Franz–Keldysh effect

${\displaystyle H={1 \over 2m}(\mathbf {p} +e\mathbf {A} )^{2}+V(r)}$ (A: vector potential, V(r): periodic potential)

${\displaystyle A={1 \over 2}A_{0}e[e^{i(k_{p}\cdot r-\omega t)}+e^{-i(k_{p}\cdot r-\omega t)}]}$ (k_p and e are the wave vector of em field and unit vector.)

Neglecting the square term ${\displaystyle A^{2}}$ and using the relation ${\displaystyle A\cdot p=p\cdot A}$ within the Coulomb gauge ${\displaystyle \nabla \cdot A=0}$, we obtain

${\displaystyle H\sim {p^{2} \over 2m}+V(r)+{e \over m}A\cdot p}$

Then using the Bloch function ${\displaystyle |jk\rangle =e^{ik\cdot r}u_{jk}(r)}$ (j= v, c that mean valence band, conduction band)

the transition probability can be obtained such that

${\displaystyle w_{cv}={2\pi \over \hbar }|\langle ck'|{e \over m}A\cdot p|vk\rangle |^{2}\delta [\mathrm {E} _{c}(k')-\mathrm {E} _{v}(k)-\hbar \omega ]}$ ${\displaystyle ={\pi e^{2} \over 2\hbar m^{2}}A_{0}^{2}|\langle ck'|exp(ik_{p}\cdot r)e\cdot p|vk\rangle |^{2}\delta [\mathrm {E} _{c}(k')-\mathrm {E} _{v}(k)-\hbar \omega ]}$(${\displaystyle k_{p}}$ means wave vector of light) ${\displaystyle e\cdot p_{cv}={1 \over V}\int _{v}e^{i(k_{p}+k-k')\cdot r}{u^{*}}_{ck'}(r)e\cdot (p+\hbar k)u_{vk}(r)d^{3}r}$

Power dissipation of the electromagnetic waves per unit time and unit volume gives rise to following equation

${\displaystyle \hbar \omega w_{cv}={1 \over 2}\omega \kappa _{2}\epsilon _{0}{E_{0}}^{2}}$

From the relation between the electric field and the vector potential, ${\displaystyle {E}=-{{\partial A} \over {\partial t}}}$, we may put ${\displaystyle E_{0}=\omega A_{0}}$

And finally we can get the imaginary part of the dielectric constant and surely the absorption coefficient. ${\displaystyle \kappa ={{\pi e^{2}} \over {\epsilon _{0}m^{2}\omega ^{2}}}\sum _{k,k'}|e\cdot p_{cv}|^{2}\delta [\mathrm {E} _{c}(k')-\mathrm {E} _{v}(k)-\hbar \omega ]\delta _{kk'}}$

An electron in the valence band(wave vector k) is excited by photon absorption into the conduction band(the wave vector at the band is ${\displaystyle k'=k_{e}}$) and leaves a hole in the valence band (the wave vector of the hole is ${\displaystyle k_{h}=-k}$). In this case, we include the electron-hole interaction.(${\displaystyle V(r_{e}-r_{h})}$)

Thinking about the direct transition, ${\displaystyle |k_{e}|,|k_{h}|}$ is almost same. But Assume the slight difference of the momentum due to the photon absorption is not ignored and the bound state- electron-hole pair is very weak and the effective mass approximation is valid for the treatment. Then we can make up the following procedure, the wave function and wave vectors of the electron and hole

${\displaystyle \Psi _{ij}(r_{e},r_{h})=\psi _{ik_{e}}(r_{e})\psi _{jk_{h}}(r_{h})}$ (i, j are the band indices, and r_e, r_h, k_e, k_h are the coordinates and wave vectors of the electron and hole respectively)

And we can take a total wave vector K such that

${\displaystyle K=k_{e}+k_{h}.}$ ${\displaystyle H=H_{e}+H_{h}+V(r_{e}-r_{h})}$

Then, Bloch functions of the electron and hole can be constructed with the phase term ${\displaystyle A_{cv}^{n,K}}$ ${\displaystyle \Psi ^{n,K}(r_{e},r_{h})=\sum _{c,k_{e},v,k_{h}}A_{cv}^{n,K}(k_{e},k_{h})\psi _{ck_{e}}(r_{e})\psi _{vk_{h}}(r_{h})}$

If V varies slowly over the distance of the integral, the term can be treated like following.

${\displaystyle [\mathrm {E} _{c}(k_{e})+\mathrm {E} _{h}(k_{h})+V(r_{e}-r_{h})-\epsilon ]A_{c,V}^{n,K}(k_{e},k_{h})=0(*)}$

here we assume that the conduction and valence bands are parabolic with scalar masses and that at the top of the valence band ${\displaystyle \mathrm {E} _{v}=0}$, i.e. ${\displaystyle \mathrm {E} _{c}(k_{e})={{\hbar ^{2}k_{e}^{2}} \over {2m_{e}}}+\mathrm {E} _{G},\mathrm {E} _{h}(k_{h})={{\hbar ^{2}k_{h}^{2}} \over {2m_{h}}}}$ (${\displaystyle \mathrm {E} _{G}}$ is the energy gap)

Now, The Fourier transform of ${\displaystyle A_{cv}^{n,K}(k_{e},k_{h})}$ and above (*), the effective mass equation for the exciton may be written as

