Rewriting Feynman amplitudes and the Dirac equation

[JD_PM]

TL;DR

I want to understand how to get the named equations below from Mandl & Shaw book.

I was studying the photon polarization sum process (second edition QFT Mandl & Shaw,https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf) and got stuck in how to get certain equations.

We work in a gauge in which the polarization vectors of the external photons are of the form

$$\epsilon = (0, \vec \epsilon), \ \ \ \ \epsilon' = (0, \vec \epsilon') \tag 1$$

The 4 products $\epsilon k$ and $\epsilon'k'$ are

$$\epsilon k=-\vec \epsilon \cdot \vec k =0, \ \ \ \ \epsilon' k'=-\vec \epsilon' \cdot \vec k'=0 \tag 2$$

Let us work in the LAB frame.Then $p=(m, 0, 0, 0)$ Thus we have

$$p \epsilon = p \epsilon'=0 \tag 3$$

Given the anticommutation relation $\{\gamma^{\alpha}, \gamma^{\beta}\}=2g^{\alpha \beta}$ and the Dirac equation $(\not{\!p}-m)u(\vec p)=0$ we get

$$\not{\!p} \not{\!\epsilon}u=-m\not{\!\epsilon} u, \ \ \ \ \not{\!p} \not{\!\epsilon'}\not{\!u}=-m\not{\!\epsilon'} u \tag 4$$

Given the Feynman Amplitude $\mathscr{M}=\mathscr{M}_a+\mathscr{M}_b$, where

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon'} (\not{\!p}+\not{\!k}+m) \not{\!\epsilon} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = i e^2 \frac{\bar u' \not{\!\epsilon} (\not{\!p}-\not{\!k'}+m) \not{\!\epsilon'} u}{2(pk')} \tag 5$$

$\mathscr{M}_a$ and $\mathscr{M}_b$ simplify to

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon} \not{\!k'} \not{\!\epsilon'} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = -i e^2 \frac{\bar u' \not{\!\epsilon} \not{\!k'} \not{\!\epsilon'} u}{2(pk')} \tag 6$$

My questions are:

1) How to get Eqs $(4)$

I almost get $\not{\!p} \not{\!\epsilon}u=-m\not{\!\epsilon} u$; based on the Dirac equation we get $\not{\!p}u(\vec p)=mu(\vec p)$, so we simply multiply by $\not{\!\epsilon'}$ (on the left side of the equation). However, note I do not get the negative sign.

How to get $\not{\!p} \not{\!\epsilon'}\not{\!u}=-m\not{\!\epsilon'} u$? I guess we have to use $\{\gamma^{\alpha}, \gamma^{\beta}\}=2g^{\alpha \beta}$, but how? I mean, all we can get out of the anticommutation relation is that $\Big(\gamma^0\Big)^2=1$ and $\Big(\gamma^i\Big)^2=-1$, where $i=1,2,3$.

2) How to get Eqs $(6)$

Based on $(1), (2)$ and $(3)$ I get $\not{\!\epsilon'} \not{\!p}=0, \not{\!\epsilon} \not{\!p}=0, \not{\!\epsilon'} m=0, \not{\!\epsilon} m=0$. So I end up getting

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon'} \not{\!k} \not{\!\epsilon} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = -i e^2 \frac{\bar u' \not{\!\epsilon} \not{\!k'} \not{\!\epsilon'} u}{2(pk')} \tag 7$$

Note I get the same $\mathscr{M}_b$ but not the same $\mathscr{M}_a$. Is it a typo on the book or I am wrong?

Any help is appreciated.

PS: I asked thie question https://math.stackexchange.com/questions/3703823/rewriting-feynman-amplitudes-and-the-dirac-equation but got not response.

 

[nrqed]

Summary:: I want to understand how to get the named equations below from Mandl & Shaw book.

I was studying the photon polarization sum process (second edition QFT Mandl & Shaw,https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf) and got stuck in how to get certain equations.

We work in a gauge in which the polarization vectors of the external photons are of the form

$$\epsilon = (0, \vec \epsilon), \ \ \ \ \epsilon' = (0, \vec \epsilon') \tag 1$$

The 4 products $\epsilon k$ and $\epsilon'k'$ are

$$\epsilon k=-\vec \epsilon \cdot \vec k =0, \ \ \ \ \epsilon' k'=-\vec \epsilon' \cdot \vec k'=0 \tag 2$$

Let us work in the LAB frame.Then $p=(m, 0, 0, 0)$ Thus we have

$$p \epsilon = p \epsilon'=0 \tag 3$$

Given the anticommutation relation $\{\gamma^{\alpha}, \gamma^{\beta}\}=2g^{\alpha \beta}$ and the Dirac equation $(\not{\!p}-m)u(\vec p)=0$ we get

$$\not{\!p} \not{\!\epsilon}u=-m\not{\!\epsilon} u, \ \ \ \ \not{\!p} \not{\!\epsilon'}\not{\!u}=-m\not{\!\epsilon'} u \tag 4$$

