relaxation time with single isotropic band model
Jiang
27 March 2017
isotropic band
Back to the expression of the scattering rate within the first-order perturbation theory \[ W(\vec{k}) = \int \frac{2 \pi}{\hbar} | \left< f | H | i \right> |^2 \delta(E_f - E_i) dS_f \] where the integral is carried out over the final states. The number of final states in an elementary volume of k-space is \[ dS_f = dk~ V/(2\pi)^3 \] where V is the volume of the unit cell.
So we can express this integral in the wave vector \(k\) space with the polar coordinate
\[ W(\vec{k}) = \frac{V}{\hbar(2 \pi)^2} \int_{0}^{\infty} d k' \int_0^{\pi} d\theta \sin(\theta) k' \int_0^{2 \pi} d\phi k' \frac{2 \pi}{\hbar} \left| \left< f | H | i \right> \right|^2 \delta(E_{k'} - E_{k} + \hbar \omega_q) \]
Figure 1 - relation between \(k\) and \(k'\)
Because of the isotropic condition, the problem reduces to 2D. We can simply: \(\int_0^{2 \pi} d \phi = 2\pi\). We can write the matrix element as following \[ \left| \left< f | H | i \right> \right|^2 = \frac{\hbar}{2NM'} \frac{C^2_{\vec{q},b}I^2(\vec{k},\vec{k}')}{\omega_{q,b}}[n(\omega_{q,b}) + 1/2 \mp 1/2] \] where the electron and phonon wave vectors should satisfy the momentum conservation \[ \vec{k}' = \vec{k} \pm \vec{q} \]
\[ W(k) =\frac{V}{4 \pi N M'} \int_{-1}^{1} d(\cos\theta) \int_0^\infty dk' ~ k'^2 \frac{(C_{q,\theta,b}I_{k,k',\theta})^2}{\omega_{q,b}} \delta(E_{k'} - E_{k} + \hbar \omega_q) \delta_{\vec{k} \pm \vec{q} - \vec{k}'} [n(\omega_{q,b}) + 1/2 \mp 1/2] \]
With \(\delta_{\vec{k} \pm \vec{q} - \vec{k}'}\), we can express q as function of k, k’ and \(\theta\), so \(C_{q,\theta,b}I_{k,k',\theta}\) is a function of just \(\theta\) and \(k'\).
The integration variable is changed from \(k'\) to \(E_{k'}\) in order to get rid of the delta function of the energy conservation. \[ W(k) = \frac{V}{4 \pi N M'} \int_{-1}^{1} d(\cos\theta) \int_0^\infty dE_{k'} \frac{dk(E_{k'})}{dE} ~ k'^2 \frac{(C_{q,\theta,b}I_{k,k',\theta})^2}{\omega_{q,b}} \delta(E_{k'} - E_{k} + \hbar \omega_q) (n(\omega_{q,b}) + 1/2 \mp 1/2) \] Because of the correspondance between \(k'\) and \(E_{k'}\) together with the momentum conservation condition, the terms in the inner integral is a function of \(E_{k'}\). Since \(\int_{-\infty}^{\infty} dx f(x) \delta(x) = f(0)\), the inner integral is equal to evaluating the terms on \(E_{k'}\). Therefore
\[ W(k) = \frac{V}{4 \pi N M'} \int_{-1}^{1} d(\cos\theta)~ \left[ \frac{dk(E_{k'})}{dE} k'^2 \frac{(C_{q,\theta,b}I_{k,k',\theta})^2}{\omega_{q,b}} (n(\omega_{q,b}) + 1/2 \mp 1/2) \right]_{E_{k'}=E_{k} - \hbar \omega_q } \]
Here, the energy conservation and momentum conservation conditions define the electron wave vector \(k'\) and the phonon wave vector \(q\), by satisfying the equation system
\[ \begin{cases} E_{k'} = E_k - \hbar \omega_q \\ q^2 = k^2 + k'^2 - 2 k k' \cos \theta \end{cases} \]
\[ E_{k'} = E_k - \hbar \omega_q \bigg\vert_{q=\sqrt{k^2 + k'^2 - 2 k k' \cos \theta}} \]
Here the phonon frequency is a function of the phonon vector \(q\). In the special case of acoustic branch for small \(q\), \(\omega_q = v_s q\) with \(v_s\) the sound velocity. In the case of optical branch for small \(q\), \(\omega_q = \omega_0\), a constant.
