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

  • #1

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!
 

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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