# Showing Feynman-amplitudes' gauge invariance (for Compton Scattering)

• JD_PM
In summary, the Feynman amplitude for Compton scattering is gauge invariant while the individual contributions are not. This is proven by considering the gauge transformations and using Ward's identity. It is important to write the propagator with all the matrices in the numerator in order to manipulate it correctly.
Gaussian97 said:
Well, this actually makes no sense. You are adding a spinnor with its adjoint, this operation is not defined at all.

My bad, I tend to overcomplicate things.

Gaussian97 said:
I recommend you to take a look at the definition of the gamma matrices, that's all you need.

OK, recalling that

$$\{ \gamma_{\mu}, \gamma_{\nu} \} = 2 \eta_{\mu \nu}$$

We get

$$p\!\!\!/ k\!\!\!/ = p^{\mu} \gamma_{\mu} \gamma_{\nu} k^{\nu} = p^{\mu} ( 2\eta_{\mu \nu} - \gamma_{\nu} \gamma_{\mu}) k^{\nu} = -p^{\mu} \gamma_{\nu} \gamma_{\mu} k^{\nu} + 2 \eta_{\mu \nu}p^{\mu}k^{\nu} = -k\!\!\!/ p\!\!\!/ + 2 p_{\nu} k^{\nu}$$

Where ##p^{\mu}## and ##k^{\nu}## commute with ##\gamma_{\nu}## and ##\gamma_{\mu}## respectively as the former are not matrices.

But this does not lead to the desired Dirac solution; recalling that we are working with the relation

$$\left(p\!\!\!/_i + m \right)k\!\!\!/_i \frac{u(\vec p_i)}{2m}$$

And plugging what we have just obtained into it yields

$$\Big[ 2 p_{\nu} k^{\nu} + k\!\!\!/_i \left(-p\!\!\!/_i + m\right) \Big] \frac{u(\vec p_i)}{2 m}$$

Which does not give zero...

I know I must be missing something trivial here...

Yes, that's exactly what we wanted, you related ##\not{\! k}\not{\! p}## with ##\not{\! p}\not{\! k}##. Now, you find that
$$\Lambda^+ \not{\!k} u \neq 0$$
but remember we didn't want this to be zero.
JD_PM said:
And plugging what we have just obtained into it yields

$$\Big[ 2 p_{\nu} k^{\nu} + k\!\!\!/_i \left(-p\!\!\!/_i + m\right) \Big] \frac{u(\vec p_i)}{2 m}$$

Which does not give zero...
Now, what Dirac equation tells you about the term
$$\left(-p\!\!\!/_i + m\right) u(\vec p_i)$$
?
Also you can do a similar analysis to find an expresion for
$$\bar{u} \not{\!k} \Lambda^+$$

JD_PM
Gaussian97 said:
but remember we didn't want this to be zero.

Ahhh I was thinking each term should equal zero by means of Dirac's solution! I was wrong.

Gaussian97 said:
Now, what Dirac equation tells you about the term
$$\left(-p\!\!\!/_i + m\right) u(\vec p_i)$$
?

It tells us that it is zero. Thus we end up with

$$p_{\mu} k^{\mu} \frac{u(\vec p_i)}{ m}$$

Gaussian97 said:
Also you can do a similar analysis to find an expresion for
$$\bar{u} \not{\!k} \Lambda^+$$

By a completely analogous procedure (as you suggested) we get

$$\frac{\bar u(\vec p_f)}{2 m}\Big[ 2 p_{\nu} k^{\nu} + \left(-p\!\!\!/_i + m\right) k\!\!\!/_i \Big]=\frac{\bar u(\vec p_f)}{m} p_{\nu} k^{\nu}$$

Plugging these two into your top equation at #29 we get

$$p_{\nu} k^{\nu} \bar u(\vec p_f) \gamma^{\mu} u(\vec p_i) \left( \frac{1}{2p_i \cdot k_i + i \varepsilon} - \frac{1}{2p_i \cdot k_f - i \varepsilon} \right)$$

We know that the above term must be zero. The propagators are different so the only way is that ##\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)## vanishes.

