Interacting Fermi Gases
Mike Hermele
(Dated: February 11, 2010)
Notes on Interacting Fermi Gas for Physics 7450, Spring 2010
I.
FERMI GAS WITH DELTA-FUNCTION INTERACTION
Since it is easier to illustrate certain techniques and ideas, we will start by considering a Fermi gas with short-range
(in fact, delta function) interactions. We consider spin-1/2 fermions. The Hamiltonian is
Z
h ∇2 i
XZ
ψσ (r) + V0 d3 r n̂(r)n̂(r).
(1)
H=
d3 rψσ† −
2m
σ
This what we get from a potential V (r) = 2V0 δ(r − r ′ ). There is no external potential U (r).
II.
BASIC IDEA OF MEAN-FIELD THEORY, AND THE HARTREE APPROXIMATION
Our basic tool for understanding interacting problems in this course is mean-field theory. One way to express
the basic idea of mean-field theory is that we try to approximate the effect of interactions on a single particle, by
replacing the effect of all the other particles with some kind of averaged background field, or mean field. Another way
to say it is that we shall try to replace our interacting problem with an effective non-interacting problem. This new
non-interacting problem will not in general be the same as just dropping the interactions altogether. It will instead
incorporate them in some relatively simple way.
How do we do this in practice? By far the simplest way is to pick some physical observable that we hope behaves
reasonably well as an average field felt by all the particle. Let’s consider the fermi gas with delta function interactions.
A good guess might be to pick the density operator, which we write in the following form:
n̂(r) = n + (n̂(r) − n).
(2)
Here n is the average density n = hψgs |n̂(r)|ψgs i. In the present case, n is our mean field. The term in parentheses
describes the fluctuations of the density about its average. What we will assume – without justification – is that
the fluctuations are small. (We’ll see what this assumption means in practice shortly.) Depending on whether you’re
an optimist or a pessimist, you might think about this as an inspired guess, or an uncontrolled approximation. One
good point of view is that this is a “first-cut” technique to study a system, when you have some guess about what it
might do – this guess could come from some experimental data, or perhaps just intuition. After studying the system
in mean-field theory, you can try to do something more sophisticated. But there are plenty of cases where mean-field
theory alone does an excellent job of explaining an experiment or set of experiments – superconductivity is a great
example of this, which we will come to later in the course.
To proceed, we write the interaction term in the Hamiltonian as
Z
Z
Z
i
h
V0 d3 r n̂(r)n̂(r) = V0 d3 r n + (n̂(r) − n) n + (n̂(r) − n) = V0 d3 r n2 + 2n(n̂(r) − n) + (n̂(r) − n)2 (3)
Z
h
i
≈ V0 d3 r n2 + 2n(n̂(r) − n) .
(4)
The approximation is to drop the term quadratic in fluctuations – however, we do keep the linear term. If we don’t
keep the linear term, then we just replace the interaction with a constant, and there is no chance for the mean field to
have any influence on the fermions. So we have to keep at least the linear term to get any nontrivial answers. What
we have just done is mean-field theory where the mean field is the particle density – this is often called the Hartree
approximation.
It turns out that the Hartree approximation is kind of trivial for the Fermi gas. To see why, let’s look again at the
expression for the approximated interaction term:
Z
Z
h
i
V0 d3 r n2 + 2n(n̂(r) − n) = V0 V n2 + 2nV0 d3 r n̂(r) − n
(5)
(6)
= V0 V n2 + 2nV0 N̂ − N
= V0 V n2 .
(7)
2
R
R
Here, we used the fact that d3 r n̂(r) = N̂ , and d3 r n = N , and also that the number of particles is a constant N ,
so we can replace N̂ → N . The term linear in fluctuations completely disappeared! All that happened is that the
total ground state energy picked up a constant shift – this is an effect, but it is not a very remarkable one, because
the fermions do not feel this shift and they move around exactly as before, under the influence of exactly the same
non-interacting Hamiltonian we started with. Even though the Hartree approximation didn’t do much (for this system
– sometimes it is more interesting), it does illustrate the basic idea of mean field theory. Let’s next move on to a
different mean field that gives a more physically interesting result.
III.
