# Frame-independent probability amplitude for Compton scattering in scalar QED

In scalar QED, the matrix element for compton scattering is the sum of three diagrams: the s-channel, t-channel and the seagull diagram. After simplifcation, making use of the orthogonality between photon polarization vectors and photon momenta, we get the following expression for the matrix element. \begin{align} i\mathcal{M} &= -ie^{2} \left[ \frac{(2p+k)_\mu (2p'+k')_\nu}{(p+k)^2-m^2}+\frac{(2p'-k)_\mu (2p-k')_\nu}{(p-k')^2-m^2} -2\eta_{\mu\nu} \right]\epsilon^\mu \epsilon'^{\nu *} \\ &=-i2e^{2} \left[ \frac{p_\mu p'_\nu}{p.k}-\frac{p'_\mu p_\nu}{p.k'} -\eta_{\mu\nu} \right]\epsilon^\mu \epsilon'^{\nu *} \end{align} where $$p$$ and $$p'$$ are the four momenta of the incoming and outgoing (scalar) electrons, while $$k$$ and $$k'$$ are the incoming and outgoing photons four momenta, which correspond with the polarization vectors $$\epsilon$$ and $$\epsilon'$$. $$\eta_{\mu \nu}$$ is the metric tensor. I want to find an expression for the probability amplitude, corresponding with this matrix element for unpolarized photons and which is frame independent. So I want to calculate $$\frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4}\left[ \frac{p_\mu p'_\nu}{p.k} - \frac{p'_\mu p_\nu}{p.k'} -\eta_{\mu \nu} \right]\left[ \frac{p_\alpha p'_\beta}{p.k}-\frac{p'_\alpha p_\beta}{p.k'} -\eta_{\alpha\beta} \right]\sum_{\epsilon \epsilon'}\epsilon^\mu \epsilon^{\alpha *} \epsilon'^{\nu *} \epsilon'^{\beta}$$ The polarization vector sum $$\sum_{\epsilon \epsilon}\epsilon^\mu \epsilon^{\alpha *}$$ is equal to $$-\eta^{\mu \alpha}$$ (up to a term that does not contribute to the amplitude due to the Ward identity). Working out this expression and using $$\eta_{\mu \nu}\eta^{\mu \nu}=4$$ and eliminating $$p'$$ using $$p'=p+k-k'$$ I get the following result ($$m$$ is the electron mass). $$\frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ m^4 \left(\frac{1}{p.k}-\frac{1}{p.k'} \right)^2 +2m^2\left(\frac{1}{p.k}-\frac{1}{p.k'} \right)+4 \right]$$ However, the last term of this expression, i.e. the 4, does not seem right to me. If I apply this formula in the LAB frame, I get $$\frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ 3 + \cos^2 \theta \right]$$ with $$\theta$$ the deviation angle of the photon. However, there is another way to calculate this probability amplitude, i.e. by calculating $$\vert \mathcal M \vert ^2$$ in the LAB frame for each polarization state separately and taking the sum afterwards. This leads to $$\frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ 1 + \cos^2 \theta \right]$$ which is more credible, as it is proportional to the (classical) result for Thomson scattering. So, it seems that the factor between brackets should be $$1+ \cos^2 \theta$$ and that the term 4 in the frame independent expression should be 2. Can somebody help me out and tell me where I made a mistake?

Details of the calculation below. • Do you have a reference for the claim that it should be a 1 not a 3? – gmarocco Dec 28 '19 at 1:09
• Doing the calculation I also got a 4. A 2 is obtained here: cds.cern.ch/record/1114396/files/p293.pdf but they dont show the whole calculation and I think there may be some mistakes. E.g. there's a -1 missing in Eq. (5). – Zarathustra Dec 28 '19 at 14:49
• @Zarathustra Yes I saw that, but it is littered with mistakes/typos as you say. Also equation (13), which they say they compare with the literature, is wrong on dimensional grounds, so would not trust it. – gmarocco Dec 28 '19 at 17:19
• @gmarocco. Instead of adding a reference for my claim, I added an alternative way of calculating, which leads to the expected result (1 instead of 3). – jac Dec 30 '19 at 13:08
• So I did the calculation as well and I also get a 3. I suspect that the algebra is right, but somehow the procedure of calculating the spin-sum via a metric substitution is wrong. If we calculate in the lab frame, the two terms containing momenta vanish. If we were then to use $\sum \epsilon \epsilon^* \rightarrow -\eta$, we would surely get the wrong answer. But I could not tell you why, as the expression does indeed still satisfy the Ward identity (?). Anyway, sorry that this is not really useful to anyone but I'm putting this all down mainly for myself... – gmarocco Dec 30 '19 at 23:51