Damn, I think I got it! (I may not celebrate beforehand though...)

The happy idea is this. We make the following observation: ##\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)## is a ##1 \times 1## matrix, which is multiplied by another ##1 \times 1## matrix (i.e. by a scalar in our case). We learned that the product of two ##1 \times 1## matrices equals to its trace, so we get

$$p_{\nu} k^{\nu} tr(\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)) \left( \frac{1}{2p_i \cdot k_i + i \varepsilon} - \frac{1}{2p_i \cdot k_f - i \varepsilon} \right)$$

We also learned that if the trace contains an odd number of gamma matrices then it yields zero (M&S page 453).

If this is correct, I will proceed with the other 2 terms.

Well, we're on the right way, but not quite there yet. Let me recapitulate a little, to not forget our aim.

We want to prove the invariance of the amplitude
$$\mathcal{M}= - 2ime^2 \bar u(\vec p_f) \varepsilon_{\mu} (\vec k_f) \left[ \gamma^{\mu} \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\gamma^{\nu} - \gamma^{\nu} \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}\gamma^{\mu} \right] \varepsilon_{\nu} (\vec k_i)u(\vec p_i)$$
Under the gauge transformation
$$\varepsilon\!\!\!/ \left(\vec k_i\right) \rightarrow \varepsilon\!\!\!/ \left(\vec k_i\right) + \lambda \not{\!k}_i, \qquad \varepsilon\!\!\!/ \left(\vec k_f\right) \rightarrow \varepsilon\!\!\!/ \left(\vec k_f\right) + \lambda' \not{\! k}_f$$

In our way to prove that, we want to prove that the quantity
$$\bar u(\vec p_f)\left[ \gamma^{\mu} \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\not{\!k}_i - \not{\!k}_i \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}\gamma^{\mu} \right] u(\vec p_i) \\$$
is zero.

And now you have computed
$$\Lambda^{+}(\vec p_i + \vec k_i)\not{\!k}_i u(\vec p_i) = \frac{p_i\cdot k_i}{m} u(\vec p_i)$$
You still need
$$\bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_i - \vec k_f) = \bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_f - \vec k_i)$$
Notice that this is NOT what you have done in the previous post.

But let me comment on some things now:
JD_PM said:
By a completely analogous procedure (as you suggested) we get

$$\frac{\bar u(\vec p_f)}{2 m}\Big[ 2 p_{\nu} k^{\nu} + \left(-p\!\!\!/_i + m\right) k\!\!\!/_i \Big]=\frac{\bar u(\vec p_f)}{m} p_{\nu} k^{\nu}$$
I don't agree whith that, first of all, the LHS is not what we need to compute. But even if it was, you are using that
$$\bar{u}(\vec p_f) (-\not{\!p}_i + m) = 0$$
which is not true at all. Of course then what follows is not true, but let me still comment some things:

JD_PM said:
The happy idea is this. We make the following observation: ##\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)## is a ##1 \times 1## matrix, which is multiplied by another ##1 \times 1## matrix (i.e. by a scalar in our case). We learned that the product of two ##1 \times 1## matrices equals to its trace, so we get

$$p_{\nu} k^{\nu} tr(\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)) \left( \frac{1}{2p_i \cdot k_i + i \varepsilon} - \frac{1}{2p_i \cdot k_f - i \varepsilon} \right)$$

We also learned that if the trace contains an odd number of gamma matrices then it yields zero (M&S page 453).

I agree that the trace of a number is itself and that the trace of an odd number of gamma matrices is zero. But you don't have just gamma matrices in the trace, you also have a ##\bar{u}## and a ##u##, which completely ruins your argument.

So no, unfortunately, we're not done yet, we still need to find the correct expresion for
$$\bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_i - \vec k_f) = \bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_f - \vec k_i)$$

JD_PM
@Gaussian97, thanks for the recap.

