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I'm working through problem 2 from Weinberg's QFT vol. 1 chapter 2. The problem statement is simple. For one observer, a photon is travelling in the $y$-direction with linear polarization in the $z$-direction. In Weinberg's notation, they describe the state as:

$$\Psi_p=\frac{1}{\sqrt{2}}\left[\Psi_{p,+}+\Psi_{p,-}\right]$$

where $p\equiv p^{\mu}=(\vec{p},p^0)=(0,p,0,p)$, and the $\pm$ subscripts correspond to $\pm 1$ helicity photon states. Of course, I am using the natural units convention $\hbar=c=1$.

Now consider another observer, travelling at a velocity $v$ in the $z$ direction relative to the first. How does this new observer describe the same state?

I have already gone through the corresponding sections in the chapter which covers the general/formal solution to the problem. What we want to calculate is:

$$U(\Lambda)\Psi_{p}=\frac{1}{\sqrt{2}}\sqrt{\frac{(\Lambda p)^0}{p^0}}\left[e^{i\theta}\Psi_{p,+}+e^{-i\theta}\Psi_{p,-}\right]$$

where $\Lambda$ is the Lorentz transformation of interest - a simple boost in the $z$-direction - and $\theta=\theta(\Lambda,p)$ rotation around the $z$-axis corresponding to the general Wigner-rotation.

So this problem essentially amounts to calculating $\theta$. This is straightforward: a Wigner-rotation $W(\Lambda,p)$ for a massless particle can always be expressed as $S(\alpha,\beta)R(\theta)$, where $R(\theta)$ is the rotation around the $z$-axis that we're interested in, so simply calculate the explicit Wigner-rotation $W=L^{-1}(\Lambda p) \Lambda L(p)$ , and look at the $x-y$ block, which we know will look like a rotation.

$$W=\begin{bmatrix}\cos\theta & \sin\theta & . & . \\ -\sin\theta & \cos\theta & . & . \\ . & . & . & .\\ . & . & . & .\end{bmatrix}$$

So, I begin with the calculation. The standard four-momentum I will use is $k^{\mu}=(\vec{k},k^0)=(0,0,\kappa,\kappa)$. Using this, I begin by calculating $L(p)$.

$$\begin{align} L(p)&=R(\hat{k}\rightarrow \hat{p}) B(\kappa \rightarrow |p|)\\ &=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos(-\pi/2) & \sin(-\pi/2) & 0\\ 0 & -\sin(-\pi/2) & \cos(-\pi/2) & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & A & B\\ 0 & 0 & B & A \end{bmatrix}\\ &=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & -A & -B \\ 0 & 1 & 0 & 0\\ 0 & 0 & B & A \end{bmatrix} \end{align}$$

where I have shortened the notation via:

$$A=\frac{p^2+\kappa^2}{2 p\kappa}, \,\,\,B=\frac{p^2-\kappa^2}{2 p\kappa}$$

I now calculate the other part of the Wigner-rotation, $L^{-1}(\Gamma p)$.

$$\begin{align} \Lambda &= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & \gamma & v\gamma \\ 0 & 0 & v\gamma & \gamma \end{bmatrix}, \,\,\, \gamma\equiv \frac{1}{\sqrt{1-v^2}}\\ &\\ \implies \Lambda p &=(0, p, pv\gamma, p\gamma )\\ &\\ L^{-1}(\Lambda p)&=B(|\Lambda p|\rightarrow \kappa )R(\hat{\Lambda p}\rightarrow \hat{k})\\ &=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & A' & B'\\ 0 & 0 & B' & A' \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos(\phi) & \sin(\phi) & 0\\ 0 & -\sin(\phi) & \cos(\phi) & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\\ &=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos\phi & \sin\phi & 0\\ 0 & -A'\sin\phi & A'\cos\phi & B'\\ 0 & -B'\sin\phi & B'\cos\phi & A' \end{bmatrix} \end{align}$$

where I have shortened the notation via:

$$A'=\frac{\kappa^2+\gamma^2p^2}{2\gamma p\kappa}, \,\,\,B'=\frac{\kappa^2-\gamma^2p^2}{2\gamma p\kappa},\,\,\, \tan\phi=\frac{1}{v\gamma}$$

So the final Wigner-rotation matrix is:

$$W=L^{-1}(\Lambda p) \Lambda L(p)=\left[ \begin{array}{cc|cc} 1 & 0 & . & . \\ 0 & 1 & . & . \\ \hline . & . & . & .\\ . & . & . & . \end{array} \right] $$

So $\theta=0$, i.e. the polarization doesn't change, in the end? I have no other intuition for this result so I don't know any sanity checks. It seems odd that Weinberg would include a problem for which the final answer is boring.

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That result seems intuitive to me. The meaning of the initial polarization is that it doesn't get through an $x$ polarizing filter. The other observer will agree that the photon is blocked, and they will also see the polarizing filter as an $x$ polarizing filter, so they'll say the polarization is orthogonal to that. And that's exactly what your final result says.

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