In the first expression that you write for $$i\mathcal{M}$$, you have simplified to the second line using a gauge condition (Lorentz gauge) $$\epsilon_\mu k^\mu=0$$. This is a legal choice but it is incompatible with replacing both polarizations sums with $$-\eta_{\mu\nu}$$, since you have partially fixed the gauge in this way. It's simple to check that squaring the expression in the first line for $$i\mathcal{M}$$ and replacing in there the polarization sums with the metric tensors yields the correct answer. Alternatively, you get the correct answer squaring the second line after having used $$\epsilon_\mu k^\mu=0$$ but then including consistently the general expression for (at least one of the) two polarizations sums. You can find a discussion on this point, as well as the explicit calculation of the squared amplitude in the Lorentz gauge, in some hand written notes here, from page 11 to the last page.

• Nevermind, sorry. I now realise upon typing this why you are right: the $k' \cdot \bar{k'}$ term will not vanish. – gmarocco Jan 3 at 11:41
• @TwoBs. Going through the notes, everything is clear to me up to the point where you say (p.13) "However since we gaugefixed, we have seen that the gauge transformation leaves the amplitude invariant only because $\epsilon' . k'=0$.This means we cannot replace both polarization sums by the metric tensor." I do not see how you get to this conclusion. The gauge transformation indeed leaves the amplitude invariant because of the Lorentz condition $\epsilon' . k'=0$, so what?Can you please elaborate a bit more? – jac Jan 7 at 20:45
• @TwoBs. By the way, I kind of came to the same conclusion following a slightly different reasoning. I posted it as a separate answer for reasons of readability. Does this make sense to you? – jac Jan 7 at 20:45
• Yes this is a true and often overlooked step in the calculation - the matrix element must be transverse for the replacement of the sum over polarisation with the metric and after gauge fixing will generally fail to be so. This is also discussed by jac below – lux Jan 7 at 22:22
• @jac I think the point is that we can replace one of the spin sums by the metric, as it vanishes due to the gauge condition. However, when we try to do the second spin substitution, there are no more polarization vectors. Hence the gauge condition can no longer guarantee invariance, and we just have to do one of the spin sums by hand. You can try doing these sum substitutions one at a time dotted into what you call $\mathcal{M}_{(1)}$ - you'll find one extra term left over. – gmarocco Jan 9 at 9:48

Just to fill in a few of the blanks of TwoBs answer:

Using \begin{align*} \sum_{s} \epsilon^s(k)^{\mu}\epsilon^s(k)^{\nu *} = -\eta^{\mu\nu} + \frac{1}{2E^2}(k^{\mu}\bar{k}^{\nu}+k^{\nu}\bar{k}^{\mu}), \end{align*} from TwoBs notes, we would like to calculate \begin{align*} \sum_{s,s'} |\mathcal{M}|^2 = 4e^4 \left( \frac{p_{\mu}p'_{\nu}}{p\cdot k}-\frac{p'_{\mu}p_{\nu}}{p\cdot k'}-\eta_{\mu\nu}\right) \left( \frac{p_{\alpha}p'_{\beta}}{p\cdot k}-\frac{p'_{\alpha}p_{\beta}}{p\cdot k'}-\eta_{\alpha\beta}\right) \epsilon^{s}(k)^{\mu}\epsilon^{s}(k)^{\alpha *} \epsilon^{s'}(k')^{\beta } \epsilon^{s'}(k')^{\nu *}. \end{align*} We can see that the term in brackets in the spin sum involving $$k$$ always vanish. You can just do it once at a time: under $$\epsilon^{s}(k)^{\mu} \rightarrow k^\mu$$ the expression is invariant, and similarly under $$\epsilon^{s}(k)^{\nu *} \rightarrow k^\nu$$. Both of these are thanks to $$\epsilon(k)\cdot k=0$$. So now we know that we can replace one of the spin sums with just the metric, as the momentum terms vanish. If we do the next spin sum, we don't have the luxury of dotting into the polarization vectors anymore - they're all gone. You can show that under $$\epsilon(k') \rightarrow k'$$, you are left with a term that is now $$k' \cdot \bar{k}'$$, which is not 0.