By solving the previous non-linear equation, we can obtain the energy \(E_{k'}\) and also \(k'\). By substituting all the corresponding terms in the integral, we will then only need to integrate over the \((\cos\theta)\) from -1 to 1.
To solve this non-linear equation we can use the Newton-Raphson method, by defining a function \[ F(k') = E_{k'} - E_k + \hbar \omega_q \bigg\vert_{q=\sqrt{k^2 + k'^2 - 2 k k' \cos \theta}} \] It’s equivalent to find the root of \(F(x) = 0\)
\[ F' = \frac{dF}{dk'} = \frac{dE}{dk} \bigg\vert_{k=k'} + \hbar \frac{d \omega }{dq} \bigg\rvert_{q = \sqrt{k^2 + k'^2 - 2 k k' \cos \theta}} \times \frac{1}{ \sqrt{k^2 + k'^2 - 2 k k' \cos \theta}} ( k' - k \cos \theta) \]
Then update iteratively the \(k'\) until the convergence
\[ k'_{n+1} = k'_n - \frac{F(k'_n)}{F'(k'_n)} \]
momentum relaxation time for acoustic phonon
We calculate the momentum relaxation time similarly to the scattering rate.
One thing to note here is that the relaxation time approximation (RTA) can only be employed in the elastic scattering case. So the phonon energy \(\hbar \omega_q\) should be small enough, otherwise the RTA cannot work.
For absorbtion: \[ \frac{1}{\tau_m} \bigg\vert_\text{abs} = \frac{V}{4 \pi N M'} \int_{-1}^{1} d(\cos\theta)~ (1 - \cos \theta) \left[ \frac{dk(E_{k'})}{dE} k'^2 \frac{(C_{q,\theta,b}I_{k,k',\theta})^2}{\omega_{q,b}} (n(\omega_{q,b}) + 1/2 \mp 1/2) \right]_{E_{k'}=E_{k} - \hbar \omega_q } \]
Let’s assume that the phonons are in equilibrium and follow the Bose distribution \[ n(\omega_{q,b}) = \frac{1}{e^{\hbar \omega_{q,b}/k_B T}-1} \] For very small \(\omega_{q,b}\) \[ n(\omega_{q,b}) \approx \frac{k_B T}{\hbar \omega_{q,b}} >> 1 \]
\[ n(\omega_{q,b}) \approx n(\omega_{q,b}) +1 \]
For acoustic phonons under the deformation potential approximation \[ C_{q,b} = \Xi_d q \] and \[ \omega_{q,b} = v_s q \]
\[ \frac{1}{\tau_m} \bigg\vert_\text{abs} = \frac{V k_B T \Xi^2_d}{4 \pi \hbar N M' v_s^2} \int_{-1}^{1} d(\cos\theta)~ (1 - \cos \theta) \left[ \frac{dk(E_{k'})}{dE} k'^2 I_{k,k',\theta}^2 \right]_{E_{k'}=E_{k} - \hbar \omega_q } \]
\[ \begin{cases} E_{k'} = E_k - \hbar v_s q \\ q^2 = k^2 + k'^2 - 2 k k' \cos \theta \end{cases} \]
\[ \begin{cases} F(k') = E_{k'} - E_k + \hbar v_s {\sqrt{k^2 + k'^2 - 2 k k' \cos \theta}}. \\ F' = \frac{dF}{dk'} = \frac{dE}{dk} \bigg\vert_{k=k'} + \hbar v_s \times \frac{1}{ \sqrt{k^2 + k'^2 - 2 k k' \cos \theta}} ( k' - k \cos \theta) \end{cases} \]
\[ \begin{cases} k'_0 = k_\text{guess} \\ k'_{n+1} = k'_n - \frac{F(k'_n)}{F'(k'_n)} \\ \lim_{n \to \infty} k'_n = k' \bigg\vert_{E_{k'}=E_k \pm \hbar\omega_q} \end{cases} \]
Now we should be able to evaluate \[ \frac{dk(E_{k'})}{dE} k'^2 I_{k,k',\theta}^2 ~, \] then by multiplying \((1-\cos\theta)\) and finally integrating over \(\cos\theta\) gives the momentum relaxation time.
The emission is the same as absorbtion, so \[ \frac{1}{\tau_m} = \frac{1}{\tau_m} \bigg\vert_\text{abs} + \frac{1}{\tau_m} \bigg\vert_\text{ems} = \frac{2}{\tau_m} \bigg\vert_\text{abs} \]
momentum relaxation time for polar LO phonon
Since \(\omega_{q,b}\) may be quite large for LO phonon, is the RTA still applicable ?