Let me go very slowly here.

We now want to compute

$$\bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_i - \vec k_f) = \bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_f - \vec k_i)$$

We agreed on the following

\begin{align*}

&\bar{u}(\vec p_f)\not{\!k}_i\Lambda^+(\vec p_i - \vec k_f=\vec p_f - \vec k_i)= \bar{u}(\vec p_f)\not{\!k}_i \frac{p\!\!\!/_f - \vec k_i + m}{2 m}\\

&= \bar{u}(\vec p_f) \frac{k\!\!\!/_i p\!\!\!/_f }{2 m} + \frac 1 2 \bar{u}(\vec p_f) \not{\!k}_i \\

&= \frac{\bar u(\vec p_f)}{2m} k\!\!\!/_i \left(p\!\!\!/_f + m \right)

\end{align*}

We know that

$$k\!\!\!/_i p\!\!\!/_f = 2 p_f \cdot k_i - p\!\!\!/_f k\!\!\!/_i$$

Thus we have

$$\bar{u}(\vec p_f) \frac{2 p_f \cdot k_i - p\!\!\!/_f k\!\!\!/_i}{2 m} + \frac 1 2 \bar{u}(\vec p_f) \not{\!k}_i$$

The following was indeed wrong

JD_PM said:
By a completely analogous procedure (as you suggested) we get

$$\frac{\bar u(\vec p_f)}{2 m}\Big[ 2 p_{\nu} k^{\nu} + \left(-p\!\!\!/_i + m\right) k\!\!\!/_i \Big]=\frac{\bar u(\vec p_f)}{m} p_{\nu} k^{\nu}$$

Instead, I think the successive is OK

$$\frac{\bar u(\vec p_f)}{2 m}\Big[ 2 p_f \cdot k_i + \left(-p\!\!\!/_f + m\right) k\!\!\!/_i \Big]= \frac{p_f\cdot k_i}{m} \bar u(\vec p_f)$$

Where I used the Dirac solution

$$\bar{u}(\vec p_f) (-\not{\!p}_f + m) = 0$$

Thus we see that

$$\bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_i - \vec k_f) = \bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_f - \vec k_i) = \frac{p_f\cdot k_i}{m} \bar u(\vec p_f)$$

Are we on the same page so far?

Yes, now it's perfect!

Great, so following the exact same procedure we indeed see that the following is incorrect

$$p_{\nu} k^{\nu} \bar u(\vec p_f) \gamma^{\mu} u(\vec p_i) \left( \frac{1}{2p_i \cdot k_i + i \varepsilon} - \frac{1}{2p_i \cdot k_f - i \varepsilon} \right)$$

And the successive is OK

$$\frac{\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)}{m} \Big[ \frac{p_i \cdot k_i}{2p_i \cdot k_i + i\varepsilon}-\frac{p_i \cdot k_f}{2p_i \cdot k_f - i\varepsilon}\Big]$$

You seem to suggest that the trick of taking the trace and using that the trace contains an odd number of gamma matrices does not work because we do not have an odd number of gamma matrices... Any hint to unveil the right trick then? ;)

Well, first of all, one question. You have proved
$$\bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_i - \vec k_f) = \frac{p_f\cdot k_i}{m} \bar u(\vec p_f)$$
But then, when substituting you have used
$$\bar u(\vec p_f)\not{\!k}_i \Lambda^{+}(\vec p_i - \vec k_f) = \frac{p_i\cdot k_f}{m} \bar u(\vec p_f)$$
Is this a typo? Or have you proved this two things are equal?

JD_PM
That is a typo indeed, I meant

$$\frac{\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)}{m} \Big[ \frac{p_i \cdot k_i}{2p_i \cdot k_i + i\varepsilon}-\frac{p_f \cdot k_i}{2p_i \cdot k_f - i\varepsilon}\Big]$$

Gaussian97 said:
Or have you proved this two things are equal?

