Take the 2-minute tour ×

In this paper on the Quantum Hall effect the authors refer to something called the correlation energy of electrons. It is defined at the top of page 5 as

$E=\frac{n}{2}\int (g(r)-1)V(r)dA\ ,$

where $n$ is electron density, $g(r)$ is the pair correlation function, $V(r)$ is the potential and $dA$ is the area element (area because the system is 2D).

I'm trying to understand where this expression comes from, but searching around for "correlation energy" all I can find is connected to the Hartee-Fock approximation, which is unrelated I think.

Testing the expression out it gives the same results as other ways of computing the energy per particle, with background charge included.

Does anyone have an explanation for this expression?

EDIT: prompted by an answer below I'll expand a bit on the mentioned "other ways of computing the energy."

First we have the interaction between the electrons, which gives the following energy per particle:

$E_{el}=\displaystyle\frac{\langle V\rangle}{N}=\frac{1}{N}\sum_{i<j}^N\int\prod_{k=1}^NdA_k\psi^*V(r_{ij})\psi=\frac{N-1}{2}\int\prod_{k=1}^NdA_k|\psi|^2V(r)\ , \ \ \ \ \ \ \ \ \ \ (1)$

where $\psi(r)=\langle r|\psi\rangle$ is the wavefunction and we've used the fact that it is an eigenfunction of $r$ and antisymmetric in electron exchange.

At this point it is convenient to mention that all this takes place on a sphere; that is the electrons live on a spherical shell at the center of which we have a magnetic Dirac monopole. I didn't include this in the original post because I don't think it's relevant to the question of the expression for the correlation energy, but it explains the next term:

To also include the effect of the positive background charge we can place it all in the center. The energy per electron resulting from the interaction between the background chance and the electrons, and the background charge with itself, will then be:

$E_{bg}=-\frac{N}{2R}\ ,\ \ \ \ \ \ \ \ \ (2)$

where $R$ is the radius of the sphere (as explained better in e.g. Jain's book "Composite Fermions").

Returning to the expression for the correlation energy in the original post, we note that the definition of the pair correlation function for an isotropic system (so that we only need the relative coordinate, e.g. $r=r_1-r_2$), is:


The term including $g(r)$ is therefore:

$\frac{n}{2}\displaystyle\int dA\ g(r)V(r)=\frac{N-1}{2}\int\prod_{k=1}^NdA_k|\psi|^2V(r)\ ,\ \ \ \ \ \ \ \ \ (3)$

where we've used the fact that $g(r)$ refers to any of the electrons to introduce an additional integration, over e.g. electron $1$, and cancel it's contribution with $1/A$, and the fact that $g(r)$ is isotropic to change an integration over a relative coordinate (the one in the original integral) to e.g. over electron $2$, bringing the integration to all electrons instead of just number $3$ to $N$.

Lastly we look at the last term of the original expression:

$-\frac{n}{2}\displaystyle\int dA\ V(r)=-\frac{n}{2}\int_0^R\frac{4\pi r\ dr}{r}=-\frac{N}{2R}.\ \ \ \ \ \ \ (4)$

We see that (1) corresponds to (3) and (2) to (4), so that this way of calculating the energy gives the same answer as the "correlation energy." The reason I'm interested in finding out about the latter is that this expression seems to be valid in more general cases, or at least easier to calculate, than the first method above. One example of this is when in the second Landau level and using an effective interaction in the first Landau level, so that $V(r)\neq \frac{1}{r}$ and it is less clear how to handle the background charge.

share|improve this question
I found a paper which apparently explains it (see Formula 3.25 page 32, in Chapter 3.4 pages 31,32). Not sure this is the simplest paper... –  Trimok Sep 13 '13 at 15:53
@Trimok Thank you! That might be very helpful; I'm digging in. Also, maybe there is a connection to Hartree-Fock after all.. –  jorgen Sep 13 '13 at 17:24
Don't know, see also Electronic correlation –  Trimok Sep 13 '13 at 17:40
So basically the difference between our interpretations comes down to whether the LHS of your (4) is supposed to correspond to the uniform background charge, or the energy an electron would have if the electrons were distributed uniformly (i.e. uncorrelated). Of course these two things are the same which is what makes it confusing. Either way, we agree that the correlation energy is equal to the difference of a single electron energy in a positionally correlated vs uncorrelated system in this case. –  NowIGetToLearnWhatAHeadIs Sep 16 '13 at 19:45
You may be right, but I don't immediately understand how these are the same thing: The way I understand you one is the difference in energy of the electrons when we turn the spatial correlation on and off, while the other is the total interaction energy per particle (with spatial correlations) of electron-electron, electron-background and background-background. –  jorgen Sep 16 '13 at 19:52

2 Answers 2

Looking at it, I would say it is the energy caused by turning on positional correlations.

Let's say that there aren't spatial correlations. Then the no correlation energy is $E_{n.c.} = \frac{1}{2} \int n V(r) dV$, where the quantities are defined as in your question.

Now lets suppose we put an electron on the origin and turn on correlations. The density of the other electrons will no longer be uniform because of attraction and repulsion with the electron at the origin. But by symmetry, we expect the electron density function to depend only on the distance from this electron we have put in. One can express this radial dependence with a function $g(r)$ which has the following properties: $g(r) \to 1$ as $r \to \infty$, and $g(r)$ is proportional to the electron density. Another way of saying this is that $g(r)$ is the electron density perturbed by correlations, normalized by the unperturbed density (since the perturbed and unperturbed densities should be equal at infinity where no one knows about the electron at the origin). With this in place, the correlated electron density is $\rho_c(r) = n g(r)$. The energy of this electron configuration is $E_c = \frac{1}{2} \int \rho_c(r) V(r) dV = \frac{1}{2} \int n g(r) V(r) dV$.

Now lets find the energy from turning on positional correlation. After turning on correlations, there is an energy $E_c$, but before turning on correlations, there was an energy $E_{n.c.}$. Thus there was an energy difference $E_c - E_{n.c.} =\frac{1}{2} \int n g(r) V(r) dV - \frac{1}{2} \int n V(r) dV = \frac{n}{2} \int (g(r) -1)V(r)dV $. This is just a guess though. Tell me if it makes sense.

share|improve this answer
Thanks for the insight! It has helped me think further about this. At first sight I don't agree completely, I think it's best to expand my original post to show what I mean, so I'll do that. –  jorgen Sep 16 '13 at 18:33

In case someone else ends up here, here's what I think is going on:

As shown in the edit of the question the term containing $g(r)$ gives $E_{el-el}$, the Coulomb energy per particle of the electron-electron interaction. Then one assumes a homogeneous positive background of charge density $n=N/A$ to keep the system neutral. Placing an electron at the north pole the Coloumb energy of this with the background is

$E_{el-bg}=-n\displaystyle\int V(r)dA\ ,$

and this will be the same for each electron.

Then the interaction between the background and itself should be $E_{bg-bg}=-\frac{1}{2}E_{el-bg}$ (half as many pairs), and the three terms together give the original expression in the question. Two comments:

I think it is a coincidence that this is essentially the same expression that occurs in connection with the Hartree-Fock approximation in (3.25) in the paper that Trimok refers to in the comments and also that NowIGetToLearnWhatAHeadIs derives in another answer. The quantities being calculated are quite different.

It seems some authors use the same expression, with an effective potential, in the 2nd Landau level and other situations, thus assuming that the background feels the same, altered potential. This was what threw me off in the first place (together with the fact that it was referred to as "correlation energy," which seems to point to the Hartree-Fock interpretation).

share|improve this answer

Your Answer