If so let’s calculate it. First the phonon distribution is quite different for absorbtion and emission
absorbtion \[ n_\text{a} = n(\hbar \omega_0) = \frac{1}{e^{\hbar \omega_0/k_B T}-1} \]
emission \[ n_\text{e} = n(\hbar \omega_0) +1 \]
\[ \begin{aligned} \frac{1}{\tau_{m,LO}} = \frac{V}{4 \pi N M'} \int_{-1}^{1} d(\cos\theta)~ (1 - \cos \theta) \left\{ \left[ \frac{dk(E_{k'})}{dE} k'^2 \frac{(C_{q,\theta,b}I_{k,k',\theta})^2}{\omega_0} ( n(\hbar \omega_0) +1) \right]_{E_{k'}=E_{k} - \hbar \omega_q } \\ + \left[\frac{dk(E_{k'})}{dE} k'^2 \frac{(C_{q,\theta,b}I_{k,k',\theta})^2}{\omega_0} n(\hbar \omega_0) \right]_{E_{k'}=E_{k} + \hbar \omega_q } \right\} \end{aligned} \] \[ C_q^2 = \left( \frac{ee^*}{V_0 \epsilon_0} \right)^2 \frac{q^2}{(q^2+q_0^2)^2} \]
Let’s not considering the screening for the moment, and insert the formula of effective charge \[ e^{*2}= M V \omega_0^2\epsilon_0^2 \left(\frac{1}{\epsilon_\infty} - \frac{1}{\epsilon} \right) . \]
\[ \begin{aligned} \frac{1}{\tau_{m,LO}} = \frac{e^2 \omega_0^2}{4 \pi \epsilon_p } \int_{-1}^{1} d(\cos\theta)~ (1 - \cos \theta) \left\{ \left[ \frac{dk(E_{k'})}{dE} \frac{k'^2}{q^2} \frac{I_{k,k',\theta}^2}{\omega_0} ( n(\hbar \omega_0) +1) \right]_{E_{k'}=E_{k} - \hbar \omega_q } \\ + \left[ \frac{dk(E_{k'})}{dE} \frac{k'^2}{q^2} \frac{I_{k,k',\theta}^2}{\omega_0} n(\hbar \omega_0) \right]_{E_{k'}=E_{k} + \hbar \omega_q } \right\} \end{aligned} \]
\[ q^2 = k^2 + k'^2 - 2 k k' \cos \theta \]
And finally, \[ \begin{aligned} \frac{1}{\tau_{m,LO}} = \frac{e^2 \omega_0}{4 \pi \epsilon_p } \int_{-1}^{1} d(\cos\theta)~ (1 - \cos \theta) \left\{ \left[ \frac{dk(E_{k'})}{dE} \frac{k'^2}{k^2 + k'^2 - 2 k k' \cos \theta} I_{k,k'}^2 (n(\hbar \omega_0) +1) \right]_{E_{k'}=E_{k} - \hbar \omega_0 } \\ + \left[ \frac{dk(E_{k'})}{dE} \frac{k'^2}{k^2 + k'^2 - 2 k k' \cos \theta} I_{k,k'}^2 n(\hbar \omega_0) \right]_{E_{k'}=E_{k} + \hbar \omega_0 } \right\} \end{aligned} \]
Verify this with Ridley’s formular
But Ridley used a different factor \((q/k)\cos\theta\) instead of \((1-\cos\theta)\), where \(\theta\) is the angle between \(q\) and \(k\). Therefore, with my notation, it is \((k-k'\cos\theta)/k\).
His formular corresponds to \(\hbar^2 k^2/2m = \gamma(E)\) situation.