STONER FERROMAGNETISM
Some metals such as Fe, Ni and Co are ferromagnetic. At low enough temperature, the electron spins spontaneously
align along some axis, and the system has a net magnetic moment. It is thought that this effect can be caused by
repulsive interactions – this is the Stoner mechanism of ferromagnetism. Now, our model is certainly not adequate
to describe a metal like Fe. For one thing, so far we’re considering short-range interactions. Moreover, effects of the
crystal lattice are very important in transition metals – the electron band structure is not free-electron-like, and the
Fermi surface has a non-spherical shape and often even a non-spherical topology. What we will see is that, at least in
mean-field theory, our model system becomes ferromagnetic if the interaction is strong enough.
It is actually not known with any certainty whether this mean field result can really be trusted. But we can make a
reasonable physical argument, so let’s start with that. Suppose we start from the non-interacting Fermi gas, with equal
populations of up and down spins, and start turning up the strength of the interactions by increasing V0 . The ground
state energy is going to go up, of course, and at some point the system will want to find a way to make the energy not
go up so much. Now, one thing we can observe is that the Pauli principle forbids two up spins from occupying the
same point in space – so two up spins don’t feel the delta function interaction. Only an up spin and a down spin can
feel the interaction. So one way to reduce the interaction contribution to the energy, is to start polarizing the system,
so that (say) more of the spins are up – the more of the spins that are up, the less the contribution of the interaction
to the ground state energy. Now it’s important to keep in mind that there is a cost to doing this – namely, the total
kinetic energy goes up when we take some of the down spins and convert them to up spins. So there is a competition
between the kinetic energy and the interaction energy. But if the interactions are very large, it’s probably okay to
pay some kinetic energy in order to reduce the contribution of the interaction energy. This line of reasoning says it is
natural to expect a ferromagnetic ground state once the interaction is strong enough.
Now let’s address the possibility of ferromagnetism in mean-field theory. We need to allow for the density of spin-up
and spin-down electrons to be different. This will be all we need to do, since we will assume that the total spin points
along the z-axis. (The axis is arbitrary, so we might as well choose the z-axis.) Now, we could try to do the Hartree
approximation as before, but using
n̂↑ (r) = n↑ + (n̂↑ (r) − n↑ )
n̂↓ (r) = n↓ + (n̂↓ (r) − n↓ ).
(8)
(9)
However, we would get exactly the same answer as before, namely that the electrons don’t feel the mean field. (Here,
note that n = n↑ + n↓ is the total density.)
Instead, we can rearrange the creation and annihilation operators in the interaction term, which suggests a different
mean field we can try. Let’s see how this goes:
Z
XZ
3
V0 d r n̂(r)n̂(r) = V0
d3 r ψσ† (r)ψσ (r)ψσ† ′ (r)ψσ′ (r).
(10)
σ,σ′
In the Hartree approximation, we grouped the first two operators together, and the second two operators together
(ψσ† (r)ψσ (r) and ψσ† ′ (r)ψσ′ (r)). But there is no reason we have to do it that way. Instead it’s worth seeing what
happens if we group the first and third operators (and so also the second and fourth operators). This will allow us
to try a different mean-field theory, and it will give us ferromagnetism. Let’s move the operators around to get the
desired grouping. First we want to exchange the positions of the last two creation and annihilation operators, which
3
gives
V0
XZ
d3 r ψσ† (r)ψσ (r)ψσ† ′ (r)ψσ′ (r) = −V0
σ,σ′
XZ
d3 r ψσ† (r)ψσ (r)ψσ′ (r)ψσ† ′ (r) + 2δ(0)V0
= −V0
d3 r ψσ† (r)ψσ (r)
σ
σ,σ′
XZ
XZ
d3 r ψσ† (r)ψσ (r)ψσ′ (r)ψσ† ′ (r) + 2δ(0)V0 N
(11)
σ,σ′
= V0
XZ
d3 r ψσ† (r)ψσ′ (r)ψσ (r)ψσ† ′ (r) + 2δ(0)V0 N .
