Agenda for week 3: Fermi gas

Learning goals

• Learning the idea of one- and two-particle Fock-space operators
• Understanding the concept of the Fermi sea and learning how to compute some of its properties
• Particle-hole formalism
• Interacting Fermion systems: Hartree-Fock approximation

Reading assignment:

Notes for week 3: Many-fermion systems

More details:

• Tuominen, Sec. 7.2, 7.5, 7.6

Preliminary exercises,

Do these during/after reading the assignment work. Will be discussed in class March 23rd.

1. Write the free-fermion Hamiltonian (nonrelativistic kinetic energy) in terms of the Fock-space operators in
   1. In terms of arbitrary single-particle basis states
   2. Momentum representation
   3. Position representation

2. By considering an eigenstate of the total momentum operator
   show that the hole creation operator increases momentum by

3. For the spin-independent 2-body interaction as in the section "Two-body interaction" in the notes, calculate its expectation value in states where
   1. Two particles are in the same momentum state but different spin states
   2. Two particles are in the same spin state but different momentum states
   Identify the "direct" and "exchange" interactions. Draw Feynman diagrams for all possible spin and momentum configurations, and mark the spin and momentum in the diagram.

Homework exercises, third week

Will be discussed in the tutorial session on Thursday March 25th. Return a scanned pdf with your solution by Friday March 26th, at 9 pm using the form below. Then check and grade your solution with the help of the model solutions and resubmit your graded solutions by Tuesday March 30th at 9 am.

Notes for week 3: Fermi gas

One- and two-particle Fock-space operators

One-particle operators

A general 1-particle operator is a sum of identical operators, each of which operate on one-particle states where acts solely on the th particle. An example of such a 1-particle operator is the kinetic energy operator which is diagonal in spin.

We should now find the corresponding operator in the -particle Fock space. Let us write the operators in a diagonal basis so that where is the expectation value of the single-particle operator. Next, we want to consider the action of to a -particle Fock state with . Therefore we have to go back to description of the -particle Fock state in terms of the single-particle states. For example, take a bosonic state In the second to last step, we exchanged the order of summation, since the sum over is the same for any permutation (as we sum over all indices). Then, we can use the definition of the number operator, and since the above relation holds for any state (derivation is the same for a fermionic state), we have More generally, we can work in a non-diagonal basis by using and which gives 1-particle operator in the Fock space For fermions, simply change bosonic to fermionic .

In terms of the field operators and , we can write the 1-particle operator (assuming local operators ) for fermionic system as An example of a local operator is the kinetic energy operator for which .

Action of one-particle operator on a Fock space state

Now you should be able to go back and do Question 1 in the preliminary exercises

Two-particle operators

2-particle operators describe pairwise interactions between particles. In many cases, these pairwise interactions are described by a pair-potential, e.g. Coulomb interaction between electrons. Thus, it is convenient to work in the position-basis, i.e. with the field operators . For brevity, we neglect spin in this section.

Consider a symmetric 2-particle interaction . Now, we want to find in the second quantized form so that it matches the position-basis result Note how the sums go over all the possible interaction pairs once. Since we know the form of 1-particle operator, we can guess a possible generalization For example, an equally plausible integrand does not work, so the guess must be fairly good. In any case, the prefactor is still to be determined.

We can now apply the ansatz to the above equation. For a fermionic system, we have by the use of the anticommutation relations. By integrating over and we find that . The minus sign we may absorb to the ordering of the field operators giving in the end (including spin)

This interaction can be illustrated with the Feynman diagram

in which the annihilation and creation operators are represented by the incoming and outgoing lines, respectively. Solid lines are fermions, and the interaction is represented by the wiggly line connecting the two electron lines. Feynman diagrams will become more familiar (and they will be given a more rigorous meaning) in the quantum field theory course, but for now, we just use them to give an intuitive description for the interaction operators.

As before, we can then move to a different basis to obtain a more general form 2-particle operator in the Fock space where . Generally, . This result holds for bosonic system similarly.

Two-body interaction in a translationally invariant system

We concentrate especially on two-body interactions that depend only on the distance between the fermions. In this case Then where the second line is a transformation from the field operators to those specified by the quantum numbers . In particular, for definite-momentum states the wave functions are plane waves, Now let us define and . We can write the above Hamiltonian as Now define , , and (for those momenta coupled by the -function in the above equation). Moreover, the Fourier transformation of the two-body potential is With these, we can write the most general form of the two-body Hamiltonian in the translationally invariant case as This interaction can be represented with the following Feynman diagram, in which the spin of the individual particles is conserved. Momentum conservation in the diagram shows up so that at every vertex the sum of incoming momenta equals the sum of outgoing momenta.

Think carefully how the momentum and spin assignments correspond to the creation and annihilation operators!

Non-interacting Fermi gas

Let us consider a gas of nonrelativistic, noninteracting spin- fermions, a Fermi gas, for which I.e. for 1-particle states and wavefunctions

In order to have wavefunctions normalized to 1, we put the particles in a box of volume , and require periodic boundary conditions: In a box , the allowed values of momentum are , for integer .

