- #1
geoduck
- 257
- 2
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.
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.