This is from Mahan’s book 1.
Section 1.3. of this book contains a nice discussion of the electron-phonon interactions.
Let’s start from the Hamiltonian
\[ H = H_p + H_e + H_{ei} \] with the phonon Hamiltonian \[ H_p = \sum_{q \lambda} \omega_{q \lambda} a^\dagger_{q \lambda} a_{q \lambda}, \] and the electron Hamiltonian \[ H_e = \sum_i \left[ \frac{p_i^2}{2m} + \frac{e^2}{2}\sum_{i \neq j} \frac{1}{r_{ij}} \right] \] summing over all the electrons, and the electron-ion interaction \[ H_{ei} = \sum_{ij} V_{ei}(r_i - R_j) \] is the potential between electron at \(r_i\) and ion at \(R_j\).
with small atomic displacements \(Q_j\) \[ R_j = R^0_j + Q_j \]
\[ V_{ei}(r_i - R_j) = V_{ei}(r_i - R^0_j - Q_j) = V(r_i - R^0_j) - Q_j \cdot \nabla V_{ei}(r_i - R^0_j) + O(Q^2) \] then we neglect the \(O(Q^2)\) term and consider only the linear electron-phonon interaction.
The constant term can be solved for the equilibrium Hamiltonian, which gives the Bloch states of the solids. Reformulating the Hamiltonian into \(H = H_0 + V\), where \(H_0\) is an equilibrium Hamiltonian that can be solved and \(V\) is a perturbation.
the electron-phonon interaction is \[ V_{ep}(r) = \sum_j Q_j \cdot \nabla V_{ei}(r - R^0_j) \]
Assume that the electron-ion potential \(V_{ei}\) possesses a Fourier transform \[ V_{ei}(r) = \frac{1}{N} \sum_{q} V_{ei}(q) e^{i q \cdot r} \]
\[ \nabla_r V_{ei}(r) = \frac{i}{N} \sum_{q} q V_{ei}(q) e^{i q \cdot r} \]
then we combine the two \[ V_{ep}(r) = \frac{i}{N} \sum_{q} V_{ei}(q) e^{i q \cdot r} q \cdot (\sum_j Q_j e^{-i q \cdot R^0_j}) \]
Using an earlier definition (1.85) \[ \frac{i}{N} \sum_j Q_j e^{-i q \cdot R^0_j} = \frac{i}{\sqrt{N}} \sum_G Q_{q+G} = - \sum_G ( \frac{\hbar}{2 M N \omega_q})^{1/2} \xi_{q+G} (a_q + a_q^\dagger) \] \(\xi\) is phonon polarization vector, \(MN = \rho v\) remember to sum over the phonon modes \(\lambda\) too, here is omitted only for the sake of simple notation.
Here, the \(V_{ep}\) is defined as the unscreened electron-atom potential. The electron-electron interaction (screening) can reduce a lot this potential in metals. \(V_{ep}\) can be obtained from first-principle or pseudo-potential calculations.
back to Ridley’s book 2.
Consider the adiabatic electron-phonon interaction:
the vibration of the ions about their equilibrium position ==> instantaneous changes in the energy of electrons and thus introduce a time-dependent component into the adiabatic one-electron Schroedinger equation.
The scattering rate is given by \[ W = \int \frac{2 \pi}{\hbar} | \left< f|V_{ep}|i \right> |^2 \delta(E_f - E_i) dS_f \]
In the first order approximation, the delta function is replaced by \[ \sin[(E_f - E_i)t / \hbar] / [(E_f - E_i) \pi] \] where \(t\) is time, and approximate to a delta function when \(t \rightarrow \infty\)
\[
W(k) = \frac{V}{8 \pi^2 N M'} \int dk' \frac{C^2_{q,b} I^2(k,k')}{\omega_{q,b}} \delta_{k \pm q - k', K}
\times (n(\omega_{q,b})+ 1/2 \mp 1/2) \delta(E_{k'}-E_{k} \mp \hbar \omega_{q,b}) ,
\] with the coupling coefficient
\[
C_{q,b} I(k,k') = \int_{\text{cell}} dr \psi^*_{n'k'}(r) H_{q,b}(r) \psi_{nk}(r) ,
\] and the \(I\) is the overlap integral \[
I(k,k') = \int_{\text{cell}} dr \psi^*_{n'k'}(r) \psi_{nk}(r) .
