# Action of a Scalar Field in Path Integral Formulation Peskin & Schroeder (Pag. 285)

I'm really confused on the discretization stuff on this chapter of P&S. My question is related to the computation of the Action in scalar field theory done in page 285. When they compute the action of the real scalar field, they get $$\int d^4x\left(\frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2\right)=-\frac{1}{V}\sum_n\frac{1}{2}(m^2-k_n^2)|\phi(k_n)|^2.\tag{p.285}$$

I'm stuck on this equation. Here is what I tried to do:

Following from the left-hand side, I can use the properties of the (continuous) Fourier transform and get $$\int d^4x\left(\frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2\right)=\int d^4x \frac{1}{2}\left(k^2-m^2\right)(\phi(x))^2.$$

The field is real, therefore $$(\phi(x))^2 = |\phi(x)|^2$$. Now I think Parseval's theorem should be used, $$\int d^4x \frac{1}{2}\left(k^2-m^2\right)(\phi(x))^2 = \dfrac{1}{2}(k^2-m^2)\int d^4x |\phi(x)|^2 =\dfrac{1}{2}(k^2-m^2)\int \dfrac{d^4k}{(2\pi)^4}|\phi(k)|^2.$$

Changing the integral in $$d^4k$$ to a series in $$n$$, (I'm assuming I can use eq. 9.22 backwards $$\int \frac{d^4k}{(2\pi)^4} \to \frac{1}{V}\sum_n$$), $$\dfrac{1}{2}(k^2-m^2)\int \dfrac{d^4k}{(2\pi)^4}|\phi(k)|^2 = \dfrac{1}{2V}(k^2-m^2)\sum_n|\phi(k_n)|^2.$$

Here is a problem, I have a continuous variable $$k$$ and a discrete one $$k_n$$ on the same equation instead of just $$k_n$$. What I'm doing wrong? Is the eq. 9.22 backwards the problem?

Here's is my go at this problem. I think its easier if no steps are skipped so that mistakes aren't made with fourier transforming: $$$$\int d^{4}x\:\frac{1}{2}(\partial_{\mu}\phi\partial^{\mu}\phi-m^{2}\phi^{2}) \\\overset{\text{Fourier}}{=}\int d^{4}x\int \frac{d^{4}k_{1}}{(2\pi)^{4}}\int\frac{d^{4}k_{2}}{(2\pi)^{4}}\frac{1}{2}(-k_{1}\cdot k_{2}-m^{2})\hat{\phi}(k_{1})\hat{\phi}(k_{2})e^{-ix\cdot (k_{1}+k_{2})} \\ = \int \frac{d^{4}k_{1}}{(2\pi)^{4}}\int\frac{d^{4}k_{2}}{(2\pi)^{4}}\frac{1}{2}(-k_{1}\cdot k_{2}-m^{2})\hat{\phi}(k_{1})\hat{\phi}(k_{2})(2\pi)^{4}\delta^{(4)}(k_{1}+k_{2}) \\ = \int \frac{d^{4}k_{1}}{(2\pi)^{4}}\frac{1}{2}(k_{1}\cdot k_{1}-m^{2})\hat{\phi}(k_{1})\hat{\phi}(-k_{1}) \\ = \int \frac{d^{4}k_{1}}{(2\pi)^{4}}\frac{1}{2}(k_{1}\cdot k_{1}-m^{2})|\hat{\phi}(k_{1})|^{2}\rightarrow\frac{1}{V}\sum_{n}\frac{1}{2}(k_{n}^{2}-m^{2})|\hat{\phi}(k_{n})|^{2}$$$$
• Thank you so much! I keep forgetting to insert different integration variables $k_1$ and $k_2$ in this cases. Correct me if I'm wrong, but I think that P&S doesn't swap the metric. When we take the derivative of the Fourier transform, we get $$\partial_\mu \phi(x) = \int \dfrac{d^4k}{(2\pi)^4}(\partial_\mu e^{-ik\cdot x})\hat{\phi}(k)= \int \dfrac{d^4k}{(2\pi)^4}(-ik_\mu) e^{-ik\cdot x}\hat{\phi}(k)$$ Hence we get a factor of $(-i)^2$ in the inner product, and when we take $k_1 = -k_2$ by the delta function condition the minus sign vanishes and we keep the same metric signature.