# Weinberg's QFT -- Two moving observers see a W-boson differently....

• Fedor Indutny

## Homework Statement

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

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

$$U(\Lambda) \Psi_{p,\sigma} = \sqrt{\frac{(\Lambda p)^0} {p^0}} \sum_{\sigma'}D_{\sigma' \sigma}^{(j)}(W(\Lambda,p))\Psi_{\Lambda p,\sigma'} \\ W(\Lambda,p) = L^{-1}(\Lambda p)\Lambda L(p)$$

Where $L(p)$ for mass positive-definite particle is chosen to be:

$$L^i_k(p) = \delta_{ik} + (\gamma - 1)\hat{p}_i \hat{p}_k \\ L^i_0(p) = L^0_i(p) = \hat{p}_i \sqrt{\gamma^2 - 1} \\ L^0_0(p) = \gamma \\ \hat{p}_i \equiv p_i /|\textbf{p}| \\ \gamma \equiv \sqrt{\textbf{p}^2 + M^2}/M$$

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

$$L(p) = \begin{bmatrix} \gamma_p & 0 & \beta_p & 0 \\ 0 & 1 & 0 & 0 \\ \beta_p & 0 & \gamma_p & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Where $\beta_p = \sqrt{\gamma_p^2 - 1}$.

$\Lambda$ is then:

$$\Lambda = \begin{bmatrix} \gamma & 0 & 0 & \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \beta & 0 & 0 & \gamma \end{bmatrix}$$

Again $\beta$ is defined similarly.

$L(\Lambda p)$ is:

$$L(\Lambda p) = \begin{bmatrix} \gamma_{\Lambda p} & 0 & \beta_{\Lambda p} \frac{p_y}{\tau} & \beta_{\Lambda p} \frac{\beta E}{\tau} \\ 0 & 1 & 0 & 0 \\ \beta_{\Lambda p} \frac{p_y}{\tau} & 0 & (\gamma_{\Lambda p} - 1) (\frac{p_y} {\tau})^2 + 1 &(\gamma_{\Lambda p} - 1) \frac{p_y \beta E}{\tau^2} \\ \beta_{\Lambda p} \frac{\beta E}{\tau} & 0 & (\gamma_{\Lambda p} - 1) \frac{p_y \beta E}{\tau^2} & (\gamma_{\Lambda p} - 1) (\frac{\beta E}{\tau})^2 + 1 \end{bmatrix}$$

Where:

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

Inverting this matrix should be the same thing as changing the sign of $\Lambda p$, 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 $W_{00} = 1$ and $W_{0i}=W_{i0}=0$, since it should leave the $\vec{k}$ 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!