Second quantization (What does "strongly correlated" mean?)

In the early days of quantum mechanics, the standard description of a many-particle problem was via the wave function for all the particles. You are all familiar, I'm sure, with Slater determinants for states of multiple identical fermions. A convenient way to write the overall wavefunction of three electrons in states $\psi\_1, \psi\_2, \psi\_3$ is
$$ \Psi( {\bf r}_1, {\bf r}_2, {\bf r}_3) = \left| \matrix{\psi_1( {\bf r}_1)&\psi_1( {\bf r}_2)&\psi_1( {\bf r}_3)\cr \psi_2( {\bf r}_1)&\psi_2( {\bf r}_2)&\psi_2( {\bf r}_3) \cr \psi_3( {\bf r}_1)&\psi_3( {\bf r}_2)&\psi_3( {\bf r}_3)} \right|. \label{slatdet} $$
This satisfies the requirement of asymmetry; note that the wavefunction vanishes if any two of the ${\bf r}\_i$ or $\psi\_i$ are equal.

An example of when we might want to use Slater determinants is when we study a many-body system in the Hartree approximation. Suppose we want to approximate the many-body state of $N$ electrons moving in some constant background potential $V_b$ (for example, from the ions of a solid). We seek $N$ lowest-energy eigenstates of the time-independent Schr\"odinger equation in the form
$$ H \psi = \left(-{\hbar^2 \nabla^2 \over 2 m} + V(r) + V_b(r)\right) \psi_i = E_i \psi_i, $$
where the self-consistent potential $V(r)$ is determined through
$$ V(r) = \int\,d{\bf r}_2 {e^2 n( {\bf r}_2) \over | {\bf r}_2 - {\bf r}|} = \int\,d {\bf r}_2 \sum_i {e^2 |\psi_i( {\bf r}_2)|^2 \over | {\bf r}_2 - {\bf r}|}. $$
Here $n({\bf r})$ is the electron number density.

Now the Hartree approximation is to find a self-consistent solution of these equations. However, the natural resulting many-body wavefunction that gives the potential above is the symmetric combination of the $\psi_i$, which does not satisfy the requirement of asymmetry under particle exchange. To be more precise, the Hartree equations can be justified as minimizing the total energy over all wavefunctions of the form
$$ \Psi = \psi_1( {\bf r}_1) \psi_2( {\bf r}_2) \ldots $$
which is not properly asymmetric.

The main error in the Hartree description, as you probably know, is that it ignores the "exchange" interaction between identical particles. We can make a better approximation by taking the Slater determinant of some orbitals $\psi_i$ as the many-body ground state. Note that the many-body state still can be thought of as $N$ independent orbitals, which is not true for a general state of $N$ electrons, as we shall see. Minimizing over Slater determinants of independent orbitals gives the famous Hartree-Fock equation,
$$ -{\hbar^2 \over 2m} \nabla^2 \psi_i({\bf r}) + (V({\bf r}) + V_b({\bf r})) \psi_i({\bf r}) -\sum_j \int\,d{\bf r}_2 {e^2 \over |{\bf r}-{\bf r}_2|} \psi_j^* ({\bf r}_2) \psi_i({\bf r}_2) \psi_j(r) \delta_{s_j s_i} = E_i \psi_i. $$

Note that the exchange term (the last term on the left side) is now an integral operator, rather than just acting on $\psi_i({\bf r})$. Here we have put back a spin delta-function as a reminder that the exchange term only occurs for equal spin (so that the particles are identical). It is important to remember that the Hartree-Fock description, which can be improved in various ways that are discussed in the 240 series, is qualitatively quite good for many materials.

This course will concentrate, however, on the minority of materials for which the true quantum-mechanical many-body state is completely different from the simple Hartree picture. Let us first introduce a compact notation for Slater determinants. First, focus on problems with translational invariance, so that the single-particle states have well-defined momentum and can be labeled by their momentum ${\bf k}$. We assume for now that the particles are spinless, to simplify the notation.
You may have seen a similar notation before for raising and lowering operators of harmonic oscillators, which we connect below to creation and annihilation operators of bosons. For now we stick to fermions, which will be created by $c^\dagger$ and annihilated by $c$.

Write $|0\rangle$ for the state of zero particles (sometimes this notation will also be used for the ground state of a many-particle system, so be careful). Then the state of one particle of momentum ${\bf k}$ is written
$$ c^\dagger_k |0\rangle. $$
The adjoint of $c^\dagger_k$ we write as $c_k$. The zero-particle state is annihilated by all the $c_k$:
$$ c_k |0\rangle = 0. $$
Another requirement is that trying to put two fermions into the same single-particle state should also give 0, as should trying to get rid of two fermions:
$$ c^\dagger_k c^\dagger_k = c_k c_k = 0, $$
where this notation means that the operators give 0 applied to any many-body state.

