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Momentum cutoff equivalent to discretizing space?

  1. Aug 9, 2014 #1
    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 [itex]k=\frac{2\pi n}{L}[/itex] and [itex]\lambda=\frac{L}{n}[/itex]

    If you consider your fields in momentum space, and put it in a box of size [itex]\Lambda[/itex], is space discretized? Would the relationship be [itex]x=\frac{2\pi n}{\Lambda}[/itex] and
    [itex]\lambda=\frac{\Lambda}{n}[/itex], where [itex]\lambda[/itex] 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., [itex]\int \phi(x)^2 d^dx[/itex], then Fourier transform the action to an integral in momentum space [itex]\int \phi(-k)\phi(k) d^dk[/itex], then impose a momentum cutoff? Because if you had a momentum cutoff to begin with, [itex]\int \phi(x)^2 d^dx[/itex] 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 [itex]\phi(k)[/itex] such that [itex]\phi(x)=\int \phi(k) e^{-ikx} dk [/itex] satisifies the differential equation in position space and the boundary conditions [itex]\phi(x=0)=\phi(x=L) [/itex]? There seems to me a big difference between having a discrete [itex]\phi_k[/itex] rather than a continuous [itex]\phi(k) [/itex], 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.
     
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  3. Aug 10, 2014 #2

    atyy

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