(12)
σ,σ′
The last term, while it is infinite because of δ(0), is just a constant – so we can drop it. Next, we again exchange the
last two creation and annihilation operators. This generates another infinite constant that we also drop, and we are
left with the interaction term
XZ
−V0
d3 r ψσ† (r)ψσ′ (r)ψσ† ′ (r)ψσ (r).
(13)
σ,σ′
This form suggests that we might try instead ψσ† (r)ψσ′ (r) as a mean field. Now, in a ferromagnetic state with the
spin polarized along the z-axis, we must have that hψσ† (r)ψσ′ (r)i vanishes if σ 6= σ ′ . The reason is that the total spin
raising operator is given by
Z
S + = d3 r ψ↑† (r)ψ↓ (r),
(14)
and S + = Sx + iSy , where Sx and Sy are the x- and y-components of the system’s total spin, respectively. So if
hS + i =
6 0 (or, equivalently, if hS − i =
6 0), this means the spin is not completely polarized along the z-axis and has
some component along x or y. Therefore we can make our mean-field guess as follows:
ψσ† (r)ψσ′ (r) = δσσ′ nσ + (ψσ† (r)ψσ′ (r) − δσσ′ nσ ).
(15)
If we plug this into our interaction term, and neglect terms quadratic in the fluctuations, after a little algebra we
find:
Z
h
i
XZ
−V0
d3 r ψσ† (r)ψσ′ (r)ψσ† ′ (r)ψσ (r) ≈ −V0 V (n2↑ + n2↓ ) − 2V0 d3 r n↑ (n̂↑ (r) − n↑ ) + n↓ (n̂↓ (r) − n↓ ) . (16)
σ,σ′
It is instructive to express this, rather than in terms of nσ , in terms of n = n↑ + n↓ , and sz = (n↑ − n↓ )/2. sz is the
density of spin angular momentum pointing along the z-axis – it’s often just referred to as “the z-component of the
spin density.” Some algebra shows that
Z
Z
Z
h
h
i
1
n2 + 2s2z − V0 n d3 r n̂(r) − 2V0 sz d3 r n̂↑ (r) − n̂
−V0 V (n2↑ + n2↓ ) − 2V0 d3 r n↑ (n̂↑ (r) − n↑ ) + n↓ (n̂↓ (r) − n↓ ) = V0 V
2
Z
h
i
1
2
2
= V0 V − n + 2sz − 2V0 sz d3 r n̂↑ (r) − n̂↓ (r) .
2
Our mean-field Hamiltonian is then
Z
h
i 1
h ∇2 i
XZ
2
3
†
ψσ (r) + 2V0 V sz − 2V0 sz d3 r n̂↑ (r) − n̂↓ (r) − V0 V n2 .
H ≈ HMF =
d rψσ −
2m
2
σ
(18)
The most remarkable thing here is that the fermions feel their own spin density sz as a Zeeman magnetic field! This
is really the essence of mean-field theory. We never applied a magnetic field to the system, but, in mean-field theory,
the fermions feel the spin density of all the other fermions as a magnetic field. Since this effective magnetic field itself
leads to a spin density, this is a sort of feedback effect, that allows the system to develop a spontaneous magnetic
moment.
We’re not done yet, though. The problem with HMF is that it has the additional parameter sz – really, we should
be able to calculate sz in terms of the basic parameters characterizing our system, like interaction strength V0 and
density n. We can do this by observing that sz will choose itself in order to minimize the ground state energy obtained
from HMF . Moreover, we can calculate the ground state energy of HMF without too much trouble – there will be N↑
4
up-spin fermions forming a fermi surface, with Fermi wavevector kF ↑ , and similarly for the down-spin electrons. The
total fermion number is fixed at N = N↑ + N↓ . The total energy then has the form
1
EMF = EK↑ + EK↓ + 2V0 V s2z − 2V0 sz (N↑ − N↓ ) − V0 V n2
2
1
= EK↑ + EK↓ + 2V0 V s2z − 4V0 V s2z − V0 V n2
2
1
= EK↑ + EK↓ − 2V0 V s2z − V0 V n2 .
