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

[JD_PM]

Homework Statement

Show that the Feynman amplitude for Compton scattering $\mathcal{M} = \mathcal{M}_a + \mathcal{M}_b$ is gauge invariant while the individual contributions $\mathcal{M}_a$ and $\mathcal{M}_b$ are not, by considering the gauge transformations



$$\varepsilon^{\mu} (\vec k_i) \rightarrow \varepsilon^{\mu} (\vec k_i) + \lambda k^{\mu}, \ \ \ \ \varepsilon^{\mu} (\vec k_f) \rightarrow \varepsilon^{\mu} (\vec k_f) + \lambda' k^{'\mu}$$

This is exercise 8.7 in Mandl & Shaw

Relevant Equations

Please see below

Show that the Feynman amplitude for Compton scattering $\mathcal{M} = \mathcal{M}_a + \mathcal{M}_b$ is gauge invariant while the individual contributions $\mathcal{M}_a$ and $\mathcal{M}_b$ are not, by considering the gauge transformations

$$\varepsilon^{\mu} (\vec k_i) \rightarrow \varepsilon^{\mu} (\vec k_i) + \lambda k^{\mu}, \ \ \ \ \varepsilon^{\mu} (\vec k_f) \rightarrow \varepsilon^{\mu} (\vec k_f) + \lambda' k^{'\mu}$$

The Feynman amplitudes for Compton Scattering by electrons are (let us ignore helicity and polarization indices for simplicity)

\begin{equation*}
\mathcal{M}_a = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i)
\end{equation*}

\begin{equation*}
\mathcal{M}_b = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i)
\end{equation*}

Given the gauge transformations

$$\varepsilon\!\!\!/ (\vec k_i) \rightarrow \varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/, \ \ \ \ \varepsilon\!\!\!/ (\vec k_f) \rightarrow \varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/'$$

We get

\begin{align*}
\mathcal{M}'_b &= -e^2 \bar u(\vec p_f) \left(\varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/ \right) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \left(\varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/' \right) u(\vec p_i) \\
&= -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i) \\
&- \lambda \lambda' e^2 \bar u(\vec p_f) k\!\!\!/ \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/' u(\vec p_i) \\
&-\lambda' e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/' u(\vec p_i) \\
&-\lambda e^2 \bar u(\vec p_f) \lambda k\!\!\!/ \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_f) u(\vec p_i) \\
\end{align*}

Analogously

\begin{align*}
\mathcal{M}'_a &= -e^2 \bar u(\vec p_f) \left(\varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/' \right) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \left(\varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/ \right) u(\vec p_i) \\
&= -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i) \\
&- \lambda \lambda' e^2 \bar u(\vec p_f) k\!\!\!/' \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/ u(\vec p_i) \\
&-\lambda' e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) k\!\!\!/ u(\vec p_i) \\
&-\lambda' e^2 \bar u(\vec p_f) \lambda k\!\!\!/' \left( \frac{i}{p\!\!\!/ - m + i\varepsilon}\right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i) \\
\end{align*}

OK. I am not sure if what follows is correct but there we go. Next we should use Ward's identity but I am not sure how to apply it in this problem. I thought of $k\!\!\!/ =0$ and $k\!\!\!/' =0$ but that is not correct, as it would imply that the individual Feynman amplitudes are invariant by themselves, which is stated to not be the case.

Any hint is appreciated.

Thank you

 

[JD_PM]

Hi @Gaussian97 I just wanted to ping you in case you missed it. If you do not have time I understand of course

 

[Gaussian97]

Hello, well basically the Ward identity is what you want to prove here, so is not a good idea try to use it to prove what you want.
What I would do is, first of all, what is $p$? Write $p$ as a function of the external 4-momenta. And then write the propagator with all the matrices in the numerator, otherwise, it will be impossible to manipulate it.

 

[JD_PM]

What I would do is, first of all, what is $p$? Write $p$ as a function of the external 4-momenta.

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

$$\mathcal{M}_a = -e^2 \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) iS_F \left(q=p_i +k_i = p_f + k_f \right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i)$$

And then write the propagator with all the matrices in the numerator, otherwise, it will be impossible to manipulate it.