You seem to suggest that they are equal. I guess that it can be proven using energy-momentum conservation i.e. ##p_i - k_f = p_f - k_i##. But I cannot show it off the top of my head.

If you wish, let us focus on the current trick to advance :)

JD_PM said:
Any hint to unveil the right trick then? ;)

We know that

$$\frac{tr(\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i))}{m} \Big[ \frac{p_i \cdot k_i}{2p_i \cdot k_i + i\varepsilon}-\frac{p_f \cdot k_i}{2p_i \cdot k_f - i\varepsilon}\Big]$$

Let us focus on the trace term. We know that

$$tr(\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)) = tr(\gamma^{\mu} u(\vec p_i) \bar u(\vec p_f))$$

Argh, the first relation that comes to mind is

$$\frac{p\!\!\!/ + m}{2 m}=\Lambda^+ (\vec p) = \sum_{r=1}^2 u_r(\vec p) \bar u_r(\vec p)$$

But it cannot be directly applied because the momentum vectors associated to the spinors are not equal...

JD_PM said:
That is a typo indeed, I meant

$$\frac{\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)}{m} \Big[ \frac{p_i \cdot k_i}{2p_i \cdot k_i + i\varepsilon}-\frac{p_f \cdot k_i}{2p_i \cdot k_f - i\varepsilon}\Big]$$
You seem to suggest that they are equal. I guess that it can be proven using energy-momentum conservation i.e. ##p_i - k_f = p_f - k_i##. But I cannot show it off the top of my head.

If you wish, let us focus on the current trick to advance :)
Yes, they are indeed equal, which follows quite easy from conservation of 4-momentum, indeed this is the next step.

JD_PM said:

We know that

$$\frac{tr(\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i))}{m} \Big[ \frac{p_i \cdot k_i}{2p_i \cdot k_i + i\varepsilon}-\frac{p_f \cdot k_i}{2p_i \cdot k_f - i\varepsilon}\Big]$$

Let us focus on the trace term. We know that

$$tr(\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)) = tr(\gamma^{\mu} u(\vec p_i) \bar u(\vec p_f))$$

Argh, the first relation that comes to mind is

$$\frac{p\!\!\!/ + m}{2 m}=\Lambda^+ (\vec p) = \sum_{r=1}^2 u_r(\vec p) \bar u_r(\vec p)$$

But it cannot be directly applied because the momentum vectors associated to the spinors are not equal...
No, unfortunately, that trace is not zero. Even if both spinors had the same momentum (which is not the case) they need to have the same spin, which is also not the case, and we should have a sum over the spins, which again is not the case. But even if you had all these conditions, notice how the product ##u \bar{u}## would give you another gamma matrix, so you would have a trace of two gamma matrices, which is not zero at all.

JD_PM
Gaussian97 said:
Yes, they are indeed equal, which follows quite easy from conservation of 4-momentum, indeed this is the next step.

I see. I will show it explicitly, but let me first ask: how would this lead us to show that the whole term is zero? I thought that the key was in the ##\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)## term...

Nop, actually the key is to prove
$$\frac{p_i \cdot k_i}{2p_i \cdot k_i + i\varepsilon}-\frac{p_f \cdot k_i}{2p_i \cdot k_f - i\varepsilon}=0$$
To do that is convenient to prove what I told you, and also to remember that ##\varepsilon## is just a parameter used to make the integrals convergent, at the end of the day one must take the limit ##\varepsilon \to 0##

JD_PM
OK I finally got it! (once again, I want to appreciate your patience )