2
(19)
(20)
(21)
Here, EKσ is the kinetic energy of the spin-σ fermions. The kinetic energies can be expressed in terms of the Fermi
wavevector using the standard results from the theory of the non-interacting electron gas. One has
EKσ =
V
V (6π 2 )5/3 5/3
5
k
=
n ,
F
σ
20π 2 m
20π 2 m σ
(22)
where for the second equality we used nσ = kF3 σ /6π 2 . So the total energy is
EMF =
i
1
V (3π 2 )5/3 h
5/3
5/3
(n
+
2s
)
+
(n
−
2s
)
− 2V0 V s2z − V0 V n2 .
z
z
20π 2 m
2
(23)
For sufficiently small V0 this function is minimized when sz = 0, but once V0 increases past some critical value, it is
minimized for a nonzero spin density. To understand why this happens, you either look at this function in detail, or
consider the simplified function f (x) = (1 + x)5/3 + (1 − x)5/3 − αx2 , in the range |x| < 1. For α = 0, this function is
minimized at x = 0 and increases as |x| increases. But, once α is large enough, the second derivative of the function
at x = 0 goes negative, and the minimum starts moving away from x = 0.
Returning to the behavior of EMF , it is worth noting that |sz | ≤ n/2, and |sz | = n/2 means the system is fully
polarized (either all spins up or all spins down). Because of this fact there are two important values of V0 . As V0 is
increased from 0, first there is a value at which sz becomes nonzero. Then, as V0 is increased further, there is another
value of V0 above which |sz | = n/2. Finding these values is left as an exercise.
IV.
VARIATIONAL APPROACH
There are many different ways to do mean-field calculations. In the present case, it turns out we can get the
same results as above, but using an approach that gives some additional insight. The idea is to use a variational
wavefunction approach. Consider some state |ψi, and evaluate the expectation value of the energy Evar = hψ|H|ψi.
The variational principle of quantum mechanics tells us that Evar ≥ E0 , where E0 is the exact ground state energy
of H. Also, if Evar = E0 , then |ψi is an exact ground state. This by itself is too general and doesn’t help us that
much. But suppose we have some set of variational wavefunctions for which it is not too hard to evaluate Evar . Then
suppose we vary parameters, moving around in this set of wavefunctions to minimize Evar as best as possible. In
doing so, we will get a variational approximation to the ground state.
The class of wavefunctions we will use will be Slater determinants, in which case it is easy to evaluate Evar using a
result called Wick’s theorem. The idea is to minimize Evar to find the best Slater determinant approximating the true
ground state – since Slater determinants are ground states of non-interacting Hamiltonians, this essentially means we
try to find the non-interacting system that best approximates the interacting system. This is more less the same idea
as the mean-field theory we’ve been describing – we’re trying to reduce the interacting problem to some approximate
non-interacting one.
In practice, it is still very hard to find the best Slater determinant among all possible Slater determinants, so usually
we just look at some simple class of Slater determinant wavefunctions, and find the best wavefunction among this
class. The wavefunctions we will look at are the same as the ground states of HMF above – N↑ up-spin fermions form
a Fermi surface, and N↓ down-spin fermions form a Fermi surface.
We will state Wick’s theorem without proof and prove it later. The theorem says the following: Suppose |ψi is a
Slater determinant wavefunction. Suppose further that the operators A, B, C and D are all linear combinations of
annihilation operators, then
hψ|A† B † CD|ψi = hψ|A† D|ψihψ|B † C|ψi − hψ|A† C|ψihψ|B † D|ψi.
(24)
5
Now we can evaluate Evar for our Fermi gas with delta-function interaction. First we have to put the operators in
the right order in the interaction term, so we can use Wick’s theorem. We have, for the interaction term
Z
XZ
Hint = V0 d3 r n̂(r)n̂(r) = V0
d3 r ψσ† (r)ψσ (r)ψσ† ′ (r)ψσ′ (r)
(25)
σ,σ′
= −V0
XZ
d3 r ψσ† (r)ψσ† ′ (r)ψσ (r)ψσ′ (r) + δ(0)V0 N .
