Green's function & Klein Gordon

  • Thread starter Sekonda
  • Start date
  • #1
207
0

Main Question or Discussion Point

Hello,

I was wondering what the use in the Green's function for the Klein-Gordon equation was, I have listed it below:

[tex]\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}e^{ip\cdot(x-x')}[/tex]

We find this gives an infinite result when the Klein gordon equation is applied to it and if x=x', what does this mean, i.e.

[tex](\frac{\partial^2 }{\partial t^2}-\nabla^2+m^2)\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}e^{ip\cdot(x-x')}=i\delta^{4})(x-x')[/tex]

Does this mean the particle described by the green's function in the first equation can only be found in one place when x=x'?

Also if we set x=x', the Green's function above turns to

[tex]\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}[/tex]

Which is alike to the 2 point function first order correction of form

[tex]-\frac{i\lambda}{2}\int \frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}[/tex]

What is the relationship here? What use are the Green's function to the above equation? p is the four vector :

[tex]p=(E,\mathbf{p})[/tex]

Thanks,
SK
 

Answers and Replies

  • #2
207
0
I'm now aware that the first equation in my previous post is also a position space propagator for the Klein-Gordon equation; would anyone be able to explain to me why for this diagram

2pointfunction.png


The correction to the 2 point function for this first order can be given as an integral over the 3 position space propagators associated with (x',x'), (x,x') and (x',y)?

Along with the -iλ/2 from the vertex (/symmetry factor = 1/2)

Thanks
 

Related Threads on Green's function & Klein Gordon

Replies
4
Views
3K
Replies
1
Views
985
Replies
5
Views
4K
Replies
4
Views
343
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
581
Top