I'm wondering how would one go on about to calculate the static structure function with the ground state being $|\phi_0\rangle$:

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

for the case $\vec{q}\neq 0$, knowing the operator $\hat{n}_\vec{q}$ is defined as follows:

$\hat{n}_\vec{q}=\sum_{\vec{k},\sigma} a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma}$

$\Rightarrow \hat{n}_\vec{q}\hat{n}_{-\vec{q}}=\sum_{\vec{k},\sigma} a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma}\sum_{\vec{k}',\sigma'} a^\dagger_{\vec{k}'\sigma'}a_{\vec{k}'-\vec{q}\sigma'}\\ =\sum_{\vec{k},\sigma} a^\dagger_{\vec{k}\sigma}a_{\vec{k}+\vec{q}\sigma} a^\dagger_{\vec{k}\sigma}a_{\vec{k}-\vec{q}\sigma}$

All that I understand is that this chain of operators will remove a particle with momentum $\vec{k}$, such as $|\vec{k}|\leq k_F$, where $k_F$ is the Fermi momentum, create (and later destroy) a particle with momentum $(\vec{k}-\vec{q})$, which lies outside the Fermi sphere due to the Pauli exclusion principle, and then put that particle back to its original position in the momentum sphere with radius $k_F$. For this reason, summing over the spins would just yield a factor of 2, since $\sigma\neq\sigma'$ is not allowed. And that taking the continuum limit in the sum should allow to solve the problem. However, I don't have the faintest clue as to how to solve for $S_\vec{q}$...

EDIT: The $\vec{q}=0$ case just gives 1; we were asked to solve this explicitly for $\vec{q}\neq0$ but we did not cover Wick's theorem, which, unfortunately, would have helped a lot...

  • $\begingroup$ Have you tried to apply Wick's theorem? $\endgroup$
    – Meng Cheng
    Oct 1, 2015 at 21:33
  • $\begingroup$ We did not cover Wick's theorem; this question was actually assigned in an advanced quantum mechanics class. $\endgroup$ Oct 1, 2015 at 23:16

1 Answer 1


With Wick's theorem it would be very straightforward. But in this case it is not too difficult to directly work out the expectation value:

$ \displaystyle S_q=\langle \phi_0|n_qn_{-q}|\phi_0\rangle=\sum_{k,k'}\langle \phi_0 |c_{k'}^\dagger c_{k'+q}c_k^\dagger c_{k-q}|\phi_0\rangle $

The string of four operators, when applied to the Fermi sea on the right( $|\phi_0\rangle$), does the following: first it kicks out a fermion below the Fermi sea, so $|k-q|<k_F$, and excite it to momentum k (so $|k|>k_F$). Then the next two operators should bring this state back to the ground state to get an overlap with $\langle\phi_0$. Therefore, it is easy to convince yourself that one must have $k'+q=k$. Therefore we really only have one sum:

$\displaystyle S_q=\sum_{k}\langle \phi_0 |c_{k-q}^\dagger c_{k}c_k^\dagger c_{k-q}|\phi_0\rangle = \sum_{k}\langle \phi_0 |c_{k-q}^\dagger c_{k-q}c_{k}c_k^\dagger|\phi_0\rangle = \sum_{k}\langle \phi_0 |\hat{n}_{k-q}(1-\hat{n}_k)|\phi_0\rangle=\sum_{|k-q|<k_F,|k|>k_F}1 $

So in the end it boils down to evaluating the volume of the Fermi sea that satisfies the kinematic constraint.


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