〉 ∫ dte e 0 i
where ω = (E21 – E10). The integral yields 1 / (γ − iω ) so that the absolute square of C(1s2p) is  〈φ 210  z  φ100 〉  P(1s → 2 p) = e E 2 (ω 2 + γ 2 )
2
2
2 0
15
We may use  〈φ 210  z  φ100 〉  = 2
2 2 a to complete the calculation. 310 0
2. Here we need to calculate the absolute square of 1 T iω 21t 2 a 2πx a πx dt e sinωt × λ ∫ dx sin ( x − )sin ∫ 0 0 i a a 2 a Let us first consider the time integral. We will assume that at t = 0 the system starts in the ground state. The time integral then becomes
∫
∞
0
dte
iω 21 t
sin ωt =
1 ∞ ω i(ω +ω )t i(ω −ω )t dt{e 21 − e 21 } = 2 ∫ 2i 0 ω − ω 221
We have used the fact that an finitely rapidly oscillating function is zero on the average. In the special case that ω matches the transition frequency, one must deal with this integral in a more delicate manner. We shall exclude this possibility. The spatial integral involves
2 a 2π x π x a dx sin sin (x − ) = ∫ 0 a a a 2 a 3π x a 1 πx cos − cos (x − ) ∫ 0 2 a a a =
1 a d a 3π x 3πx πx a πx a a a dx sin − sin sin − sin (x − ) − ∫ a 0 dx π a 3π a a 3π a 2 π a
1 a2 3π x a 8 πx a 2 = 2 cos − 2 cos = −2 2 π 9 a π a 9π a 0 The probability is therefore
2
2
2
ω λ 16a P12 = 2 2 9π (ω 21 − ω 2 ) 2 2
(b) The transition from the n = 1 state to the n = 3 state is zero. The reason is that the eigenfunctions for all the odd values of n are all symmetric about x = a/2, while the potential (x – a/2) is antisymmetric about that axis, so that the integral vanishes. In fact, quite generally all transition probabilities (even even) and (odd odd) vanish. (c) The probability goes to zero as ω 0.
3. The only change occurs in the absolute square of the time integral. The relevant one is
∫
∞
−∞
dte
iω 21 t − t 2 /τ
e
2
= πe
−ω 2τ 2 / 4
which has to be squared. When τ ∞ this vanishes, showing that the transition rate vanishes for a very slowly varying perturbation.
4. The transition amplitude is Cn→ m =
λ i
〈m 
= − iλ
∞ (A + A + )  n 〉∫ dte iω ( m −n )te −αt cosω1 t 0 2Mω
1 α − iω (m − n) δ m,n −1 n + δ m,n +1 n + 1 2Mω (α − iω (m − n)) 2 + ω12
(
)
(a) Transitions are only allowed for m = n ± 1. (b) The absolute square of the amplitude is, taking into account that (m – n)2 = 1,
λ2 2Mω
(nδ m,n −1 + (n + 1)δm ,n+1)
α 2 + ω2 (α 2 + ω12 − ω 2 )2 + 4α 2ω 2
When ω1 ω, nothing special happens, except that the probability appears to exceed unity when α2 gets to be small enough. This is not possible physically, and what this suggests is that when the external frequency ω 1 matches the oscillator frequency, we get a resonance condition as α approaches zero. Under those circumstances first order perturbation theory is not applicable. When α 0, then we get a frequency dependence similar to that in problem 2.
5. The two particles have equal and opposite momenta, so that
3
E i = ( pc)2 + mi2 c 4 The integral becomes ∞ 1 2 2 6 ∫ dΩ∫0 p dpδ (Mc − E1 ( p) − E 2 (p)) (2π )
and it is only the second integral that is of interest to us. Let us change variables to u = E1(p) + E2(p) then 2
du =
2
pdp pc pc dp + dp = (E1 + E 2 ) E1 E2 E1 E 2
and the momentum integral is
∫
∞
0
p 2 dpδ(Mc 2 − E1 ( p) − E 2 ( p)) = ∫
=p
∞
(m1 + m2 )c 2
p
E1 E 2 du δ (Mc 2 − u) 2 uc
E1 E 2 4 Mc
To complete the expression we need to express p in terms of the masses. We have (m 2 c 2 )2 + p2 c 2 = (Mc 2 − (m1c 2 ) 2 + p 2c 2 )2 = (Mc 2 ) 2 − 2Mc 2 E1 ( p) + (m1c 2 ) 2 + p 2c 2 This yields (Mc ) + (m1c ) − (m 2 c ) 2Mc 2 2 2
E1 (p) =
2 2
2 2
and in the same way (Mc ) + (m2 c ) − (m1c ) E 2 ( p) = 2Mc 2 2 2
2 2
2 2
By squaring both sides of either of these we may find an expression for p2. The result of a short algebraic manipulation yields
4
2
c p = (M − m1 − m 2 )(M − m1 + m 2 )(M + m1 − m 2 )(M + m1 + m2 ) 4M 2 2
6. The wave function of a system subject to the perturbing potential
λ V(t) = V f(t) where f(0) = 0 and Limf (t) = 1, with df(t)/dt << ω f(t), is given by t→ ∞
 ψ (t)〉 = ∑ Cm (t)e
−iE 0m t /
φ m 〉
m
and to lowest order in V, we have Cm (t) =
1 t iωt' dt'e f (t')〈φ m  V  φ0 〉 i ∫0
