I am working through Weinberg's QFT book, and in problem 1 in chapter 2 I ran into copious amounts of algebra, so I am trying to "cheat" a little by using some assumptions, but am unsure of their validity. The problem goes like this: "Observer $O$ sees a W boson with momentum $p$ along the y-axis and spin z component $\sigma$. How does observer $O'$ describe the state when she is traveling along the z-axis with velocity $v$ relative to $O$?"

So, we need to work out the Wigner rotation to find the angle of rotation:

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

Where $L(p)$ is a boost that takes the standard momentum $(M,0,0,0)$ to four momentum $(p_0,\bf{p})$. However, the matrix $L^{-1}(\Lambda p)$ is really nasty. So I thought about going about it this way. We see that the momentum transforms to

$$\Lambda p= (\gamma p_0,0,p,-\gamma vp_0) $$

in the $O'$ frame. I took this as meaning that the Wigner rotation should be a rotation about the x-axis by some angle $\theta$. Is this assumption correct? From here I rearranged the Wigner rotation expression to

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

Which has a relatively simple form as a function of $\theta$. I then used the fact that $ L(\Lambda p)$ should be a symmetric matrix to get an expression for $\theta$ in terms of $p$ and $v$. Once we have $\theta$ I think the rest of the computation is pretty straightforward as the Wigner D-matrix elements can be calculated pretty easily. I'm mainly worried about my approach to find the Wigner rotation...is my logic sound? If not, how would you approach this problem? The algebra in the direct computation seems pretty unmanageable!


After a few days with no activity, I am wondering if there is anything I can do to improve this question?

  • $\begingroup$ You could add a bounty to draw attention. $\endgroup$
    – Nephente
    Mar 30, 2016 at 19:37

1 Answer 1


The boost matrix can be chosen in block form as (in $c=1$ units):

$$L(\mathbf{p}) = \begin{bmatrix} \frac{E}{M}& \frac{\mathbf{p}^t}{M}\\ \frac{\mathbf{p}}{M}& 1_{(3\times3)}+\frac{\mathbf{p}\mathbf{p}^t}{M(E+M)} \end{bmatrix}$$

where $\mathbf{p}$ is the 3-momentum and $ E = \sqrt{\mathbf{p}^2+M^2}$.

It can be easily checked that this matrix boosts the momentum from its rest frame to its laboratory frame, Also, it can be easily checked that:

$$L(\mathbf{p})^{-1} = L(-\mathbf{p})$$

by matrix multiplication. The inverse boost acts on the momentum in the laboratory frame and brings it back to its rest frame value.

The remaining work is just to substitute the special case in the exercise.

  • $\begingroup$ Thanks for the response! Do you think that the rotation will end up being about the X-axis for this particular case? $\endgroup$ Apr 3, 2016 at 16:22
  • $\begingroup$ As this is what I was trying to avoid...;). $\endgroup$ Apr 3, 2016 at 23:33
  • $\begingroup$ @TylerHG I don't think that the rotation will end up along the x-axis, since it depends on both $p$ and the boost speed $v$. The expression you wrote for $\Lambda p$ is correct. So, I don't see how one can avoid performing the full computation by multiplying three 4-matrices. It is not so terrible because they are quite sparse. $\endgroup$ Apr 4, 2016 at 10:22
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    $\begingroup$ Thank you for your responses! I guess this time my laziness has lost the battle $\endgroup$ Apr 4, 2016 at 16:24

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