Translational invariance implying diagonal representation in momentum space

Question

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!

Answer 1

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 $q=-p$.

Answer 2

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.

Answer 3

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

Answer 4

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.