Doing one of these, say $$\epsilon(k)^{\mu} \rightarrow k^{\mu}$$, we have \begin{align*} (\frac{p\cdot k}{p\cdot k}p'_{\nu} - \frac{p'\cdot k}{p\cdot k'}p_{\nu}-k_{\nu})\epsilon(k')^{\nu*}... &= -k'\cdot \epsilon(k')^* = 0. \end{align*}

• The first spin sum has (as the term in brackets linear in $k$) indices $\mu \alpha$ and these do not contract with the polarization vector components with indices $\beta$ and $\nu$. So, I cannot see how you can apply the gauge condition, which requires a contraction between between an upper index and the same lower index (i.e. a dot product). Or did I miss something? – jac Jan 10 at 19:53
• @jac Does what I added help? – gmarocco Jan 10 at 20:20
• Yes!...I got it. Thanks a million :) – jac Jan 10 at 22:45

Another reasoning leading to the conclusion that one cannot just simplify the expression for $$\mathcal{M}$$ (first equation) using the Lorentz gauge condition $$\epsilon . k=0$$ and subsequently replace in the simplified expression (second equation) $$\sum \epsilon_\mu \epsilon_\nu^*$$ by $$-\eta_{\mu \nu}$$ is the following. The matrix element of the first equation can be written as follows $$\mathcal{M}= \mathcal{M}^{\mu \nu} \epsilon_\mu \epsilon_{'\nu}^*=\left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right) \epsilon_\mu \epsilon^{'*}_\nu$$ where $$\mathcal{M}^{\mu \nu}_{(2)}$$ represents the terms for which in the Lorentz gauge $$\mathcal{M}^{\mu \nu}_{(2)}\epsilon_\mu \epsilon^{'*}_\nu=0$$, i.e. the terms that we dropped in simplifying the first equation. The Ward identity states that $$\mathcal{M}^{\mu \nu} k_\mu =0$$ and that $$\mathcal{M}^{\mu \nu}k^{'}_\nu=0$$ If we calculate the probability amplitude, we get $$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2= \left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right)\left( \mathcal{M}^{\alpha \beta }_{(1)} + \mathcal{M}^{\alpha \beta}_{(2)}\right) \sum \limits_{\epsilon \epsilon'}\epsilon_\mu \epsilon_\alpha^* \epsilon^{'*}_\nu\epsilon^{'}_\beta$$ We know that the polarization sum can be written as follows $$\sum \limits_{\epsilon}\epsilon_\mu \epsilon_\alpha^*=\left( -\eta_{\mu \alpha}+W_{\mu \alpha}(k) \right)$$where $$W_{\mu \alpha}(k)$$ is a term that is linear in $$k_\mu$$ and $$k_\alpha$$. Applying now the previous expression we get $$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2= \left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right)\left( \mathcal{M}^{\alpha \beta }_{(1)} + \mathcal{M}^{\alpha \beta}_{(2)}\right) \left( -\eta_{\mu \alpha}+W_{\mu \alpha}(k) \right)\left( -\eta_{\nu \beta}+W_{\nu \beta}(k') \right)$$ The Ward indentity allows us to rewrite this as follows.$$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2= \left( \mathcal{M}^{\mu \nu }_{(1)} + \mathcal{M}^{\mu \nu}_{(2)}\right)\left( \mathcal{M}^{\alpha \beta }_{(1)} + \mathcal{M}^{\alpha \beta}_{(2)}\right) \left( -\eta_{\mu \alpha} \right)\left( -\eta_{\nu \beta} \right)$$ However, if we first simplify the expression for the matrix element using the Lorentz gauge condition we get $$\frac{1}{2}\sum \limits_{\epsilon \epsilon'} \vert \mathcal{M}\vert^2= \mathcal{M}^{\mu \nu }_{(1)} \mathcal{M}^{\alpha \beta }_{(1)} \left( -\eta_{\mu \alpha }+W_{\mu \alpha}(k) \right)\left( -\eta_{\nu \beta} +W_{\nu \beta}(k')\right)$$ which is an expression that cannot further be simplified using the Ward identity.

Conclusion of all this is that when first using the Lorentz gauge condition to simplify the expression for the matrix element, we cannot anymore do the replacement $$\sum \limits_{\epsilon}\epsilon_\mu \epsilon_\alpha^* \rightarrow -\eta_{\mu \alpha}$$ Correction: as explained in the answers from TwoBs and gmarocco, you can do the replacement for one of the two spin sums, but not for the second.

So, when doing the calculation of the probability amplitude using the very first expression for the matrix element in the initial question and applying the aforementioned replacement, I get after quite some algebra $$\frac{1}{2}\sum_{\epsilon \epsilon'} \vert \mathcal{M} \vert^2=2e^{4} \left[ m^4 \left(\frac{1}{p.k}-\frac{1}{p.k'} \right)^2 +2m^2\left(\frac{1}{p.k}-\frac{1}{p.k'} \right)+2 \right]$$ which is the expected result.