Given the Feynman Amplitude $\mathscr{M}=\mathscr{M}_a+\mathscr{M}_b$, where

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon'} (\not{\!p}+\not{\!k}+m) \not{\!\epsilon} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = i e^2 \frac{\bar u' \not{\!\epsilon} (\not{\!p}-\not{\!k'}+m) \not{\!\epsilon'} u}{2(pk')} \tag 5$$

$\mathscr{M}_a$ and $\mathscr{M}_b$ simplify to

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon} \not{\!k'} \not{\!\epsilon'} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = -i e^2 \frac{\bar u' \not{\!\epsilon} \not{\!k'} \not{\!\epsilon'} u}{2(pk')} \tag 6$$

My questions are:

1) How to get Eqs $(4)$

I almost get $\not{\!p} \not{\!\epsilon}u=-m\not{\!\epsilon} u$; based on the Dirac equation we get $\not{\!p}u(\vec p)=mu(\vec p)$, so we simply multiply by $\not{\!\epsilon'}$ (on the left side of the equation). However, note I do not get the negative sign.

How to get $\not{\!p} \not{\!\epsilon'}\not{\!u}=-m\not{\!\epsilon'} u$? I guess we have to use $\{\gamma^{\alpha}, \gamma^{\beta}\}=2g^{\alpha \beta}$, but how? I mean, all we can get out of the anticommutation relation is that $\Big(\gamma^0\Big)^2=1$ and $\Big(\gamma^i\Big)^2=-1$, where $i=1,2,3$.

Because of the anticommutation relation of the Dirac matrices, we have for arbitrary four vectors a and b,
$\not{\!a} \not{\!b} = -\not{\!b} \not{\!a} + 2 a \cdot b$
I will get back to your other questions a bit later.

 

[nrqed]

Summary:: I want to understand how to get the named equations below from Mandl & Shaw book.

I was studying the photon polarization sum process (second edition QFT Mandl & Shaw,https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf) and got stuck in how to get certain equations.

We work in a gauge in which the polarization vectors of the external photons are of the form

$$\epsilon = (0, \vec \epsilon), \ \ \ \ \epsilon' = (0, \vec \epsilon') \tag 1$$

The 4 products $\epsilon k$ and $\epsilon'k'$ are

$$\epsilon k=-\vec \epsilon \cdot \vec k =0, \ \ \ \ \epsilon' k'=-\vec \epsilon' \cdot \vec k'=0 \tag 2$$

Let us work in the LAB frame.Then $p=(m, 0, 0, 0)$ Thus we have

$$p \epsilon = p \epsilon'=0 \tag 3$$

Given the anticommutation relation $\{\gamma^{\alpha}, \gamma^{\beta}\}=2g^{\alpha \beta}$ and the Dirac equation $(\not{\!p}-m)u(\vec p)=0$ we get

$$\not{\!p} \not{\!\epsilon}u=-m\not{\!\epsilon} u, \ \ \ \ \not{\!p} \not{\!\epsilon'}\not{\!u}=-m\not{\!\epsilon'} u \tag 4$$

Given the Feynman Amplitude $\mathscr{M}=\mathscr{M}_a+\mathscr{M}_b$, where

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon'} (\not{\!p}+\not{\!k}+m) \not{\!\epsilon} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = i e^2 \frac{\bar u' \not{\!\epsilon} (\not{\!p}-\not{\!k'}+m) \not{\!\epsilon'} u}{2(pk')} \tag 5$$

$\mathscr{M}_a$ and $\mathscr{M}_b$ simplify to

$$\mathscr{M}_a = -i e^2 \frac{\bar u' \not{\!\epsilon} \not{\!k'} \not{\!\epsilon'} u}{2(pk)}, \ \ \ \ \mathscr{M}_b = -i e^2 \frac{\bar u' \not{\!\epsilon} \not{\!k'} \not{\!\epsilon'} u}{2(pk')} \tag 6$$

My questions are:

1) How to get Eqs $(4)$

I almost get $\not{\!p} \not{\!\epsilon}u=-m\not{\!\epsilon} u$; based on the Dirac equation we get $\not{\!p}u(\vec p)=mu(\vec p)$, so we simply multiply by $\not{\!\epsilon'}$ (on the left side of the equation). However, note I do not get the negative sign.

How to get $\not{\!p} \not{\!\epsilon'}\not{\!u}=-m\not{\!\epsilon'} u$? I guess we have to use $\{\gamma^{\alpha}, \gamma^{\beta}\}=2g^{\alpha \beta}$, but how? I mean, all we can get out of the anticommutation relation is that $\Big(\gamma^0\Big)^2=1$ and $\Big(\gamma^i\Big)^2=-1$, where $i=1,2,3$.

2) How to get Eqs $(6)$

Based on $(1), (2)$ and $(3)$ I get $\not{\!\epsilon'} \not{\!p}=0, \not{\!\epsilon} \not{\!p}=0, \not{\!\epsilon'} m=0, \not{\!\epsilon} m=0$. So I end up getting

These are not correct. For example, $\not{\!\epsilon} \not{\!p} = - \not{\!\epsilon} \not{\!p} +2 \epsilon \cdot p$ which reduces to $- \not{\!\epsilon} \not{\!p}$ in your case. Using this to move around the $\not{\!p}$ so that it acts on the spinor, you should recover their result.