Let us prove that ##p_i \cdot k_f = p_f \cdot k_i##

For that, let us come back to the COM frame and recall what I wrote at #12 (spoiler message)

$$p_i=(E_1=\sqrt{\vec p^2 + m_e^2}, \vec p)$$

$$k_i=(E_2=\sqrt{\vec k^2 + (m_p=0)^2}, \vec k)$$

$$p_f=(E_1=\sqrt{\vec p^2 + m_e^2}, -\vec p)$$

$$k_f=(E_2=\sqrt{\vec k^2 + (m_p=0)^2}, -\vec k)$$

Thus we have

\begin{align*}
&p_i \cdot k_f = \sqrt{\vec p^2 + m_e^2}\sqrt{\vec k^2 + (m_p=0)^2} -\vec p \cdot (-\vec k) \\
&= \sqrt{\vec p^2 + m_e^2}\sqrt{\vec k^2 + (m_p=0)^2} -(-\vec p) \cdot \vec k \\
&= p_f \cdot k_i
\end{align*}

Thus, taking the limit ##\varepsilon \to 0##, we end up with

$$\frac{p_i \cdot k_i}{2p_i \cdot k_i}-\frac{p_i \cdot k_f}{2p_i \cdot k_f}=\frac{\cancel{p_i \cdot k_i}}{2\cancel{p_i \cdot k_i}}-\frac{\cancel{p_i \cdot k_f}}{2\cancel{p_i \cdot k_f}} =0$$

If you agree on my result, I will proceed with the other two terms.

Yeah, we finally got our result. Although the proof is much nicer just squaring the equation ##p_i - k_f = p_f - k_i##
$$(p_i - k_f)^2 = (p_f - k_i)^2 \Longrightarrow p^2 + k^2 - 2 p_i \cdot k_f = p^2 + k^2 - 2 p_f \cdot k_i \Longrightarrow p_i \cdot k_f = p_f \cdot k_i$$

JD_PM
Great, let us prove that the last term of the master equation at #27 vanishes (the procedure is completely analogous to what we have done so far). Its structure is

$$\bar u(\vec p_f)\left[k\!\!\!/_f \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\gamma^{\nu} - \gamma^{\nu} \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}k\!\!\!/_f \right] u(\vec p_i) \\$$

Let's work out first ##\bar u(\vec p_f) k\!\!\!/_f \Lambda^{+}(\vec p_i + \vec k_i)##

\begin{align*}
&\bar{u}(\vec p_f)\not{\!k}_f\Lambda^+(\vec p_i + \vec k_i=\vec p_f + \vec k_f)= \bar{u}(\vec p_f)\not{\!k}_f \frac{p\!\!\!/_f + k\!\!\!/_f + m}{2 m}\\
&= \bar{u}(\vec p_f) \frac{k\!\!\!/_f p\!\!\!/_f }{2 m} + \frac 1 2 \bar{u}(\vec p_f) \not{\!k}_f \\
&= \frac{\bar u(\vec p_f)}{2 m}\Big[ 2 p_f \cdot k_f + \left(-p\!\!\!/_f + m\right) k\!\!\!/_f \Big] \\
&= \frac{p_f\cdot k_f}{m} \bar u(\vec p_f)
\end{align*}

Where we used

$$p\!\!\!/ k\!\!\!/ = p^{\mu} \gamma_{\mu} \gamma_{\nu} k^{\nu} = p^{\mu} ( 2\eta_{\mu \nu} - \gamma_{\nu} \gamma_{\mu}) k^{\nu} = -p^{\mu} \gamma_{\nu} \gamma_{\mu} k^{\nu} + 2 \eta_{\mu \nu}p^{\mu}k^{\nu} = -k\!\!\!/ p\!\!\!/ + 2 p \cdot k$$

\begin{equation*}
k\!\!\!/_f k\!\!\!/_f = k^2 = 0
\end{equation*}

Analogously we compute ##\Lambda^{+}(\vec p_i - \vec k_i)k\!\!\!/_f u(\vec p_i)##

\begin{equation*}
\Lambda^{+}(\vec p_i - \vec k_i)k\!\!\!/_f u(\vec p_i)= \frac{p_i\cdot k_f}{m} u(\vec p_i)
\end{equation*}

Thus we get

$$\frac{\bar u(\vec p_f) \gamma^{\mu} u(\vec p_i)}{m} \Big[ \frac{p_f \cdot k_f}{2p_i \cdot k_i + i\varepsilon}-\frac{p_i \cdot k_f}{2p_i \cdot k_f - i\varepsilon}\Big]$$

By energy-momentum conservation we know that ##p_f \cdot k_f = p_i \cdot k_i##. Thus, by taking the limit
##\varepsilon \to 0##, we see that the above term vanishes.

