note on polar LO-phonon
Jiang
20 February 2017
electron-phonon Hamiltonian
\[ H_{ep} = \int d^3R ~ \rho(R) ~ \phi(R) \] where \(\rho\) is the charge density of the electron (= free charge density) and \(\phi\) the potential created by the bound charges.
Gauss’s law states \[ \nabla \cdot D(R) = \rho(R) \] where \(D \equiv \epsilon_0 E + P\) is a vector field called the electric displacement field.
Then the Hamiltonian is \[ H_{ep} = \int d^3R ~ \nabla \cdot D(R) ~ \phi(R) \]
\[ \int d^3R ~ \nabla \cdot [D(R) \phi(R)] = \int d^3R ~ \nabla \cdot D(R) ~ \phi(R) + \int d^3R ~ D(R) \cdot \nabla \phi(R) \] The left term is zero because the potential field goes to zero at \(R \rightarrow \infty\). Therefore the Hamiltonian becomes
\[ H_{ep} = - \int d^3R ~ D(R) \cdot \nabla \phi(R). \]
Because \(\phi\) is the potential field created by the bounded charge, the corresponding electric displacement field should be zero \[ \nabla \phi(R) = E(R) = - P(R) / \epsilon_0 \]
So the Hamiltonian becomes \[ H_{ep} = - \frac{1}{\epsilon_0} \int d^3R ~ D(R) \cdot P(R) . \]
In the present of screening caused by the motion of charges, we can assume a Thomas-Fermi potential
\[ D(R) = - \nabla [ \frac{e}{4 \pi |r -R|} e^{- q_0 |r-R|} ] = \frac{e}{4 \pi} \left( \frac{1}{|r-R|^2} + \frac{q_0}{|r-R|} \right) e^{- q_0 |r-R|} s(r-R) ~ u_R \] where \(s\) is the unit step function, \(u_R\) is unit direction vector.
\[ H_{ep} = \frac{e e^*}{\epsilon_0 V_0 4 \pi \sqrt{N}} \int d^3R ~ \left( \frac{1}{|r-R|^2} + \frac{q_0}{|r-R|} \right) e^{- q_0 |r-R|} s(r-R) ~ u_R \cdot \sum_q Q_q ~ a_q e^{i q \cdot R} \] Change of variable \((r-R) \rightarrow R\), and transform into polar coordinate \[ H_{ep} = \frac{e e^*}{\epsilon_0 V_0 4 \pi \sqrt{N}} \int_0^\infty dR ~R^2 \int_0^\pi d\theta \sin{\theta} \int_0^{2 \pi} d\phi ~ \left( \frac{1}{R^2} + \frac{q_0}{R} \right) e^{- q_0 R} ~ u_R \cdot \sum_q Q_q ~ a_q e^{i q \cdot (r-R)} \] \[ H_{ep} = \frac{e e^*}{\epsilon_0 V_0 4 \pi \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \int_0^\infty dR ~R^2 \int_0^\pi d\theta \sin{\theta} \int_0^{2 \pi} d\phi ~ \left( \frac{1}{R^2} + \frac{q_0}{R} \right) e^{- q_0 R} \cos{\theta} ~ e^{-i q \cos(\theta) R} \] \[ H_{ep} = \frac{e e^*}{\epsilon_0 V_0 2 \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \int_0^\infty dR ( 1 + q_0 R ) ~ e^{- q_0 R} \int_0^\pi d\theta \sin{\theta}\cos{\theta} ~ e^{-i q \cos(\theta) R} \] \[ H_{ep} = \frac{e e^*}{\epsilon_0 V_0 2 \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \int_0^\infty dR ( 1 + q_0 R ) ~ e^{- q_0 R} \left( \frac{2 i (qR \cos(qR) - \sin(qR))}{q^2 R^2} \right) \] \[ H_{ep} = \frac{i~ e e^*}{\epsilon_0 V_0 \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \int_0^\infty dR ( 1 + q_0 R ) \frac{ qR \cos(qR) - \sin(qR)}{q^2 R^2} ~ e^{- q_0 R} \] First for the case without screening (\(q_0=0\)) \[ H_{ep} = \frac{i~ e e^*}{\epsilon_0 V_0 \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \int_0^\infty dR \frac{ qR \cos(qR) - \sin(qR)}{q^2 R^2} = \frac{i~ e e^*}{\epsilon_0 V_0 \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \left( \frac{-1}{q} \right) \]
Then for the case with screening \[ H_{ep} = \frac{i~ e e^*}{\epsilon_0 V_0 \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \int_0^\infty dR ( 1 + q_0 R ) \frac{ qR \cos(qR) - \sin(qR)}{q^2 R^2} ~ e^{- q_0 R} = \frac{i~ e e^*}{\epsilon_0 V_0 \sqrt{N}} \sum_q Q_q e^{i q \cdot r} \left(- \frac{q}{q^2 + q^2_0} \right) \] Here we obtain the equation (3.139) in Ridley’s book 1.
