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

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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

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

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

\begin{equation}
    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
\end{equation}

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 

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

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

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...

asked Jun 19 in Applied Physics by leppavirta (5 points) [ no revision ]

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