where ω = (E m − E 0 ) / and at time t = 0 the system is in the ground state. The time integral is 0
∫
t
0
dt'e
iωt '
0
t
f (t') = ∫ dt' f (t') 0
iωt'
d e 1 = dt' iω iω
t
d
∫ dt' dt' (e 0
iωt '
f (t')) −
1 t iω t' dt'e df (t') / dt' i ω ∫0
The second term is much smaller than the term we are trying to evaluate, so that we are iω t left with the first term. Using f(0) = 0 we are left with e / iω, since for large times f(t) = 1. When this is substituted into the expression for Cm(t) we get iωt
Cm (t) = −
e 〈φ  V  φ 0 〉 0 (E m − E 00 ) m
m≠ 0
Insertion of this into the expression for ψ(t)> yields
 ψ (t)〉 = φ0 〉 + e
− iE 00 t /
〈φm  V  φ 0 〉 φm 〉 0 0 m ≠ 0 E0 − Em
∑
On the other hand the ground state wave function, to first order in V is 〈φ n  V  φ0 〉 0 0  φn 〉 n≠ 0 E 0 − E n
 w0 〉 = φ 0 〉 + ∑ It follows that 〈w 0  ψ (t)〉 = 1+ e
− iE00 t /
〈φ 0  V  φm 〉〈φ m V  φ 0 〉 (E 00 − E m0 )2 m ≠0
∑
5
Thus to order V the right side is just one. A fuller discussion may be found in D.J.Griffiths Introduction to Quantum Mechanics.i
7. The matrix element to be calculated is M fi = −
e2 4πε 0
* 3 3 3 3 ∫ d r1 ∫ d r2...∫ d rA Φ f (r1,r2,..rA )∫ d r Z
1
∑ r − r i =1
e −ip.r /
V
ψ 100 (r )Φ i (r1,r2 ,..rA )
i

The summation is over I = 1,2,3,..Z , that is, only over the proton coordinates. The outgoing electron wave function is taken to be a plane wave, and the Φ are the nuclear wave functions. Now we take advantage of the fact that the nuclear dimensions are tiny compared to the electronic ones. Since rI  << r , we may write 1 1 r •r = + 3 i + ...  r − ri  r r The 1/r term gives no contribution because 〈Φ f  Φ i 〉 = 0. This is a shorthand way of saying that the initial and final nuclear states are orthogonal to each other, because they have different energies. Let us now define Z
d = ∑ ∫ d r1 ∫ d r2 ..∫ d rA Φ f (r1,r2,..)r j Φ i (r1,r2 ,..) 3
3
3
*
j =1
The matrix element then becomes M fi = −
e
2
4πε 0
∫d r 3
e
−ip . r /
d• r 3 ψ 100 (r ) V r
The remaining task is to evaluate this integral. First of all note that the free electron energy is given by 2
p = ∆E +  E100  2m where ∆E is the change in the nuclear energy. Since nuclear energies are significantly larger than atomic energy, we may take for p the value p = 2m∆E . To proceed with the integral we choose p to define the z axis, and write p / = k . We write the r coordinate in terms of the usual angles θ and φ . We thus have
6
∫ d re
−ip.r /
∫ dΩ∫
∞
3
0
d.r 3 ψ 100 (r ) = r 3/ 2
dre
−ikr cosθ
2 Z − Zr / a 0 e (dx sin θ cosφ + dy sin θ sinφ + dz cos θ ) 4π a0
The solid angle integration involves
∫
2π
0
dφ , so that the first two terms above disappear.
We are thus left with 3/2
1 ∞ 1 Z 2π dz d(cosθ ) dr cosθe − ikr cosθ e −Zr /a 0 = ∫ ∫ −1 0 π ao 3/2
1 cosθ 1 Z 2π (d.pˆ )∫ d(cosθ ) −1 π ao (Z / a0 + ik cos θ)
The integral, with the change of variables cosθ = u becomes
∫
1
∫
1
−1
−1
du
u = Z / a0 + iku
du
u(Z / a0 − iku) = (Z / a0 )2 + k 2 u 2 2
u −ik ∫−1 du (Z / a0 )2 + k 2 u 2 1
−i k w2 2i a0 ak arctan( 0 ) 2 ∫−k dw 2 2 =− 2 k− k Z (Z / a0 ) + w Z k
ka0 ∆E k 2∆E 1 = = . If Z is not too large, then the 2 2 2 = Z mcZα Z mc α Z (13.6eV ) factor is quite large, because nuclear energies are in the thousands or millions of electron volts. In that case the integral is simple: it is just
Note now that
3/2
π Z d• p 1 Z (2π ) 2 (−2i) 1− 2a0 p p π a0 We evaluate the rate using only the first factor in the square bracket. We need the absolute square of the matrix element which is 3
Z (d.p) 2 (− ) 16π a0 p 4 4πε 0 V The transition rate per nucleus is e
2
2
2
7
2π d pV p 2 R fi = − ∆E)  M fi  ∫ 3 δ( (2π) 2m 3
2
Z (d • p)2 2π d 3 pV p2 1 e 2 16π 2 = δ ( − ∆E) ∫ (2π) 3 2m p4 V 4πε 0 a0 2
3
In carrying out the solid angle integration we get
∫ dΩ(d • p)
2
=
4π 2 2 d  p 3
so that we are left with some numerical factors times
∫ dpδ (p
2
/ 2m − ∆E ) =
Putting all this together we finally get 2
16 3 d R fi = (Zα ) 2 a0 3
2
mc mc 2∆E
2
We write this in a form that makes the dimension of the rate manifest.
