Does first order QED matrix elements vanish?

[center o bass]

Hi! I'm trying to learn the Feynman rules trough Wick's theorem right now and I'm focusing on QED.

Here the first order term of the S-operator can be written as

$$-ie\int d^4x :\bar \psi(x) \gamma^\mu \psi(x) A^\mu(x):$$

but the author of the book I'm reading (Greiner Reinhart) claims that all the matrix elements vanish due to the fact that they are not kinematically allowed. That is alright. I agree that they are not kinematically allowed...

So I wondered do they actually vanish if you try to compute them in QED or is this kinematical requirement just imposed (as an additional postulate)?

I tried to verify the for one of the terms in the first order term when expanding the field operators out as positive and negative frequency terms. One of the terms will certainly be
$$\bar \psi^{(-)} A^{(-)} \psi^{(+)}$$ which is interpreted as a photon emission from an electron. Now since $\psi^{(+)}$ annhilates an electron we need an electron in the initial state
$$|i\rangle = b^\dagger_{p_1 s_1} |0\rangle$$
and since $\psi^{(-)}$ and $A^{(-)}$ creates an electron and a photon we need an electron and a photon in the final state
$$|f\rangle = b^\dagger_{p_2 s_s} a^\dagger_{k_2 \lambda_2}. |0\rangle$$

Now it does not seem to me that the corresponding matrix element for this process obviously vanishes. Since $\psi^{(-)} \sim b_{ps}^\dagger \ \ A^{(-)}\sim a^\dagger_{k \lambda} \ \ \psi^{(+)} \sim b_{ps}$ the matrix element will go as

$$\langle f| \bar \psi^{(-)} A^{(-)} \psi^{(+)}|i\rangle \sim \langle 0 |b_{p_2 s_2} a_{k_2 \lambda_2} b^\dagger_{p s} a_{k \lambda}^\dagger b_{p's'} b^\dagger_{p_1 s_1}|0\rangle.$$

Every annhilation operator here have to be commuted past a creation operator before it can do any damage.. Will this matrix element become zero anyhow? Any easy way to see that it will or does one have to embark on the calculation?

 

[Bill_K]

Whoa, too many creation/annihilation operators! You can calculate S matrix elements in one of two ways. Either directly take the overlap between the in and out states: S = <out|in>. Or, "reduce" the in and out states by replacing them with creation operators acting on the vacuum. This second way expresses the S matrix as a vacuum expectation value, S = <0|ψγψA|0>.

The kinematic constraint comes as a consequence when you go from x space to p space.

∫ψ(x)A(x)ψ(x) d⁴x ∝ ∫∫ψ(p)e^-ipx d⁴p ∫ψ(p')e^ip'x d⁴p' ∫A(k)e^ikxd⁴k d⁴x

If you do the x integral first, you get a delta function that conserves momentum.

∫e^-ipxe^ip'xe^ikx d⁴x ∝ δ⁴(p' + k - p)

 

[center o bass]

Whoa, too many creation/annihilation operators! You can calculate S matrix elements in one of two ways. Either directly take the overlap between the in and out states: S = <out|in>. Or, "reduce" the in and out states by replacing them with creation operators acting on the vacuum. This second way expresses the S matrix as a vacuum expectation value, S = <0|ψγψA|0>.

The kinematic constraint comes as a consequence when you go from x space to p space.

∫ψ(x)A(x)ψ(x) d⁴x ∝ ∫∫ψ(p)e^-ipx d⁴p ∫ψ(p')e^ip'x d⁴p' ∫A(k)e^ikxd⁴k d⁴x

If you do the x integral first, you get a delta function that conserves momentum.

∫e^-ipxe^ip'xe^ikx d⁴x ∝ δ⁴(p' + k - p)

Why are there too many creation and annhilation operators? They come from the initial and final states + the interaction.

 

[Bill_K]

They come from the initial and final states + the interaction.

This is doing it twice. The |in> state is an eigenstate of the full Hamiltonian including the interaction. In the infinite past it was an incoming electron, that's the boundary condition, but in the present it evolves, because the interaction Hamiltonian is busy turning it into other things. The |out> state is also an eigenstate of the full Hamiltonian, with a different boundary condition: it's the outgoing photon and electron in the infinite future.

|in> can turn into many things as it evolves, but the probability amplitude for |in> to turn specifically into the |out> state we're interested in is just the overlap, <out|in>. The interaction is already in there, in the states!

The LSZ reduction formulas build the |in> and |out> states from creation operators acting on the vacuum, so that all one ever needs to calculate in order to get the S-matrix elements is vacuum expectation values.

 

[center o bass]

This is doing it twice. The |in> state is an eigenstate of the full Hamiltonian including the interaction. In the infinite past it was an incoming electron, that's the boundary condition, but in the present it evolves, because the interaction Hamiltonian is busy turning it into other things. The |out> state is also an eigenstate of the full Hamiltonian, with a different boundary condition: it's the outgoing photon and electron in the infinite future.

|in> can turn into many things as it evolves, but the probability amplitude for |in> to turn specifically into the |out> state we're interested in is just the overlap, <out|in>. The interaction is already in there, in the states!

The LSZ reduction formulas build the |in> and |out> states from creation operators acting on the vacuum, so that all one ever needs to calculate in order to get the S-matrix elements is vacuum expectation values.

I sort of agree with what you're saying here, but I have not yet reached the LSZ reduction formulaes. I'm reading 'Greiner Reinhart - Field Quantization' and I'm in the section where they present the Feynman rules for QED. As far as I understand it, it is the S matrix which does the evolving and in order to evaluate the matrix elements one has to expand it in powers of the interaction. The author uses the method I illustated explicitly at pages 245, 246 etc..

I.e.

$$S_{fi} = \langle \phi_f|\hat S|\phi_i\rangle$$