The most important condition on the $c_k$ is the anticommutation relation:
$$ \{ c^\dagger_k, c_k\} = c^\dagger_k c_k + c_k c^\dagger_k = 1. $$
This essentially says that the probability of single-particle state $k$ being empty in many-body state $|\Psi\rangle$,
which is $\langle \Psi | c^\dagger_k c_k | \Psi \rangle$, plus the probability of its being occupied, which is $\langle \Psi | c^\dagger_k c_k | \Psi \rangle$, should sum to 1. Recall that this anticommutation relation holds only for fermions; clearly something different will be required for bosons.

Now we need to add the spin of the fermions and ask how the operators of different states $k$ alter each other. For instance, is the state $c^\dagger_{k_1} c^\dagger_{k_2} |0\rangle$ the same as $c^\dagger_{k_1} c^\dagger_{k_2} |0\rangle$?

There is an element of choice here, but the simplest way to think about it is to consider the operator $c^\dagger_{k_1}$ as "adding on" a new row to the Slater determinant (\ref{slatdet}). Then, since interchanging two rows of a determinant changes the sign of a determinant, we should have
$$ c^\dagger_{k_1} c^\dagger_{k_2} |0\rangle = -c^\dagger_{k_1} c^\dagger_{k_2} |0\rangle, $$
which we summarize as
$$ \{ c^\dagger_{\sigma_1 k_1}, c^\dagger_{\sigma_2 k_2}\} = \{ c_{\sigma_1 k_1}, c_{\sigma_2 k_2}\} = 0. $$
so that the two sequences create the same many-body state but with a sign difference.
As you might expect, this sign difference is only there for fermions, and will not be present for bosons. Finally, restoring the spin variable $\sigma$, we write the full anticommutation relation as
$$ \{ c^\dagger_{\sigma_1 k_1}, c_{\sigma_2 k_2}\} = \delta_{k_1,k_2} \delta_{\sigma_1,\sigma_2}. $$
Hence these operators just anticommute unless both the momentum and the spin are the same, in which case there is a number 1 on the right side, expressing the idea that the fermionic state should have probability 1 of being occupied or empty.

Now let's see how the above need to be modified for bosons. The main examples of bosonic operators that will appear in this course are the modes of the electromagnetic field, or of phonon excitations in a solid. For bosonic operators, we have a commutation relation instead of an anticommutation relation:
$$ [b_{k_1}, b^\dagger_{k_2}] = \delta_{k_1,k_2}. $$
From this you can show that the number of quanta in mode $k$, i.e., the expectation value of the number operator $n_k \equiv b^\dagger_k b_k$, is increased by 1 by the creation operator $b^\dagger_k$, and decreased by 1 by the annihilation operator

We are now in a position to understand the simplest example of coherent states, which are useful in taking the classical limit of a harmonic oscillator or in setting up Feynman path integrals over classical configurations.

Let's return to fermions. We can write the filled Fermi sea as
$$ \prod_{|k|\leq k_F} c^\dagger_{k\uparrow} c^\dagger_{k\downarrow} |0\rangle. $$
Here all states of momentum below the Fermi momentum are filled by the creation operators, while all those above the Fermi momentum are empty. We could equally well have written this state as a Slater determinant, of course.

As motivation for next time, consider the following famous example of a strongly correlated wavefunction:
$$ \Psi_{BCS} = \prod_k (u_k + v_k c^\dagger_{k \downarrow} c^\dagger_{-k \uparrow}) |0\rangle,\quad |u_k|^2 + |v_k|^2 = 1. $$
Here $u_k$ and $v_k$ are some $k$-dependent complex numbers satisfying the normalization constraint above. Now this will just give the filled Fermi sea above for a particular choice of the $u_k, v_k$: the filled Fermi sea results from
$$ u_k = \cases{0&if $k\leq k_f$\cr1&if $k>k_f$},\quad v_k = \cases{1&if $k\leq k_f$\cr0&if $k>k_f$}. $$ However, suppose we smear out the sharp boundary at the Fermi level by an energy $\Delta$, so that now over some interval, $u_k$ and $v_k$ are both between 0 and 1. This might seem to be similar to the Fermi gas at finite temperature, because then the occupancy is also smeared out over a distance $kT$ near the Fermi level. However, the smeared BCS state is very different on a fundamental level because it has perfect pair correlations. For every momentum $k$, even in the smeared state, if the spin-up orbital is occupied then the spin-down orbital at momentum $-k$ is automatically also occupied. This does not occur in the Fermi gas at finite temperature, where the spin-up and spin-down orbitals are independent.