m 2∆E
CHAPTER 16. 1. The perturbation caused by the magnetic field changes the simple harmonic oscillator Hamiltonian H0 to the new Hamiltonian H H = H0 +
q B• L 2m
If we choose B to define the direction of the z axis, then the additional term involves B Lz. When H acts on the eigenstates of the harmonic oscillator, labeled by nr, l, ml >, we get
3 qB H  nr,l,m l 〉 = ω(2nr + l + + m  n ,l,m l 〉 2 2m l r Let us denote qB/2m by ωB . Consider the three lowest energy states: nr = 0, l = 0, the energy is 3ω / 2 . nr = 0, l = 1 This threefold degenerate level with unperturbed energy 5ω / 2 , splits into three nondegenerate energy levels with energies
1 E = 5ω / 2 + ω B 0 −1 The next energy level has quantum numbers nr = 2, l = 0 or nr = 0, l = 2. We thus have a fourfold degeneracy with energy 7ω / 2. The magnetic field splits these into the levels according to the ml value. The energies are
2 1 E = 7ω / 2 + ω B 0,0 −1 −2
nr = 1,0
2, The system has only one degree of freedom, the angle of rotation θ. In the absence of torque, the angular velocity ω = dθ/dt is constant. The kinetic energy is 2 2 2 2 1 1 (M v R ) 1 L 2 = E = Mv = 2 2 MR 2 2 I
where L = MvR is the angular momentum, and I the moment of inertia. Extending this to a quantum system implies the replacement of L2 by the corresponding operator. This suggests that 2
H=
L 2I
(b) The operator L can also be written as p x R. When the system is placed in a constant magnetic field, we make the replacement 1 q p → p − qA = p − q(− r × B) = p + r × B 2 2 The operator r represents the position of the particle relative to the axis of rotation, and this is equal to R. We may therefore write q q 2 L = R × p → R × (p + R × B) = L + (R (R • B) − R B) 2 2 If we square this, and only keep terms linear in B , then it follows from (R.B) = 0, that 2
2
1 2 L q L qB 2 − L •B = − L H = (L − qR L • B)= 2I 2I 2M 2I 2M z The last step is taken because we choose the direction of B to define the z axis. The energy eigenvalues are therefore E=
2 l(l + 1) qB − m 2I 2M l
where ml = l,l − 1,l − 2,...− l . Note that the lowest of the levels corresponds to ml = l. 3. In the absence of a magnetic field, the frequency for the transition n = 3 to n = 2 is determined by
2π ν =
1 1 2 2 1 mc α − 2 4 9
so that mc α 5 4π 36 2
ν=
2
The lines with ∆ml = ± 1 are shifted upward (and downward) relative to the ∆ml = 0 (unperturbed ) line. The amount of the shift is given by h∆ν =
eB 2mc
so that
∆ν =
eB 4π mc
Numerically ν = 0.4572 x 1015 Hz and with B = 1 T, ∆ν = 1.40 x 1010 Hz. Thus the frequencies are ν and ν(1 ± ∆ ν / ν ) . Thus the wavelengths are c /ν and (c / ν )(1∓ ∆ν / ν ) . This leads to the three values λ = 655.713 nm, with the other lines shifted down/up by 0.02 nm.
4. The Hamiltonian is H=
1 2 ( p − qA ) − qE • r 2m
Let us choose E = (E, 0, 0 ) and B = ( 0, 0, B), but now we choose the gauge such that A = (0, Bx, 0). This leads to
(
)
1 2 2 2 px + (py − qBx) + pz − qEx = 2m 1 ( p2x + p2y + p2z − 2qBpy x + q 2 B 2 x 2 − 2mqEx) = 2m
H=
Let us now choose the eigenstate to be a simultaneous eigenstate of H, pz (with eigenvalue zero) and py (with eigenvalue k ). Then the Hamiltonian takes the form 2 2
k 2 1 2 1 1 2 H= + px + ( qBx − k − mE / B ) − k + mE / B) ) ( 2m 2m 2m 2m This is the Hamiltonian for a shifted harmonic oscillator with a constant energy added on. We may write this in the form 2 2 2 kE mE 1 q B k − mE / B H=− − m 2 x − 2 + B 2 m qB 2B
Thus the energy is
2
1 kE mE 2 qB (n + E=− − ) 2 + 2 B 2B m with n = 0,1,2,3,… 5. We first need to express everything in cylindrical coordinates. Since we are dealing with an infinite cylinder which we choose to be aligned with the z axis,, nothing depends on z, and we only deal with the ρ and φ coordinates. We only need to consider the Schrodinger equation in the region a ≤ ρ ≤ b.
