Hello PF :)(adsbygoogle = window.adsbygoogle || []).push({});

Let me for the moment consider just <0|[tex]\varphi(y)\varphi(x)[/tex]|0> as a propagator (instead of commutator of the fields)... and so in this expression evolves only <0|aa[tex]^{+}[/tex]|0> part.

Now my question is:

1) We can consider this expression as <0|a vector multiplied by a[tex]^{+}[/tex]|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 dont 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[tex]^{+}[/tex]|0> vector where aa[tex]^{+}[/tex]|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 cant apply the same logic to my 1) treatment.

Thanks alot

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Interpretation of Propagator

**Physics Forums | Science Articles, Homework Help, Discussion**