How to Calculate the Static Structure Factor for Noninteracting Fermions?

[Catria]

Homework Statement

Calculate the static structure factor for noninteracting fermions

$S(\vec{q})=\frac{1}{N}\langle \phi_0|\hat{n}_{\vec{q}}\hat{n}_{-\vec{q}} | \phi_0\rangle$

where $\hat{n}_\vec{q}=\sum_{\vec{k},\sigma} a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma}$ is the particle density operator in the momentum representation and $|\phi_0\rangle$ is the ground state. Take the continuum limit $\sum_{\vec{k},\sigma} \to 2V\int \frac{d^3 k}{(2\pi)^3}$ and calculate $S(\vec{q})$ explicitly.

Hint: Consider the $\vec{q}=0$ and $\vec{q}\neq 0$ separately.

Homework Equations

The spin-independent one-particle operator: $\langle\vec{q}|O|\vec{k}\rangle =\frac{1}{V}\int d\vec{r} e^{-i(\vec{q}-\vec{k})\cdot\vec{r}} O(\vec{r})$

The Attempt at a Solution

The case $\vec{q}=0$:

In that case, the operator $\hat{n}_\vec{q}\hat{n}_{-\vec{q}}=\sum_{\vec{k},\sigma} a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma}a^\dagger_{\vec{k}\sigma}a_{\vec{k}-{\vec{q}}\sigma} =\sum_{\vec{k},\sigma} a^\dagger_{\vec{k}\sigma}a_{\vec{k}\sigma}a^\dagger_{\vec{k}\sigma}a_{\vec{k}\sigma}$

means that the integral reads

$S(0)=\frac{2V}{N}\int \frac{d^3 k}{(2\pi)^3} n^2_{\vec{k}\sigma} = 1$.

The physical process, in the $\vec{q}\neq 0$ case, is as follows: there is a particle at location $\vec{k}$ in momentum space within the Fermi sphere of radius $k_F$ that gets annihilated, and is created back with momentum $\vec{k}-\vec{q}$, which lies outside of the Fermi sphere, due to Pauli's exclusion principle. Said particle is annihilated and created back at the hole it first left behind.

Due to this constraint, if there was some spherical-coordinate integral to evaluate in the $\vec{q}\neq 0$ case, the radial integration limits would probably be $k_F-|\vec{q}|,k_F$. But this is as far as I have taken the integral:

$S(\vec{q})=2\int \frac{d^3 k}{(2\pi)^3} e^{-i\vec{k}\cdot\vec{q}} a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma}a^\dagger_{\vec{k}\sigma}a_{\vec{k}-{\vec{q}}\sigma}$

I'd like to think I am getting stuck at a rather advanced stage...

 

[mark726]

The exponential term can be expanded into a Fourier series, and the operators can be expressed as the spin-independent one-particle operator. But I am not sure how to proceed from here, so I am hoping someone can give me some hints.

One approach to solving this problem is to use the Wick's theorem, which allows us to express the product of four fermionic operators in terms of two-particle operators. In this case, we have four fermionic operators: a^\dagger_{\vec{k}\sigma}, a_{\vec{k}+\vec{q}\sigma}, a^\dagger_{\vec{k}\sigma}, and a_{\vec{k}-{\vec{q}}\sigma}. Using Wick's theorem, we can write the product of these operators as a sum of two-particle operators:

a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma}a^\dagger_{\vec{k}\sigma}a_{\vec{k}-{\vec{q}}\sigma} = \langle a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma}\rangle a^\dagger_{\vec{k}\sigma}a_{\vec{k}-{\vec{q}}\sigma} + \langle a^\dagger_{\vec{k}\sigma}a_{\vec{k}-{\vec{q}}\sigma}\rangle a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma} + \langle a^\dagger_{\vec{k}\sigma}a^\dagger_{\vec{k}\sigma}\rangle a_{\vec{k}+\vec{q}\sigma}a_{\vec{k}-{\vec{q}}\sigma} + \langle a_{\vec{k}+\vec{q}\sigma}a_{\vec{k}-{\vec{q}}\sigma}\rangle a^\dagger_{\vec{k}\sigma}a^\dagger_{\vec{k}\sigma}

The first two terms are the ones that contribute to the calculation of the static structure factor. The third and fourth terms do not contribute because they involve two-particle operators, which do not contribute to the expectation value of a one-particle operator.

Using the spin-independent one-particle operator, we can write the first two terms as:

\langle a^\dagger