From my professor lecture:

The Photon field is defined as: $$A_{\mu}(x)= \sum_{s=1,2}\int \frac{d^3 \vec{p}}{{(2 \pi)^{3}\sqrt{2 E_{p}}}}\left(\epsilon^{s}_{\mu}(p) a^s_{\hspace{0.05cm}\vec{p}}e^{-ip^{\nu}x_{\nu}}+\epsilon^{*s}_{\mu}(p)a^{\dagger s}_{\hspace{0.05cm}\vec{p}}e^{ip^{\nu}x_{\nu}}\right)$$ with the commutation relations: $$[a^s_{\hspace{0.05cm}\vec{p}},a^{\dagger r}_{\hspace{0.05cm}\vec{k}}]=(2 \pi)^{3}\delta^{sr}\delta{\left(\vec{p}-\vec{k}\right)}.$$ The Feynman rules for external lines can be obtained by acting the field $A_{\mu}$ on the initial and final state particles:

Incoming photon ~~~~• :$\hspace{2.5 cm}$ $A_{\mu}|\gamma(k,r)\rangle=\epsilon^{r}_{\mu}(k)$

Outgoing photon •~~~~ :$\hspace{2.5 cm}$ $\langle\gamma(k,r)|A_{\mu}=\epsilon^{*r}_{\mu}(k)$

Where we used the commutation relations, the expression of the field and the relation: $$|\gamma(k,r)\rangle = \sqrt{2E_{k}} a^{\dagger r}_{\hspace{0.05cm}\vec{k}}|0\rangle$$

Then the lecture follows with the same thing for a fermionic and a scalar field, so far so good, but there are three things that I don't understand:

  1. From what I see we use just "half" of the expression of the field:

$$A_{\mu}(x)|\gamma(k,r)\rangle= \sum_{s=1,2}\int \frac{d^3 \vec{p}\sqrt{2E_{k}}}{{(2 \pi)^{3}\sqrt{2 E_{p}}}}\epsilon^{s}_{\mu}(p)e^{-ip^{\nu}x_{\nu}}a^s_{\hspace{0.05cm}\vec{p}} a^{\dagger r}_{\hspace{0.05cm}\vec{k}}|0\rangle$$

But not the additional term with $a^{\dagger s}_{\hspace{0.05cm}\vec{p}} a^{\dagger r}_{\hspace{0.05cm}\vec{k}}$.

And the same for the outgoing photon for the term with both annihilation operators.

I think this is because the fist term ($a^{\dagger s}_{\hspace{0.05cm}\vec{p}}$) of the field creates a photon while the second ($a^{s}_{\hspace{0.05cm}\vec{p}}$) annihilates a photon, but then the notation $A_{\mu}|\gamma(k,r)\rangle$ shouldn't be wrong? And the same for the premise "The Feynman rules for external lines can be obtained by acting the field $A_{\mu}$ on the initial and final state particles" ?

  1. Redoing the same calculation I actually obtain and additional prefactor:


$\hspace{0.75cm}$ And $e^{ik^{\mu}x_{\mu}}$ for the outgoing particle. Why these prefactors are neglected?

  1. When we consider the S-matrix element we write: $\langle i|S|f \rangle$ (with i the initial stante and f the final state).

Now, since the incoming photon ~~~~• is an initial state why we use the ket $|\gamma\rangle$ and not the bra $\langle\gamma|$? , that would lead to: ~~~~•$=\langle\gamma(k,r)|A_{\mu}=\epsilon^{*r}_{\mu}(k)$ (and viceversa for the outgoing one).


1 Answer 1

  1. What you are really supposed to calculate is $\langle 0 |A_\mu(x) | \gamma(k,\epsilon) \rangle$. This gets rid of your second term of the form $a^\dagger a^\dagger$.

  2. The pre-factor $e^{-ik\cdot x}$ has been absorbed into the remaining parts of the Feynman diagram. To see how this works you should really properly work out the Feynman rules starting from position space, then do all the Wick contractions, then Fourier transform to momentum space. Through this process keep track of the $e^{-ik\cdot x}$ and see what happens to it.

  3. The $S$-matrix is $\langle f | S | i \rangle$ NOT $\langle i | S | f \rangle$.

  • $\begingroup$ Thanks a lot, it is finally more clear. For 1) but it is common to use my professor’s notation and omit the $⟨0|$ because I didn’t find it in any other books. For 2) I did it and it should cancel out with the same prefactor for the outgoing particle right (considering momentum conservation at the vertex)? @PraharMitra $\endgroup$
    – Aleph12345
    Commented Jul 6, 2021 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.