# Momentum cutoff equivalent to discretizing space?

1. Aug 9, 2014

### geoduck

Does having a momentum cutoff require space to be discretized?

As an analogy, suppose you put your fields in a box of size L, and require periodic boundary conditions. Then momentum is discretized. The relationship is $k=\frac{2\pi n}{L}$ and $\lambda=\frac{L}{n}$

If you consider your fields in momentum space, and put it in a box of size $\Lambda$, is space discretized? Would the relationship be $x=\frac{2\pi n}{\Lambda}$ and
$\lambda=\frac{\Lambda}{n}$, where $\lambda$ here is the wavelength in momentum space?

If this is the case, then would it make sense to start from the action as an integral in position space, i.e., $\int \phi(x)^2 d^dx$, then Fourier transform the action to an integral in momentum space $\int \phi(-k)\phi(k) d^dk$, then impose a momentum cutoff? Because if you had a momentum cutoff to begin with, $\int \phi(x)^2 d^dx$ should really be a sum over discrete values of x rather than an integral.

Also, in general, just because the interval on which a function is defined is finite, must this imply the Fourier transform is discrete? Can you still use a continuous Fourier transform (i.e., integral) and consider all $\phi(k)$ such that $\phi(x)=\int \phi(k) e^{-ikx} dk$ satisifies the differential equation in position space and the boundary conditions $\phi(x=0)=\phi(x=L)$? There seems to me a big difference between having a discrete $\phi_k$ rather than a continuous $\phi(k)$, and just saying that your system is in a box of length L doesn't tell you whether to use discrete or continuous Fourier transforms to momentum space.

2. Aug 10, 2014