Interacting Fermi Gases Mike Hermele (Dated: February 11, 010) Notes on Interacting Fermi Gas for Physics 7450, Spring 010 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/ fermions. The Hamiltonian is H = d 3 rψ ψ (r) + V 0 d 3 r ˆn(r)ˆn(r). (1) m This what we get from a potential V (r) = V 0 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: ˆn(r) = n + (ˆn(r) n). () Here n is the average density n = ψ gs ˆn(r)ψ gs. 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 V 0 d 3 r ˆn(r)ˆn(r) = V 0 d 3 r n + (ˆn(r) n) n + (ˆn(r) n) = V 0 d 3 r n + n(ˆn(r) n) + (ˆn(r) n) (3) V 0 d 3 r n + n(ˆn(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: V 0 d 3 r n + n(ˆn(r) n) = V 0 V n + nv 0 d 3 r ˆn(r) n (5) = V 0 V n + nv 0 ˆN N (6) = V 0 V n. (7)
Here, we used the fact that d 3 r ˆn(r) = ˆN, and d 3 r n = N, and also that the number of particles is a constant N, so we can replace ˆN 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 V 0. 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 ˆn (r) = n + (ˆn (r) n ) (8) ˆn (r) = n + (ˆn (r) n ). (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: V 0 d 3 r ˆn(r)ˆn(r) = V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r). (10) In the Hartree approximation, we grouped the first two operators together, and the second two operators together (ψ (r)ψ (r) and ψ (r)ψ (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 V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r) = V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r) + δ(0)v 0 d 3 r ψ (r)ψ (r) = V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r) + δ(0)v 0N (11) = V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r) + δ(0)v 0N. (1) 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 V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r). (13) This form suggests that we might try instead ψ (r)ψ (r) as a mean field. Now, in a ferromagnetic state with the spin polarized along the z-axis, we must have that ψ (r)ψ (r) vanishes if. The reason is that the total spin raising operator is given by S + = d 3 r ψ (r)ψ (r), (14) and S + = S x + is y, where S x and S y are the x- and y-components of the system s total spin, respectively. So if S + 0 (or, equivalently, if S 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: ψ (r)ψ (r) = δ n + (ψ (r)ψ (r) δ n ). (15) If we plug this into our interaction term, and neglect terms quadratic in the fluctuations, after a little algebra we find: V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r) V 0V (n + n ) V 0 d 3 r n (ˆn (r) n ) + n (ˆn (r) n ). (16) It is instructive to express this, rather than in terms of n, in terms of n = n + n, and s z = (n n )/. s z 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 V 0 V (n + n ) V 0 d 3 r n (ˆn (r) n ) + n (ˆn (r) n ) ) V 0 n ( 1 = V 0 V n + s z ( = V 0 V 1 ) n + s z V 0 s z d 3 r ˆn(r) V 0 s z d 3 r ˆn (r) ˆn (r). d 3 r ˆn (r) ˆn Our mean-field Hamiltonian is then H H MF = d 3 rψ ψ (r) + V 0 V s z m V 0s z d 3 r ˆn (r) ˆn (r) 1 V 0V n. (18) The most remarkable thing here is that the fermions feel their own spin density s z 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 H MF is that it has the additional parameter s z really, we should be able to calculate s z in terms of the basic parameters characterizing our system, like interaction strength V 0 and density n. We can do this by observing that s z will choose itself in order to minimize the ground state energy obtained from H MF. Moreover, we can calculate the ground state energy of H MF without too much trouble there will be N
