Correlation functions in position and momentum space

[spaghetti3451]

What is the relation between the correlators ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle## and ##\langle 0 | T\phi(p)\phi(x=0) | 0 \rangle##?

I can derive the momentum space Feynman rules for ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle##. Are the momentum space Feynman rules for ##\langle 0 | T\phi(p)\phi(x=0) | 0 \rangle## the same?

 

[eljose79]

The correlators ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle## and ##\langle 0 | T\phi(p)\phi(x=0) | 0 \rangle## are related through the Fourier transform. In momentum space, the correlator ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle## can be written as ##\int \frac{d^{4}p}{(2\pi)^{4}}e^{-ip\cdot(x_{1}-x_{2})}\langle 0 | T\phi(p)\phi(-p) | 0 \rangle##.

The momentum space Feynman rules for ##\langle 0 | T\phi(p)\phi(x=0) | 0 \rangle## will be similar to those for ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle##, but with some key differences. The main difference is that the momentum space Feynman rules for ##\langle 0 | T\phi(p)\phi(x=0) | 0 \rangle## will only involve one external momentum, while the Feynman rules for ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle## involve two external momenta. This is because in the latter case, we are correlating two fields at different points in space, while in the former case, we are correlating a field at a point in space with itself.

Another important difference is that the momentum space Feynman rules for ##\langle 0 | T\phi(p)\phi(x=0) | 0 \rangle## will involve the propagator for the field at a point, while the Feynman rules for ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle## will involve the propagator for the field at two different points.

In summary, while the momentum space Feynman rules for ##\langle 0 | T\phi(p)\phi(x=0) | 0 \rangle## may have some similarities to those for ##\langle 0 | T\phi(x_{1})\phi(x_{2}) | 0 \rangle##, they are not exactly