(26)
σ,σ′
The last term is a constant which we ignore. Dropping this term, we have
Evar = EK↑ + EK↓ + hψ|Hint |ψi
h
i
XZ
= EK↑ + EK↓ − V0
d3 r hψσ† (r)ψσ′ (r)ihψσ† ′ (r)ψσ (r)i − hψσ† (r)ψσ (r)ihψσ† ′ (r)ψσ′ (r)i
(27)
(28)
σ,σ′
= EK↑ + EK↓ − V0
XZ
σ,σ′
= EK↑ + EK↓ − V0
XZ
σ,σ′
i
h
d3 r (δσσ′ nσ )(δσσ′ nσ ) − n2
(29)
i
h
d3 r n2↑ + n2↓ − n2
(30)
1
= EK↑ + EK↓ − 2V0 V s2z + V0 V n2 .
2
(31)
Except for the last term, this is exactly what we found before for EMF . So, in particular, by minimizing this we find
the same spin density sz .
V.
INTERACTING ELECTRON GAS AND THE FOCK APPROXIMATION
The Hamiltonian of the interacting electron gas is
Z
h ∇2 i
XZ
e2
[n̂(r) − n][n̂(r ′ ) − n]
3
†
H=
ψσ (r) +
d r ψσ −
.
d3 rd3 r ′
2m
2
|r − r ′ |
σ
(32)
The interaction potential is the Coulomb potential V (r) = e2 /|r|. (We use cgs units here and throughout for all
formulas of electromagnetism, as is conventional in solid state physics.) We suppose that the density of electrons is n
– note that the interaction term has been modified to include a uniform positive background charge, of charge density
en. The (average) charge density from the electrons is −en, so overall the system is electrically neutral. This very
simple model is often referred to as the “jellium” model of a solid – the pointlike positive charges of the ion cores
have been smeared out into a uniform positive background.
We can apply our mean-field techniques to this model, too. But first, say some simple things that don’t require any
calculation, and that will be helpful in discussing the mean-field results. In the discussion below I keep the factors of
~. A important length scale in the problem is a, the typical spacing between particles. It is related to the density by
a ∝ n−1/3 . The way a is usually defined more precisely in this context is to imagine that each particle occupies its
own sphere of equal volume, and a is the radius of these spheres. Since 1/n is the volume per particle, we have
4
1
= πa3 ,
n
3
(33)
3 1/3
n−1/3 .
4π
(34)
and so
a=
Another important length is the Bohr radius a0 = ~2 /me2 . The ratio of these lengths is rs = a/a0 . Now, by crude
dimensional analysis, the kinetic energy per electron is EK ∼ ~2 /ma2 . And, the Coulomb energy per electron is on
the order of EC ∼ e2 /a. The ratio is
EC
a
∼
= rs .
EK
a0
(35)
6
So this says that, when the gas is very dense, rs is small and the Coulomb energy is small compared to the kinetic
energy. This means that the fermions only interact weakly. In this limit we expect that the system will be similar
to non-interacting fermions – that is, a degenerate Fermi gas. On the other hand, at low density, rs is large and the
Coulomb energy dominates. To minimize the total energy, it is more important for the electrons to minimize their
Coulomb energy – it is believed that the best way to do this is for the electrons to form a regular crystal lattice, which
is called a Wigner crystal. The reason this does a good job at minimizing Coulomb energy is that the crystal lattice
allows electrons to always stay pretty far away from their neighbors. So at large rs we have a Fermi gas (with maybe
some corrections to the behavior due to interactions), and at small rs we have a Wigner crystal. What happens in
between is an open problem, and a very challenging one at that. Something we know empirically from metals is that
the Fermi-gas-like behavior can persist up to pretty large values of rs , maybe rs ∼ 10 or more. The fact that such
a strongly interacting system can behave more or less like a Fermi gas is remarkable – this situation is described by
Landau’s theory of Fermi liquids. If there is time, we might have some more to say about this later in the course.
This discussion is useful for doing mean-field theory, since it tells us that most likely mean-field theory will be
good for small rs . That’s because the interactions are weak in that limit, so a description in terms of an effective
non-interacting problem is more likely to be valid.
The first thing we can see is that the Hartree approximation doesn’t do anything interesting here, just like for the
gas with delta function interactions. That’s because the interaction term is already quadratic in the fluctuating part
of the density, (n̂(r) − n), so the Hartree approximation just tells us to drop the interactions all together.