So I think you want to work with the following form of the fermion propagator (?)

$$S_F = \frac{q\!\!\!/ + mc}{p^2 -m^2c^2 + i \varepsilon}$$

 

[Gaussian97]

Yes, but notice that the momentum $p$ or $q$ is different for both amplitudes.

So I think you want to work with the following form of the fermion propagator (?)

$$S_F = \frac{q\!\!\!/ + mc}{p^2 -m^2c^2 + i \varepsilon}$$

Yes, assuming that $p=q$.

 

[JD_PM]

Yes, but notice that the momentum $p$ or $q$ is different for both amplitudes.

Indeed. Let us explicitly work with $\mathcal{M}_a$ as an example; $\mathcal{M}_b$ should follow once I understand the procedure.

Yes, assuming that $p=q$.

Oops typo.

OK so we have

$$\mathcal{M}_a = \frac{-ie^2}{
q^2 -m^2c^2 + i \varepsilon} \bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \left(q\!\!\!/ + mc \right) \varepsilon\!\!\!/ (\vec k_i) u(\vec p_i)$$

... to manipulate it.

Once here, what kind of manipulations do you have in mind?

 

[Gaussian97]

Well, there are two parts here, one, to prove that $\mathscr{M}_a$ and $\mathscr{M}_b$ are not invariant, and second to prove that $\mathscr{M}_a+\mathscr{M}_b$ is invariant.
I think the first ones are not really important, but 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.
For the second part, proving that $\mathscr{M}$ is invariant, you cannot simply work with one of the amplitudes, since you'll need both together to make it invariant.
I recommend you to write the expression for $\mathscr{M}$ (without the gauge transformation for now) and try to simplify it.

 

[JD_PM]

OK, let's focus on showing that $\mathcal{M}=\mathcal{M}_a+\mathcal{M}_b$ is invariant.

We have

\begin{align*}
&\mathcal{M}=\mathcal{M}_a+\mathcal{M}_b \\
&- \frac{ie^2 \bar u(\vec p_f)}{q^2 -m^2c^2 + i \varepsilon} \left[ \varepsilon\!\!\!/ (\vec k_f) \left(q\!\!\!/ + mc \right) \varepsilon\!\!\!/ (\vec k_i) + \varepsilon\!\!\!/ (\vec k_i) \left(q\!\!\!/ + mc \right) \varepsilon\!\!\!/ (\vec k_f)\right] u(\vec p_i)
\end{align*}

We recall how the projector operators are defined

\begin{equation*}
\Lambda^{\pm} (\vec q) := \frac{\pm q\!\!\!/ + mc}{2 mc}
\end{equation*}

Thus we get

\begin{align*}
&\mathcal{M}=\mathcal{M}_a+\mathcal{M}_b \\
&- \frac{ie^2 \bar u(\vec p_f)}{q^2 -m^2c^2 + i \varepsilon} \left[ \varepsilon\!\!\!/ (\vec k_f) \left(q\!\!\!/ + mc \right) \varepsilon\!\!\!/ (\vec k_i) + \varepsilon\!\!\!/ (\vec k_i) \left(q\!\!\!/ + mc \right) \varepsilon\!\!\!/ (\vec k_f)\right] u(\vec p_i) \\
&= -ie^2\frac{\bar u(\vec p_f)}{q^2 -m^2c^2 + i \varepsilon}2mc \Lambda^+(\vec q) \left[ \varepsilon\!\!\!/ (\vec k_f) \varepsilon\!\!\!/ (\vec k_i) + \varepsilon\!\!\!/ (\vec k_i) \varepsilon\!\!\!/ (\vec k_f)\right] u(\vec p_i)
\end{align*}

I would say we should go for the gauge transformations at this point. But I am not sure how to deal with the product of the polarization states.

I've been reading a bit and found the following relation (Mandl & Shaw, page 135)

Is this the approach you were picturing in your mind?

