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Using Feynman rules for QED, we can write the Feynman amplitude of a typical electromagnetic scattering process, for example, $$e^-(k)\mu^-(p)\to e^-(k')\mu^-(p'),$$ at the lowest order, in terms of the spinors $u_s(p),\bar{u}_s(p)$, and the photon propagator $-ig_{\mu\nu}/q^2$.

However, it turns out that the same amplitude can also be written in terms of the matrix elements of the electromagnetic currents as: $$\langle e^-(k')|(-e)\bar\psi\gamma^\mu\psi|e^-(k)\rangle \frac{1}{q^2} \langle \mu^-(p')|(-e)\bar{\psi'}\gamma_\mu\psi'|\mu^-(p)\rangle $$ where $\psi,\psi'$ fields destroy electron and muon respectively. How do we arrive at this expression?

I start from the perturbation theory. The relevant expression for the amplitude at the second-order in the perturbation theory is $$ \langle e^-(k')\mu^-(p')|\left[-\frac{1}{2}\int d^4x\int d^4y ~~\mathscr{T}\left[(-e\bar{\psi}\gamma^\mu\psi A_\mu)_x (-e\bar{\psi'}\gamma^\nu\psi' A_\nu)_y\right]\right]|e^-(k) \mu^-(p)\rangle$$ where $\mathscr{T}$ is the time-ordering operator. Retaining the only term that leads to a nonzero contribution, I find, $$ \langle e^-(k')\mu^-(p')|\left[-\frac{e^2}{2}\int d^4x\int d^4y ~~\underbrace{A_\mu(x)A_\nu(y)} :\bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}^{'}(y)\gamma^\nu\psi^{'}(y):\right]|e^-(k) \mu^-(p)\rangle$$ where $::$ represents normal ordering.

Now, how do we manipulate the last expression to make it a product of the matrix elements of two electromagnetic currents $j_\mu=\bar\psi\gamma_\mu\psi$ and $j^\prime_\mu=\bar{\psi'}\gamma_\mu{\psi'}$ in the form as shown in the second expression? Am I doing it right?

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  • $\begingroup$ Why did you use normal ordering? $j_{\mu} = \bar{\psi} \gamma_\mu \psi$ is not normal-ordered. $\endgroup$
    – QuantumAI
    Commented Feb 21, 2022 at 13:44
  • $\begingroup$ A similar question: physics.stackexchange.com/q/214023 $\endgroup$
    – MrQ
    Commented Feb 24, 2022 at 10:21

2 Answers 2

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I think there is something wrong with the spinor fields you are using in your Lagrangian. You should have two families of spinor fields: one that creates/annihilates electrons and another one that creates/annihilates muons. The spinors originating from the two families should not interact in any way and therefore the two "sub-amplitudes" should factor out the way you describe they factor out.

EDIT: Well, I am not 100% sure of the explanation I will provide, but I think it can justify what I understand to be your concern. You can define two vacuum states: one will be with respect to the electrons/positrons and the other will be with respect to the muon/anti-muon. The product of those vacuum states will comprise the vacuum we already know and cherish! Namely, $|0\rangle=|0\rangle_e|0\rangle_{\mu}$. Electron creation/annihilation operators (obtained from expanding $ψ(x)$) act on the former vacuum state, whereas muon creation/annihilation operators (obtained from expanding $ψ′(x)$) act on the latter.

So, having that in mind, you can write the in and out states as products of states belonging in the "electron Hilbert space" with states belonging in the "muon Hilbert space". After doing that, then it is only natural that the creation/annihilation operators regarding muons will act on the latter and the creation/annihilation operators regarding the electrons will act on the former and the factorization property emerges in this way

I hope this helps. If something is not clear, please let me know.

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    $\begingroup$ Yes, that is true. For that, I have used two different fields: $\psi$ and $\psi^\prime$. $\endgroup$ Commented Jun 13, 2022 at 15:02
  • $\begingroup$ Sorry, I hadn't notice. Is it still not clear how to reach to the electromagnetic current, written in terms of two matrix elements, given that you have two families of spinor fields?? $\endgroup$
    – schris38
    Commented Jun 13, 2022 at 15:06
  • $\begingroup$ Having problems in separating it into the electron current and the muon current. $\endgroup$ Commented Jun 13, 2022 at 20:46
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    $\begingroup$ Well, I am not 100% sure of the explanation I will provide, but I think it can justify what I understand to be your concern. You can define two vacuum states: one will be with respect to the electrons/positrons and the other will be with respect to the muon/anti-muon. The product of those vacuum states will comprise the vacuum we already know and cherish! Namely, $|0>=|0>_e|0>_{\mu}$. Electron creation/annihilation operators (obtained from expanding $\psi(x)$) act on the former vacuum state, whereas muon creation/annihilation operators (obtained from expanding $\psi'(x)$) act on the latter. $\endgroup$
    – schris38
    Commented Jun 14, 2022 at 7:29
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    $\begingroup$ So, having that in mind, you can write the in and out states as products of states belonging in the "electron Hilbert space" with states belonging in the "muon Hilbert space". After doing that, then it is only natural that the creation/annihilation operators regarding muons will act on the latter and the creation/annihilation operators regarding the electrons will act on the former and the factorization property emerges in this way $\endgroup$
    – schris38
    Commented Jun 14, 2022 at 7:37
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If I understand the problem correctly and if I am not mistaken, what is missing is apply the LSZ reduction formula. This leaves the spinors and the $A_\mu A_\nu$ contraction.

So integration of $A_\mu A_\nu$ contraction leads $(2\pi)^4\delta^{(4)}(k-k' -p +p')\frac{-i\eta_{\mu\nu}}{(k-k')^2}$.

As for the spinors:

$\bar{u}^{s'}_{e^-}(k')\gamma^\mu u^{s}_{e^-}(k) \cdot \bar{u}^{r'}_{\mu^-}(p')\gamma^\nu u^{r}_{\mu^-}(p) = j^\mu_{e^-} \cdot j^\nu_{\mu^-}$

I hope this answer the question.

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