There is a different way of doing the mean-field theory which is called the Fock, or exchange, approximation.
Essentially, the idea is to proceed like we did for Stoner ferromagnetism, but we do not allow for a ferromagnetic
ground state. (We could also study Stoner ferromagnetism in this case, but we’ll keep it simpler to start out.) We
will encounter an instructive problem here in the behavior of the mean-field Hamiltonian, so our focus will just be
on getting HMF . In particular, we will drop all the constant terms in HMF and just focus on the terms actually
involving the fermion operators.
Our first step is to rearrange the fermion operators as we did for Stoner ferromagnetism. There are a bunch of
constant terms in the interaction term of the electron gas, having to do with the background positive charge density.
We also generate a bunch of constant terms, like before, when we rearrange the fermion operators. We drop all these
constant terms, and only focus on
Hint
e2
→−
2
Z
d3 rd3 r ′
ψσ† (r)ψσ′ (r ′ )ψσ† ′ (r ′ )ψσ (r)
|r − r ′ |
(36)
This suggests that we make the mean-field guess
where
ψσ† (r)ψσ′ (r ′ ) = Fσσ′ (r − r ′ ) + ψσ† (r)ψσ′ (r ′ ) − Fσσ′ (r − r ′ ) ,
(37)
Fσσ′ (r − r ′ ) = hψσ† (r)ψσ′ (r ′ )i.
(38)
What we are going to assume is that the ground state of the interacting Fermi gas keeps all the symmetries of
the non-interacting problem. This means, for example, we assume no ferromagnetism (since ferromagnetism breaks
rotational symmetry of the spin). And we also assume no breaking of spatial symmetries, etc. This assumption is
already implicit in the definition of Fσσ′ (r − r ′ ), since we take it to depend only on the difference r − r ′ , which is a
consequence of translation symmetry. If we plug this mean-field guess, drop the term quadratic in fluctuations, and
keep only the terms involving the fermion operators, we find
Z
i
e2 X
1 h
Fσσ′ (r − r ′ )ψσ† ′ (r ′ )ψσ (r) + Fσ′ σ (r ′ − r)ψσ† (r)ψσ′ (r ′ ) .
(39)
Hint → −
d3 rd3 r ′
′
2
|r − r |
′
σ,σ
Therefore the mean-field Hamiltonian is
Z
i
h ∇2 i
XZ
e2 X
1 h
ψσ (r)−
Fσσ′ (r − r ′ )ψσ† ′ (r ′ )ψσ (r)+ Fσ′ σ (r ′ − r)ψσ† (r)ψσ′ (r ′ ) .
d3 rd3 r ′
HMF =
d3 r ψσ† −
′
2m
2
|r − r |
′
σ
σ,σ
(40)
The issue here is how we are supposed to calculate the function F . If we guess a form for F , that will give us a
mean-field Hamiltonian and hence a ground state wavefunction. Then we can use this wavefunction to re-calculate
F by Fσσ′ (r − r ′ ) = hψσ† (r)ψσ′ (r ′ )i. We had better get the same answer that we started with! This is the notion of
self-consistency, which crops up in mean-field theory all the time. There is some important physics behind it, which
7
is that the mean-field is generated by all the particles together, but it also then influences all the particles. So there
is a kind of feedback effect. Because of this, we are not allowed to plug in just any function F into HMF – it has to
be self-consistent. In complete generality, the problem of finding self-consistent solutions can be tricky. However, in
many cases there is a way around it, which often consists of making an inspired guess.
In this case, the easiest way to guess is not to guess the form for F , but to guess the ground state wavefunction.
Remember that we wanted to assume our interacting Fermi gas preserves all the symmetries of the original noninteracting Fermi gas. The only Slater determinant I can think of that satisfies these properties is actually the ground
state of the non-interacting problem, |ψ0 i. (This is just the state we construct by filling up electrons into plane wave
states starting from zero energy, until we reach the Fermi energy.) So we are guessing that HMF has the same ground
state wavefunction as the non-interacting problem – it’s important to note that this does not mean that HMF has
all the same properties as the non-interacting problem; indeed, we will see there is a big difference. We will use the
non-interacting ground state to calculate F , and then we will check that indeed |ψ0 i is really the ground state of
HMF .
