Understanding the Propagator in Quantum Field Theory

[Neitrino]

Hello PF :)
Let me for the moment consider just <0|$$\varphi(y)\varphi(x)$$|0> as a propagator (instead of commutator of the fields)... and so in this expression evolves only <0|aa$$^{+}$$|0> part.
Now my question is:

1) We can consider this expression as <0|a vector multiplied by a$$^{+}$$|0> which is <1|1> so this is a transition aplitude that one "one-particle state" will go to another "one-particle state" but I don't understad the idea of propagation in such treatment...
a)One-particle is created at x position - this is one quantum state
b)One-particle is created at y position - this is another quantum state
and multiplication of these quantum states I appreciate as a propagator?

2) We can consider above expression as multiplication of <0 vector by aa$$^{+}$$|0> vector where aa$$^{+}$$|0> is creation of particle at x position =|1> and death of this one-particle state at y position again giving me the vacuum.
So since the annihilation of already born one-particle state happens at y position I should assume that this one-particle state SHOULD TRAVEL to y position where it is annihilated by "a" (basically if "something" is annihilated "somewhere" this "something" should first reach that "somwhere" place)and this travel corresponds to propagation of particle from x to y?

Is my undersyanding correct ? If my understanding is correct I can't apply the same logic to my 1) treatment.

Thanks a lot

 

[Neitrino]

no ideas ... :(

 

[Neitrino]

? :(

 

[kanato]

Both of your perspectives are equivalent. A maybe easier to understand definition is this:
$$G(x,t;x',t') = \langle 0 | \psi(x,t) \psi^\dagger(x',t') |0 \rangle$$

The operator $$\psi^\dagger(x',t')$$ creates a particle in the vacuum state at position x' and time t'. Then the operator $$\psi(x,t)$$ attempts to destroy a particle at position x and time t. So what the propagator really measures is that if a particle at x',t' is allowed to time evolve for time t-t', what is its probability amplitude for being at x? The idea being that the time evolution operator may cause the particle which starts localized at x' to "spread out."

Your confusion might be because you've left off the time arguments. If you have your operators at the same time, then $$\langle 0 | \varphi(y) \varphi^\dagger(x) |0 \rangle$$ is the projection of your state x onto state y, and is really just testing their orthogonality.