We start with H =
(
1 2 2 Π + Πy 2m e x
)
where Π x = −i
∂ ∂ + eAx ; Π y = −i + eAy ∂x ∂y
To write this in cylindrical coordinates we use Eq. (1633) and the fact that for the situation at hand Ax = −sinϕ Aϕ ; Ay = cosϕ Aϕ ; Aϕ =
Φ 2πρ
where Φ is the magnetic flux in the interior region. When all of this is put together, the equation Hψ ( ρ,ϕ ) = E ψ (ρ,ϕ ) takes the form
Φ 1 ∂ψ e 2 Φ 2 ∂ 2ψ 1 ∂ψ 1 ∂ 2ψ − 2ie + + + ψ 2π ρ 2 ∂ϕ ρ 2 2π 2me ∂ρ 2 ρ ∂ρ ρ 2 ∂ϕ 2 2
Eψ = −
To solve this, we use the separation of variables technique. Based on previous experience, we write
ψ (ρ,ϕ ) = f (ρ )e imϕ The singlevaluedness of the solution implies that m = 0,±1,±2,±3,… With the notation k = 2m e E / the equation for f(ρ) becomes 2
2
2
2
d f 1 df eΦ −k f (ρ ) = − m+ f 2 + dρ ρ dρ 2π 2
If we now introduce z = kρ and ν = m +
eΦ the equation takes the form 2π
d 2 f (z) 1 df (z) ν 2 + + 1− 2 f (z) = 0 z z dz dz2 This is Bessel’s equation. The most general solution has the form
f ( ρ) = AJ ν ( kρ) + BN ν (k ρ) If we now impose the boundary conditions f ( ka) = f (kb) = 0 we end up with
AJν (ka) + BN ν ( ka) = 0 and
AJν (kb) + BN ν ( kb) = 0 The two equations can only be satisfied if
Jν (ka )Nν (kb) − Jν ( kb)Nν (ka) = 0 This is the eigenvalue equation, and the solution k clearly depends on the order ν of the Bessel functions, that is, on the flux enclosed in the interior cylinder.
CHAPTER 17 1. We start with Eq. (1719) . We define k as the z axis. This means that the polarization vector, which is perpendicular to k has the general form
ε (λ ) = ˆi cosϕ + ˆjsinϕ This leads to
B = ∇ × A = −i
2ε 0ωV
kkˆ × (ˆi cosϕ + ˆj sinϕ ) = B0 ( ˆjcosϕ − ˆi sinϕ )
We are now interested in gp − gn ) ( p) (n) m X 0 {(σ (y p ) − σ (n y )cosϕ − (σ x − σ x )sinϕ}X1 2 2
M = B0
The operators are of the form sinϕ 0 −icosϕ 0 0 σ y cosϕ − σ x sinϕ = − = icosϕ 0 sinϕ 0 ie iϕ It is simple to work out the “bra” part of the scalar product
0 1 χ (+p) χ −(n) − χ −( p) χ +(n) ) iϕ ( ie 2
−iϕ
0 − iϕ 0 p ie
−ie
0 n
−ie
−iϕ
with the help of 0
χ+
ie
iϕ
− iϕ
0 = (1 0) iϕ 0 ie
−ie
−iϕ
−iϕ − iϕ = (0 −ie ) = −ie χ − 0
−ie
and 0 χ − iϕ ie
− iϕ
0 = (0 1) iϕ 0 ie
−ie
This implies that the “bra” part is
−iϕ
iϕ = (ie 0
−ie
0)= ie χ + iϕ
− iϕ
0
−ie
0
1 2
(χ (+p) χ−(n) − χ−( p) χ+(n) )
ie
= − 2i(e
−iϕ
iϕ
= 0 n
−iϕ
0 − iϕ 0 p ie
−ie
−ie
−iϕ
iϕ ( p ) (n ) χ (−p) χ (n) − + e χ + χ+ )
= − 2i(e −iϕ X1−1 + e iϕ X11 )
−1
For the “ket” state we may choose X triplet = α X1 + β X1 + γX1 , and then the matrix element is 1
M = −i 2B0
0
g p − gn iϕ − iϕ (e α + e γ ) 2 2
2. We are interested in finding out for what values of l, m, the matrix element 1 〈,m  (ε .p)(k.r) + (ε .r)(p.k)  0,0〉 2 does not vanish. We use the technique used in Eq. (1722) to rewrite this in the form
1 ime 〈,m  [H 0,ε .r]( k.r) + (ε .r)[H0 ,k.r]  0.0〉 = 2 ime 〈,m  H0 (ε .r)(k.r) − (ε .r)H 0 (k.r) + (ε .r)H 0 (k.r) − (ε .r)(k.r)H 0  0,0〉 = 2 ime (E − E 0,0 )〈,m  (ε .r)(k.r)  0,0〉 2 ,m Let us now choose k to define the z axis, so that k = ( 0,0,k). Since ε is perpendicular to k , we may choose it to be represented by ε = (cosα, sinα, 0). Then, with the usual polar coordinates, we have
(ε .r)(k.r) = k(cosαsinθcosφ + sinαsinθsinφ )cosθ = = ksinθ cosθ cos(φ − α ) This is a linear combination of Y21(θ ,φ ) and Y2,−1(θ,φ ) . Thus the angular integral is of the form
δ ,2 .
∫ dΩY
* l , m 2,±1 0,0
Y Y
, and since Y0,0 is just a number, the integral is proportional to
There is also a selection rule that requires m = ± 1. This comes about because of our choice of axes.