We are only left to show that the following term vanishes

$$\bar u(\vec p_f)\left[k\!\!\!/_f \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}k\!\!\!/_i - k\!\!\!/_i \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}k\!\!\!/_f \right] u(\vec p_i) \\$$

Which looks a bit harder because we've got an extra ##k\!\!\!/## to deal with. Working on it...

Mmm so it was not hard at all. I get

$$\frac{\bar u(\vec p_f) k\!\!\!/_i u(\vec p_i)}{m} \Big[ \frac{p_f \cdot k_f}{2p_i \cdot k_i + i\varepsilon}-\frac{p_i \cdot k_f}{2p_i \cdot k_f - i\varepsilon}\Big]\underbrace{=}_{\varepsilon \to 0} \frac{\bar u(\vec p_f) k\!\!\!/_i u(\vec p_i)}{m} \left[\frac{\cancel{p_i \cdot k_i}}{2\cancel{p_i \cdot k_i}}-\frac{\cancel{p_i \cdot k_f}}{2\cancel{p_i \cdot k_f}} \right] =0$$

Mmm I suspect I missed something, don't you think? What I mean is that I expected to derive Ward's identity ##k_{\mu} \mathcal{M}^{\mu}=0## at some point but I did not (at least I do not see it).

Your post #51 is essentially correct, except for some typos with the indices.
Then for #52 you're right, there was nothing more difficult in proving this "double term". In fact, if you take the first term we proved to be zero, and contracting by ##k_f## you get this double term. The same is true if you take the one you proved zero in #51, contracting with ##k_i## you get this last one. Since of course anything multiplied by zero is zero this last term was far from difficult, was actually trivially zero.

And yes you have proven Ward's identity, look that the first term you proved zero was, in fact, ##k_\mu \mathscr{M}^{\mu}##, the second was ##k'_\mu \mathscr{M}^{\mu}## and this last one ##k_{\mu}k'_\nu \mathscr{M}^{\mu\nu}##
So all these three terms were zero, which is essentially Ward's identity.

JD_PM
Thank you very much @Gaussian97. The key concepts are clear and the main problem solved. To finish it off though (you may have just sighed at this point. The following is just extra discussion so I would perfectly understand if you were to stop it here).

Gaussian97 said:
proving something is not invariant is not trivial in general. I would simply choose arbitrary values for the momenta and spin, choose a representation for the Dirac matrices and compute the amplitude numerically, seeing that is really not invariant.

Let us work out only with one the two Feynman Amplitudes

\begin{align*}
\mathcal{M}_a = - 2ime^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left[ \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon} \right] \varepsilon\!\!\!/ (\vec k_i)u(\vec p_i)
\end{align*}

Under the gauge symmetries

\begin{equation*}
\varepsilon\!\!\!/ \left(\vec k_i\right) \rightarrow \varepsilon\!\!\!/ \left(\vec k_i\right) + \lambda \not{\!k}_i, \qquad \varepsilon\!\!\!/ \left(\vec k_f\right) \rightarrow \varepsilon\!\!\!/ \left(\vec k_f\right) + \lambda' \not{\! k}_f
\end{equation*}

