Computing the density operator commutation relations (Atland & Simons) I'm trying to work through Altland and Simons' example of interacting fermions in one dimension. It's in chapter 2, page 70 (you can find it here).
They define fermionic operators
$$
a_{sk}^\dagger
$$
where $s=L/R$. $a_{Lk}^\dagger$ is an operator that creates an electron going to the left with momentum $(-k_F+k)$, and $a_{Rk}^\dagger$ is an operator that creates an electron going to the right with momentum $(k_F+k)$. So basically, $a_{Lk}^\dagger=a_{-k_F+k}^\dagger$, $a_{Rk}^\dagger=a_{k_F+k}^\dagger$. These operators are restricted to exist only for small $k$.
Then, they define density operators
$$
\rho_{sq}=\sum_k a^\dagger_{sk+q}a_{sk}
$$
They go on to show that the commutation relations for the density operators are
$$
[\rho_{sq},\rho_{s'q'}]=\delta_{s,s'}\sum_k (a^\dagger_{sk+q}a_{sk-q'}-a^\dagger_{sk+q+q'}a_{sk})
$$
Now, here's the part I don't understand. They say they want to replace the right side of the equation with its ground state expectation value. They define the ground state of the theory by $|\Omega\rangle$. Then they claim that 
$$
\langle\Omega|a^\dagger_{sk}a_{sk'}|\Omega\rangle = \delta_{kk'}
$$
Why should this be true? I understand that in the noninteracting theory, $a^\dagger_{sk}a_{sk'}|\Omega\rangle$ is orthogonal to $|\Omega\rangle$ unless $k=k'$. But in the interacting theory, the ground state could be in a superposition of states that means $\langle\Omega|a^\dagger_{sk}a_{sk'}|\Omega\rangle\neq0$.
They ultimately use this to prove
$$
\langle\Omega|[\rho_{sq},\rho_{s'q'}]|\Omega\rangle = \delta_{s,s'}\delta_{q,-q'}\sum_k\langle\Omega|(a^\dagger_{sk+q}a_{sk+q}-a^\dagger_{sk}a_{sk})|\Omega\rangle
$$
and I don't see any other way to prove this.
What am I missing?
 A: $\def\kk{\mathbf{k}}
\def\ii{\mathrm{i}}
\def\qq{\mathbf{q}}$
You can prove this using translation invariance, without any other assumptions on the nature of the interacting ground state. I am going to give the argument in $D$ dimensions, which obviously holds in $D=1$ as a special case.
Spatial translations of the system as a whole are generated by the centre-of-mass momentum ($\hbar = 1$)
$${\bf P} = \sum_{\kk,s} \kk a_{\kk s}^\dagger a_{\kk s},$$
where boldface type denotes a vector in $D$ dimensions. The unitary operator $T_{\bf r} = \mathrm{e}^{\ii {\bf P} \cdot {\bf r}}$ describes a translation of the coordinate system by an amount ${\bf r}$. Now, the crucial assumption is that the interacting ground state satisfies the condition ${\bf P} \lvert \Omega \rangle = 0$, i.e. $T_{\bf r} \lvert \Omega \rangle = \lvert \Omega \rangle $. This holds for any system with translation-invariant interactions, since in that case 1) the Hamiltonian $H$ commutes with ${\bf P}$, so that eigenstates of $H$ are also eigenstates of ${\bf P}$, and 2) eigenstates of ${\bf P}$ gain a positive contribution ${\bf P}^2/2M$ to their energy (in non-relativistic approximation), with $M$ the total mass of the system. Therefore, the ground state $\lvert \Omega \rangle$ lies in the ${\bf P}=0$ sector. Of course, the above formal arguments prove what should already be obvious: a system in its ground state has no overall motion in the lab frame.
The rest of the argument follows from some simple algebra. One readily proves the commutation relation
$$ [\mathbf{P},a_{\kk s}] = -\kk a_{\kk s}, $$
which means physically that the ladder operator $a_{\kk s}$ reduces the total momentum of the system by an amount $\kk$, and further implies the translation property $$T_{\bf r} a_{\kk s}T_{\bf r}^\dagger = \mathrm{e}^{-\ii {\bf k} \cdot {\bf r}}a_{\kk s}.$$
It follows that
\begin{align}
 \langle \Omega \rvert a^\dagger_{\kk s}a_{\kk' s} \lvert \Omega \rangle & = \langle \Omega \rvert \left(T_{\bf r}^\dagger T_{\bf r}\right)  a^\dagger_{\kk s} \left(T_{\bf r}^\dagger T_{\bf r}\right) a_{\kk' s} \left(T_{\bf r}^\dagger T_{\bf r}\right) \lvert \Omega \rangle \\ 
& =  \langle \Omega \rvert \left(T_{\bf r}a^\dagger_{\kk s}T_{\bf r}^\dagger \right)\left(T_{\bf r} a_{\kk' s}T_{\bf r}^\dagger \right)\lvert \Omega \rangle  \\ 
& = \mathrm{e}^{\ii (\kk - \kk')\cdot {\bf r}}\langle \Omega \rvert a^\dagger_{\kk s}a_{\kk' s} \lvert \Omega \rangle,
\end{align}
where the first equality follows from unitarity of $T_{\bf r}$, the second from the translation-invariance of the interacting ground state, and the third using the translation property of the ladder operators. Noting that the above equality holds for all ${\bf r}$, we conclude that both sides must vanish unless $\kk = \kk'$.
Finally, it is worth mentioning that this argument does not rely in any way on fermionic statistics, and the same holds true in a bosonic system. Indeed, a similar property holds true for any operator which increases the momentum of the system by a definite amount (thus satisfying the translation property above). This includes the density Fourier components themselves,
$$ \rho_{\qq s} = \sum_\kk a^\dagger_{\kk+\qq s}a_{\kk s},$$
which satisfy
$$\langle \Omega \rvert \rho_{\qq s}^\dagger  \rho_{\qq' s}\lvert \Omega \rangle \propto \delta_{\qq\qq'}.$$