up-spin fermions forming a fermi surface, with Fermi wavevector k F, and similarly for the down-spin electrons. The total fermion number is fixed at N = N + N. The total energy then has the form E MF = E K + E K + V 0 V s z V 0s z (N N ) 1 V 0V n (19) = E K + E K + V 0 V s z 4V 0V s z 1 V 0V n (0) = E K + E K V 0 V s z 1 V 0V n. (1) Here, E K is the kinetic energy of the spin- 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 E K = where for the second equality we used n = k 3 F /6π. So the total energy is V 0π m k5 F = V (6π ) 5/3 0π m n5/3, () E MF = V (3π ) 5/3 0π (n + s z ) 5/3 + (n s z ) 5/3 V 0 V s z 1 m V 0V n. (3) For sufficiently small V 0 this function is minimized when s z = 0, but once V 0 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 αx, 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 E MF, it is worth noting that s z n/, and s z = n/ means the system is fully polarized (either all spins up or all spins down). Because of this fact there are two important values of V 0. As V 0 is increased from 0, first there is a value at which s z becomes nonzero. Then, as V 0 is increased further, there is another value of V 0 above which s z = n/. Finding these values is left as an exercise. 4 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 ψ, and evaluate the expectation value of the energy E var = ψhψ. The variational principle of quantum mechanics tells us that E var E 0, where E 0 is the exact ground state energy of H. Also, if E var = E 0, then ψ 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 E var. Then suppose we vary parameters, moving around in this set of wavefunctions to minimize E var 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 E var using a result called Wick s theorem. The idea is to minimize E var 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 H MF 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 ψ is a Slater determinant wavefunction. Suppose further that the operators A, B, C and D are all linear combinations of annihilation operators, then ψa B CDψ = ψa Dψ ψb Cψ ψa Cψ ψb Dψ. (4)
Now we can evaluate E var 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 H int = V 0 d 3 r ˆn(r)ˆn(r) = V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r) (5) The last term is a constant which we ignore. Dropping this term, we have = V 0 d 3 r ψ (r)ψ (r)ψ (r)ψ (r) + δ(0)v 0 N. (6) E var = E K + E K + ψh int ψ (7) = E K + E K V 0 d 3 r ψ (r)ψ (r) ψ (r)ψ (r) ψ (r)ψ (r) ψ (r)ψ (r) (8) = E K + E K V 0 = E K + E K V 0 d 3 r (δ n )(δ n ) n (9) d 3 r n + n n (30) = E K + E K V 0 V s z + 1 V 0V n. (31) Except for the last term, this is exactly what we found before for E MF. So, in particular, by minimizing this we find the same spin density s z. 5 V. INTERACTING ELECTRON GAS AND THE FOCK APPROXIMATION The Hamiltonian of the interacting electron gas is H = d 3 r ψ ψ (r) + e m d 3 rd 3 r ˆn(r) nˆn(r ) n r r. (3) The interaction potential is the Coulomb potential V (r) = e /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 and so a = 1 n = 4 3 πa3, (33) ( 3 ) 1/3n 1/3. (34) 4π Another important length is the Bohr radius a 0 = /me. The ratio of these lengths is r s = a/a 0. Now, by crude dimensional analysis, the kinetic energy per electron is E K /ma. And, the Coulomb energy per electron is on the order of E C e /a. The ratio is E C E K a a 0 = r s. (35)
So this says that, when the gas is very dense, r s 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, r s 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 r s we have a Fermi gas (with maybe some corrections to the behavior due to interactions), and at small r s 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 r s, maybe r s 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 r s. 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, (ˆn(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 H MF. In particular, we will drop all the constant terms in H MF 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 H int e This suggests that we make the mean-field guess where d 3 rd 3 r ψ (r)ψ (r )ψ (r )ψ (r) r r ψ (r)ψ (r ) = F (r r ) + ψ (r)ψ (r ) F (r r ), (37) 6 (36) F (r r ) = ψ (r)ψ (r ). (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 F (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 H int e d 3 rd 3 r 1 r r F (r r )ψ (r )ψ (r) + F (r r)ψ (r)ψ (r ). (39) Therefore the mean-field Hamiltonian is H MF = d 3 r ψ ψ (r) e d 3 rd 3 r 1 m r r F (r r )ψ (r )ψ (r)+f (r r)ψ (r)ψ (r ). (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 F (r r ) = ψ (r)ψ (r ). 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