 

[Gaussian97]

We have

\begin{align*}
&\mathcal{M}=\mathcal{M}_a+\mathcal{M}_b \\
&- \frac{ie^2 \bar u(\vec p_f)}{q^2 -m^2c^2 + i \varepsilon} \left[ \varepsilon\!\!\!/ (\vec k_f) \left(q\!\!\!/ + mc \right) \varepsilon\!\!\!/ (\vec k_i) + \varepsilon\!\!\!/ (\vec k_i) \left(q\!\!\!/ + mc \right) \varepsilon\!\!\!/ (\vec k_f)\right] u(\vec p_i)
\end{align*}

No, this is not correct, that's why I told you to write $q$ in terms of the initial and final momenta. Not only because you'll need to know this relation to prove the invariance, but also because you are using the same letter for two different things, and this always leads you to mistakes.
Even if this were okay, you could still factor the photon polarizations, which will simplify the expression a lot when you do the transformation.

I've been reading a bit and found the following relation (Mandl & Shaw, page 135)

View attachment 274508

Is this the approach you were picturing in your mind?

No, I don't think this is necessary. You are not summing over polarizations, the amplitude must be invariant for each polarization.

 

[JD_PM]

No, this is not correct, that's why I told you to write $q$ in terms of the initial and final momenta.

Absolutely, thanks for your patience

So we instead have

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

Should we proceed by applying the gauge transformations
$\varepsilon\!\!\!/ (\vec k_i) \rightarrow \varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/, \ \ \ \ \varepsilon\!\!\!/ (\vec k_f) \rightarrow \varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/'$?

 

[Gaussian97]

Absolutely, thanks for your patience

So we instead have

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

Ok, now it's almost correct, the only thing you should care about is that $\Lambda$ is a matrix and therefore doesn't commute with other matrices.
Even then, you can still simplify it a little before doing the gauge transformation, first of all, you have common terms that you can factor out, this is especially useful for the polarizations. Second, the denominator of the propagators can be also simplified using $k^2=0$ and $p^2=m^2$.

 

[JD_PM]

OK let me go slowly. We have

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

Sorry this may be trivial for you but I want to see if I really understand it.

Spoiler: Let's prove $k^2 = 0$

(i.e. why this is a physical process)

To do so, let us chose a particular Lorentz frame: the center of mass (COM) frame. For one of the vertices of $\mathcal{M}_a$ we have

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

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

Taking the square of $k_i$

$$k_i^2 = k_i \cdot k_i = E_2^2 -\vec k \cdot \vec k = \vec k \cdot \vec k - \vec k \cdot \vec k = 0$$

So indeed this looks like a physical photon. Let us check the other vertex to be sure. We should of course get $k_i^2 = 0$ as well

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

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

Taking the square of $k_f$

$$k_f^2 = k_f \cdot k_f= E_2^2 -\vec k \cdot \vec k = \vec k \cdot \vec k - \vec k \cdot \vec k = 0$$

We conclude that this photon is indeed physical.

Spoiler: Let's prove $p^2 = m^2$

On the COM frame, we indeed get

$$p^2 = p_i^2 = p_f^2 = E_2^2 - \vec p^2= \vec p^2 + m_e^2 - \vec p^2 = m_e^2$$

Note: I have switched to natural units.

 

[JD_PM]

Back to the original problem:

When you suggest to factor terms out of our result do you mean that we should be able to simplify the propagators so that they are the same and we get a term of the form (?)
$$\text{consts.} \times \text{propagator} \times u(\vec p_f)\left(\varepsilon\!\!\!/ (\vec k_f) \Lambda^{+}(\vec p_i + \vec k_i) \varepsilon\!\!\!/ (\vec k_i) + \varepsilon\!\!\!/ (\vec k_i) \Lambda^{+}(\vec p_i - \vec k_f) \varepsilon\!\!\!/ (\vec k_f) \right) u(\vec p_i)$$

I do not get the above structure. The most simplified form I could obtain is the following

\begin{align*}

&\mathcal{M}=\mathcal{M}_a+\mathcal{M}_b \\

&= - ie^2\frac{2m}{2 p_i \cdot k_i + i \varepsilon} \left[\bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \Lambda^{+}(\vec p_i + \vec k_i) \varepsilon\!\!\!/ (\vec k_i)u(\vec p_i) \right]

