# Pauli Jordan Green's function

Yoran91
Hi everyone,

I'm going through some lecture notes on Quantum Field Theory and I came across a derivation of an explicit form of the Pauli Jordan Green's function for the Klein-Gordon field.

The equations used in my lecture notes are equivalent to the ones in http://www.physics.byu.edu/faculty/berrondo/wt752/Invariant%20Functions.pdf
.

My question is actually quite simple: how does equation 4.144 follow from 4.143 in the above pdf? (the same equations are in my lecture notes as well).

Homework Helper
Gold Member
It's just due to the Euler identity

$$e^{i\theta} = \cos \theta + i \sin\theta.$$

This identity shows up in a lot of places, so it is a good one to remember!

It's just due to the Euler identity
There's a bit more to it than that, isn't there? Looks to me like he pulled a switcheroo in notation. In Eq 4.143, p·x is a 4-dimensional dot product, but in Eq. 4.144, the same p·x is 3-dimensional.

Yoran91
I see that I'd need to use $\sin(x) = 1/2i (e^{ix}-e^{-ix})$, but it seems like the factor $\exp(i\vec{p}\cdot \vec{x})$ is factored out, which can't be true. They way I see it is

$\exp(-ip\cdot x)-\exp(ip\cdot x)=\exp(-iEt+i\vec{p}\cdot\vec{x})-\exp(iEt-i\vec{p}\cdot\vec{x})$, in which it is not possible to just factor $\exp(i\vec{p}\cdot \vec{x})$ out and be left with $\sin(Et)$ (with some factors)

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dauto
Yes, the p and x on equation 4.144 should be bold to make that point clear

dauto
I see that I'd need to use $sin(x) = 1/2i (e^{ix}-e^{-ix})$, but it seems like the factor $\exp(i\vec{p}\cdot \vec{x})$ is factored out, which can't be true.

why not?

dauto
I see your problem. You're forgetting that the integral is over a symmetric region. Make the change of variables $\vec{p}' = - \vec{p}$ in one of the integrals and all is well

andrien
the spatial momentum goes from -infinity to infinity.so you can change the sign in one of them without any effect

Yoran91
But wouldn't that introduce a factor -1 (Jacobian) as well, so that the term left over would be a cosine rather than a sine?

dauto
There is another -1 factor from the reversal of the integration limits

Yoran91
Ah of course! Thanks!