Solving QFT Problem: Deriving <k'|(\partial_\mu \phi^\dag)\phi|k>

[PJK]

Somehow I have problems with figuring out the following problem:
I know that the scalar field is obeying the follwoing equations:
$$<0|\phi(x)|k> = e^{ikx}$$
$$<0|\phi(x)^\dag|k> = 0$$
$$<k'|\phi(x)^\dag|0> = e^{-ik'x}$$
$$<k'|\phi(x)|0> = 0$$
And I was told that I can deduce the following result from the equations above:
$$<k'|(\partial_\mu \phi^\dag)\phi|k> = - i k'_\mu e^{-i(k'-k)x}$$
I can 'derive' this when I sandwich a vacuum projector in the lhs:
$$<k'|(\partial_\mu \phi^\dag)\phi|k> = (\partial_\mu <k'|\phi^\dag|0>)<0|\phi|k>$$
But I do not understand why I am allowed to do this?

 

[Physics Monkey]

Hi PJK,

A good way to think about this problem is to insert a complete set of states (which you are always allowed to do) and ask which states contribute to the matrix element. As you will see, many don't.

Hope this helps.

 

[PJK]

Thanks for your answer Physics Monkey!
This is what I originally did, but I do not understand why the overlap with the two-particle states vanishes?

 

[Physics Monkey]

Hi PJK,

What do you know about the action of the field operator on a one particle state? It might be useful to think about the charge of the resulting state (I'm assuming $$\phi(x)$$ is a complex field).

 

[PJK]

Hey Physics Monkey,

ok I think I understand what you mean:
$$\phi \approx b + c^\dag$$ So it destroys phi particles and creates phi antiparticles
$$\phi^\dag \approx b^\dag + c^$$ So it destroys phi antiparticles and creates phi particles

Thus in a process phi -> phi where the momenta k_i of the incoming and outgoing particle are not equal the phi operator has to destroy the incoming particle and the phi^\dag operator has to destroy the outgoing. So it is guaranteed that after these operators acted on the its corresponding state to right or to the left the vacuum is obtained. Is that correct? But this argument doesn't hold when the momenta are equal, or?

Furthermore I thought about another proof, but I am not sure if it is correct: I could sandwitch in a complete set of momentum operator states:
$$\int d\tilde{p} <k'|\phi(x)^\dag|p><p|\phi(x)|k>$$
The only surviving state of the complete set of eigenstates is the vacuum.
Is this proof correct?

Thank you very much for your hints!

 

[Finbar]

Is it some thing like

$$\left|p\rangle \langle p \right|= 1$$

Is the identity matrix?

So you can insert it any where.

 

[Demystifier]

Is it some thing like

$$\left|p\rangle \langle p \right|= 1$$

Is the identity matrix?

So you can insert it any where.

It is identity in the 1-particle sector, but not on the whole QFT Hilbert space.

 

[Physics Monkey]

Hi PJK,

I haven't checked carefully, but it does look like if the external momenta are equal then one has the option of creating and destroying an anti-particle instead of messing with the particle. Of course, the particle process remains valid as well.

Also, the single particle momentum states don't by themselves form a complete basis, so you can't just insert them here (Edit: as Demystifier points out). Of course, the eigenstates of the full momentum operator do span the entire Hilbert space, but there one must consider multiparticle states as well.