The Hamiltonian, written in terms of creation and annihilation operators, is where is summed over all allowed values, as determined from the boundary conditions and spin is summed over the set .

The total number of particles is given by the expectation value of

We observe that , so the total number of particles is conserved in the noninteracting Fermi gas. This is understandable as there are no interactions in the system which could change the number of particles.

Ground state

The next task is find the ground state of the Fermi gas. If the particles were bosons, they would all occupy the lowest energy state in the ground state. Now, for the case of identical fermions, there can be at most one particle occupying a state , i.e. two fermions for each .

Thus, for fermions, the ground state is such that all states with energies below some highest energy are occupied by the particles in the system. is the Fermi energy. In the momentum space the occupied states form a sphere of radius , where is known as the Fermi momentum.

The -particle ground state can be expressed as

and are fixed by the number of particles. To determine them, we apply the total number operator to the ground state. At the thermodynamic limit we find a particle density where the integral was evaluated by noticing that it describes a three dimensional sphere in momentum space. The surface of the sphere (the states ) is the Fermi surface.

Fig: Dispersion relation of free fermions (blue line). The Fermi energy is marked with the red dashed line. The occupied states for -particle ground state are marked with the bold blue line.

Fig: Occupied states (blue) of the Fermi gas form a sphere in in the momentum space.

Let us also calculate the ground state energy. It is In other words, the ground state energy is This can also be expressed in terms of the number of particles by using the relation between the Fermi momentum and the particle density : This result is used below.

Correlation functions

Single-particle correlation function

The single-particle correlation function of the Fermi gas is defined by This correlation function describes the probability amplitude that annihilating a particle at and creating another one at (in the same spin state) yields the Fermi gas. Alternatively, it can be regarded as the probability amplitude of the transition from state to the state . The prefactor (where is the particle density) makes sure that . This can be seen by a direct calculation (exercise) or identifying as the operating yielding the local density for particles with spin .

We can express the single-particle correlation function of the Fermi gas in terms of the spherical Bessel function : Note that the size of the correlation function depends only on the distance between the two points, not on their position or the orientation of the vector connecting them. This reflects the translation and rotation symmetry of the Fermi gas.

We can plot it:

The single-particle correlation function of the Fermi gas oscillates within a scale of the order of the Fermi wavelength and tends to zero for .

Pair distribution function

We can also define a pair distribution function of a Fermi gas via The pair distribution function describes the conditional probability amplitude of finding a particle of spin at , if there is another particle of spin at .

In the exercises, you will show that

• If ,
• If , .

Plotting :

shows that , whereas . This is a signature of an effective repulsion between pairs of equal-spin fermions even in a non-interacting gas! This repulsion originates from the Pauli exclusion principle.

Particle and hole excitations

So far we have only discussed the ground state properties of the Fermi gas. To describe dynamics, we must consider the other states of the system, the excited states.

In the Fermi gas, the simplest possible excited state with particles is with and . This state is obtained by taking a single particle () from a ground state and moving it to a state () above the Fermi surface. We say this state has a particle excitation with () and a hole excitation with ().

The energy of such an excited state is (show this as an exercise!), which is always higher than the ground state energy . We can now choose to measure the single particle energies not from the band bottom, but relative to the Fermi energy. We define a new dispersion relation which describes the energy of a particle and hole excitations. Both excitations have a positive energy. In terms of , the energy of the excited state relative to GS is then a sum of positive quantities, i.e. the excited state energy is a sum of particle and hole excitation energies. This also applies to more complicated excited states of the Fermi gas.

Fig:Particle-hole excitation spectrum in the direction.

The creation and annihilation operators for particles are the same as the original operators. For holes, the annihilation and creation operators switch their roles and we also invert the momentum and spin, so that always increases the total momentum and spin of the system by and , respectively.

The new operators have the same anticommutation relations as the original operators:

The vacuum has the property , i.e. it is quenched by all original annihilation operators. Let us check how the new annihilation operators act on the Fermi gas ground state :

• For particles we have , which annihilates the ground state because modes with are not occupied
• For holes we have which also annihilates the ground state because modes with are occupied and (the Pauli principle: you cannot create a particle in a state that is already occupied). Thus

The ground state of the Fermi gas is an effective vacuum. It is annihilated by both particle and hole annihilation operators for all and . This has practical consequences, as it makes the calculation of expectation values of particle-hole operator products easy. For example, instead of an expectation value -operator product, we might only need to evaluate an expectation value of -operator product.

Hamiltonian in particle-hole basis

The Hamiltonian of the Fermi gas also needs to be transformed to the particle-hole basis. To do this, we first divide it into two parts, of which one will describe particles and the other holes: The two parts can then be separately expressed in terms of the new dispersion and particle-hole operators. We get where and are the particle and hole number operators. The particle number operator of the original particles is Thus, the Hamiltonian is For a system with constant number of particles, the second term is 0, and the last term is a constant energy shift which can be scaled away. We end up with a simple Hamiltonian After dropping the term, the expectation value of the Hamiltonian corresponds to the energy of the system relative to the effective vacuum , and not relative to the original vacuum .