\]
For the polar, optical and piezoelectric interactions the macroscopic fields produce long-range fields that varies slowly over a unit cell ==>
This section is from [Cardona’s fundamental of semiconductors][cardona]
The variation in the Hamiltonian is approximated by the shifting of the electronic energies \(E_{nk}\)
\[ \left( \frac{\partial H_e}{\partial R_j} \right) |_{R^0_j} \approx \frac{\partial E_{nk}}{\partial R_j} |_{R^0_j} \]
In the infinite wave length situation, the movement becomes a global displacement of the lattice, so no phonon energies in the deformation potential approximation. In order to couple electrons and phonons, we have to consider small wave vector but not completely zero.
we consider long wave length (small wave vector) acoustic phonon situation.
\[ d_{ij} = \frac{\partial \delta R_i}{\partial R_j} \]
\(d_{ij}\) can be decomposed into two parts
the antisymmetric part corresponds to a rotation of the crystal, why? and the symmetric part describes a strain (known as the strain tensor) ==> \(S_{ij}\)
back to Ridley’s book 3.
The displacement in the plane wave expansion is \[ u(R) = \frac{1}{\sqrt{N}} \sum_q Q_q a_q e^{iq \cdot R} + c.c. \] where \(a_q\) is the unit polarization vector.
\[ H_{ep} = \sum_{ij} \Xi _{ij} S_{ij} \]
\(S_{ij}\) is the strain tensor elements. Inserting the displacement expression, we obtain \[ S_{ij} = \frac{1}{2}(\frac{\partial u_i}{\partial R_j} + \frac{u_j}{\partial R_i}) = \frac{1}{2 \sqrt{N}} \sum_q \left[ i Q_q (a_i q_j + a_j q_i) e^{i q \cdot R} +c.c. \right] \]
so the Hamiltonian is \[ H_{ep} = \frac{1}{2 \sqrt{N}} \sum_q \left[ i Q_q e^{i q \cdot R} \sum_{ij} \Xi_{ij} (a_i q_j + a_j q_i) +c.c. \right] \] and the coupling coefficient is \[ C^2_{q} = \left\{ \sum_{ij} \Xi_{ij} (a_i q_j + a_j q_i) \right\}^2 \] where \(i,j\) corresponds to the position vector \((x_1,x_2,x_3)\) index (in the 2D material case, only two instead of three), and this is a \(q\) dependent function. Also the deformation potential \(\Xi\) is \(q\) dependent.
strictly speaking, we should consider each direction of \(q\) separately in the sum (which is discussed in Herring’s paper 4). Here we apply an implicit averaging over the azimuthal angle of \(q\), regarding elastic anisotropy as small.
The \(\Xi_{ij}\) is listed in the table 1.7 (p.39) for \(\Gamma\), L and X valleys. For \(\Gamma\) valley the \(\Xi\) is reduced to just \(\Xi_d\) times an identity matrix.
In the case of L and X valleys it is convenient to consider transverse and longitudinal modes separately. with an implicit averaging over the azimuthal angle of \(q\). (condition is elastic anisotropy is small)
\[ \Xi_L (\theta) = \Xi_d + \Xi_u \cos^2{\theta_q} \]
\[ \Xi_T (\theta) = \Xi_u \sin{\theta_q} \cos{\theta_q} \]
\(\theta_q\) is the angle between \(q\) and the principal axis of the spheroidal valley. why? still need to read Herring’s paper… This derivation allows us to change the dependence of the deformation potential from \(q\) to the angle \(\theta_q\).