\\

&+ie^2\frac{2m}{2 p_i \cdot k_i - i \varepsilon} \left[\bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \Lambda^{+}(\vec p_i - \vec k_f) \varepsilon\!\!\!/ (\vec k_f)u(\vec p_i) \right]

\end{align*}

 

[Gaussian97]

Well, no, I wasn't talking about factoring the propagator, because they are clearly different. In fact in your expression

\begin{align*}

&\mathcal{M}=\mathcal{M}_a+\mathcal{M}_b \\

&= - ie^2\frac{2m}{2 p_i \cdot k_i + i \varepsilon} \left[\bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_f) \Lambda^{+}(\vec p_i + \vec k_i) \varepsilon\!\!\!/ (\vec k_i)u(\vec p_i) \right]

\\

&+ie^2\frac{2m}{2 p_i \cdot k_i - i \varepsilon} \left[\bar u(\vec p_f) \varepsilon\!\!\!/ (\vec k_i) \Lambda^{+}(\vec p_i - \vec k_f) \varepsilon\!\!\!/ (\vec k_f)u(\vec p_i) \right]

\end{align*}

The second term should have $\frac{1}{p_i\cdot k_f}$ instead of $\frac{1}{p_i\cdot k_i}$.
But notice that there are lots of common things in both expressions, like the $-2ime^2$ or the $u$ spinnors and the polarizations $\varepsilon$.

 

[JD_PM]

The second term should have $\frac{1}{p_i\cdot k_f}$ instead of $\frac{1}{p_i\cdot k_i}$.

Typo, my bad.

... and the polarizations $\varepsilon$.

But they are different: notice that the first term is left-multiplied by $\varepsilon\!\!\!/ (\vec k_f)$ while the second one is left-multiplied by $\varepsilon\!\!\!/ (\vec k_i)$. I think we cannot factor them out.

 

[Gaussian97]

Yes, but in both terms you have both polarizations, the order is not a big deal, since the non-commutability is due to the fact that they are contracted with Dirac matrices, but the polarizations themselves are number and therefore commute, so we don't care in which order appear.

 

[JD_PM]

Ahhh I think I get it!

Based on

$$\varepsilon\!\!\!/ := \gamma^{\mu} \varepsilon_{\mu}$$

We get

\begin{align*}

&\mathcal{M}=\mathcal{M}_a+\mathcal{M}_b \\

&= - 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^{\mu} \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}\gamma^{\nu} \right] \varepsilon_{\nu} (\vec k_i)u(\vec p_i)

\end{align*}

What about now?

 

[Gaussian97]

Yeah, that's the idea. Just be careful because now, if you follow the contractions, $\varepsilon_{\mu}(k_f)$ is always on the left. Or, in other words, you could also factor out $\gamma^\mu$, which is equivalent to factor the original $\not{\!\varepsilon}(k_f)$ which we already discussed is not possible.

After fixing that we can start doing the gauge transformations, but first of all the ones you give here are quite ambiguous

Should we proceed by applying the gauge transformations
$\varepsilon\!\!\!/ (\vec k_i) \rightarrow \varepsilon\!\!\!/ (\vec k_i) + \lambda k\!\!\!/, \ \ \ \ \varepsilon\!\!\!/ (\vec k_f) \rightarrow \varepsilon\!\!\!/ (\vec k_f) + \lambda' k\!\!\!/'$?

what is $k$ and $k'$? And what is $\lambda$ and $\lambda'$?

 

[JD_PM]

Just be careful because now, if you follow the contractions, $\varepsilon_{\mu}(k_f)$ is always on the left. Or, in other words, you could also factor out $\gamma^\mu$, which is equivalent to factor the original $\not{\!\varepsilon}(k_f)$ which we already discussed is not possible.

Absolutely, I think the way to fix it is to make $\varepsilon$ index independent! There is only one way I could think of to achieve that: introducing the Kronecker-delta distribution. I have been trying different ways to get it but it does not seem to work. Should I insist or this is not the right approach?

