Random potential disorder average

[gonadas91]

Hi,

I have a basic question concerning disorder average in random potentials. Suppose we have a hamiltonian (in second quantised notation) in the form:
$$H=H_{0}+\int d\vec{r}\psi^{\dagger}(\vec{r})V(\vec{r})\psi(\vec{r})$$
with $V(\vec{r})$ some random potential satisfying $\langle V(\vec{r})\rangle=0$ and $\langle V(\vec{r})V(\vec{r}')\rangle = D\delta(\vec{r}-\vec{r}')$, that is, a gaussian-like random potential that is uncorrelated as a function of the distances. In fact, the total number of scatterers is important and one defines $V$ on a microscopic level (for points scatterers):
$$V(\vec{r})=\sum_{j=1}^{N}V(\vec{r}-\vec{r}_{j})\to V(\vec{r}-\vec{r}_{j})=V_{0}\delta(\vec{r}-\vec{r}_{j})$$

Usually, one works in the momentum basis if $H_{0}$ is the standard free hamiltonian and expands the fields in that basis:

$$ \begin{eqnarray} H_{0}=\frac{\vec{p}^{2}}{2m}\to \psi(\vec{r})=\frac{1}{\sqrt{V}}e^{i\vec{p}.\vec{r}}\psi(\vec{p})\end{eqnarray} $$

Then, when it comes to the potential $V(\vec{r})$ one gets in momentum space:
$$V(\vec{p}-\vec{p}') \equiv V(\vec{q})= \int d\vec{r} e^{i\vec{q}.\vec{r}}V(\vec{r})$$
Then, to take the disorder average:
$$\begin{align} \langle V(\vec{q})\rangle = 0\\
\langle V(\vec{q})V(\vec{q}')\rangle = D\delta_{\vec{q}+\vec{q}'}
\end{align}$$
I am not 100% sure about the first equation above, which states that if $\langle V(\vec{r})\rangle=0\to \langle V(\vec{q})\rangle=0$, is that correct? Secondly, imagine we don't have the simple free-plane wave hamiltonian, but a more general $H_{0}$ from which a complete set of eigenstates is known so:
$$H_{0}\phi_{\lambda}(\vec{r})=\varepsilon_{\lambda}\phi_{\lambda}(\vec{r})$$
Then we expand the fields in this set of eigenstates, and we get the projected $V_{\lambda\lambda'}$ in this new basis:
$$V_{\lambda\lambda'}=\int d\vec{r}\phi_{\lambda}^{*}(\vec{r})V(\vec{r})\phi_{\lambda}(\vec{r})$$
Assuming that the $\langle V(\vec{r})\rangle$ and $\langle V(\vec{r})V(\vec{r}')\rangle$ do NOT change, so they are still representing uncorrelated, point scattering potentials, how does this change the above considerations of the model for the potential in this new basis? I mean, do we have:

$$\begin{align} \langle V_{\lambda\lambda'}\rangle = 0\\
\langle V_{\lambda_{1}\lambda_{1'}}V_{\lambda_{2}\lambda_{2'}}\rangle = \int d\vec{r} \phi_{\lambda_{1}}^{*}(\vec{r})\phi_{\lambda_{2}}^{*}(\vec{r})\langle V(\vec{r})V(\vec{r}')\rangle\phi_{\lambda_{1'}}(\vec{r})\phi_{\lambda_{2'}}(\vec{r})
\end{align}$$
Are these equations correct here? Thanks in advance!

 

[eljose79]

Hello,

Thank you for your question. Yes, your first equation is correct. Since the disorder average is taken over an ensemble of random potentials, the average of the potential itself will be zero.

As for your second set of equations, they are also correct. When expanding the fields in a new basis, the disorder average of the potential will still be zero and the correlation function will still represent uncorrelated, point scattering potentials. The only difference is that the potential will now be expressed in terms of the new basis functions, as shown in your equations.

I hope this helps clarify your understanding. Let me know if you have any further questions.