Random potential disorder average

In summary, the disorder average of a hamiltonian with a random potential is taken over an ensemble of potentials, resulting in a zero average of the potential and a correlation function that represents uncorrelated, point scattering potentials. This remains true even when expanding the fields in a new basis.
  • #1
gonadas91
80
5
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!
 
Physics news on Phys.org
  • #2


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.
 

Related to Random potential disorder average

What is random potential disorder average?

Random potential disorder average refers to the average value of a random potential distribution in a given system. This type of disorder is often used in statistical mechanics and condensed matter physics to model the effects of random fluctuations on a system.

How is random potential disorder average calculated?

The calculation of random potential disorder average involves taking the average value of the potential distribution over all possible configurations in a given system. This can be done using various mathematical techniques, such as Monte Carlo simulations or analytical methods.

What is the significance of random potential disorder average?

Random potential disorder average is important because it helps us understand how random fluctuations affect the behavior of a system. It can provide insights into the properties of materials and their response to external stimuli, such as temperature or pressure.

How does random potential disorder average impact the behavior of physical systems?

Random potential disorder average can have a significant impact on the behavior of physical systems. It can lead to phenomena such as localization, where particles are confined to a specific region, or Anderson localization, where particles are completely trapped. It can also affect the transport properties of materials, such as electrical conductivity.

What are some real-world applications of random potential disorder average?

Random potential disorder average has many real-world applications, particularly in the fields of materials science and engineering. It is used to study the behavior of disordered materials, such as glasses or polymers, and to design new materials with specific properties. It is also used in the development of electronic devices, such as transistors, where disorder can have a significant impact on performance.

Similar threads

Replies
1
Views
602
  • Quantum Physics
Replies
5
Views
675
  • Quantum Physics
Replies
4
Views
2K
  • Quantum Physics
Replies
6
Views
872
Replies
1
Views
1K
  • Quantum Physics
Replies
4
Views
1K
Replies
3
Views
613
  • Quantum Physics
Replies
4
Views
787
Replies
4
Views
1K
  • Quantum Physics
Replies
6
Views
1K
Back
Top