the scattering rate
The coupling coefficient is therefore \[ C_q^2 = \left( \frac{e e^*}{V_0 \epsilon_0} \right)^2 \frac{q^2}{(q^2 + q^2_0)^2}, \] and the scattering rate is \[ W(k) = \frac{V_0}{8 \pi^2 M \omega_0} \left( \frac{e e^*}{V_0 \epsilon_0} \right)^2 \int_0^{q_{ZB}} dq \int_{-1}^{1} d(\cos{\theta}) \int_0^{2 \pi} d\phi ~ \frac{q^4}{(q^2 + q^2_0)^2} \delta_{k \pm q - k',0} \delta(E_{k'}-E_k \mp \hbar \omega_0) \times (n(\omega_0) + 1/2 \mp 1/2) , \] \[ W(k) = \frac{V_0}{4 \pi M \omega_0} \frac{1}{\hbar v} \left( \frac{e e^*}{V_0 \epsilon_0} \right)^2 \left[ n(\omega_0) \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} + \{n(\omega_0)+1 \} \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} \right], \] where \(v=\sqrt{2 E_k/m^*}\) denotes the group velocity of electrons, and \[ \int dq \frac{q^3}{(q^2 + q^2_0)^2} = \frac{1}{2} \left[ \frac{q_0^2}{q_0^2+q^2} + \log(q_0^2 + q^2) \right] + const. \] and the \(q_{\min}\) and \(q_{\max}\) is given by energy conservation condition for absorbsion \[ q_{\min} = k(x^+_k -1); q_{\max} = k(x^+_k +1), \text{where } x_k^+=(1+ \hbar \omega_0/E_k)^{1/2} \] for emission \[ q_{\min} = k(1 - x^+_k); q_{\max} = k(1 + x^+_k), \text{where } x_k^- =(1- \hbar \omega_0/E_k)^{1/2} \]
For the case of no screening (\(q_0 = 0\)), so not for highly doped semiconductors, then the previous integrant reduces to \(\log(q)\). We can obtain directly the following relation for the scattering rate \[ W(k) = \frac{V_0}{4 \pi M \omega_0} \frac{1}{\hbar v} \left( \frac{e e^*}{V_0 \epsilon_0} \right)^2 \left[ n(\omega_0) \log \frac{\sqrt{1+ \hbar \omega_0/E_k}+1}{\sqrt{1+ \hbar \omega_0/E_k}-1} + \{n(\omega_0)+1 \} \log \frac{1+\sqrt{1-\hbar \omega_0/E_k}}{1-\sqrt{1-\hbar \omega_0/E_k}} \right], \]
The result given in Ridley’s book is the following \[ W(k) = \frac{V_0}{2 \pi M \omega_0} \frac{1}{\hbar v} \left( \frac{e e^*}{V_0 \epsilon_0} \right)^2 \left[ n(\omega_0) \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} \right)^{1/2} + \{n(\omega_0)+1 \} \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} -1 \right)^{1/2} \right], \] but what exactly is his definition for sinh? <== it’s OK, it’s the usual definition of sinh. I try to “reverse engineer” his result here. Change of variable in order to ease the writing \(\left( \frac{E_k}{\hbar \omega_0} \right)^{1/2} \rightarrow x\) \[ 2 \sinh^{-1} x \equiv 2 \log (x + \sqrt{1+x^2}) = \log (x + \sqrt{1+x^2})^2 =\log \frac{(x + \sqrt{1+x^2})^2}{( \sqrt{1+x^2} +x)( \sqrt{1+x^2}-x)} = \log \frac{\sqrt{1+x^2} +x}{ \sqrt{1+x^2}-x} = \log \frac{\sqrt{1+x^{-2}} +1}{ \sqrt{1+x^{-2}}-1} \] this gives the first term. \[ 2 \sinh^{-1} \sqrt{x^2-1} \equiv 2 \log(\sqrt{x^2-1} + \sqrt{1+x^2-1})= \log(\sqrt{x^2-1} + x)^2 = \log\frac{(\sqrt{x^2-1} + x)^2 }{(x + \sqrt{x^2-1})(x - \sqrt{x^2-1})} = \log\frac{x + \sqrt{x^2-1} }{x - \sqrt{x^2-1}} = \log\frac{1 + \sqrt{1-x^{-2}} }{1 - \sqrt{1-x^{-2}}} \] this gives the second term.