##\mathcal{M}_a## becomes

\begin{align*}
&\mathcal{M}_a'= -2ime^2 \bar u(\vec p_f) \left( \varepsilon\!\!\!/ \left(\vec k_f\right) + \lambda' \not{\! k}_f \right) \left[ \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon} \right] \left( \varepsilon\!\!\!/ \left(\vec k_i\right) + \lambda \not{\! k}_i \right) u(\vec p_i) \\
&= -2ime^2 \bar u(\vec p_f)\varepsilon\!\!\!/ \left(\vec k_f\right)\left[\frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\right]\varepsilon\!\!\!/ \left(\vec k_i\right) u(\vec p_i) \\
&- 2\lambda \lambda' ime^2 \bar u(\vec p_f)k\!\!\!/_f \left[\frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\right]k\!\!\!/_i u(\vec p_i) \\
&- 2 \lambda ime^2 \bar u(\vec p_f)\varepsilon\!\!\!/ \left(\vec k_f\right)\left[\frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\right]k\!\!\!/_i u(\vec p_i) \\
&- 2 \lambda' ime^2 \bar u(\vec p_f) k\!\!\!/_f \left[\frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\right]\varepsilon\!\!\!/ \left(\vec k_i\right) u(\vec p_i) \\
\end{align*}

We clearly see that the last three terms above will not vanish (as the negative counterpart coming from ##\mathcal{M}_b## are missing), so ##\mathcal{M}_a## is indeed changing under the given gauge symmetries.

However, you seemed to suggest we should explicitly show that each of those 3 terms yield a non-zero term through a particular representation of the gamma matrices (though I have to say I always try to avoid them while doing proofs).

OK let's show it for one of those 3, say (let me only write the structure)

\begin{equation*}
\bar u(\vec p_f)\varepsilon_{\mu} \left(\vec k_f\right)\left[ \gamma^{\mu} \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\gamma^{\nu} \right]k_{\nu} u(\vec p_i)
\end{equation*}

The following follows from M&S, pages 460 and 461

First off, let us pick the Dirac-Pauli representation i.e.

\begin{equation*}
\gamma^{0}=\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix},
\ \ \ \ \ \gamma^{j}= \begin{pmatrix}
0 & \sigma_j \\
-\sigma_j & 0 \\
\end{pmatrix}
\end{equation*}

Where ##\sigma_j## stand for Pauli matrices i.e.

\begin{equation*}
\sigma_x=\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix},
\ \ \ \ \ \sigma_y= \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix},
\ \ \ \ \ \sigma_z= \begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}
\end{equation*}

Next we need to write down a complete set of plane wave states. Defining two non-relativistic spinors i.e.

\begin{equation*}
\chi_1 := \chi_2' := \begin{pmatrix}
1 \\
0 \\
\end{pmatrix}, \ \ \ \ \chi_2 := \chi_1' := \begin{pmatrix}
0 \\
1 \\
\end{pmatrix}
\end{equation*}

We see that the relativistic Dirac spinor gets the form (please let me omit the prove; a plane-wave-solutions discussion would deserve a thread on its own)

\begin{equation*}
u_r(\vec 0) =
\begin{pmatrix}
\chi_r \\
0 \\
\end{pmatrix},
\ \ \ \ u_r(\vec p) = \frac{p\!\!\!/ + m}{(2m E_{\vec p} + 2m^2)^{1/2}}u_r(0)
\end{equation*}

Where ##r=1,2##

For our particular choice of gamma-matrices, we get

\begin{equation*}
\large u_1(\vec p)= \left(\frac{E_{\vec p} + m}{2m}\right)^{1/2}
\begin{pmatrix}
1 \\
0 \\
\frac{p^z}{E_{\vec p}+m} \\
\frac{p^x + ip^y}{E_{\vec p}+m}
\end{pmatrix}, \ \ \ \
u_2(\vec p) = \left(\frac{E_{\vec p} + m}{2m}\right)^{1/2}
\begin{pmatrix}
0 \\
1 \\
\frac{p^x - ip^y}{E_{\vec p}+m} \\
\frac{-p^z}{E_{\vec p}+m}
\end{pmatrix}
\end{equation*}

Besides, we must pick a particular representation of the polarization vectors as well. It is costumary to choose the following (M&S page 77, EQ 5.21)

\begin{equation*}
\varepsilon_{0}^{\mu} (\vec k) := (1,0,0,0)
\end{equation*}
\begin{equation*}
\varepsilon_{j}^{\mu} (\vec k) := (0, \vec \varepsilon_{j} (\vec k))
\end{equation*}

Before proceeding with the computation: is the above procedure what you meant?

