Rate of Photon Absorption in a Semiconductor - Fermi's Golden Rule

+ 1 like - 0 dislike
484 views

In this paper, the rate of photon absorption in a semiconductor is given by Fermi's golden rule:

$$w_{1,2} = \frac{2\pi}{\hbar}\frac{e^{2}}{m^{2}c^{2}}\frac{2V}{(2\pi)^{3}}\int_{B.Z.}|\mathbf{A}\cdot\mathbf{p}_{1,2}|^{2} \delta (E_{1}(k)-E_{2}-h\nu)dk$$

$$w_{total} = \sum_{1,2}w_{12}$$

where $\textbf{A}$ is the vector potential of the light, $V$ is the volume. A factor of 2 for spin is included and the integration goes over one Brillouin zone (B.Z). and transitions are considered between bands with energy $E_{i}(k)$. The index 1 runs over all the empty bands, and 2 runs over all the filled bands. $h\nu$ is photon energy.

I am struggling to derive this equation, in particular the volume, $V$, element.

My attempt:

In the case when a photon interacts with the electron the hamiltonian in Schrodinger equation is modified to become

\hat{H} = H_{0} + U(t),

where $U(t)$ is describing the perturbation caused by the incident photon. The electromagnetic perturbation is of the form

U = - \frac{i\hbar e}{2mc} \nabla \cdot \mathbf{A} - \frac{i\hbar e}{mc} \mathbf{A} \cdot \nabla + \frac{e^{2}}{2mc^{2}}|\mathbf{A}|^{2} - e\phi

where $\mathbf{A}$ is the vector potential and $\phi$ is the scalar potential. We define the vector potential and the scalar potential such that  $\nabla \cdot \mathbf{A} = 0$ and  $\phi = 0$ so that the time dependent potential reduces to

U = -\frac{i\hbar e}{m_{e}c} \mathbf{A}\cdot \nabla + \frac{e^{2}}{2m_{e}c^{2}}|\textbf{A}|^{2}.

Noting that the momentum operator, $\textbf{p} = -i\hbar\nabla$, and assuming the $\textbf{E}$ field of the photon is negligible, so that $|\textbf{A}|^{2}\approx 0$, the potential further reduces to

U = \frac{e}{m_{e}c} \textbf{A}\cdot \textbf{p}.

The probability per unit time $w$ that a photon makes a transition at a given $k$ in the BZ is given by

$$w \approx \frac{2\pi}{\hbar} | \langle v|\hat{H}|c \rangle|^{2} \delta (E_{1}(k)-E_{2}-h\nu)$$

Recognising that the perturbation matrix elements $| \langle v|\hat{H}|c \rangle|$ and the joint of the density of state are $k$ dependent, we obtain upon the integration of the above equation over the total transition probability per unit time to be

$$W = \frac{2\pi}{\hbar} \frac{2}{(2\pi)^{3}} \int | \langle v|\hat{H}|c \rangle|^{2} \delta (E_{1}(k)-E_{2}(k)-h\nu) d^{3}k$$

Substituting our expression for $\hat{H}$ becomes

$$W = \frac{2\pi}{\hbar} \frac{2}{(2\pi)^{3}} \frac{e^{2}}{m^{2}c^{2}} \int | \langle v|\textbf{A}\cdot\textbf{p}|c \rangle|^{2} \delta (E_{1}(k)-E_{2}(k)-h\nu) d^{3}k$$

But this doesn't seem to be leading anywhere useful...

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.