Pauli Jordan Green's function

In summary, the equation 4.144 follows from 4.143 in the pdf due to the Euler identity. The notation in equation 4.143 is changed to make the equation easier to remember, but this change has no effect on the result.
  • #1
Yoran91
37
0
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).
 
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  • #2
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!
 
  • #3
fzero said:
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.
 
  • #4
I see that I'd need to use [itex]\sin(x) = 1/2i (e^{ix}-e^{-ix})[/itex], but it seems like the factor [itex]\exp(i\vec{p}\cdot \vec{x})[/itex] is factored out, which can't be true. They way I see it is

[itex]\exp(-ip\cdot x)-\exp(ip\cdot x)=\exp(-iEt+i\vec{p}\cdot\vec{x})-\exp(iEt-i\vec{p}\cdot\vec{x})[/itex], in which it is not possible to just factor [itex]\exp(i\vec{p}\cdot \vec{x})[/itex] out and be left with [itex]\sin(Et)[/itex] (with some factors)
 
Last edited:
  • #5
Yes, the p and x on equation 4.144 should be bold to make that point clear
 
  • #6
Yoran91 said:
I see that I'd need to use [itex]sin(x) = 1/2i (e^{ix}-e^{-ix})[/itex], but it seems like the factor [itex]\exp(i\vec{p}\cdot \vec{x})[/itex] is factored out, which can't be true.

why not?
 
  • #7
I see your problem. You're forgetting that the integral is over a symmetric region. Make the change of variables [itex] \vec{p}' = - \vec{p} [/itex] in one of the integrals and all is well
 
  • #8
the spatial momentum goes from -infinity to infinity.so you can change the sign in one of them without any effect
 
  • #9
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?
 
  • #10
There is another -1 factor from the reversal of the integration limits
 
  • #11
Ah of course! Thanks!
 

1. What is Pauli Jordan Green's function?

Pauli Jordan Green's function is a mathematical tool used in quantum mechanics to describe the behavior of a quantum system in terms of its energy levels and wave functions.

2. How is Pauli Jordan Green's function used?

Pauli Jordan Green's function is used to calculate the probability amplitude for a particle to transition from one state to another in a quantum system. This information is crucial for understanding the dynamics and properties of the system.

3. Who discovered Pauli Jordan Green's function?

Pauli Jordan Green's function was first introduced by the physicists Wolfgang Pauli and Pascual Jordan in 1927 as a solution to the Schrödinger equation.

4. What are the applications of Pauli Jordan Green's function?

Pauli Jordan Green's function is widely used in many areas of physics, including quantum mechanics, condensed matter physics, and statistical mechanics. It is also used in engineering and chemistry to describe the behavior of complex systems.

5. Can Pauli Jordan Green's function be extended to include interactions between particles?

Yes, Pauli Jordan Green's function can be extended to include interactions between particles through the use of perturbation theory. This allows for a more accurate description of real-world systems that may have interactions between particles.

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