I'm reading section 9.2 of Peskin and Schroeder, specifically where they begin the explicit computation of the two point correlation function for a free scalar field using the path integral (p.285). They use a finite (size $L$ along each dimension) square lattice (lattice spacing $\epsilon$ along each dimension) to do this, then take the continuum limit in the end.

They begin by using the discrete Fourier transform to do a change of variables $$\phi(x_{i})=\frac{1}{V}\sum_{n}e^{-ik_{n}\cdot x_i}\phi(k_n)\tag{9.21}$$ where $V=L^4$, $k^\mu_n=2\pi n^\mu/L $ and the four sums are over all integers such that $|k^{\mu}_n|<\pi/\epsilon$, so they are finite sums. I think this also means that the range of the summation indices includes the integer zero. More precisely (I think), the lattice is assumed to have $(2M+1)$ points and each of the four sums runs from $-M$ to $M$. Here is where my confusion comes in. Since the $\phi(x_i)$ are real then $\phi^*(k_n)=\phi(-k_n)$ so P&S then consider the real and imaginary parts of the $\phi(k_n)$ with $k_n^0>0$ as the independent variables in the path integral. My question is what about all the $\phi(k_n)$ that have $k_n^0=0$? it seems to me there should be $d\phi(k_n)$ measures in the path integral that have $k_n^0=0$ but the restriction they state just ignores them which does not make sense to me. I'm not sure where I'm going wrong with my reasoning so any clarification is appreciated.

  • $\begingroup$ Those states have no energy, never vibrating in time, and thus cannot be physical. $\endgroup$ Commented May 18, 2023 at 10:17
  • $\begingroup$ Do you mind explaining in more detail? I don't understand where energy comes in exactly and in what context? I thought this was just a change in the integration variables in the path integral. @naturallyInconsistent $\endgroup$
    – Function
    Commented May 18, 2023 at 15:37
  • $\begingroup$ $k^0=0=\omega$ is the energy of a wave, the angular frequency of the wave in time. $\endgroup$ Commented May 18, 2023 at 15:39
  • $\begingroup$ Yes but only in a specific context. For example when said plane waves are solutions to the Schrödinger equation in NRQM then your statement is true. But I'm struggling to see how it relates to the context of my question. To my mind it seems P&S are merely changing variables in the path integral and here $k^\mu_n=2\pi n^\mu/L$ by definition for the discrete Fourier transform, there is no physical interpretation for them like the one you put forth. @naturallyInconsistent $\endgroup$
    – Function
    Commented May 18, 2023 at 16:46
  • $\begingroup$ Related: physics.stackexchange.com/q/43312/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented May 19, 2023 at 5:00

1 Answer 1


I think I figured it out. The key thing I forgot is the constraint on the field configuration at the initial and final time in the definition of the path integral. Once this is taken into account then the counting seems to work.

First, a mathematical note, the form of the discrete Fourier transform for which the reality of the function $\phi(x^\mu_m)$ being transformed translates into the requirement that $\phi^*(k^\mu_n)=\phi(-k^\mu_n)$ is precisely given by (e.g. see appendix B in Stone and Goldbart p.780)

$$\phi(x_m)=\frac{1}{V}\sum_{n^0=-M}^{+M}\sum_{n^1=-M}^{+M}\sum_{n^2=-M}^{+M}\sum_{n^3=-M}^{+M}e^{-ik_n\cdot x_m}\phi(k_n)$$


$$\phi(k_n)=\frac{V}{(2M+1)^4}\sum_{m^0=0}^{2M}\sum_{m^1=0}^{2M}\sum_{m^2=0}^{2M}\sum_{m^3=0}^{2M}e^{+ik_n\cdot x_m}\phi(x_m)$$

where $V=L^4=(2M+1)^4\epsilon^4$ and $k^\mu_n=2\pi n^\mu/L$. Note that $|k^\mu_n|=2\pi |n^\mu|/L\le2\pi M/L=\frac{\pi}{\epsilon}\frac{2M}{2M+1}<\pi/\epsilon$.

Therefore, P&S are implicitly assuming a lattice where the spacetime points are given by $$x^\mu_m=(m^0\epsilon,m^1\epsilon,m^2\epsilon,m^3\epsilon)$$ where $$m^\mu=0,...,2M.$$ Now, one would naively think that this means there are $(2M+1)^4$ integration measures $d\phi(x^0_m,x^1_m,x^2_m,x^3_m)$ in our discretized path integral, however, recall that there is a constraint on the field configuration when $x^0=0$, so we do not integrate over $d\phi(x^0_m,x^1_m,x^2_m,x^3_m)$ that have $x^0_m=0$ (the integrands in eq. 9.18 in P&S will of course still contain factors with such $\phi(x_m)$'s and they will cancel with the corresponding factors in the denominator of that formula). Therefore, in reality we have $(2M)(2M+1)^3$ integration variables. Note that the field configuration at the final time is naturally excluded already since $x^0=L=(2M+1)\epsilon$ is the total time yet the $m^0$ index above only runs up to $2M$.

After the change of variable, we should have the same number of integration variables, so we need to similarly exclude some of the $d\phi(k^0_n,k^1_n,k^2_n,k^3_n)$. The range of the indices $n^\mu=-M,...,M$ in the sum in eq. 9.21 again furnishes a naive counting of $(2M+1)^4$ possible integration measures $d\phi(k^0_m,k^1_m,k^2_m,k^3_m)$. We can choose to not integrate over all the $d\phi(k^0_n,k^1_n,k^2_n,k^3_n)$ that have $k^0_n=0$ (again the integrands in eq. 9.18 in P&S will contain such factors and they will cancel with the corresponding factors in the denominator of that formula). Therefore, we have $(2M)(2M+1)^3$ integration variables. Now we remember that the $\phi(k_n)$ are actually complex so this is really $2(2M)(2M+1)^3$ real integration variables. We can restrict to $k^0_n>0$ then we are back to $(2M)(2M+1)^3$ real integration variables. Therefore, we are integrating over the same number of discrete variables in the path integral as we did before Fourier transforming.


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