3. In the transition under consideration, the radial part of the transition rate is unchanged. The only change has to do with the part of the matrix element that deals with the dependence on the polarization of the photon emitted in the transition. 2 2 2 Eq. (1744), for example shows that δm ,1 is multiplied by εx + εy = 1− εz and this factor carries some information about the direction of the photon momentum, even though that does not appear explicitly in the matrix element. We proceed as follows: The direction of the polarization of the initial atomic state defines the z axis. Let the photon momentum direction be given by
dˆ = ˆi sinΘcosΦ + ˆjsinΘsinΦ + kˆ cosΘ We may define two unit vectors perpendicular to this. For the first one we take dˆ × kˆ , which, after being divided by the sine of the angle between these two vectors, i.e. by sinΘ , yields εˆ1 = − ˆi sinΦ + ˆjcosΦ The other one is εˆ2 = dˆ × εˆ1 (two vectors perpendicular to each other), which leads to
εˆ2 = ˆi cosΘcosΦ + ˆj cosΘsinΦ − kˆ sinΘ In the coordinate system in which dˆ represents the z axis, the ει vectors represent the x and y axes, and since the photon polarization must lie in that new x – y plane, we see that the polarization vector has the form
ε = cos χ εˆ1 + sin χ εˆ2 Thus
εz = kˆ • ε = − sin χ sinΘ , εx = ˆi • ε = cos χ sinΦ + sin χ cosΘcosΦ,
εy = ˆj • ε = − cos χ cosΦ + sin χ cosΘsinΦ and
εx2 + εy2 = 1− ε z2 = 1− sin2 χ sin 2 Θ Thus the final answer (using Eq. (1744) is
dΓ =
α ω 3 215 1 δ (1− sin2 χ sin 2 Θ)d(cosΘ)dΦ 2π c 2 310 2 m,1
The dependence on the polarization appears in the sin χ term. 2
4. First of all, we need to recognize what 2p 1s means for the harmonic oscillator in three dimensions. The numbers “2” and “1” usually refer to the principal quantum number, e.g n = nr + + 1 for the hydrogen atom. Here the energy spectrum is characterized by 2n r + + 1 , and it is this combination that we call the principal quantum number. Thus we take the 2p 1s transition to mean (n r = 0, = 1) → ( nr = 0, = 0) . To solve this problem we recognized that nothing changes in the angular integration that was done for the 2p 1s transition in hydrogen. The only change in the matrix element involves the radial functions. In hydrogen we calculated
∫
∞
0
3
r R21(r ) R10 (r) dr
using the radial functions for hydrogen. Here the same integral appears, except that the radial functions are those of the threedimensional harmonic oscillator. Here, the properly normalized eigenfunctions are 2 mω R10( r) = 1/ 4 π
3/ 2
e
− mω r 2 / 2
and
R21(r) =
8 3
1/ 2
1 mω π 1/4
5/4
re
− mω r 2 / 2
Note that these functions appear in the solution to problem 813. Given these, the integral that yields the matrix element is straightforward. We have
M=
8 3
1/2
2 mω π 1/ 2 1/ 2
2
∫
∞ 0
4 −mω r 2 /
drr e
4 2 m ω = 1/2 π 3 m ω
5/2
1 ∞ 3/2 −x dxx e ∫ 0 2
4 2 m ω = 1/2 π 3 m ω
5/2
1 3π 1/ 2 2 4
1/ 2
2
=
2
3 . We check that this has the dimensions of a (length)2 as 2mω required. To get the decay rate, we just take the hydrogen result and make the substitution
The square of this is
15
 M hydrogen
2 2 3 2  = 9 a0 → M  = 3 2mω 2
This then leads to the rate
4 ω3 2α ω R = α 2  M 2 = ω 3 mc 2 9 c
CHAPTER 19 1. We have 1 3 −i∆.r d re V (r) ∫ V
M fi =
If V(r) = V(r), that is, if the p9otential is central, we may work out the angular integration as follows:
M fi =
2π π 1 ∞ 2 −i∆r cosθ r V ( r )dr ∫ dφ ∫ sinθdθ e ∫ 0 0 0 V
with the choice of the vector ∆ as defining the z axis. The angular integration yields
∫
2π
0
π
1
0
−1
dφ ∫ sin θdθe − i∆r cosθ = 2π ∫ d(cos θ )e −i∆r cosθ =
4π sin∆ r ∆r
so that
M fi =
1 4π V ∆
∫
∞ 0
rdrV (r )sin∆ r
Note that this is an even function of ∆ that is, it is a function of ∆ = (p f − p i ) / 2
2
2
2. For the gaussian potential
M fi =
2 2 1 4π V0 ∞ rdr sin ∆re −r /a ∫ 0 V ∆
Note that the integrand is an even function of r. We may therefore rewrite it as
∫
∞
0
rdr sin ∆re
− r2 / a 2
=
1 ∞ − r2 / a 2 rdr sin ∆re ∫ 2 −∞
The integral on the right may be rewritten as