Above, we calculated the energy of the simple excited state where and . By using the transformed Hamiltonian and Wick's theorem, we find which agrees with the earlier calculation.

Now you should be able to go back and do Question 2 in the preliminary exercises

Hartree-Fock (mean field) method

Let us consider the ideal Fermi gas together with the two-body interactions in using the form we arrived at earlier here, so that the total Hamiltonian is of the form In the case where the single-particle eigenstates are spin degenerate and described by some momentum , and when the two-body interaction is spin-independent and translation-invariant, we may simplify this to the form

Finding the ground state of this Hamiltonian for a generic -particle system (with a large ) turns out to be very hard if not impossible. This generic problem is discussed for example in Ch. 1 of this book by Xiao-Gang Wen. When there are only a few particles, such exact ground states may be found computationally. However, in the case of a large number of particles majority of approaches at least start from an approximative scheme introduced below: that of a Hartree-Fock or mean field method, whose idea is to try to pose the interacting Hamiltonian in terms of the closest analogue non-interacting Hamiltonian, i.e., in terms of an effective Fermi gas, where the "other" particles act as an effective potential term in the single-particle problem. In some cases this emerging effective potential spontaneously breaks some symmetry of the system. Such symmetry breaking can be associated with phase transitions. Below we describe two such models, related with spontaneous symmetry breaking in the spin sector (and hence describing magnetism) or one where we have to redefine the fermionic states as in the BCS theory of superconductivity.

First step: Slater determinant ansatz

The Hartree-Fock method is perhaps the most often used approach to finding an approximate ground state for a system of a large number of pair-wise interacting fermions. The main idea is to assume that the ground state can still be written in terms of some Slater determinant, where the quantum numbers that are included in this product are chosen such that the energy is minimized. The approach is thus similar to the one in the variational principle discussed in earlier quantum mechanics courses.

In the Ansatz, we assume that the quantum numbers are organized such that the states correspond to occupied states, and the others () to unoccupied states.

Second step: expectation value of energy with ansatz

Now suppose our Hamiltonian is

Let us compute the expectation value of the energy : These can be simplified under the assumption that is a Slater determinant. and We also notice that the latter term is non-zero only if (i) and or (ii) and (and all are ). In other words When showing this, it is enough to assume as for we would be removing two fermions from the same state. Then we simply need to shift the annihilation operator next to the corresponding creation operator and notice that we are calculating the expectation value of . At this point you can probably infer the reason for the sign change?

Plugging this result to the expression for the energy and using the Kronecker -functions to get rid of some of the sums, the result is The two pairwise interaction terms are frequently called the "direct" and "exchange" terms or the "Hartree" and the "Fock" terms.

Third step: minimize energy to find states

Minimizing in general leads to a set of non-linear equations for the Ansatz states, which have to be solved as a (complicated) problem of coupled Schrödinger-like equations. However, the simplfication we have achieved with the Hartree-Fock approximation is that these are now separate coupled single-particle equations, not one huge -particle equation. We do not write the equations down here (see e.g. Tuominen). In stead, we will use this approach in a couple of concrete cases next week. However, the generic Hartree-Fock method is widely used for example for the

• Atomic shell model
• To describe an interacting electron gas (or an "electron liquid")
• Nuclear shell model
• and many other systems

Alternative formulation: self-consistent mean field method

An alternative formulation of the Hartree-Fock theory can be made so that we replace the two-body interaction term by an effective one-body potential that depends on the state of the system. In other words, we approximate in the two-body interaction The two-body interaction term is hence of the form Since for indistinguishable particles the two-body interaction satisfies we can interchange the dummy indices and in the second and the fourth term. As a result, we get an effective single-particle Hamiltonian This procedure defines two "potentials" denoted by operators and corresponding to the Hartree and Fock interaction terms. In principle the problem thus reduces to finding the single-particle spectrum under such potentials. The ground state then corresponds to filling lowest-energy eigenstates of this new problem. However, now the potentials and themselves depend on . The usual procedure for solving this problem is that of a self-consistency iteration:

• Start from an Ansatz
• Compute and (or their representation in a chosen basis)
• Find the spectrum of the resulting single-particle Hamiltonian
• Choose the lowest-energy states as a new Ansatz
• Iterate until convergence

System with translational invariance

Let us consider the case where the system contains translational invariance and the two-body interaction is spin-conserving, as discussed above. In that case the effective one-body Hamiltonian is (note that strictly speaking the following assumes that spin is a good quantum number - this assumption can also be lifted)

where we defined two self-consistent fields, containing the constant Hartree term and the momentum and spin dependent Fock term Next week, we will use this approach to show that the spin dependence in the Fock term may lead to a spontaneous symmetry breaking to a state with non-vanishing spin polarization. This gives us a microscopic model for ferromagnetism.

Now you should be able to go back and do Question 3 in the preliminary exercises