Klein-Gordon Field: Understanding Eq. (1)

[grimx]

Hi everyone! Im' a new member and I'm studying Quantum Field Theory.

I read this:

"The interpretation of the real scalar field is that it creates a particle (boson) with momentum p at the point x."

and :

[itex]\phi[/itex][itex]\left(x\right)[/itex] [itex]\left|0\right\rangle[/itex] = [itex]\int \frac{d^3p}{(2\pi)^3(2\varpi_p)}[/itex] [itex]e^{-ipx} |p\rangle[/itex] (1)

but I didn't understand this equality... i know that:

[itex]\phi (x) = \int \frac{d^3p}{(2\pi)^3(2\varpi_p)} (a_p e^{ipx} + a^+_p e^{-ipx})[/itex] (2)

So... where it goes the term [itex]a_p e^{ipx}[/itex] in the expression (1) ?

Can someone kindly show me all the steps?
I know it's a stupid question, but I can not understand.

thank you very much!

 

[WannabeNewton]

The vacuum is annihilated by ##a_p## by definition.

 

[grimx]

Thank you for your reply.
But in theory... [itex]a_p[/itex] it should not destroy the particle created by [itex]a^+_p[/itex]??

What am I doing wrong?

Thank you.

 

[WannabeNewton]

We don't have ##a_p a^{\dagger}_p## in the free KG field. We have ##a_p## attached to the negative frequency modes and ##a^{\dagger}_p## attached to the positive frequency modes so they act independently of one another.

As such ##\phi(x)|0\rangle## simply creates a particle at ##x##.

 

[grimx]

Thanks! :)