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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!

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4 Answers 4

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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 $q=-p$.

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  • $\begingroup$ interesting that you defined a translationally invariant correlation function as one that depends only on the combination (x-y). Of course it seems reasonable, as any translation will automatically be cancelled by the subtraction $\endgroup$ Sep 26, 2015 at 16:47
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    $\begingroup$ I clarified where the dependence on only $x-y$ comes from. $\endgroup$
    – josh314
    Sep 26, 2015 at 18:30
  • $\begingroup$ Nick, glad I could help. If you believe this has addressed your question sufficiently, would you be so kind as to accept it as the answer? $\endgroup$
    – josh314
    Sep 27, 2015 at 1:19
  • $\begingroup$ I did accept the answer I believe? I just click the up button, correct? Just for the sake of clarification, a three point correlation function would then have to depend on some combination like x+y-2z, making it two dimensional in momentum space, correct? $\endgroup$ Sep 27, 2015 at 8:10
  • $\begingroup$ Nick, notice that some questions will have a green check mark next to the top answer, for example, physics.stackexchange.com/questions/20813/…. Such answers have been "accepted" by the poster of the question. Only the original poster (OP) can accept an answer although any user with sufficient privileges can upvote or downvote an answer. Upvoting/downvoting is done by clicking the up/down arrows next to answer. To accept an answer there should be a greyed-out checkmark you can click on. You can only accept one answer per question. $\endgroup$
    – josh314
    Sep 27, 2015 at 13:36
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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.

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  • $\begingroup$ ok I see this. But I don't see how this implies that the correlation function is diagonal. Maybe you are saying that the correlation function is itself an inner product of hamiltonian eigenstates, implying that each is itself an eigenstate of the momentum operator? $\endgroup$ Sep 26, 2015 at 11:14
  • $\begingroup$ @NickMurphy : Think about an operator's representation in its own eigenspace representation !! $\endgroup$
    – user35952
    Sep 26, 2015 at 13:07
  • $\begingroup$ @user35952 Hey ok I think I see what you are saying is that of course any operator is diagonal in a basis of its own eigenstates. But this seems to prove to me that one CAN choose a momentum basis that diagonalizes the correlation function. Not that it is necessarily so $\endgroup$ Sep 26, 2015 at 13:14
  • $\begingroup$ @NickMurphy : Yes. But the choice is always there when, especially so, when Hamiltionian commutes with the translation operator and hence with the momentum operator as well !! $\endgroup$
    – user35952
    Sep 26, 2015 at 13:16
  • $\begingroup$ @user35952 Ok nice, I think I'm following. Thanks! $\endgroup$ Sep 26, 2015 at 13:17
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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

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  • $\begingroup$ I'm still not quite sure why we are talking about the hamiltonian here. I am talking purely about the correlation function, which does not involve the hamiltonian, but quantum fields. $\endgroup$ Sep 26, 2015 at 13:28
  • $\begingroup$ @NickMurphy : My bad, I was solving something else and it got mixed up !! $\endgroup$
    – user35952
    Sep 26, 2015 at 14:38
  • $\begingroup$ No problem, thanks a lot for the help and answers! $\endgroup$ Sep 26, 2015 at 15:12
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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.

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  • $\begingroup$ Ok so this seems ok, but the correlation function is only defined to on the vacuum state. So extrapolating to all states seems invalid. And even then, it seems you have proven that the correlation function can be made diagonal in momentum space, but isn't necessarily so, as all we have shown is that there exists one such basis, but not that every momentum basis diagonlizes both $\endgroup$ Sep 26, 2015 at 12:55
  • $\begingroup$ In Peskin in schroder, on page 381, he states that on the translationally invariant vacuum, the correlation function MUST be diagonal in momentum space, i.e more than it is just possible to choose such a momentum basic. $\endgroup$ Sep 26, 2015 at 13:30
  • $\begingroup$ Yes, I found it. I believe the correct answer is the one provided by @josh314 physics.stackexchange.com/q/209231 $\endgroup$
    – gented
    Sep 26, 2015 at 16:58

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