To calculate W, we also need \(n(\omega_0)\) \[ n(\omega_0) = \frac{1}{e^{\hbar \omega_0 / k_B T} - 1} \]
effective charge
In polar materials, the permittivity can be different at low and high frequencies, which is related to the effective charge on the ions. The electric field causes the ionic motion, which then relates to the polarisation vector of the ions. This allows us to express the effective charge on the atoms \[ e^{*2}= M V_0 \omega_0^2\epsilon_0^2 \left(\frac{1}{\epsilon_\infty} - \frac{1}{\epsilon} \right) . \] This simplifies the expression of scattering rate, and also reduces the number of parameters. The scattering rate is \[ W(k) = \frac{\omega_0 e^2}{ 2 \pi \hbar \epsilon_p v} \left[ n(\omega_0) \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} \right)^{1/2} + \{n(\omega_0)+1 \} \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} -1 \right)^{1/2} \right], \] where \[ \frac{1}{\epsilon_p} = \frac{1}{\epsilon_\infty} - \frac{1}{\epsilon} . \]
Example: Plot the W(k) as a function of the energy for room temperature T = 300 K:
source("scatteringRate.R")
hbar = 6.58211899E-16 # [eV s]
temp = 300 #[K] temperature
epsinf = 10.92 #High-frequency dielectric constant
eps0 = 12.9 #Static dielectric constant
epsp = 1/(1/epsinf - 1/eps0) #effective permittivity
w0 = 8.75e12 #[hz] frequency
me = 0.0655 #effective mass
Emax=4*hbar*w0*2*3.14 #[ev] max value of energy grid
E = seq(0,Emax,by=0.001*Emax) #ev energy grid
W0 = 2.0*w0
W = scatteringRate_LOpolar_isotropic_woscreening_effcharge(permittivity = epsp,
freq=w0,
eff_mass=me,
temperature=temp,
energy=E)
plot(E/(hbar*w0*2*3.14),(W/W0), type="l")
energy relaxation rate
the energy relaxation is easier to obtain from the scattering rate obtained just previously. \[ \frac{dE_k}{dt} = \frac{e^2 \omega_0^2}{2 \pi \epsilon_p v} \left[ n(\omega_0) \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} \right)^{1/2} - \{n(\omega_0)+1 \} \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} -1 \right)^{1/2} \right] . \]
momentum relaxation rate
