Hi! I'm trying to learn the Feynman rules trough Wick's theorem right now and I'm focusing on QED.(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex] -ie\int d^4x :\bar \psi(x) \gamma^\mu \psi(x) A^\mu(x):[/tex]

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

[tex] \bar \psi^{(-)} A^{(-)} \psi^{(+)}[/tex] which is interpreted as a photon emission from an electron. Now since [itex]\psi^{(+)}[/itex] annhilates an electron we need an electron in the initial state

[tex]|i\rangle = b^\dagger_{p_1 s_1} |0\rangle[/tex]

and since [itex]\psi^{(-)}[/itex] and [itex] A^{(-)}[/itex] creates an electron and a photon we need an electron and a photon in the final state

[tex]|f\rangle = b^\dagger_{p_2 s_s} a^\dagger_{k_2 \lambda_2}. |0\rangle[/tex]

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

[tex]\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. [/tex]

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?

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# Does first order QED matrix elements vanish?

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