(
)
2 2 2 2 1 ∞ 1 ∞ rdr sin ∆re − r / a = ∫ rdr e −r /a +i∆ r − c .c ) ∫ 2 −∞ 4 i −∞
Now
1 ∂ ∞ ∂ ∆a π − a 2 ∆2 / 4 1 ∞ − r 2 / a 2 + i∆r − r 2 / a 2 + i∆r −a 2 ∆2 /4 rdre = dre = −i a π e =i e ∫ ∫ 4i −∞ i ∂∆ −∞ ∂∆ 2 3
Subtracting the complex conjugate and dividing by 4i gives
M fi =
(
)
3 1 −a 2 ∆2 /4 a π V0 e V
The comparable matrix element for the Yukawa potential is ∞ 1 4π 1 b − r/ b VY b ∫ dre sin∆r = 4π VY 0 V ∆ V 1+ b 2 ∆2 3
M fi =
We can easily check that the matrix elements and their derivatives with respect to ∆2 at ∆ = 0 will be equal if a = 2b and VY = 2 πV0 . The differential cross section takes its simplest form if the scattering involves the same particles in the final state as in the initial state. The differential cross section is dσ µ 2 = 2 4  U (∆)  dΩ 4π 2
where µ is the reduced mass and U(∆) = VM fi . We are interested in the comparison
(dσ / dΩ)gauss e −2b ∆ 2 −2X = 2 2 −2 = (1+ X ) e (dσ / dΩ)Yukawa (1+ b ∆ ) 2 2
where we have introduced the notation X = b2x2. This ratio, as a function of X, starts out at X = 0 with the value of 1, and zero slope, but then it drops rapidly, reaching less than 1% of its initial value when X = 4, that is, at ∆ = 2/b. 3. We use the hint to write
dσ µ b p dσ = 2 2= 2 4 4π V0 dΩ π d∆ 4π 1 + b2 ∆2 2
2
3
2
The total cross section may be obtained by integrating this over ∆2 with the range given 2 2 2 by 0 ≤ ∆ ≤ 4 p / , corresponding to the values of cosθ between –1 and + 1.. The 2 2 2 integral can actually be done analytically. With the notation k = p / the integral is
∫
4k 2
0
d∆2
1 1 4 k 2 b 2 dx 4k 2 = = (1 + b2 ∆2 ) 2 b 2 ∫0 (1+ x) 2 1 + 4k 2b 2
This would immediately lead to the cross section if the particles were not identical. For identical particles, there are symmetry problems caused by the Pauli Exclusion Principle and the fact that the protons have spin 1/2. The matrix elements are not affected by the
spin because there is no spinorbit coupling or any other spin dependence in the potential. However: In the spin triplet state, the spatial wave function of the proton is antisymmetric, while for the spin singlet state, the spatial wave function is symmetric. This means that in the original Born approximation we have
e ik .r ∓ e −ik .r e −ik '.r ∓ e ik '.r V (r ) = ∫d r 2 2 3
∫ d rV (r)e 3
−i( k ' −k ).r
∓ ∫ d 3 rV (r )e − i( k +k'). r
The first term has the familiar form
b3 b3 4π V0 = 4π V0 1+ b 2 ∆2 1 + 2b 2 k 2 (1− cosθ ) and the second term is obtained by changing cosθ to  cosθ.. Thus the cross section involves 1 1 d(cosθ ) ∓ 2 2 2 2 2 2 2 2 1+ 2b k − 2b k cosθ 1+ 2b k + 2b k cosθ
1
∫
−1
1
→ ∫ dz −1
=
1 1 ∓ 1 + a − az 1 + a + az
2
2
4 2 ∓ ln(1+ 2a) 1+ 2a a(1 + a)
where a = 2b2k2. Thus the total cross section is 8πµ b 2 1 4 2 2 V0 ln(1 + 4 k b ) ∓ 2 2 4 2 2 2 2 1+ 4 k b k b (1 + 2k b ) 2 6
σ=
The relation to the center of mass energy follows from E = p / 2µ = k / 2µ , so that 2
2 2
2µE (1.67 × 10−27 kg)(100 × 1.6 × 10−13 J ) k = 2 = (1.054 × 10−34 J.s) 2 2
With b = 1.2 x 101`5 m, we get (kb)2 = 3.5, so that σ = 4.3 x 1028 m2 = 4.3 x 1024 cm2 = 3.4 barns.
4. To make the table, we first of all make a change of notation: we will represent the proton spinors by χ ± and the neutron spinors by η± . To work out the action of
σ p • σ n = σ pzσ nz + 2(σ p +σ n − + σ p −σ n + ) on the four initial combinations, we will use σ + χ + = σ − χ − = 0; σ + χ − = χ +; σ − χ + = χ − and similarly for the neutron spinors. Thus [σ pzσ nz + 2(σ p +σ n − + σ p −σ n + )]χ +η + = χ +η+ [σ pzσ nz + 2(σ p +σ n − + σ p −σ n + )]χ +η − = − χ +η− + 2 χ −η + [σ pzσ nz + 2(σ p +σ n − + σ p −σ n + )]χ −η + = − χ −η+ + 2 χ +η − [σ pzσ nz + 2(σ p +σ n − + σ p −σ n + )]χ −η − = χ −η− From this we get for the matrix A + Bσ p • σ n , with rows and columns labeled by (++), (+),(+). ()the following
A+B 0 A + Bσ p • σ n = 0 0
0 0 A − B 2B 2B A−B 0 0
0 0 0 A + B
The cross sections will form a similar matrix, with the amplitudes replaced by the absolute squares, i.e. A+B2, 2B2, and AB2.
