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).

 

[fzero]

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!

 

[Bill_K]

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)

 

[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!