 

[Gaussian97]

No, is much easier, you simply associate the index $\mu$ with $\varepsilon(k_f)$, so in the first term, since $\varepsilon(k_f)$ was on the left, you write $\gamma^\mu$ on the left, but since in the other term was in the right you need to write $\gamma^\mu$ on the right. And the other way around for $\gamma^\nu$.
So the correct expression is
$$\mathcal{M}= - 2ime^2 \bar u(\vec p_f) \varepsilon_{\mu} (\vec k_f) \left[ \gamma^{\mu} \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\gamma^{\nu} - \gamma^{\nu} \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}\gamma^{\mu} \right] \varepsilon_{\nu} (\vec k_i)u(\vec p_i)$$
Take a moment to check that this is equivalent to what you had when we contract the indices.

 

[JD_PM]

No, is much easier, you simply associate the index $\mu$ with $\varepsilon(k_f)$, so in the first term, since $\varepsilon(k_f)$ was on the left, you write $\gamma^\mu$ on the left, but since in the other term was in the right you need to write $\gamma^\mu$ on the right. And the other way around for $\gamma^\nu$.
So the correct expression is
$$\mathcal{M}= - 2ime^2 \bar u(\vec p_f) \varepsilon_{\mu} (\vec k_f) \left[ \gamma^{\mu} \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\gamma^{\nu} - \gamma^{\nu} \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}\gamma^{\mu} \right] \varepsilon_{\nu} (\vec k_i)u(\vec p_i)$$

$\times 2$

Take a moment to check that this is equivalent to what you had when we contract the indices.

I would not say equivalent, as my expression (as you noticed) yielded $\not{\!\varepsilon}(k_f)$ to the left on both terms which is incorrect. So you are right and my equation #17 is wrong

 

[JD_PM]

Great, let's tackle gauge invariance. The transformations I provided are those presented by M&S themselves and I agree they are not clear. What I understand is that they really meant 4-vectors i.e. (we note that gamma matrices do not transform under gauge transformations so no necessity of including them)

$$\varepsilon^{\mu} (\vec k_i) \rightarrow \varepsilon^{\mu} (\vec k_i) + \partial^{\mu} f(k_i), \ \ \ \ \varepsilon^{\nu} (\vec k_f) \rightarrow \varepsilon^{\nu} (\vec k_f) + \partial^{\nu} g(k_f)$$

But what are $f(k), g(k)$? Not clear, but based on page 134 we could guess that

$$f(k), g(k) \sim \exp(\pm ikx)$$

 

[JD_PM]

Let me provide you with more info

Besides, I have been reading M&S pages 134, 135

 

[Gaussian97]

Ok, my question was essentially to see if you knew what they were or not. I see you don't have it very clear. The answer is that $k$ and $k'$ must be the 4-momenta of the photons, while $\lambda$ and $\lambda'$ can be any two real numbers.

Knowing that, we are ready to do the transformation of $\mathscr{M}$.

 

[JD_PM]

Mmm but why we don't have the regular gauge transformation structure i.e.

$$\varepsilon^{\mu} (\vec k_i) \rightarrow \varepsilon^{\mu} (\vec k_i) + \partial^{\mu} f(k_i), \ \ \ \ \varepsilon^{\nu} (\vec k_f) \rightarrow \varepsilon^{\nu} (\vec k_f) + \partial^{\nu} g(k_f)$$

?

 

[Gaussian97]

You can look at section 8.3 of M&S, there you can see the reason, if you have doubts then we can discuss it here if you want.

 

[JD_PM]

You can look at section 8.3 of M&S, there you can see the reason, if you have doubts then we can discuss it here if you want.

If it is not clear after re-reading 8.3 I will indeed comment on it, thank you.

Good, let us assume it for now and proceed with the explicit computation of the gauge transformations.