To calculate F we go to k-space:
Fσσ′ (r − r ′ ) = hψ0 |ψσ† (r)ψσ′ (r ′ )|ψ0 i
1 X −ik·r ik·r
=
e
e hψ0 |ψσ† (k)ψσ′ (k′ )|ψ0 i.
V
′
(41)
(42)
k,k
When we learned about second quantization, we discussed how to evaluate the expectation value hψ0 |ψσ† (k)ψσ′ (k′ )|ψ0 i
for spinless fermions. The only difference with spin is that the result should be proportional to δσσ′ – if we remove
a fermion of one spin, we need to put back a fermion of the same spin – otherwise we get a state orthogonal to the
ground state and the expectation value vanishes. Moreover, we get the same answer for removing and putting back
an up spin or a down spin (we don’t have any ferromagnetism, for example). Therefore
hψ0 |ψσ† (k)ψσ′ (k′ )|ψ0 i = δσσ′ δk,k′ Θ(kF − |k|),
(43)
where ǫk = k2 /2m and Θ(x) = 1 for x > 0 and Θ(x) = 0 for x < 0. Plugging this into our expression for F we find
Fσσ′ (r − r ′ ) ≡ δσσ′ F (r − r ′ ) =
δσσ′ X −ik·(r−r′ )
e
Θ(kF − |k|).
V
(44)
k
We could go ahead and evaluate this expression for F (by turning the sum into an integral over k), but it is more
int
useful to plug it into the interaction part of the mean-field Hamiltonian, HMF
, given in Eq. (39). We can first simplify
this a bit:
Z
i
e2 X
1 h
†
int
′
′
′
†
′
′ (r − r )ψ ′ (r )ψ (r) + Fσ ′ σ (r − r)ψ (r)ψ ′ (r )
= −
HMF
F
(45)
d3 rd3 r ′
σσ
σ
σ
σ
σ
2
|r − r ′ |
σ,σ′
Z
i
e2 X
1 h
F (r − r ′ )ψσ† (r ′ )ψσ (r) + F (r ′ − r)ψσ† (r)ψσ (r ′ )
(46)
= −
d3 rd3 r ′
′
2 σ
|r − r |
Z
i
F (r − r ′ ) h † ′
e2 X
†
′
ψ
(r
)ψ
(r)
+
ψ
(r)ψ
(r
)
(47)
d3 rd3 r ′
= −
σ
σ
σ
σ
2 σ
|r − r ′ |
i
XZ
F (r − r ′ ) h †
2
′
= −e
d3 rd3 r ′
ψ
(r)ψ
(r
)
.
(48)
σ
σ
|r − r ′ |
σ
Above we used the fact that F (r) = F (−r).
This form is still not so useful – we really want to express HMF in a form where it is a sum of a bunch of number
operators, each multiplied by a single-particle energy. Since we’ve assumed that the system is completely homogeneous
and translation-invariant, we expect that the single-particle eigenstates will still be plane waves. So we should try to
Fourier transform and go to momentum space. To do that, we express the fermion creation/annihilation operators in
8
terms of their Fourier transforms:
int
HMF
= −e2
XZ
d3 rd3 r ′
σ
i
F (r − r ′ ) h †
′
ψ
(r)ψ
(r
)
σ
σ
|r − r ′ |
Z
e2 X X
F (r − r ′ ) −ik·r ik′ ·r′ †
= −
e
e
ψσ (k)ψσ (k′ )
d3 rd3 r ′
V σ
|r − r ′ |
′
k,k
#
"Z
2 XX
′
′ ′
e
F
(r
−
r
)
= −
e−ik·r eik ·r .