For clarity, let’s seperate the absorbsion part and the emission part \[ \frac{1}{\tau_{m,LO}} \bigg\vert_\text{abs} = \frac{e^2 \omega_0}{4 \pi \epsilon_p } n(\hbar\omega_0) \int_{-1}^{1} d(\cos\theta) \left\{ \frac{k-k'\cos\theta}{k} I_{k,k',\theta}^2 \frac{k'^2}{k^2+k'^2 - 2kk'\cos\theta} \frac{1}{2}\frac{\sqrt{2m}}{\hbar}\frac{1}{\sqrt{\gamma(E_{k'})}}\frac{d\gamma(E_{k'})}{dE} \right\}_{E_{k'} = E_k+\hbar\omega_0} \]
Since \(E_{k'} = E_k+\hbar\omega_0\) is constant w.r.t. \(\theta\), \(k'\) is also constant w.r.t. \(\theta\). Therefore
\[ \frac{1}{\tau_{m,LO}} \bigg\vert_\text{abs} = \frac{e^2 \omega_0}{4 \pi \epsilon_p } n(\hbar\omega_0) \frac{1}{2}\frac{\sqrt{2m}}{\hbar} \left\{ \frac{1}{\sqrt{\gamma(E_{k'})}} \frac{d\gamma(E_{k'})}{dE} \right\}_{E_{k'} = E_k+\hbar\omega_0} \int_{-1}^{1} d(\cos\theta)(1- \frac{k'}{k}\cos\theta) \frac{k'^2}{k^2+k'^2 - 2kk'\cos\theta} I_{k,k',\theta}^2 \]
Ridley simplified this express by taking out the \(I_{k,k',\theta}^2\) out of the integral, thus assuming that \(I\) is not very sensible to \(\theta\).
\[ \frac{1}{\tau_{m,LO}} \bigg\vert_\text{abs} = \frac{e^2 \sqrt{2m} \omega_0}{8 \pi \hbar \epsilon_p } n(\hbar\omega_0) \left\{ I_{k,k'}^2\frac{1}{\gamma(E_{k'})}\frac{d\gamma(E_{k'})}{dE} \right\}_{E_{k'} = E_k+\hbar\omega_0} \times \left[ \frac{k'^2-k^2}{2k^2} \log\frac{k'+k}{k'-k} + \frac{k'}{k} \right] \]
\[ \begin{aligned} \frac{1}{\tau_{m,LO}} \bigg\vert_\text{abs} = \frac{e^2 \sqrt{2m} \omega_0}{8 \pi \hbar \epsilon_p } n(\hbar\omega_0) I_{k,k'}^2 \bigg\vert_{E_{k'} = E_k+\hbar\omega_0} \times \frac{1}{\gamma(E_k+\hbar\omega_0)}\frac{d\gamma(E_k+\hbar\omega_0)}{dE} \\ \times \left[ \frac{\gamma(E_k+\hbar\omega_0)-\gamma(E_k)}{2\gamma(E_k)} \log\frac{\sqrt{E_k+\hbar\omega_0}+\sqrt{E_k}}{\sqrt{E_k+\hbar\omega_0}-\sqrt{E_k}} + \frac{\sqrt{E_k+\hbar\omega_0}}{\sqrt{E_k}} \right] \end{aligned} \]
Equivalently the emission part is \[ \begin{aligned} \frac{1}{\tau_{m,LO}} \bigg\vert_\text{ems} = \frac{e^2 \sqrt{2m} \omega_0}{8 \pi \hbar \epsilon_p } ( n(\hbar\omega_0) +1) I_{k,k'}^2 \bigg\vert_{E_{k'} = E_k-\hbar\omega_0} \times \frac{1}{\gamma(E_k-\hbar\omega_0)}\frac{d\gamma(E_k-\hbar\omega_0)}{dE} \\ \times \left[ \frac{\gamma(E_k) - \gamma(E_k-\hbar\omega_0)}{2\gamma(E_k)} \log\frac{\sqrt{E_k} + \sqrt{E_k-\hbar\omega_0}}{\sqrt{E_k} - \sqrt{E_k-\hbar\omega_0}} + \frac{\sqrt{E_k-\hbar\omega_0}}{\sqrt{E_k}} \right] \end{aligned} \]
We can do this differently by inserting the good formula of \(I_{k,k',\theta}^2\) into the integral \[ I_{k,k',\theta} = (a_k a_{k'} + c_k c_{k'} \cos\theta), \] this will make the integral even more complicated. It may not worth the time to develop this analytically. Fawcett gave the final formula of the scattering rate in his paper in 1970 1, using simple non-parabolic band \(\hbar^2k^2/2m=E(1+\alpha E)\). We will do this integral numerically using the true k.p-8 band structure and the overlap integral and then compare to his expression.
summary of the numerical procedure
- for a given \(\cos\theta\), solve the non-linear equation system of energy and momentum conservation condition, obtaining \(k'\) and \(E_{k'}\)
- evaluate a function with \(\cos\theta\) and \(k'\) (this function contains derivative of the inverse of \(E(k)\) and some complicated terms)
- repeat step 1 and 2 for many \(\cos\theta\) values from -1 to 1 and sum them up to obtain the integral
This procedure is general for acoustic and LO phonon, and also general for scattering rate and relaxation time.