Let's go slowly. We have

\begin{align*}
&\mathcal{M}'= - 2ime^2 \bar u(\vec p_f) \left( \varepsilon_{\mu} (\vec k_f) + \lambda' k_{\mu}'\right) \times \\
&\times \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] \times \\
&\times \left( \varepsilon_{\nu} (\vec k_i) + \lambda k_{\nu}\right) u(\vec p_i) \\
&= - 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) \\
&- 2\lambda \lambda' ime^2 \bar u(\vec p_f) k_{\mu}' \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] k_{\nu} u(\vec p_i) \\
&- 2\lambda ime^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] k_{\nu} u(\vec p_i) \\
&- 2\lambda 'ime^2 \bar u(\vec p_f) k_{\mu}' \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) \\
\end{align*}

(typing...)

 

[JD_PM]

OK I think that the best at this point is to make a guess. We have

$$\mathcal{M}' \sim \mathcal{M} - k'_{\mu}\gamma^{\mu} \gamma^{\nu} k_{\nu} - k_{\nu}\gamma^{\nu} \gamma^{\mu} k_{\mu}'-\varepsilon_{\mu} \gamma^{\mu}k_{\nu}\gamma^{\nu} - k_{\nu}\gamma^{\nu}\varepsilon_{\mu} \gamma^{\mu}-k_{\mu}' \gamma^{\mu} \varepsilon_{\nu} \gamma^{\nu} - \varepsilon_{\nu} \gamma^{\nu} k_{\mu}' \gamma^{\mu}$$

Where we cannot add consecutive terms because each carries different propagators.

But how to deal with $k_{\mu} \gamma^{\mu}$ contractions? I have been trying to get the right structure and use

$$k^{\mu}\mathcal{M}_{\mu \nu}=0$$

But I got more confused... could you please give me a hint?

 

[Gaussian97]

Well, it's true that they carry different propagators, but that doesn't mean we cannot combine them, we have split the amplitude into 4 terms, the first one is the original one, so our aim is to prove the other three terms to be zero. Let's analyse them case by case, I recommend you to start with the 3rd case. Let's forget about all the constants and polarization, the only relevant things here are the matrices (including $u$ and $\bar{u}$) and the 4-momentum. We have something of the form
$$\bar u(\vec p_f)\left[ \gamma^{\mu} \frac{\Lambda^{+}(\vec p_i + \vec k_i)}{2p_i \cdot k_i + i \varepsilon}\gamma^{\nu} - \gamma^{\nu} \frac{\Lambda^{+}(\vec p_i - \vec k_f)}{2p_i \cdot k_f - i \varepsilon}\gamma^{\mu} \right] k_{\nu} u(\vec p_i) \\$$

Now, using Dirac equation and the properties of Dirac matrices, try to simplify these two terms:
$$\Lambda^+(\vec p_i + \vec k_i)\not{\!k}_iu(\vec p_i) = \cdots$$
$$\bar{u}(\vec p_f)\not{\!k}_i\Lambda^+(\vec p_i - \vec k_f) = \cdots$$

 

[JD_PM]

Alright so

Now, using Dirac equation and the properties of Dirac matrices, try to simplify these two terms:
$$\Lambda^+(\vec p_i + \vec k_i)\not{\!k}_iu(\vec p_i) = \cdots$$
$$\bar{u}(\vec p_f)\not{\!k}_i\Lambda^+(\vec p_i - \vec k_f) = \cdots$$

Alright, you suggested to use the Dirac equation

$$(i\not{\!\partial} - m) \psi(x) = 0$$

As well as properties related to Dirac matrices. I guess those are (mostly related to $\Lambda^+$; let me omit helicity and matrix indices)

$$\Lambda^+( \vec p) u (\vec p)= u (\vec p)$$

$$\bar u (\vec p) \Lambda^+( \vec p) =\bar u (\vec p)$$

$$\Lambda^+( \vec p) = u (\vec p)\bar u (\vec p)$$

Thinking how to use them (I think all I need is rest a bit and I will see it)

 

[Gaussian97]

Alright, you suggested to use the Dirac equation

$$(i\not{\!\partial} - m) \psi(x) = 0$$

Notice that you have no derivative nor $\psi$ field in the formula you want to play with. So that form of DE doesn't look very useful. The one in momentum space is way more useful. Also, I recommend you to write also the DE for the adjoint field.