5. Consider n – p scattering again. If the initial proton spin is not specified, then we must add the cross sections for all the possible initial proton states and divide byt 2, since a priori there is no reason why in the initial state there should be more or less of upspin protons. We also need to sum over the final states. Note that we do not sum amplitudes because the spin states of the proton are distinguishable. Thus, for initial neutron spin up and final neutron spin up we have 1 2
σ (+  +) = (σ (++,++) + σ (++,−+) + σ (−+,++) + σ (−+,−+)) where on the r.h.s. the first label on each side refers to the proton and the second to the neutron. We thus get
σ (+  +) = Similarly
1 ( A + B 2 +  A − B 2 )= A 2 +  B 2 2
1 2 1 = ( 2B 2 )= 2  B 2 2
σ (−  +) = (σ (+−,++) + σ (+−,−+) + σ (−−,++) + σ (−−,−+))
Thus  A  +  B  −2  B   A  − B  P= 2 2 2 =  A  +  B  +2  B   A 2 +3  B 2 2
2
2
2
2
6. For triplet triplet scattering we have (with the notation (S,Sz) (1,1)(1,1) 〈 χ +η+  χ +η+ 〉 = A + B (1,1)(1,1) 〈 χ −η−  χ −η− 〉 = A + B (1,0) (1,0)
〈
(0,0)(0.0)
〈
(0,0)(1,0)
〈
χ +η− + χ −η+ χ +η− + χ −η+ 2

2
χ +η− − χ −η + χ +η − − χ −η+ 2

2
χ +η− − χ −η + χ +η − + χ −η+ 2

2
〉=
1 (A − B + 2B + 2 B + A − B) = A + B 2
〉=
1 (A − B − 2 B − 2 B + A − B) = A − 3B 2
1 〉 = (A − B + 2 B − 2 B − A + B) = 0 2
We can check this by noting that (in units of ,
A + Bσ p • σ n = A + 4Bs p • s n = A + 2B(S2 − s 2p − s2n ) 3 = A + 2B S(S + 1) − 2 For S = 1 this is A + B, For S = 0, it is A – B , and since 〈S = 1 S − 3 / 2  S = 0〉 = 0 by orthogonality of the triplet to singlet states, we get the same result as above. 2
7. We have, with x = kr and cosθ = u,
i d −iux e −1 −1 x dx i 1 d i 1 dg − iux = ∫ du (g(u)e − iux )− ∫ du e x −1 du x −1 du 1
1
I(x) = ∫ dug(u)e − iux = ∫ dug(u)
The first term vanishes since g(±1)=0. We can proceed once more, and using the fact that the derivatives of g(u) also vanish at u = ± 1, we find
−i I(x) = x
2
d 2 g − ixu du ∫−1 du2 e 1
and so on. We can always go beyond any predetermined power of 1/x so that I(x) goes to zero faster than any power of (1/x). 7. We proceed as in the photoelectric effect. There the rate, as given in Eq.(19111) is
R=
2π V mp e 2 dΩ ∫ 3  M fi  (2π )
Here m is the electron mass, and pe is the momentum of the outgoing.electron.The factor arose out of the phase space integral
p2 p2 p2 ∫ dpp δ 2m − E γ = ∫ d 2m mpδ 2m − Eγ = mpe 2
with pe determined by the photon energy, as shown in the delta function. In the deuteron p2 photodisintegration process, the energy conservation is manifest in δ − E γ + E B . M The delta function differs in two respects: first, some of the photon energy goes into dissociating the deuteron, which takes an energy EB ; second, in the final state two particles of equal mass move in in equal and opposite directions, both with momentum of magnitude p, so that the reduced mass Mred = M/2 appears. Thus the factor mpe will be replaced by Mp/2, where the momentum of the particle is determined by the delta function. Next, we consider the matrix element. The final state is the same as given in Eq. (19114) with pe replaced by p , and with the hydrogenlike wave function replaced by the deuteron ground state wave function. We thus have 2
2 dσ 2π (VMp / 2) V e 1 2 3 i(k − p / )• r = ( ε • p) d re ψ (r) ∫ d dΩ (2π ) 3 c M 2ε 0ωV V
ik• r
We need to determine the magnitude of the factor e . The integral is over the wave −α r function of the deuteron. If the ground state wave function behaves as e , then the −2αr probability distribution goes as e , and we may roughly take 1/2α as the “size” of the 2 2 deuteron. Note that α = ME B / . As far as k is concerned, it is given by k=
pγ
=
Eγ
c
Numerically we get, with EB =2.2 MeV, and Eγ = 10 MeV, k/2α = 0.11, which means that we can neglect the oscillating factor. Thus in the matrix element we just need 3 ik• r ∫ d re ψ d (r). The wave function to be used is
ψ d ( r) =
−α (r− r0 )
N e r 4π
r > r0
N is determined by the normalization condition 2
N 4π
∫
∞
r0
4π r dr 2
e
−2α (r− r0 )
r2
=1
So that N = 2α 2
The matrix element involves
N 4π 4π k
∫
∞
r0
e −α (r− r0 ) rdr sinkr = r
N 4π ∞ dx sink(x + r0 )e −αx ∫ 0 k N 4π ∞ = dx (sinkr0 Re(e −x (α − ik) ) + coskr0 Im(e −x (α −ik ) )) k ∫0 k N 4π α = 2 2 sin kr0 + 2 2 coskr0 k α +k α +k =
The square of this is 2 kr αr 4π N 2 r0 2 2 0 2 2 sin kr0 + 2 2 0 2 2 coskr0 2 α r0 + k r0 k α r0 + k r0
2
It follows that e 2 pr0 kr dσ αr = 2 (αr0 ) 2 2 0 2 2 sinkr0 + 2 2 0 2 2 coskr0 dΩ 4πε 0 c Mω α r0 + k r0 α r0 + k r0
2
We can easily check that this has the correct dimensions of an area. 2
For numerical work we note that αr0 = 0.52; kr0 = 0.26 E MeV and ω = E B +
p . M
9. The change in the calculation consists of replacing the hydrogen wave function 1 4π
