Translational invariance implying diagonal representation in momentum space I have just come across something in my reading of Peskin and Schroeder that claims that because a function, in this particular case a two-point correlation function, is translationally invariant, it automatically has a diagonal momentum space representation. I am not seeing this relationship, and I was hoping someone could clarify this for me! 
 A: This is just a property of Fourier transformations. If the correlation function is translational invariant then, by definition, the position space representation $D(x,y)$ transforms as $D(x+a,y+a) = D(x,y)$ for any constant $a$. Thus $D(x,y) = D(x-y,0)$ and so the correlator depends only on the difference $x-y$. For simplicity, we'll define $D(x-y) = D(x-y,0)$. Fourier transform this over both $x$ and $y$ and you'll find it diagonal in momentum space. For example, in one dimension,
\begin{eqnarray}
\tilde{D}(p,q) &\sim& \int dx dy e^{ipx + iqy} D(x-y)\\
&=& \int du dy e^{ipu} e^{i(p+q)y} D(u) \\
&\sim& \tilde{D}(p) \delta(p+q)
\end{eqnarray}
where I'm neglecting to keep track of the constants. The result is diagonal in momentum space by virtue of the delta-function, enforcing $q=-p$.
A: The fact that the system is translationally invariant implies that the translation operator commutes with the Hamiltonian. This implies that they have a basis of mutual eigenstates. Since the momentum operator generates the translations, i.e. 
$$T =e^{-\imath x p}$$
a state is an eigenstate of the translation operator if and only if it is an eigenstate of the momentum operator. 
A: I guess what is to be noted here is the fact that, the Correlation function (operator), commutes with the momentum operator, since 
$$
[D,\text e^{ixp}] = 0 \implies [D,p] = 0
$$
Having that to be the case, one can recollect that any operator represented in its own eigenspace is diagonal should answer your question.
PS : I am not completely sure about the question(answer) , so this is just my persepective
A: To expand on the answer given by @By Symmetry, let the two point function be defined as
$$
f(x_1,x_2)=\langle\Omega|\, \phi(x_1)\phi(x_2)\,|\Omega\rangle.
$$
In order the above to be translationally invariant one must require that 
$$
\langle\Omega|\,\left[P, \phi(x_1)\phi(x_2)\right]\,|\Omega\rangle = 0
$$
where $P$ is the generator of the translations as defined as a one-parameter unitary group. This does not in general require the commutator to be always zero (it just requires it to be so on the vacuum state); however if you want this relation to hold on every state $|\psi\rangle$ then the commutator always being zero implies that $P$ and $\phi(x_1)\phi(x_2)$ have a set of common eigenstates where they could be simultaneously diagonalised.
