Playing around with numerical light-like momenta $p^\mu_1,p^\mu_2$ (light-like meaning ${p_1}^\mu {p_1}_\mu={p_2}^\mu {p_2}_\mu=0$) and corresponding polarization vectors ${\epsilon^\pm_1}^\mu,{\epsilon^\pm_2}^\mu$ in Minkowski space (polarization is transversal ${\epsilon^\pm_1}^\mu {p_1}_\mu=0$ and ${\epsilon^\pm_2}^\mu {p_2}_\mu=0$ and circular ${\epsilon^\pm_1}^\mu {\epsilon^\pm_1}_\mu=0$ and ${\epsilon^\pm_2}^\mu {\epsilon^\pm_2}_\mu=0$), I noticed that the following identity always seems to hold:

$$\epsilon^\pm_1\cdot\epsilon^\mp_2=\frac{(\epsilon^\pm_1\cdot p_2)(p_1\cdot\epsilon^\mp_2)}{p_1\cdot p_2}$$

where $x\cdot y=x_\mu y^\mu$ is the inner product over the Minkowski metric. A quick search on the internet did not give any such results. At the first glance it looks a lot like the bac-cab identity in 3D Euclidean space, but I am not aware of it being true in Minkowski space also. So I wonder if the above can be proven on general grounds? Or maybe someone here knows a reference where this is explained? Thanks for any suggestion!


First of all, polarization vectors ${\epsilon^\pm_i}^\mu$ can be shifted by gauge transformations such that a quantity proportional to the corresponding Minkowski momentum is added to it:

$${\epsilon^\pm_i}^\mu \rightarrow ({\epsilon^\pm_i}^\mu+A_i p_i^\mu)$$

It is easy to show that the above equation is invariant under any such gauge transformations:

$$\epsilon^\pm_1\cdot\epsilon^\mp_2=\frac{(\epsilon^\pm_1\cdot p_2)(p_1\cdot\epsilon^\mp_2)}{p_1\cdot p_2}$$ $$\Downarrow \text{gauge trafo}$$ $$({\epsilon^\pm_1}+A_1 p_1)\cdot({\epsilon^\mp_2}+A_2 p_2)=\frac{(({\epsilon^\pm_1}+A_1 p_1)\cdot p_2)(p_1\cdot({\epsilon^\mp_2}+A_2 p_2))}{p_1\cdot p_2}$$

Expanding the brackets, we trivially see that all extra terms on LHS cancel all extra terms on RHS. From this we learn that we may consider the equation in any gauge we like and the result will be true in general.

Without loss of generality, consider $p_1^\mu$ to be a light-like Minkowski vector with only z-spatial component non-zero:


Also without loss of generality, let the corresponding polarization vector be right-polarized (which implies that the other one must be left-polarized - the reverse case is then directly given by complex conjugation). So we choose the trivial gauge:


The second momentum now can be arbitrary but light-like. We can parameterize it by the usual two angles in spatial spherical coordinates:


The corresponding left-polarized polarization vector in trivial gauge is given by:

$${\epsilon^-_2}^\mu=R_z(\phi)\cdot R_y(\theta)\cdot\left(\frac{1}{\sqrt{2}}(0,1,-i,0)\right)\\ =\frac{1}{\sqrt{2}}\left(~0~,~\cos(\phi)\cos(\theta)+i\sin(\phi)~,~\sin(\phi)\cos(\theta)-i\cos(\phi)~,~-\sin(\theta)~\right)$$

Where $R_z(\phi)$ and $R_y(\theta)$ are basic spatial rotation matrices which rotate the polarization such that it is properly oriented into the spatial surface orthogonal to $p_2^\mu$.

With this we have obtained an explicit parameterization of the most general case for two light-like Minkovski vectors with corresponding circular polarization vectors of opposite helicity. Now we have to do the multiplications under the Minkowski metric $\eta^{\mu\nu}=\text{DiagonalMatrix}(-1,1,1,1)$.

We find:

$$\epsilon^+_1\cdot\epsilon^-_2=\frac{1}{2} ((\cos (\theta )+1) \cos (\phi )+i (\cos (\theta )+1) \sin (\phi ))\\=\cos ^2\left(\frac{\theta }{2}\right) (\cos (\phi )+i \sin (\phi ))$$


$$\frac{(\epsilon^+_1\cdot p_2)(p_1\cdot\epsilon^-_2)}{p_1\cdot p_2}=\frac{E_2 \left(\frac{\sin (\theta ) \cos (\phi )}{\sqrt{2}}+\frac{i \sin (\theta ) \sin (\phi )}{\sqrt{2}}\right)\left(-E_1 \frac{\sin (\theta )}{\sqrt{2}}\right)}{ E_1 E_2 (-1+\cos (\theta ))}\\=\cos ^2\left(\frac{\theta }{2}\right) (\cos (\phi )+i \sin (\phi ))$$

Both calculations give exactly the same result, which proves that the identity is indeed true in general. By complex conjugation we can reverse the helicity of the two polarization vectors, so that the proof stays valid in the same fashion in the reversed helicity case.


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