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Prove Bloch's Theorem from Integral Schrodinger's Eqn

  1. Mar 25, 2014 #1

    king vitamin

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    Gold Member

    Prove Bloch's Theorem from Integral Schrodinger's equation (sorry! tried to make an umlaut and accidentally hit the keyboard shortcut for posting, any way to edit titles?)

    1. The problem statement, all variables and given/known data

    If a scattering potential has the translation invariance property [itex]V(\mathbf{r} + \mathbf{R}) = V(\mathbf{r})[/itex], where [itex]\mathbf{R}[/itex] is a constant vector, prove that the scattering solutions [itex]\psi_{\mathbf{k}}^{(\pm)}[/itex] of the integral form of the Schrodinger equation are Bloch wave functions, since they satisfy the relation

    \psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}+\mathbf{R}) = e^{i \mathbf{k} \cdot \mathbf{R}}\psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}).

    2. Relevant equations

    The integral form of Schrodinger's equation is given as:

    \psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}) = N e^{i \mathbf{k} \cdot \mathbf{r}} - \frac{m}{2 \hbar^2 \pi} \int d^3 r' \frac{\exp(\pm ik|\mathbf{r} - \mathbf{r}'|)}{|\mathbf{r} - \mathbf{r}'|}V(\mathbf{r}')\psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}')

    3. The attempt at a solution

    I'm don't seem to be getting anywhere here. I begin by computing:

    \psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}+\mathbf{R}) = N e^{i \mathbf{k} \cdot (\mathbf{r}+\mathbf{R})} - \frac{m}{2 \hbar^2 \pi} \int d^3 r' \frac{\exp(\pm ik|\mathbf{r}+\mathbf{R} - \mathbf{r}'|)}{|\mathbf{r}+\mathbf{R} - \mathbf{r}'|}V(\mathbf{r}')\psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}')

    The first term is fine of course. In the second term, we obviously need to use the translation property of the potential, and we need to make the integral look like the original Schrodinger's equation for [itex]\psi_{\mathbf{k}}^{(\pm)}(\mathbf{r})[/itex]. So I change variables [itex]\mathbf{r}' \rightarrow \mathbf{r}' + \mathbf{R}[/itex] (Jacobian is trivial for shifts):

    \psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}+\mathbf{R}) = N e^{i \mathbf{k} \cdot (\mathbf{r}+\mathbf{R})} - \frac{m}{2 \hbar^2 \pi} \int d^3 r' \frac{\exp(\pm ik|\mathbf{r} - \mathbf{r}'|)}{|\mathbf{r}- \mathbf{r}'|}V(\mathbf{r}')\psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}'+\mathbf{R})

    where I have used the translation property of the potential to shift it. This is where I'm stuck. I can't really work with the exact wave function, but the equation as it is seems to be impossible to work with because the arguments of the Green's function and the wave function don't match. If I could even get a hint as to the idea behind this it would really help.

    I can also write:

    \psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}+\mathbf{R}) = e^{i \mathbf{k} \cdot \mathbf{R}} \left[N e^{i \mathbf{k} \cdot \mathbf{r}} - \frac{m}{2 \hbar^2 \pi} \int d^3 r' \frac{\exp(\pm ik|\mathbf{r} - \mathbf{r}'|)}{|\mathbf{r}- \mathbf{r}'|}V(\mathbf{r}')e^{-i \mathbf{k} \cdot \mathbf{R}}\psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}'+\mathbf{R}) \right]
    where I want to show
    e^{-i \mathbf{k} \cdot \mathbf{R}}\psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}'+\mathbf{R}) = \psi_{\mathbf{k}}^{(\pm)}(\mathbf{r}')
    This would work if the wavefunction were a momentum eigenstate (replace k by p/hbar), but I don't think this is true in general (k is defined as [itex]k^2 = 2mE/\hbar^2[/itex]).
    Last edited: Mar 25, 2014
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