For any of the two modes, we have the following equation for the Hamiltonian \[ H_{ep} = \frac{1}{\sqrt{N}} \sum_q \left[ i Q_q e^{i q \cdot R} \Xi(\theta_q) q +c.c. \right] \] and the coupling coefficient is \[ C^2_q = \Xi^2(\theta_q) q^2. \]
Still with the azimuthal average approximation \[ \omega_q = v_s q \] and \[ n(\omega_q) = \frac{1}{e^{\hbar \omega_q / k_B T} - 1} \approx \frac{k_B T}{\hbar \omega_q} , \text{if } \frac{\hbar \omega_q}{k_B T} \ll 1. \]
for a spherical band, and with \(M' = \rho V/N\) where \(\rho\) is the mass density. And the overlap integral \(I(k,k')=1\). Where non-parabolicity is present this coefficient is less than unity, so more difficult. The deformation potential \(\Xi\) becomes independent of the angle and goes out of the integral for the scattering rate \[ W(k) = \frac{\Xi^2_d k_B T}{8 \pi^2 \hbar \rho v^2_{sL}} \int dk' \cdot \delta_{k \pm q - k',0} \delta(E_{k'}-E_k \mp \hbar \omega_q) , \] the two delta-functions are for momentum and energy conservation. How to treat them will be a difficulty …
Convert the k-space integral into q integral using the one-to-one relation bewteen k and q, and pass to the sperical coordinate system, \[ W(k) = \frac{\Xi^2_d k_B T}{8 \pi^2 \hbar c_L} \int_0^{q_{ZB}} dq \int_{-1}^{1} d(\cos{\theta}) \int_0^{2 \pi} d\phi \delta_{k \pm q - k',0} \delta(E_{k'}-E_k \mp \hbar \omega_q) , \] with \(c_L\) (called the average elastic constant for longitudinal modes) replaces the average \(\rho v^2_{sL}\).
After integrating over \(\phi\) and \(\cos{\theta}\), we obtain \[ W(k) = \frac{\Xi^2_d k_B T}{4 \pi \hbar^2 c_L v} \int_{q_{min}}^{q_{max}} dq ~ q \] where \(v\) denotes the group velocity of electron at energy \(E_k\). (the \(2 \pi\) just comes from the \(\int_0^{2 \pi} d\phi\)) how the integral over \(\cos{\theta}\) is done?, still need thinking…
For the situation of high temperature, where the equipartition can be assumed, it is possible to take the group velocity of electron \(v\) is much greater than the sound velocity \(v_{sL}\). ==> \(q_{min} = 0\) and \(q_{max} = 2k\)
\[ W(k) = \frac{\Xi^2_d k_B T k^2}{2 \pi \hbar^2 c_L v} \]
The relaxation time is related to the simple scattering rate \(W(k)\). The momentum and energy relaxation time are derived from the scattering rate by weighting \(W(k)\) by the appropriate change in momentum and energy. For momentum relaxation time \[ \frac{d\hbar k}{dt} = \int dk' \hbar (k'-k) W(k,k') \delta(E_{k'}-E_k \mp \hbar \omega_q) \approx - \hbar k \int dk' (1 - \cos{\theta_k}) W(k,k') , \] where \(\theta_k\) is scattering angle. \(k \approx k'\) is assumed for acoustic scattering. Conservation of energy and momentum entails \[ q^2 = 2k^2(1 - \cos{\theta_k} ). \] the factor \((1 - \cos{\theta_k} )\) just cancels out the same weighting factor, so it is equivalent to multiply the integrand by \(q^2/2k^2\). \[ 1 / \tau_m = W(k) \] for equipartition.
Example: Plot the \(\tau_m\) as a function of the energy for room temperature T = 300 K for GaAs, with parameters from Rode’s paper 5.
source("relaxationTime.R")
temp = 300 #[K] temperature
vsL = 524000 #[cm/s] Speed of sound
rho = 5.360 #[g cm-3]Material density
me = 0.0655 # effective mass
Xid = 7.0 #[ev] deformation potential
E = seq(0.01,1,by=0.01) #ev energy grid
tm = momentumRelaxationTime_acoustic_isotropic (mass_density = rho,
sound_velocity = vsL,
deformation_pot = Xid,
temperature=temp,
eff_mass=me,
energy=E)
plot(E,1/tm, type="l")
plot(E,tm, type="l")
The single particle relaxation time is related to the imaginary part of the single particle self-energy function by \[ \frac{1}{\tau_s} = \frac{2}{\hbar} \Im \Sigma(k_F,E_F). \]
… …