Solid State: Electron in periodic potential

[natugnaro]

Homework Statement

Starting from the wave function of free electron and applying first order approximation of time independent perturbation theory find wave function of electron in periodic potential.

Homework Equations

Potential energy in reciprocal space
E_{p}(\vec{r}) = \sum_{\vec{g} }E_{g}\cdot e^{i\vec{g}\vec{r}}

free electron wave function
\phi_{k}(\vec{r})=e^{i\vec{k}\vec{r}

free electron wave function in first order approximation
\Psi_{k}(\vec{r})=\phi_{k}(\vec{r})+\sum_{\vec{k}^{'}\neq\vec{k}}\frac{\int\phi^{*}_{k'}(\vec{r})\cdot E_{p}(\vec{r})\cdot\phi_{k}(\vec{r})\cdot d^{3}r}{E(k)-E(k')}\cdot\phi_{k'}(\vec{r})

The Attempt at a Solution

\int\phi^{*}_{k'}(\vec{r})\cdot E_{p}(\vec{r})\cdot\phi_{k}(\vec{r})\cdot d^{3}r=\sum_{\vec{g} }E_{g}\int e^{i(-\vec{k'}+\vec{g}+\vec{k})\cdot\vec{r}}d^{3}r=\sum_{\vec{g}}E_{g}\cdot\delta_{\vec{k'},\vec{k}+\vec{g}}

I don't understan the last part , transition from complex exponential to Kronecker delta.
I have two options:
1.) When vectors satisfy k'=k+g then I have integral of \int d^{3}r which must be 1, I am not clear about this.
2.) when vectors k' and k+g are not equal I have integral of \int [Cos(a) + i*Sin(a)] d^{3}r
(a is some real number from the dot product in exponent) ,
this integral should be zero, but I don't see how ?

 

[Mute]

Your wavefunction isn't properly normalized. It should be

\phi_k(\mathbf{r}) = \frac{1}{\sqrt{V}}e^{i\bm k \cdot \bm r}

So, what results is the integral

\frac{1}{V}\int d\mathbf{r} e^{i(-\mathbf{k}' + \mathbf{g} + \mathbf{k})\cdot \mathbf{r}}

If \mathbf{k}' = \mathbf{k} + \mathbf{g}, then you just have an integral over the volume of your crystal, so you get V/V = 1.

If \mathbf{k}' \neq \mathbf{k} + \mathbf{g}, then the integral evaluates to

\left.\frac{1}{V}\frac{e^{iq_x x}}{iq_x}\frac{e^{iq_y y}}{iq_y}\frac{e^{iq_z z}}{iq_z}\right|_{\mathcal{V}}

where it is integrated over the volume \mathcal{V} and \mathbf{q} = -\mathbf{k}' + \mathbf{k} + \mathbf{g}. Now, the wavevectors k', g and k must take values such that the wavefunction vanishes at the boundary. Since the result is proportional to your wavefunction, it vanishes at the boundary, and the result is zero.

Hence, if k' = k + g, the integral gives 1; if k' =/= k + g, the integral is zero. We write the result compactly using the kronecker delta \delta_{k',k+g}

 

[natugnaro]

Thanks for help.
The case k'=k+g is clear to me now, but I'm still having difficulties with k' =/= k + g.

Since the result is proportional to your wavefunction, it vanishes at the boundary, and the result is zero.

I can't make \left.\frac{1}{V}\frac{e^{iq_x x}}{iq_x}\frac{e^{iq_y y}}{iq_y}\frac{e^{iq_z z}}{iq_z} proportional to \phi_k(\mathbf{r}) = \frac{1}{\sqrt{V}}e^{i\bm k \cdot \bm r}
If q=k then it is proportional, but q=-k'+k+g

 

[Mute]

Thanks for help.
The case k'=k+g is clear to me now, but I'm still having difficulties with k' =/= k + g.


I can't make \left.\frac{1}{V}\frac{e^{iq_x x}}{iq_x}\frac{e^{iq_y y}}{iq_y}\frac{e^{iq_z z}}{iq_z} proportional to \phi_k(\mathbf{r}) = \frac{1}{\sqrt{V}}e^{i\bm k \cdot \bm r}
If q=k then it is proportional, but q=-k'+k+g

e^{iq_x x} e^{iq_y y} e^{iq_z z} = e^{i q \cdot r} \propto \phi_q(\mathbf{r})

k, k' and g all satisfy the boundary conditions of the problem such that the wavefunction vanishes at the boundary. Since q is just a sum of these \phi_q will also vanish at the boundary.

 

[natugnaro]

I understand transition to Kronecker delta now.
But I run into more difficulties trying to solve this problem.
I'll put it aside for now.
regards Mute.