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## Homework Statement

Suppose that observer [itex]\cal O[/itex] sees a W-boson (spin one and mass [itex]m \neq 0[/itex]) with momentum [itex]\textbf{p}[/itex] in the y-direction and spin z-component [itex]\sigma[/itex]. A second observer [itex]\cal O'[/itex] moves relative to the first with velocity [itex]\textbf{v}[/itex] in the z-direction. How does [itex]\cal O'[/itex] describe the W state?

## Homework Equations

How does one compute the Wigner rotation matrix for this? It seems that the calculations quickly become cumbersome and unwieldy, without producing any meaningful result.

## The Attempt at a Solution

I'm quite sure that the result should be rotation around x-axis in the spin space. The question is the angle of the rotation, and its dependence on the [itex]\textbf{v}[/itex].

I started it by looking up formulas from the p. 68:

[tex]

U(\Lambda) \Psi_{p,\sigma} = \sqrt{\frac{(\Lambda p)^0} {p^0}} \sum_{\sigma'}D_{\sigma' \sigma}^{(j)}(W(\Lambda,p))\Psi_{\Lambda p,\sigma'}

\\

W(\Lambda,p) = L^{-1}(\Lambda p)\Lambda L(p)

[/tex]

Where [itex]L(p)[/itex] for mass positive-definite particle is chosen to be:

[tex]

L^i_k(p) = \delta_{ik} + (\gamma - 1)\hat{p}_i \hat{p}_k

\\

L^i_0(p) = L^0_i(p) = \hat{p}_i \sqrt{\gamma^2 - 1}

\\

L^0_0(p) = \gamma

\\

\hat{p}_i \equiv p_i /|\textbf{p}|

\\

\gamma \equiv \sqrt{\textbf{p}^2 + M^2}/M

[/tex]

In the mentioned problem I took [itex]\vec{p}[/itex] to be [itex](E, 0, p_y, 0)[/itex], [itex]L(p)[/itex] is then:

[tex]

L(p) = \begin{bmatrix}

\gamma_p & 0 & \beta_p & 0 \\

0 & 1 & 0 & 0 \\

\beta_p & 0 & \gamma_p & 0 \\

0 & 0 & 0 & 1

\end{bmatrix}

[/tex]

Where [itex]\beta_p = \sqrt{\gamma_p^2 - 1}[/itex].

[itex]\Lambda[/itex] is then:

[tex]

\Lambda = \begin{bmatrix}

\gamma & 0 & 0 & \beta \\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0 \\

\beta & 0 & 0 & \gamma

\end{bmatrix}

[/tex]

Again [itex]\beta[/itex] is defined similarly.

[itex]L(\Lambda p)[/itex] is:

[tex]

L(\Lambda p) = \begin{bmatrix}

\gamma_{\Lambda p} & 0 & \beta_{\Lambda p} \frac{p_y}{\tau} & \beta_{\Lambda p} \frac{\beta E}{\tau} \\

0 & 1 & 0 & 0 \\

\beta_{\Lambda p} \frac{p_y}{\tau} & 0 & (\gamma_{\Lambda p} - 1) (\frac{p_y} {\tau})^2 + 1 &(\gamma_{\Lambda p} - 1) \frac{p_y \beta E}{\tau^2} \\

\beta_{\Lambda p} \frac{\beta E}{\tau} & 0 & (\gamma_{\Lambda p} - 1) \frac{p_y \beta E}{\tau^2} & (\gamma_{\Lambda p} - 1) (\frac{\beta E}{\tau})^2 + 1

\end{bmatrix}

[/tex]

Where:

[tex]

\gamma_{\Lambda p} \equiv \sqrt{p_y^2 + \beta^2 E^2 + M^2}/M

[/tex]

Inverting this matrix should be the same thing as changing the sign of [itex]\Lambda p[/itex], so it will flip the signs of off-diagonal matrix elements.

And here is when I gave up, the result of multiplication is so huge that I can't really figure out anything out of it. It is clear however that [itex]W_{00} = 1[/itex] and [itex]W_{0i}=W_{i0}=0[/itex], since it should leave the [itex]\vec{k}[/itex] invariant, everything else doesn't seem to help that much.

Any suggestions appreciated!

(Note that it is very likely that I made mistake somewhere in the computations above, but it doesn't seem to be making things more complex than they might be)

Thank you!