Spin-1 particle states as seen by different observers: Wigner rotation

[SpookyMulder]

Summary: Suppose that observer $\mathcal{O}$ sees a $W$ boson (spin-1 and $m > 0$) with momentum $\boldsymbol{p}$ in the $y$-direction and spin $z$-component $\sigma$. A second observer $\mathcal{O'}$ moves relative to the first with velocity $\boldsymbol{v}$ in the $z$-direction. How does $\mathcal{O'}$ describe the $W$ state?

This is, of course, the first problem from Chapter 2 of Weinberg's first volume of QFT. I should add that it is not a HW problem, I'm simply going through the book and finishing the problems I left undone. It turns out to be the only one left for me in this chapter and it's still bugging me. There was an old thread with zero activity, so I've added a new one to start afresh.

The effect of operating with a homogeneous LT $\Lambda$ represented by $U(\Lambda)$ on a one-particle state with $m>0$ and spin $j$ is given by (2.5.23):
$U(\Lambda) \Psi_{p, \sigma} = \sqrt{\frac{(\Lambda p)^0} {p^0}} \sum_{\sigma'}D_{\sigma' \sigma}^{(j)}(W(\Lambda, p)) \Psi_{\Lambda p, \sigma'}$,
where the little-group element is:
$W(\Lambda, p) = L^{-1}(\Lambda p)\Lambda L(p)$,
the standard boost $L(p)$, carrying the standard 4-momentum $k^{\mu}=(0,0,0,M)$ (in Weinberg's order of components the last one is a time component of a 4-vector) to $p^{\mu}$, is chosen to be:
$L^i_k(p) = \delta_{ik} + (\gamma - 1)\hat{p}_i \hat{p}_k \quad L^i_0(p) = L^0_i(p) = \hat{p}_i \sqrt{\gamma^2 - 1} \quad L^0_0(p) = \gamma$,
where $\hat{p}_i \equiv p_i /|\textbf{p}|$ and $\gamma \equiv \sqrt{\textbf{p}^2 + M^2}/M$,
and $D^{(j)}_{\sigma' \sigma}(R)$ are the irreducible representations of the rotation group, which can be built up from the standard matrices for infinitesimal rotations $R_{ik} = \delta_{ik} + \Theta_{ik}$, with $\Theta_{ik} = - \Theta_{ki}$ infinitesimal (Eqns. 2.5.20--22):
$D^{(j)}_{\sigma' \sigma}(1+\Theta) = \delta_{\sigma' \sigma} + \frac{i}{2} \Theta_{ik} \left(J_{ik}^{(j)}\right)_{\sigma' \sigma}$,
$\left(J_{23}^{(j)} \pm J_{31}^{(j)} \right)_{\sigma' \sigma} = \left(J_{1}^{(j)} \pm J_{2}^{(j)} \right)_{\sigma' \sigma} = \delta_{\sigma', \sigma \pm 1} \sqrt{(j \mp \sigma)(j \pm \sigma + 1)}$,
$\left(J_{12}^{(j)}\right)_{\sigma' \sigma} = \left(J_{3}^{(j)}\right)_{\sigma' \sigma} = \sigma \delta_{\sigma', \sigma}$
with $\sigma = j, j -1, \dots, -j$.

Now that I have stated everything Weinberg provides, let me say where my problem is. It's relatively easy to calculate the Wigner rotation matrix $W(\Lambda, p)$. Since $W$ is a rotation, it's supposed to be characterized by, say, Euler angles, with a general rotation given by successive rotations around fixed axes in the following order: a rotation around the 3-axis by $\gamma$, followed by a rotation around the 2-axis by $\beta$, followed by a rotation around the 3-axis by angle $\alpha$. Knowing Wigner matricies $D^{(j=1)}_{\sigma' \sigma}(\alpha, \beta, \gamma) = e^{-i(\sigma' \alpha + \sigma \gamma)} d^{(1)}(\beta)$, one can write the state as seen by $\mathcal{O}'$ as a linear combination of states with the boosted momentum $\Lambda p$ and spin-z components of $-1, 0, 1$, which, aside from the relativistic correction factor of $\sqrt{\gamma_v}$ out front, is:
$D^{(1)}_{-1, +1} \Psi_{\Lambda p, \sigma=-1} + D^{(1)}_{0, +1} \Psi_{\Lambda p, \sigma=0} + D^{(1)}_{+1, +1} \Psi_{\Lambda p, \sigma=+1}$.

