3
$\begingroup$

I'm working on the discrete Fourier series in Peskin (page 285),but I have two questions.

enter image description here

Question 1:

I tried to derive Eq.(9.21):

Consider $$ f(x)=\int \frac{d^{4} k}{(2 \pi)^{4}} e^{-i k \cdot x} \tilde{f}(k) $$ So we have \begin{align} \phi(x_i)&=\int \frac{d^{4} k}{(2 \pi)^{4}} e^{-i k \cdot x_i} \phi(k)\\&= \lim _{n \rightarrow \infty} \sum_{n} \frac{\Delta^{4} k_n}{(2 \pi)^{4}} e^{-i k_n \cdot x_i} \phi(k_n) \end{align} In finite volume we have $k_n=\frac{2\pi n}{L}$,so $$\Delta^{4} k_n=\frac{(2\pi)^4}{L^4}=\frac{(2\pi)^4}{V}$$ So the Fourier transform of finite volume is \begin{align} \phi(x_i)&=\sum_{n} \frac{\Delta^{4} k_n}{(2 \pi)^{4}} e^{-i k_n \cdot x_i} \phi(k_n)\\&= \sum_{n} \frac{(2\pi)^4}{(2 \pi)^{4}V} e^{-i k_n \cdot x_i} \phi(k_n)\\&=\frac{1}{V} \sum_{n} e^{-i k_n \cdot x_i} \phi(k_n) \end{align} This is consistent with Eq.(9.21). However, I didnʻt use $\left|k^{\mu}\right|<\pi / \epsilon$. Where does it come from?Is my derivation wrong?

Question 2:

I tried to derive this equation but I have difficulties $$ \mathcal{D} \phi(x)=\prod_{k_{n}^{0}>0} d \operatorname{Re} \phi\left(k_{n}\right) d \operatorname{Im} \phi\left(k_{n}\right). $$ Consider Eq.(9.20) \begin{align} \mathcal{D} \phi&=\prod_{i} d \phi\left(x_{i}\right)\\&=\prod_{i} d \left(\frac{1}{V} \sum_{n} e^{-i k_{n} \cdot x_{i}} \phi\left(k_{n}\right)\right) \end{align} I don't know how to go on. Now how do I derive this equation?

$\endgroup$

1 Answer 1

2
$\begingroup$

Lets start with your first question. Peskin and Schroeder are putting all of the fields in a box with volume $V = L^4$ in Euclidian Space (after Wick Rotation). This means that smallest Fourier mode the fields can have is $k^{\mu} = \frac{2\pi}{L}$. However, that box is discretized into a lattice with lattice spacing $\epsilon$. This lattice spacing brings with it a UV cutoff to all of the fields, coarse graining the fields so that the largest Fourier mode the fields can have is $k^{\mu} = \frac{\pi}{\epsilon}$. Having both a lattice spacing and a finite box means your theory have UV and IR cut offs. So the Fourier Integrals you have should really be written as

$$ \phi(x_i) = \int_{-\pi/\epsilon}^{-\pi/L}\frac{d^4k}{(2\pi)^4}e^{-ik_ix^i}\tilde{\phi}(k_i) + \int^{\pi/\epsilon}_{\pi/L}\frac{d^4k}{(2\pi)^4}e^{-ik_ix^i}\tilde{\phi}(k_i) $$ Usually you take $L\rightarrow \infty$, but that is where the $\pi/\epsilon$ comes from.

The answer to your second question I believe is found here Does "sum over all paths" in the path integral imply "sum over all paths" in momentum space when one Fourier-transforms?.

$\endgroup$
2
  • $\begingroup$ I still don't understand “This lattice spacing brings with it a UV cutoff to all of the fields”. Consider $k=\frac{2\pi n}{L}$, n=1,2,3..., $k$ does have a minimum $\frac{2\pi}{L}$, but why it also have a maximum?@user3166083 $\endgroup$
    – sky
    Mar 2, 2020 at 6:02
  • 1
    $\begingroup$ If your fields are on a discrete lattice, then, for example, you know information about two points $\phi(x_i)$ and $\phi(x_i + \epsilon)$, but not, say, $\phi(x_i + \epsilon/2)$. You cannot get arbitrarily close between two points on a lattice. This means that the fields do not have access to Fourier modes corresponding to distances shorter than the lattice spacing. $\endgroup$ Mar 2, 2020 at 16:21

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.