3/2
Z 2 e −Zr /a 0 a0
by
ψ (r) =
N sinqr 4π r
r < r0
−κ r
=
N e 4π r
r > r0
where the binding energy characteristic of the ground state of the electron determines κ as follows
κ 2 = 2me  E B  / 2 = (me cα / ) 2 with α = 1/137. The eigenvalue condition relates q to κ as follows:
qr0 cot qr0 = −κ r0 where q = 2
`
2m eV0 2 −κ 2
and V0 is the depth of the square well potential. The expression for the differential cross section is obtained from Eq. (19116) by dividing by 4(Z/a0)2 and replacing the wave function in the matrix element by the one written out above, 2
2 2 dσ 2π me pe 1 e pe 2 3 i( k − p e / ).r ˆ ˆ = ( ε . p ) d re ψ (r) ∫ dΩ (2π )3 c m e 2ε 0ω 4π
We are interested in the energydependence of the cross section, under the assumptions that the photon energy is much larger than the electron binding energy and that the potential has a very short range. The energy conservation law states that under these 3 2 assumptions ω = pe / 2m e . The factor in front varies as pe / ω ∝ pe ∝ Eγ , and thus we need to analyze the energy dependence of
∫ d re 3
form
∫ d re 3
ψ (r) =
iQ.r
4π Q
∫
∞
0
rdr sinQrψ (r)
i(k −p e / ). r
2
ψ (r) . The integral has the
2
kp p where Q = k − p e / so that Q = k + e2 − 2 e (kˆ .pˆ ) . 2 p / 2m ω 2 2 2 2 2 2 2 Now k / pe = ω / pe c = ω e 2 2 = . We are dealing with the 2m e c 2 pe c nonrelativistic regime, so that this ratio is much smaller than 1. We will therefore neglect the k –dependence, and replace Q by pe/ . The integral thus becomes 2
4π Q
2
∞ N r0 −κ r dr sin Qr sin qr + dr sin Qre ∫ ∫ 0 r 0 4π
The first integral is 1 r0 dr (cos(Q − q) r − cos(Q + q)r) = 2 ∫0 1 sin(Q − q)r0 sin(Q + q)r0 1 − ≈ − cos Qr0 sin qr0 Q+q Q 2 Q− q where, in the last step we used Q >> q. The second integral is ∞
Im ∫r dre 0
−r (κ − iQ )
e −r0 (κ −iQ ) cosQr0 −κr0 = Im ≈ e κ − iQ Q
The square of the matrix element is therefore 2 4π N 2 1 −κr 0 − sinqr0 )) 2 2 (cosQr0 (e Q Q
The square of the cosine may be replaced by 1/2, since it is a rapidly oscillating factor, 2 and thus the dominant dependence is 1/Q4 , i.e. 1 / Eγ . Thus the total dependence on the 3/2
photon energy is 1 / Eγ
3
7
or 1 / pe , in contrast with the atomic 1 / pe dependence.
10. The differential rate for process I, a + A b + B in the center of momentum frame is
dRI 1 1 2 dpb = ∑ MI 3 pb dΩ (2 j a + 1)(2J A + 1) (2π ) dE b spins
2
The sum is over all initial and final spin states. Since we have to average (rather than sum) over the initial states, the first two factors are there to take that into account. The phase factor is the usual one, written without specification of how Eb depends on pb. The rate for the inverse process II, b + B a + A is, similarly
2 dRII 1 1 2 dpa = M II ∑ 3 pa dΩ (2 j b + 1)(2J B + 1) (2π ) dE a spins
By the principle of detailed balance the sum over all spin states of the square of the matrix elements for the two reactions are the same provided that these are at the same center of momentum energies. Thus
∑M
2 I
=
spins
∑M
2
II
spins
Use of this leads to the result that (2 j a + 1)(2J A + 1) dRI (2 jb + 1)(2J B + 1) dRII = pb2 (dpb / dE b ) dΩ p2a (dpa / dE a ) dΩ Let us now apply this result to the calculation of the radiative capture cross section for the process N + P D + γ. We first need to convert from rate to cross section. This is accomplished by multiplying the rate R by the volume factor V, and dividing by the relative velocity of the particles in the initial state. For the process I, the photodisintegration γ + D N + P , the relative velocity is c, the speed of light. For process II, the value is pb/mred = 2pb/M . Thus
dσ I V dRI = ; dΩ c dΩ
dσ II MV dRII = dΩ 2 pb dΩ
Application of the result obtained above leads to
dσ II MV dRII = dΩ 2 pb dΩ MV pa (dpa / dE a ) (2 ja + 1)(2J A + 1) c dσ I = × 2 2 pb (2 jb + 1)(2J B + 1) pb (dpb / dE b ) V dΩ 2
We can calculate all the relevant factors. We will neglect the binding energy of the deuteron in our calculation of the kinematics. First (2 j γ + 1)(2 J D + 1) (2 j P + 1)(2J N + 1)
=
2× 3 3 = 2×2 2
Next, in the center of momentum frame, the center of mass energy is
pa2 pa2 W = pa c + ≈ pa c + 2 MD 4M
so that (dE a / dpa ) = c +
pa . In reaction II, 2M 2
2
p p W =2× b = b 2M M
so that (dE b / dpb ) = 2 pb / M . There is a relation between pa and pb since the values of W are the same in both cases. This can be simplified. For photon energies up to say 50 MeV or so, the deuteron may be viewed as infinitely massive, so that there is no difference between the center of momentum. This means that it is a good approximation to write 2 W = E γ = pa c = pb / M . We are thus finally led to the result that
dσ ( NP → Dγ ) 3 E γ dσ (γD → NP ) = 2 Mc 2 dΩ dΩ