- #1

gonadas91

- 80

- 5

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!