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

[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...

 

[JD_PM]

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).

 

[Gaussian97]

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).

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?

 

[Gaussian97]

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.

 

[JD_PM]

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.

 

[Gaussian97]

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.

 

[nrqed]

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.