${\displaystyle [(-{\hbar ^{2} \over 2M}\nabla ^{2})+(-{\hbar ^{2} \over 2\mu }\nabla ^{2}-{e^{2} \over 4\pi \epsilon r})]\Phi ^{n,k}(r,R)=[\mathrm {E} -\mathrm {E} _{G}]\cdot \Phi ^{n,K}(r,R)}$

${\displaystyle r=r_{e}-r_{h},R={{m_{e}r_{e}+m_{h}r_{h}} \over {m_{e}+m_{h}}},{1 \over \mu }={1 \over m_{e}}+{1 \over m_{h}},M=m_{e}+m_{h}}$

then the solution of eq is given by

${\displaystyle \Psi ^{n,K}(r,R)=\Psi _{K}(R)\psi _{n}(r)}$ ${\displaystyle \Psi ^{n,K}(r,R)={1 \over {\sqrt {V}}}exp(iK\cdot R)\phi _{n}(r)}$

${\displaystyle \phi _{n}(r)}$ is called the envelope function of an exciton. The ground state of the exciton is given in analogy to the hydrogen atom.

then, the dielectric function is

${\displaystyle \kappa _{2}(\omega )={\pi e^{2} \over {\epsilon _{0}m^{2}\omega ^{2}}}|e\cdot p_{cv}|^{2}\sum _{\lambda }|\phi _{\lambda }(0)|^{2}\delta (\mathrm {E} _{G}+\mathrm {E} _{\lambda }-\hbar \omega )(**)}$

detailed calculation is in.[5]

Franz–Keldysh effect means an electron in a valence band can be allowed to be excited into a conduction band by absorbing a photon with its energy below the band gap. Now we're thinking about the effective mass equation for the relative motion of electron hole pair when the external field is applied to a crystal. But we are not to take a mutual potential of electron-hole pair into the Hamiltonian.

When the Coulomb interaction is neglected, the effective mass equation is

${\displaystyle [-{\hbar ^{2} \over 2\mu }\nabla ^{2}-eE\cdot r]\psi (r)=\epsilon \psi (r)}$.

And the equation can be expressed,

${\displaystyle [-{\hbar ^{2} \over 2\mu }{d^{2} \over {dr_{i}^{2}}}-eE_{i}r_{i}-\epsilon _{i}]\psi (r_{i})=0}$( where ${\displaystyle \mu _{i}}$ is the value in the direction of the principal axis of the reduced effective mass tensor)

Using change of variables:

${\displaystyle \hbar \theta _{i}=({{e^{2}E_{i}^{2}\hbar ^{2}} \over {2\mu _{i}}})^{1/3},\xi _{i}={{\epsilon _{i}+eE_{i}r_{i}} \over {\hbar \theta _{i}}}}$

then the solution is

${\displaystyle \psi (\xi _{x},\xi _{y},\xi _{z})=C_{x}C_{y}C_{z}Ai(-\xi _{x})Ai(-\xi _{y})Ai(-\xi _{z})}$

where ${\displaystyle C_{i}={{\sqrt {e|E_{i}|}} \over {\hbar \theta }}}$

For example, ${\displaystyle E_{y}=E_{z}=0,E_{x}}$ the solution is given by

${\displaystyle \psi (x,y,z)=C\cdot Ai({{-eEx-\epsilon +\hbar ^{2}k_{y}^{2}/2\mu _{y}+\hbar ^{2}k_{z}^{2}/2\mu _{z}} \over {\hbar \theta _{x}}})}$

The dielectric constant can be obtained inserting this equation to the (**) (above block), and changing the summation with respect to λ to ${\displaystyle \int _{-\infty }^{\infty }d\epsilon _{x}d\epsilon _{y}d\epsilon _{z}}$

The integral with respect to ${\displaystyle d\epsilon _{x}d\epsilon _{y}}$ is given by the joint density of states for the two-D band. (the Joint density of states is nothing but the meaning of DOS of both electron and hole at the same time.)

${\displaystyle {\kappa (\omega ,E)}={{\pi ^{2}} \over {\epsilon _{0}m^{2}\omega ^{2}}}|e\cdot p_{c}v|^{2}{{e|E_{x}|} \over {\hbar \theta _{x}^{2}}}\int _{-\infty }^{\infty }{J^{2D}}_{cv}(\hbar \omega -\epsilon _{G}-\epsilon _{x})\cdot |Ai(-{{\epsilon _{x}} \over {\hbar \theta }})^{2}|d\epsilon _{x}.}$

where ${\displaystyle J_{cv}^{2D}(\hbar \omega )={(\mu _{y}\mu _{z})^{1/2} \over \pi \hbar ^{2}},\hbar \omega >\epsilon _{G}.}$

${\displaystyle =0,\hbar \omega <\epsilon _{G}.}$

Then we put ${\displaystyle \eta ={{\hbar \omega -\epsilon _{G}} \over {\hbar \theta _{x}}}}$

And think about the case we find ${\displaystyle \eta <<0}$, thus ${\displaystyle \hbar \omega <<\epsilon _{G}}$ with the asymptotic solution for the Airy function in this limit.

Finally,${\displaystyle \kappa _{2}(\omega ,E_{x})={1/2}\kappa _{2}(\omega )exp[{-4 \over 3}({{\epsilon _{G}-\hbar \omega } \over {\hbar \theta _{x}}})]}$

Therefore, the dielectric function for the incident photon energy below the band gap exist! These results indicate that absorption occurs for an incident photon.