Now, $W(\Lambda, p)$ one calculates for this problem is a devilish matrix that I won't even bother writing down. Does the problem boil down to finding the above Euler angles, despite the struggle? I thought there could be an easier approach here, especially because in the next problem (with a photon moving the same way as the W-boson in this one), everything is so much more simpler because helicity is Lorentz-invariant and the corresponding phase is also a simple one (in fact, it's zero). I just don't see Weinberg assigning this problem to his class at Austin if the calculation is so horrible, I can be wrong, of course.

Perhaps one could approach this problem by exploiting the isomorphism between the LG and $SL(2,C)/Z_2$ which Weinberg only briefly outlines in section 2.7. In this case one at least works with 2x2 matrices instead of 4x4. Another approach I see is to Taylor expand $W(\Lambda, p))$ to the first order in $\omega$ ($\Lambda = 1 + \omega$) to try to identify the angle (is this called the Wigner angle?) and axis of rotation as a function of Euler angles and the boost parameter of $\Lambda$ and then exponentiate to get the finite rotation. Could someone point out to where I could find such an expansion in literature?

 

[PeterDonis]

This is, of course, the first problem from Chapter 2 of Weinberg's first volume of QFT. I should add that it is not a HW problem,

Yes, it is a HW problem, because, even if you personally aren't assigned it right now, other PF readers will be. So it needs to be in the homework forum, and I have moved it there.

 

[malawi_glenn]

Yes, it is a HW problem, because, even if you personally aren't assigned it right now, other PF readers will be. So it needs to be in the homework forum, and I have moved it there.

Very unlikely to be assigned homework problems from Weinbergs QFT books. But I do get what you mean.

 

[physicsworks]

Well, just from the fact that boosted momentum as seen by $\mathcal{O'}$ lies in the $OYZ$-plane, one can deduce that the corresponding Wigner rotation is around the $x$-axis. This may help extracting the angle of rotation from the resulting $W$ matrix, although the whole process is cumbersome and should probably be done in Mathematica. I like the idea of using spinor formalism, if you do end up solving the problem this way, I would recommend consulting this paper by O'Donnel and Visser: Elementary analysis of the special relativistic combination of velocities, Wigner rotation, and Thomas precession (section 5).
I also doubt that there is a university in the US or elsewhere that teaches standard QFT classes straight from Weinberg's texts, although I have seen his colleague from UTA Vadim Kaplunovsky occasionally using excerpts from all three volumes.

 

[SpookyMulder]

Ah, thank you, that's helpful! However, I still get a crazy matrix after multiplying $L^{-1}(\Lambda p)\Lambda L(p)$ together. How do I extract the angle that depends on both $\Lambda$ and $p$ in a complicated way? In the paper you suggested they derive a general form of a product of two boosts as a boost times a rotation, but here for $W(\Lambda, p)$ we have a product of three boosts and formulas just quickly become too complicated. I will check out spinor formalism approach though.

 

[physicsworks]

Even though $L^{-1}(\Lambda p)$ is not pretty, you can avoid dealing with it altogether by isolating this matrix on one side of your little-group defining equation $W(\Lambda, p) = L^{-1}(\Lambda p)\Lambda L(p)$. Remember that $W(\Lambda, p)$ is a pure rotation and, as was mentioned above, is around the $x$-axis, because boosts involved in the problem are parametrized by vectors lying in the $OYZ$ plane. Thus, using group multiplication properties, you can first right-multiply the above defining equation by the inverse of the Wigner rotation matrix (parametrized in the usual way), leaving the identity on the LHS, and then left-multiply both sides by $L(\Lambda p)$. This will express $L(\Lambda p)$ as a product of a rotation around the $x$-axis, followed by a boost along the $y$-axis, followed by a boost along the $z$-axis. This product is much more manageable than calculating the original triple product on the RHS and can be done by hand. Now, because the LHS $L(\Lambda p)$ is a boost, its matrix is symmetric. You can exploit this fact to extract the Wigner rotation angle and then use it to find $D^{(1)}(W)$ matrices in the linear combination of states as seen by $\mathcal{O'}$ (which you correctly provided in the first post).