- #1

Markus Kahn

- 112

- 14

- TL;DR Summary
- I'm confused about how to determine, given the definition of the spin 1/2 fields and the photon field, which Spinor belongs where in a Feynman diagram.

Consider Moller scattering, that is $$e^-(\vec p_1, \alpha)+e^-(\vec p_2, \beta) \quad\longrightarrow\quad e^-(\vec q_1, \gamma)+e^-(\vec q_2, \delta),$$

where the ##\vec{p}_i,\vec q_i## label the momenta of the in and outgoing electrons and the greek letter the spin state.

The two relevant Feynman diagrams at second order in ##e## (electric charge) are

(taken directly from the above linked Wikipedia page)

From QFT we know that we can write the photon and spinor fields as

$$

\begin{align*}

A_\mu(x) &= \int \frac{d^3p}{(2\pi)^32e(\vec p)}[c_a^\dagger(\vec p) {\varepsilon_\mu^a}^*(\vec p)e^{-ip\cdot x} + c_a(\vec p) \varepsilon_\mu^a(\vec p)e^{ip\cdot x}],\\

\psi(x) &= \int \frac{d^3p}{(2\pi)^32e(\vec p)}[a_\alpha^\dagger(\vec p) v_\alpha(\vec{p}) e^{-ip\cdot x} + b_\alpha(\vec p) u_\alpha(\vec p)e^{ip\cdot x}], \\

\bar\psi(x) &= \int \frac{d^3p}{(2\pi)^32e(\vec p)}[a_\alpha(\vec p) \bar v_\alpha(\vec{p}) e^{ip\cdot x} + b_\alpha^\dagger(\vec p) \bar u_\alpha(\vec p)e^{-ip\cdot x}],

\end{align*}

$$

where ##a^\dagger## creates

We can now write down the matrix elements of the first Feynman diagram (##t##-channel) using the Feynman rules for QED

$$M_{\alpha\beta\gamma\delta} = \left\{\bar{v}_{\alpha}(\vec p_{1})(-e\gamma^\mu)v_\gamma(\vec q_1)\right\}[-iG_{\mu\nu}(p_1-q_1)] \left\{\bar{v}_\beta (\vec p_2) (-e\gamma^\nu) v_\delta(\vec q_2)\right\}.$$

As I understand it, the expressions in the two curly brackets ##\{\hspace{2mm}\}## are ##\mathbb{C}##-numbers, so we could technically swap them without changing the value of the matrix element.

I don't really understand how the expressions in the curly brackets are supposed to be read. Let's take as an example

$$\left\{\bar{v}_{\alpha}(\vec p_{1})(-e\gamma^\mu)v_\gamma(\vec q_1)\right\}.$$

Why do we know that we have to use the spinors ##\bar v_\alpha## and ##v_\gamma## and why not ##v_\alpha## and ##\bar v_\gamma##? I do understand that we only need ##v_\sigma## and not ##u_\rho##, since we have no positrons,

I tried reading it up in Peskins book, but unfortunately, he just writes it down and says it follows directly from the Feynman rules (see section 5.1, p. 131). I don't really have access to another book at the moment...

where the ##\vec{p}_i,\vec q_i## label the momenta of the in and outgoing electrons and the greek letter the spin state.

The two relevant Feynman diagrams at second order in ##e## (electric charge) are

(taken directly from the above linked Wikipedia page)

From QFT we know that we can write the photon and spinor fields as

$$

\begin{align*}

A_\mu(x) &= \int \frac{d^3p}{(2\pi)^32e(\vec p)}[c_a^\dagger(\vec p) {\varepsilon_\mu^a}^*(\vec p)e^{-ip\cdot x} + c_a(\vec p) \varepsilon_\mu^a(\vec p)e^{ip\cdot x}],\\

\psi(x) &= \int \frac{d^3p}{(2\pi)^32e(\vec p)}[a_\alpha^\dagger(\vec p) v_\alpha(\vec{p}) e^{-ip\cdot x} + b_\alpha(\vec p) u_\alpha(\vec p)e^{ip\cdot x}], \\

\bar\psi(x) &= \int \frac{d^3p}{(2\pi)^32e(\vec p)}[a_\alpha(\vec p) \bar v_\alpha(\vec{p}) e^{ip\cdot x} + b_\alpha^\dagger(\vec p) \bar u_\alpha(\vec p)e^{-ip\cdot x}],

\end{align*}

$$

where ##a^\dagger## creates

**electrons**and ##b^\dagger## creates positrons. Note, I'm aware that this deviates from common conventions, but my lecturer wants it that way...We can now write down the matrix elements of the first Feynman diagram (##t##-channel) using the Feynman rules for QED

$$M_{\alpha\beta\gamma\delta} = \left\{\bar{v}_{\alpha}(\vec p_{1})(-e\gamma^\mu)v_\gamma(\vec q_1)\right\}[-iG_{\mu\nu}(p_1-q_1)] \left\{\bar{v}_\beta (\vec p_2) (-e\gamma^\nu) v_\delta(\vec q_2)\right\}.$$

As I understand it, the expressions in the two curly brackets ##\{\hspace{2mm}\}## are ##\mathbb{C}##-numbers, so we could technically swap them without changing the value of the matrix element.

QuestionQuestion

I don't really understand how the expressions in the curly brackets are supposed to be read. Let's take as an example

$$\left\{\bar{v}_{\alpha}(\vec p_{1})(-e\gamma^\mu)v_\gamma(\vec q_1)\right\}.$$

Why do we know that we have to use the spinors ##\bar v_\alpha## and ##v_\gamma## and why not ##v_\alpha## and ##\bar v_\gamma##? I do understand that we only need ##v_\sigma## and not ##u_\rho##, since we have no positrons,

**my problem is that I don't really understand the logic when we are supposed to choose the spinor with a bar and when the spinor without a bar.**I tried reading it up in Peskins book, but unfortunately, he just writes it down and says it follows directly from the Feynman rules (see section 5.1, p. 131). I don't really have access to another book at the moment...