ψσ† (k)ψσ (k′ )
d3 rd3 r ′
V σ
|r − r ′ |
′
(49)
(50)
(51)
k,k
Let’s evaluate the double integral in square brackets. We have, making the change of variables r → r + r ′ ,
Z
Z
F (r) ir′ ·(k′ −k) −ik·r
F (r − r ′ ) −ik·r ik′ ·r′
e
e
=
d3 rd3 r ′
e
e
d3 rd3 r ′
′
|r − r |
|r|
#" Z
"Z
#
′
′
F
(r)
=
d3 r
d3 r ′ eir ·(k −k)
e−ik·r
|r|
"Z
#
3 F (r) −ik·r
= V δkk′
d r
e
|r|
= V δkk′ G(k),
(52)
(53)
(54)
(55)
where in the last line we have defined G(k) as the Fourier transform of F (r)/|r|. Putting this back into the Hamiltonian, we have
XX
int
HMF
= −e2
G(k)ψσ† (k)ψσ (k),
(56)
σ
k
and therefore
HMF =
X X h k2
σ
k
2m
i
− e2 G(k) ψσ† (k)ψσ (k).
(57)
This tells us that the single-particle eigenstates of HMF are indeed plane-waves, but their energies are
ǫ̃k =
k2
− e2 G(k),
2m
(58)
which is not the same as for the non-interacting problem. Now, an inspection of the form of F (r) makes it clear
that F (r) only depends on |r|, and so G(k) can only depend on |k|. So ǫ̃k depends only on |k| and the energies are
spherically symmetric. As long as ǫ̃ is a monotonically increasing function of |k| (and in fact even under somewhat
weaker assumptions), then HMF indeed has the same ground state as the non-interacting problem. We just fill up
plane wave states going from small to large |k|, until we have the right number of electrons. So our guess for the
ground state wavefunction was a reasonable one.
To say anything else, we need to evaluate G(k). Plugging in the expression for F (r) in the form
Z
d3 q iq·r
F (r) =
e Θ(kF − |q|).
(59)
(2π)3
we have
F (r) −ik·r
e
|r|
Z
Z
d3 q
Θ(kF − |q|)
d3 r e−i(k−q)·r
=
(2π)3
|r|
Z
Z
3
1
d q
Θ(kF − |q|) d3 r e−i(k−q)·r
=
(2π)3
|r|
Z
d3 q Θ(kF − |q|)
= 4π
.
(2π)3 (k − q)2
G(k) =
Z
d3 r
(60)
(61)
(62)
(63)
9
In the last line we used the standard result for the Fourier transform of 1/|r|. This integral can be evaluated by going
to spherical coordinates – the details will be given in an appendix to these notes, and the result is
G(k) =
2kF
F (|k|/kF ),
π
(64)
where
F (x) =
1 1 − x2 1 + x +
ln .
2
4x
1−x
(65)
To get a feeling for what this result means, let’s express the single-particle energy ǫ̃k in a relatively simple form,
defining x = |k|/kF :
2e2 kF
k2
−
F (x)
2m
π
i
4me2
k2 h
F (x)
= F x2 −
2m
πkF
i
kF2 h 2 4 4 1/3
=
x −
rs F (x) .
2m
π 9π
ǫ̃k =
(66)
(67)
(68)
The important point here is that the interaction correction to the single-particle energy is proportional to rs , which
makes sense, since we argued that interactions have a small effect at small rs . One thing we can see right away from
this form is that ǫ̃(kF ) > ǫ̃(k) for k < kF . The way to show this is to look at the difference ǫ̃(kF ) − ǫ̃(k). This holds
even though ǫ̃k is not a monotonic function of |k|, and implies that indeed the states below kF are filled and those
above kF are empty. So the ground state of HMF is always the same as the non-interacting ground state, validating
our guess.
Something else we can see from this form is that ǫ̃k has a singularity at k = kF . In fact, the Fermi velocity diverges
there, which is not physical.
Despite the problem we found, this kind of mean-field theory can still be useful. For example, if one uses the
Hartree-Fock variational approach here to calculate the ground state energy, the result is correct in the limit of small
rs . More precisely, in the non-interacting problem the kinetic energy is proportional to 1/rs2 . Since small rs means
weak interaction, it is a good guess that the ground state energy can be expanded as a power series in rs . Using
perturbation theory techniques that are beyond the scope of this course, it can be verified that this is true, and the
first few terms in this series can be calculated (see chapter 5 Many-Particle Physics, 3rd ed., by G. D. Mahan). The
first term due to interactions goes like 1/rs , and it is this term that Hartree-Fock turns out to get correctly.