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 H MF 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, ψ 0. (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 H MF has the same ground state wavefunction as the non-interacting problem it s important to note that this does not mean that H MF 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 ψ 0 is really the ground state of H MF. To calculate F we go to k-space: F (r r ) = ψ 0 ψ (r)ψ (r )ψ 0 (41) = 1 e ik r e ik r ψ 0 ψ V (k)ψ (k )ψ 0. (4) k,k When we learned about second quantization, we discussed how to evaluate the expectation value ψ 0 ψ (k)ψ (k )ψ 0 for spinless fermions. The only difference with spin is that the result should be proportional to δ 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 ψ 0 ψ (k)ψ (k )ψ 0 = δ δ k,k Θ(k F k), (43) where ǫ k = k /m and Θ(x) = 1 for x > 0 and Θ(x) = 0 for x < 0. Plugging this into our expression for F we find F (r r ) δ F(r r ) = δ V e ik (r r ) Θ(k F k). (44) We could go ahead and evaluate this expression for F (by turning the sum into an integral over k), but it is more useful to plug it into the interaction part of the mean-field Hamiltonian, HMF int, given in Eq. (39). We can first simplify this a bit: HMF int = e d 3 rd 3 r 1 r r F (r r )ψ (r )ψ (r) + F (r r)ψ (r)ψ (r ) (45) = e = e = e k d 3 rd 3 r 1 r r F(r r )ψ (r )ψ (r) + F(r r)ψ (r)ψ (r ) d 3 rd 3 r F(r r ) r r ψ (r )ψ (r) + ψ (r)ψ (r ) d 3 rd 3 r F(r r ) r r 7 (46) (47) ψ (r)ψ (r ). (48) Above we used the fact that F(r) = F( r). This form is still not so useful we really want to express H MF 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
terms of their Fourier transforms: HMF int = e d 3 rd 3 r F(r r ) r r ψ (r)ψ (r ) (49) = e d 3 rd 3 r F(r r ) V r r e ik r e ik r ψ (k)ψ (k ) (50) k,k = e ψ V (k)ψ (k ) d 3 rd 3 r F(r r ) r r e ik r e. ik r (51) k,k Let s evaluate the double integral in square brackets. We have, making the change of variables r r + r, d 3 rd 3 r F(r r ) r r e ik r e ik r = d 3 rd 3 r F(r) e ir (k k) e ik r (5) r = d 3 r e ir (k k) d 3 r F(r) e ik r (53) r = V δ kk d 3 r F(r) e ik r (54) r = V δ kk G(k), (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 HMF int = e G(k)ψ (k)ψ (k), (56) and therefore H MF = k k k m e G(k) ψ (k)ψ (k). (57) This tells us that the single-particle eigenstates of H MF are indeed plane-waves, but their energies are ǫ k = k m e G(k), (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 ǫ k depends only on k and the energies are spherically symmetric. As long as ǫ is a monotonically increasing function of k (and in fact even under somewhat weaker assumptions), then H MF 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 we have F(r) = d 3 q (π) 3 eiq r Θ(k F q). (59) G(k) = d 3 r F(r) e ik r (60) r d 3 q = (π) 3 d 3 r e i(k q) r Θ(k F q) (61) r d 3 q = (π) 3 Θ(k F q) d 3 r e i(k q) r 1 (6) r d 3 q Θ(k F q) = 4π (π) 3 (k q). (63) 8
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 where G(k) = k F π F(k/k F), (64) F(x) = 1 + 1 x 4x ( ln 1 + x ). (65) 1 x To get a feeling for what this result means, let s express the single-particle energy ǫ k in a relatively simple form, defining x = k/k F : ǫ k = k m e k F F(x) (66) π = k F x 4me F(x) (67) m πk F = k F x 4 ( 4 ) 1/3rs F(x). (68) m π 9π The important point here is that the interaction correction to the single-particle energy is proportional to r s, which makes sense, since we argued that interactions have a small effect at small r s. One thing we can see right away from this form is that ǫ(k F ) > ǫ(k) for k < k F. The way to show this is to look at the difference ǫ(k F ) ǫ(k). This holds even though ǫ k is not a monotonic function of k, and implies that indeed the states below k F are filled and those above k F are empty. So the ground state of H MF is always the same as the non-interacting ground state, validating our guess. Something else we can see from this form is that ǫ k has a singularity at k = k F. 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 r s. More precisely, in the non-interacting problem the kinetic energy is proportional to 1/r s. Since small r s means weak interaction, it is a good guess that the ground state energy can be expanded as a power series in r s. 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/r s, and it is this term that Hartree-Fock turns out to get correctly. 9