- 969

- 3

Say you have the integral:

[tex]D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(.5+.5\frac{\vec{k}^2}{k_{0}^2})}{k^2+i\epsilon} [/tex]

Can I just set this equal to:

[tex]D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2+i\epsilon} [/tex]

since the pole structure of the denominator will force k to be on-shell?

But I thought Fourier transforms were unique? So how can two different transform functions give the same position space function?

You don't really see this type of question when using Lorentz covariant gauges, since here k_0 and vector k are split up in non-covariant way. But if you have a non-covariant gauge, you might see this.

It's just weird to me because I thought Fourier transforms were unique. Does this mean that I can use either momentum space representation of the propagator in Feynman diagrams, and I'll get the same answer?