Yes it seems ok, but I also would do the next step and fix a value for ##p_i, p_f, k_i, k_f## then is just a numerical calculation, which should be very difficult.
Probably this is not considered necessary, since you see that, to get an invariant amplitude we have used terms from A and from B, so seems plausible that is not possible to prove without all those terms.
But to prove that something is not invariant is just as easy as to find a counter-example. So a single example where the amplitude does indeed change. That's what I was saying you could do.

To be fair is not a bad-exercise, to remember what all these abstract symbols really mean. But is more a computational exercise than a physical one.

Gaussian97 said:
Yes it seems ok, but I also would do the next step and fix a value for ##p_i, p_f, k_i, k_f## then is just a numerical calculation, which should be very difficult.

It is indeed a messy computational task to plug in a particular representation of the gamma matrices, a singular choice of polarization vectors, the plane wave solutions and fixed values for ##p_i, p_f, k_i, k_f##.

I will think further whether it is possible to prove the non-invariance of ##\mathcal{M}_a## without using a particular representation of the gamma matrices.

Again, thank you.

Well, I did some computations with random numbers, in fact given:
$$p_i = (1.02, 0.18, -0.10, 0.02), \qquad k_i = (0.21, -0.18, 0.10, -0.02)$$
$$p_f = (1.02, -0.15, 0.11, 0.09), \qquad k_f = (0.21, 0.15, -0.11, -0.09)$$
and with polarizations ##r_i =2, s_i=1, r_f=1, s_f=2## we have (not sure if I'm using the same conventions as you)
$$\bar{u}(\vec p_f, r_f) = (1.42, 0,-0.06, 0.11+0.08i),\qquad u(\vec p_i, r_i) = (0, 1.42,-0.13+0.07i, 0.01)$$
$$\varepsilon_\mu(\vec k_i, s_i) = (0, -0.07, 0.04, 0.997),\qquad \varepsilon_\mu(\vec k_f, s_f) = (0, -0.58, 0.81, 0)$$
then we can compute
$$\mathscr{M}_a = -0.05 - 0.04i,\qquad \mathscr{M}_b = 0.08 + 0.05i,\qquad \mathscr{M} = 0.03 + 0.01i$$
and, choosing ##\lambda = -0.81## and ##-0.71## we can also find
$$\mathscr{M}'_a = -0.07 - 0.02i,\qquad \mathscr{M}'_b = 0.10 + 0.04i,\qquad \mathscr{M}' = 0.03 + 0.01i$$
where you can clearly see that the amplitudes by themselves are not invariant, but only their sum

JD_PM
@Gaussian97 I guess you used Mathematica or analogous (?)

I am convinced that your approach is correct but I am still wondering if we can prove it without relying on a particular representation of the gamma matrices. I may need more time to think about this.

Last edited:
JD_PM said:
Alright. Let me change notation for the internal momenta: it will be labeled as ##q##. Energy-momentum conservation at each vertex implies ##q=p_i +k_i = p_f + k_f##, so we have

$$\mathcal{M}_a = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) iS_F \left(q=p_i +k_i = p_f + k_f \right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i)$$
So I think you want to work with the following form of the fermion propagator (?)

$$S_F = \frac{q\!\!\!/ + mc}{p^2 -m^2c^2 + i \varepsilon}$$
The proof is much simpler than what you did in the rest of the thread. Once you have this form for the propagator, you just need to use commutation relations to pass the slash k_i and slash k_f past the gamma matrices in the propagator and then use Dirac's equation.

Replies
15
Views
2K
Replies
7
Views
908
Replies
1
Views
1K
Replies
8
Views
1K
Replies
7
Views
1K
Replies
1
Views
817
Replies
1
Views
720
Replies
26
Views
4K