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

[pellman]

In QFT expressions such as these hold:

real scalar:
$$\Delta_F(x-x')\propto\langle 0| T\phi(x)\phi(x')|0\rangle$$

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

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.

 

[arkajad]

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.

 

[The_Duck]

I'm only just learning this too, so take with a chunk of salt:

$$\Delta_F(x-x')$$ 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 $$\Delta_F(x-x')$$ 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 $$\langle 0| T\phi(x)\phi(x')|0\rangle$$ 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.

 

[pellman]

Thanks! That helps actually.

 

[strangerep]

In QFT expressions such as these hold:

real scalar:
$$\Delta_F(x-x')\propto\langle 0| T\phi(x)\phi(x')|0\rangle$$

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

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. :-)

 

[dextercioby]

I would recommend further reading in A.Zee's <QFT in a Nutshell> book. He has some nice, edible explanations for some advanced concepts.