similar as we did for the acoustic phonon, for the momentum relaxation rate we have to calculate the weighted average \[ \frac{d\hbar k}{dt} = \int dk' \hbar (k'-k) W(k,k') \delta(E_{k'} - E_{k} \mp \hbar \omega_q) dk' \] The change of momentum due to phonon scattering is An absorption of a phonon contribute an increase of momentum of \((q/k)\cos\theta\) in the direction of \(k\), and an emission of a phonon results in a decrease of momentum of \((q/k)\cos\theta\) too. We insert this into the momentum relaxation rate \[ \frac{d\hbar k}{dt} = \frac{\omega_0 e^2}{ 4 \pi \hbar \epsilon_p v} \left[ n(\omega_0) \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} \left( -\frac{q}{k} \cos\theta \right) + \{n(\omega_0)+1 \} \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} \left( \frac{q}{k} \cos\theta \right) \right], \]
The \(\cos\theta\) can be obtained from the energy conservation relationship \(\cos\theta = \frac{q}{2k}+\frac{m^* \omega_0}{\hbar kq}\) for the absorption and \(\cos\theta = \frac{q}{2k}-\frac{m^* \omega_0}{\hbar kq}\) for the emission. The momentum relaxation rate is \[ \frac{d\hbar k}{dt} = \frac{\omega_0 e^2}{ 4 \pi \hbar \epsilon_p v} \left[ n(\omega_0) \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} \left( \frac{q^2}{2k^2}-\frac{m^* \omega_0}{\hbar k^2} \right) + \{n(\omega_0)+1 \} \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} \left( \frac{q^2}{2k^2}+\frac{m^* \omega_0}{\hbar k^2} \right) \right], \] then \[ \frac{d\hbar k}{dt} = \frac{\omega_0 e^2}{ 4 \pi \hbar \epsilon_p v} \left[ n(\omega_0) \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} \left( \frac{q^2}{2k^2}-\frac{m^* \omega_0}{2E_k} \right) + \{n(\omega_0)+1 \} \int_{q_{\min}}^{q_{\max}} dq \frac{q^3}{(q^2 + q^2_0)^2} \left( \frac{q^2}{2k^2}+\frac{m^* \omega_0}{2E_k} \right) \right]. \] Here for no-screening, neglecting \(q_0\) we obtain \[ \frac{1}{\tau_m} = \frac{\omega_0 e^2}{ 4 \pi \hbar \epsilon_p v} \left( n(\omega_0) \left( 1 + \frac{\hbar \omega_0}{E_k} \right)^{1/2} + \{n(\omega_0)+1\} \left( 1 - \frac{\hbar \omega_0}{E_k} \right)^{1/2} + \frac{\hbar \omega_0}{E_k} \left[ -n(\omega_0) \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} \right)^{1/2} +\{n(\omega_0)+1\} \sinh^{-1} \left( \frac{E_k}{\hbar \omega_0} - 1 \right)^{1/2} \right] \right) \]
Example: Plot the \(1/\tau_m\) as a function of the energy for room temperature T = 300 K for GaAs, with parameters from Rode’s paper 2.
source("relaxationTime.R")
hbar = 6.58211899E-16 #[eV s]
temp = 300 #[K] temperature
epsinf = 10.92 #High-frequency dielectric constant
eps0 = 12.9 #Static dielectric constant
epsp = 1/(1/epsinf - 1/eps0) #effective permittivity
w0 = 8.75e12 #[hz] frequency
me = 0.0655 #effective mass
Emax=4*hbar*w0*2*3.14 #[ev] max value of energy grid
E = seq(0.0,Emax,by=0.001*Emax) #ev energy grid
tm = momentumRelaxationTime_LOpolar_isotropic_woscreening_effcharge(permittivity = epsp,
freq=w0,
eff_mass = me,
temperature=temp,
energy=E)
plot(E/(hbar*w0*2*3.14),1.0e0/tm, type="l")
plot(E/(hbar*w0*2*3.14),tm, type="l")