How to calculate the second-order pertubation in a Bose gas? I'm self-learning many body theory and right now I'm trying to solve Problem 1.3 from Quantum Theory of Many-Particle Systems by Fetter and Walecka.
Problem:

Given a homogeneous system of a spin-zero particles interacting through a potential V
show that the expectation value of the hamiltonian in the noninteracting ground state is
$$ \frac{E^{(1)}}{N} = \frac{(N − 1)V(\vec{0})}{2V}$$
where $$
V(\vec q)=\int d_3x V(\vec{x})e^{-i\vec{q}\cdot\vec{x}} $$
Then show that the second-order contribution to the ground-state energy is
$$ \frac{E^{(2)}}{N} = -\frac{(N − 1)}{2V} \int \frac{d^3q}{(2\pi)^3} \frac{m|V(\vec q)|^2}{\hbar^2\vec q^2}$$
using second order perturbation theory
$$E^{(2)}=\sum_{n\neq0}\frac{\langle0|H_1|n\rangle\langle n|H_1|0\rangle}{E_0-E_n}$$ where $|n\rangle$ is the excited state and $|0\rangle$ is the ground state.

My try:
The kinetic operator is $$ T=\sum_{m \vec k} \frac{ \hbar ^2k^2 }{2m} n_{m \vec k} = \sum_{\vec k} \frac{ \hbar ^2k^2 }{2m} n_{\vec k} $$
The sum over m goes away since spinless particles only have 1 possible m.
Moreover since they are bosons at $T=0K$ they all are in the lowest energy state.
I managed to write the interaction hamiltonian as
$$ H_1 = \sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger  c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} $$
First order perturbation theory says that
$$
E^{(1)} = \langle 0 |H_1|0 \rangle = \langle 0 |\sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger  c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} |0\rangle$$
Now I act right with the destructors and left with the creators, which will act as destructors: I need to create holes where there is an occupied state, so I can only do it at the lowest energy. That means that $ m = m', \vec p = \vec k = \vec 0 $ and $\vec q = 0$. I get:
$$\langle 0|\sum_{ \vec 0} \frac{ 1}{V} V(\vec 0) c_{\vec 0}^\dagger  c_{\vec 0}^\dagger c_{\vec 0} c_{\vec 0} |0\rangle$$
I use commutation rules to pair up the operators to get $N^2 - N$, finding finally
$$ \frac{E^{(1)}}{N} = \frac{(N − 1)V(\vec{0})}{V}$$
which lacks the $1/2$, where does that come from? Is there another double counting?
Now regarding the second order energy I need to evaluate the matrix element
$$ \langle 0|H_1|n \rangle = \langle 0 |\sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger  c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} |n\rangle $$
I know I need to transform the sum into an integral but before that I should probably find a way to eliminate the sum over $\vec k$ and $\vec p$ through some considerations on the allowed creations/destructions, but I can't manage to do so.
Edit
I try to work it all out following the suggestion in the answer, because I can't manage to obtain the right result.
So I evaluate this matrix element
$$ \langle n|H_1|0 \rangle = \langle n|\sum_{m m' \vec k \vec p\vec q} \frac{ 1}{V} V(\vec q) c_{\vec p m}^\dagger  c_{\vec k m'}^\dagger c_{\vec k + \vec q m'}c_{\vec p - \vec q m} |0\rangle $$
first operating with the destructors one the right, getting $\delta_{\vec k + \vec q, 0}\delta_{\vec p - \vec q, 0}\delta_{m, m'}\sqrt{N(N -1)}$, then with the creators on the right, which now have became $c_{\vec q m}^\dagger  c_{-\vec q m}^\dagger$, getting $ \sqrt{N(N -1)} $. Thus
$$ \langle n|H_1|0 \rangle = \sum_{\vec q} \frac{ 1}{V} V(\vec q) N(N-1)$$
I switch to the integral
$$ = \int \frac{d_3q V}{(2\pi)^3} \frac{ 1}{V} V(\vec q) N(N-1)$$
Now I need to do the modulus squared, divide it by the energy difference and put it in the sum. I don't see how this can lead me to the final result, where for example $ (N-1)$ isn't squared.
What am I missing?
 A: $H_1$ is Hermitian, so it's easier to operate first with the destruction operators on $|0\rangle$. Notice that unless their momenta is zero, they will give zero. So you will find you must destroy two zero momenta particles, and create a pair, one with $q$ and one with $-q$. These are the only $\langle n|$ states that contribute. The energy difference is then $\hbar^2 q^2/m$.
Edit with details
The second order energy term from elementary quantum mechanics is
\begin{equation}
E^{2} =
\sum_{n \neq 0} \frac{\langle 0|H_1|n\rangle \langle n|H_1|0\rangle}{E_0-E_n}
=
\sum_{n \neq 0} \frac{|\langle n|H_1|0\rangle|^2}{E_0-E_n}
\end{equation}
The only states $|n\rangle$ that have non zero matrix elements are
the normalized states with
\begin{equation}
|n(\vec q)\rangle = c^\dagger_{\vec q} c^\dagger_{-\vec q} |0'\rangle
\end{equation}
where $|0'\rangle$ is the normalized state with $N-2$ particles in
the zero momentum orbitals.
and $\vec q$ not equal to zero. Notice that since
$c^\dagger_{-\vec q} c^\dagger_{\vec q}=
c^\dagger_{\vec q} c^\dagger_{-\vec q}$, to sum over the
$|n(\vec q)\rangle$
states you should either include just one of
$|n(\vec q)\rangle$ and
$|n(-\vec q)\rangle$ or sum over all
$\vec q \neq 0$ and divide by $2$.
The energies are $E_0 = 0$, $E_{n(\vec q)} = 2\left [
\frac{\hbar^2 q^2}{2m}\right ] = \frac{\hbar^2 q^2}{m}$, so the perturbation
theory result is
\begin{equation}
E^{(2)} = -\frac{1}{2}\sum_{\vec q\ne 0}
\frac{|\langle n(\vec q)|H_1|0\rangle|^2}{\hbar^2 q^2/m}
\end{equation}
The matrix element is, as I discussed in the comments,
\begin{equation}
\langle n(\vec q)|H_1|0\rangle = \frac{\sqrt{N(N-1)}}{V}V(\vec q) \,,
\end{equation}
where the factor of 2 is canceled since there are two ways to generate
the same $|n(q)\rangle$. Putting these together gives
\begin{equation}
E^{(2)} = -\frac{N(N-1)}{2V^2}
\sum_{\vec q \ne 0} \frac{|V(\vec q)|^2}{\hbar^2 q^2/m}
= -\frac{N(N-1)}{2V}\int \frac{d^3q}{(2 \pi)^3}
\frac{|V(\vec q)|^2}{\hbar^2 q^2/m} \,.
\end{equation}