As well as properties related to Dirac matrices. I guess those are (mostly related to $\Lambda^+$; let me omit helicity and matrix indices)

$$\Lambda^+( \vec p) u (\vec p)= u (\vec p)$$

$$\bar u (\vec p) \Lambda^+( \vec p) =\bar u (\vec p)$$

$$\Lambda^+( \vec p) = u (\vec p)\bar u (\vec p)$$

Thinking how to use them (I think all I need is rest a bit and I will see it)

I refer to the properties of the gamma matrices, in general.
Also, notice that in our formulas the momenta of $\Lambda^+$ and $u$ is not the same, and also they have Dirac matrices between, so you'll need some more algebra.

 

[JD_PM]

Argh, yes! I am fresh, back at it

Notice that you have no derivative nor $\psi$ field in the formula you want to play with. So that form of DE doesn't look very useful. The one in momentum space is way more useful. Also, I recommend you to write also the DE for the adjoint field.

I see. Based on what I found, the Dirac Equation in momentum space is

$$\left(\not{\!k} - m \right) \psi (k)=0$$

One of its solutions is

$$\left(\not{\!k} - m \right) u (\vec k)=0$$

And the adjoint Dirac Equation is

$$\left(\not{\!k} + m \right) \bar \psi(k)=0$$

One of its solutions is

$$\bar u (\vec k) \left(\not{\!k} - m \right)=0$$

$$\Lambda^+(\vec p_i + \vec k_i)\not{\!k}_iu(\vec p_i) = \cdots$$

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

OK, let me go slowly. Using the definition of projection operator i.e.

$$\Lambda^{\pm} (\vec q) := \frac{\pm q\!\!\!/ + m}{2 m}$$

I get

\begin{align*}
&\Lambda^+(\vec p_i + \vec k_i)\not{\!k}_iu(\vec p_i) = \frac{p\!\!\!/_i + k\!\!\!/_i + m}{2 m} k\!\!\!/_i u(\vec p_i) \\
&= \frac{p\!\!\!/_i k\!\!\!/_i }{2 m}u(\vec p_i) + \frac 1 2 k\!\!\!/_i u(\vec p_i) \\
&= \left(p\!\!\!/_i + m \right)k\!\!\!/_i \frac{u(\vec p_i)}{2m}
\end{align*}

Where I used

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

Analogously we get

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

Once here, I feel very strongly that what we are looking for is to apply the Dirac solutions, so each term equals zero (before starting the computation, I actually expected both terms to become equal so that they would cancel each other out due to the -ive sign in one of the terms (i.e. your top equation at #29)).

Of course, the following is wrong

\begin{equation*}
\left(p\!\!\!/_i + m \right)u(\vec p_i) =0 \Rightarrow \left(p\!\!\!/_i + m \right)k\!\!\!/_i \frac{u(\vec p_i)}{2m}=0
\end{equation*}

\begin{equation*}
u(\vec p_f)\left(p\!\!\!/_f + m \right) =0 \Rightarrow \frac{\bar u(\vec p_f)}{2m} k\!\!\!/_i \left(p\!\!\!/_f + m \right) =0
\end{equation*}

Because the spinors do not commute with $k\!\!\!/_i$.

So I imagine we have to first do some algebra to get rid of the sandwiched $k\!\!\!/_i$. I do not think we are looking for a commutation/anti-commutation relation between spinors, $k\!\!\!/_i$ matrices here... What is the trick then?

PS: If the above is completely wrong, at least it was fun!

 

[Gaussian97]

Yeah, you're completely on the correct way! Just one mention here:

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

Even if this is true, your argument is not completely correct. In fact the equality $\vec p_i - \vec k_f = \vec p_f$ is false. The correct way to proceed is with the identity
$$p_i - k_f = p_f - k_i$$
i.e. conservation of momentum, then we have
$$\Lambda^+(\vec p_i - \vec k_f) = \Lambda^+(\vec p_f - \vec k_i)$$
and, when contracting with $\not{\!k}_i$ we have
$$\not{\!k}_i\Lambda^+(\vec p_i - \vec k_f) = \not{\!k}_i\Lambda^+(\vec p_f - \vec k_i)=\not{\!k}_i\frac{\not{\!p}_f-\not{\!k}_i+m}{2m}=\not{\!k}_i\frac{\not{\!p}_f+m}{2m}$$
which is indeed what you obtain.

