Why is the Green's function equal to the vacuum expectation of the field?

  • Thread starter pellman
  • Start date
  • #1
675
4
In QFT expressions such as these hold:

real scalar:
[tex]\Delta_F(x-x')\propto\langle 0| T\phi(x)\phi(x')|0\rangle[/tex]

4-spinor
[tex]S_F(x-x')]\propto\langle 0| T\psi(x)\bar{\psi}(x')|0\rangle[/tex]

where T is the time-ordering operation and the proportionality depends on the choice of normalization.

I can prove these by direct calculation against other means of deriving the Green's functions but what is the explanation as to why it holds? I don't find one in my QFT texts.

Extra credit: what the does "F" subscript denote? Seems to be a standard notation.
 

Answers and Replies

  • #2
1,444
4
F for Feynman. I looked in old good Roman's "Introduction to quantum field theory" - he does it, but it was long ago that I studied this subject, so I will stop at that.
 
  • #3
1,006
105
I'm only just learning this too, so take with a chunk of salt:

[tex]\Delta_F(x-x')[/tex] is a Green's function of the classical field's equation of motion, i.e. it represents the field evolving undisturbed except for a brief disturbance at x'. We might then reasonably think of [tex]\Delta_F(x-x')[/tex] as representing the value of the field at x that results from the disturbance created at x' (really I guess this makes more sense for the retarded Green's function). But [tex]\langle 0| T\phi(x)\phi(x')|0\rangle[/tex] represents a similar idea in the quantum theory: the amplitude for a particle (disturbance in the field) created at x' to propagate to x, or vice versa.
 
  • #4
675
4
Thanks! That helps actually.
 
  • #5
strangerep
Science Advisor
3,204
1,056
In QFT expressions such as these hold:

real scalar:
[tex]\Delta_F(x-x')\propto\langle 0| T\phi(x)\phi(x')|0\rangle[/tex]

4-spinor
[tex]S_F(x-x')]\propto\langle 0| T\psi(x)\bar{\psi}(x')|0\rangle[/tex]

where T is the time-ordering operation and the proportionality depends on the choice of normalization.

I can prove these by direct calculation against other means of deriving the Green's functions but what is the explanation as to why it holds? I don't find one in my QFT texts.

Extra credit: what the does "F" subscript denote? Seems to be a standard notation.
Take a look at my post #8 in this thread:

https://www.physicsforums.com/showthread.php?t=420953

Basically, those expressions represent the amplitude for a particle to propagate
from x' to x, subject to (a) the field satisfies relativistic (KG or Dirac) eqn of motion,
(b) the field satisfies canonical (anti)commutation relations (so that spacelike-separated
events cannot exert a causal influence on each other), and (c) boundary conditions (which
determine whether you're dealing with (say) a retarded propagator, or (more usually in
QFT) a Feynman propagator, etc.

So, in a nutshell, these relations hold because the quantum fields were constructed to
satisfy (suitably causal) relativity and also the basic principles of quantum theory. :-)
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
13,019
567
I would reccomend further reading in A.Zee's <QFT in a Nutshell> book. He has some nice, edible explanations for some advanced concepts.
 

Related Threads on Why is the Green's function equal to the vacuum expectation of the field?

Replies
7
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
11
Views
580
Replies
9
Views
2K
Replies
18
Views
3K
Replies
6
Views
733
Replies
1
Views
601
Replies
4
Views
664
Top