2
$\begingroup$

My question is quite general, but it is inspired by P&S and especially Chapter 12. In Chapter 12 they start with the generating functional, in which the functional integral is over the Fourier transform of the field configuration $\phi(k)$ $$Z[J]=\bigg(\prod_k\int d\phi(k)\bigg)e^{i\int[\mathcal{L}+J\phi]}.\tag{12.1}$$

  1. First, I would like to clear a small misunderstanding: the exponent of the exponential contains an integral but no integration measure. Integration over the momentum space is implied right? Does it even matter?

  2. And then, my real question is how do I transform from a functional integral over the field to a functional integral over the fourier transform of the field? What is going on in that functional integration measure and how do I go from $\mathcal{D}\phi(x)$ to $\mathcal{D}\phi(k)$ and vice versa??

I have read P&S and Srednicki, also did some google search, but I couldn't find anything satisfactorily rigorous.

$\endgroup$

1 Answer 1

3
$\begingroup$

Integration over the momentum space is implied right? Does it even matter?

From context, yes I would interpret the measure as an integral over momentum space. I am only speculating, but perhaps they did not include the measure to avoid confusion between the $k$ appearing in the path integral measure, and the $k$ that would appear if they wrote the measure in the action explicitly.

And then, my real question is how do I transform from a functional integral over the field to a functional integral over the fourier transform of the field? What is going on in that functional integration measure and how do I go from 𝜙(𝑥) to 𝜙(𝑘) and vice versa??

Note the integral equation \begin{equation} \tilde{\phi}(k) = \int d^4 x e^{i k x} \phi(x) \end{equation} has the form of a matrix equation \begin{equation} \tilde{\phi}_n = U_{n j} \phi_j \end{equation} if we interpret the sum over $j$ as an integral, and where $U_{n j} = e^{i k_n x_j}$ is a unitary matrix.

Therefore, when doing a change of variables in the integration measure from from $\prod_k d \tilde{\phi}(k)$ to $\prod_x d \phi(x)$, we pick up a trivial Jacobian factor $\det U = 1$.

It actually doesn't even really matter that $U$ is unitary; the key point is that $U$ is independent of the field (in other words, we've done a linear change of variables). This is because in the end, we are almost always interested in ratios of partition functions, like $Z[J]/Z[0]$, so an overall field-independent constant will drop out of the final answer.

Putting this together, we can equate the path integrals written in momentum space and real space, since they only differ by a linear change of variables \begin{equation} \left(\prod_k \int d \tilde{\phi}_k \right) e^{i \int d^4 k' \mathcal{L}} = \left(\prod_x \int d \phi_x \right) e^{i \int d^4 x' \mathcal{L}} \end{equation} The momentum space representation is usually more useful for perturbation theory.

$\endgroup$
2
  • $\begingroup$ Thanks! Can you also explain why usually physics textbooks write $\prod_k \int d\phi_k d\phi_k^*$, denoting with $z^*$ the complex conjugate of the complex number $z$. From your very clear answer I cannot understand how the complex conjugate can get out. Moreover, what about the bounds of the integral over $d\phi_k$? $\endgroup$ Commented Dec 29, 2023 at 16:22
  • $\begingroup$ @GiancarloCreanza Normally the bounds over $d\phi_k$ would run from $-\infty$ to $\infty$ (assuming $\phi_k$ is real). For your other question, I think this stack exchange answer gives a much clearer explanation than I could give in a comment: physics.stackexchange.com/a/695946/27732. Also this one: physics.stackexchange.com/questions/89002/… $\endgroup$
    – Andrew
    Commented Dec 29, 2023 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.