So I imagine we have to first do some algebra to get rid of the sandwiched $k\!\!\!/_i$. I do not think we are looking for a commutation/anti-commutation relation between spinors, $k\!\!\!/_i$ matrices here... What is the trick then?

Well, precisely that. We have $\not{\!p}\not{\!k}u(p)$ which we don't know how to deal with, but we know how to deal with $\not{\!k}\not{\!p}u(p)$. Now we just need to find some relation between $\not{\!p}\not{\!k}$ and $\not{\!k}\not{\!p}u(p)$.

 

[JD_PM]

Yeah, you're completely on the correct way! Just one mention here:
Even if this is true, your argument is not completely correct. In fact the equality $\vec p_i - \vec k_f = \vec p_f$ is false. The correct way to proceed is with the identity
$$p_i - k_f = p_f - k_i$$
i.e. conservation of momentum, then we have
$$\Lambda^+(\vec p_i - \vec k_f) = \Lambda^+(\vec p_f - \vec k_i)$$
and, when contracting with $\not{\!k}_i$ we have
$$\not{\!k}_i\Lambda^+(\vec p_i - \vec k_f) = \not{\!k}_i\Lambda^+(\vec p_f - \vec k_i)=\not{\!k}_i\frac{\not{\!p}_f-\not{\!k}_i+m}{2m}=\not{\!k}_i\frac{\not{\!p}_f+m}{2m}$$
which is indeed what you obtain.

Absolutely! Actually we indeed know that the term $\Lambda^+(\vec p_i - \vec k_f)$ is associated to

While the term $\Lambda^+(\vec p_i + \vec k_i)$ is associated to

Well, precisely that. We have $\not{\!p}\not{\!k}u(p)$ which we don't know how to deal with, but we know how to deal with $\not{\!k}\not{\!p}u(p)$.

OK at least we agree on the following idea: we need to pull $\not{\!k}$ through $\not{\!p}$.

You seem to suggest we should think of commutation/anticommutation relations. I found none...

Now we just need to find some relation between $\not{\!p}\not{\!k}$ and $\not{\!k}\not{\!p}u(\vec p)$.

At least I found one equation closed to what you asked; the following relates $\bar u(\vec p)\not{\!p}\not{\!k}$ and $\not{\!k}\not{\!p}u(\vec p)$. Left-multiplying the first Dirac solution I gave by $\not{\!k}$, right-multiplying the second Dirac solution by $\not{\!k}$ and adding them up yields

$$\bar u(\vec p) \not{\!p} \not{\!k} = m \left(\bar u(\vec p) - u(\vec p) \right) - \not{\!k}\not{\!p} u(\vec p)$$

What am I missing?

 

[Gaussian97]

At least I found one equation closed to what you asked; the following relates $\bar u(\vec p)\not{\!p}\not{\!k}$ and $\not{\!k}\not{\!p}u(\vec p)$. Left-multiplying the first Dirac solution I gave by $\not{\!k}$, right-multiplying the second Dirac solution by $\not{\!k}$ and adding them up yields

$$\bar u(\vec p) \not{\!p} \not{\!k} = m \left(\bar u(\vec p) - u(\vec p) \right) - \not{\!k}\not{\!p} u(\vec p)$$

What am I missing?

Well, this actually makes no sense. You are adding a spinnor with its adjoint, this operation is not defined at all.

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

 

[JD_PM]

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.

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...

 

[Gaussian97]

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.

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]

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.

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

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

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.

 

[Gaussian97]

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:

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:

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

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?

 

[Gaussian97]

Yes, now it's perfect!

 

[JD_PM]

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? ;)

 

[Gaussian97]

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

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

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]

Any hint to unveil the right trick then? ;)

OK, I had another idea.

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...

 

[Gaussian97]

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.

OK, I had another idea.

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]

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...

 

[Gaussian97]

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.

 

[Gaussian97]

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