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## Main Question or Discussion Point

Hello PF :)

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

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