Lorentz transformation (III)

Problem: Let $p^{\mu}\equiv[|\mathbf{p}|, p^1, p^2, p^3]^T$ be the four-momentum of a massless particle, assuming $c=1$. In some inertial frame, such four-momentum becomes to $k^{\mu}\equiv[\kappa, 0,  0,  \kappa]^T$.  The problem is to find a Lorentz transformation (matrix) $L^{\mu}_{\,\,\nu}(p)$ such that $p^{\mu} =L^{\mu}_{\,\,\nu}(p)\,k^{\nu}$.

(This problem is about the proof of Eq. (2.5.44) in Weinberg's QFT book, Volume I)

Solution:

To transform $k^{\mu}$ to $p^{\mu}$, we can first boost $k^{\mu}$ to $[|\mathbf{p}|, 0, 0, |\mathbf{p}|]^T$ followed by a spatial rotation to $p^{\mu}$. 

Recall the Lorentz boost on the energy and momentum, \begin{equation}\begin{split}E'&=&\gamma\,(E-v\,p^3)\,,\\{p'}^3&=&\gamma\,(p^3-v\,E)\,,\end{split} \end{equation} with $\gamma \equiv 1/\sqrt{1-v^2}$. In the matrix form, \begin{equation}\left[\begin{matrix}E' \\ 0 \\ 0 \\ {p'}^3\end{matrix}\right]\equiv B(v)\, \left[\begin{matrix}E \\ 0 \\ 0 \\ {p}^3\end{matrix}\right]=\left[\begin{matrix} 1/\sqrt{1-v^2} & 0 & 0 & -v/\sqrt{1-v^2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -v/\sqrt{1-v^2} & 0 &  0 & 1/\sqrt{1-v^2}\end{matrix}\right]\left[\begin{matrix}E \\ 0 \\ 0 \\ {p}^3\end{matrix}\right]\,.\end{equation}

In our case, we have \begin{equation}|\mathbf{p}| =\gamma\, \kappa (1-v)\,,\end{equation} from which we can solve \begin{equation} v= \frac{1-|\mathbf{p}|^2/\kappa^2}{1+|\mathbf{p}|^2/\kappa^2}\,.\end{equation}

As a result, the boost matrix $B(v)$ can be expressed in terms of $u\equiv |\mathbf{p}|/\kappa$ as \begin{equation}B(u)=\left[\begin{matrix} (u^2+1)/2u & 0 & 0 & (u^2-1)/2u \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ (u^2-1)/2u & 0 &  0 & (u^2+1)/2u\end{matrix}\right]\,.\tag{1}\end{equation}

Now we need to find a spatial rotation matrix $S$ that maps $[|\mathbf{p}|, 0, 0, |\mathbf{p}|]^T$ to $[|\mathbf{p}|, p^1, p^2, p^3]^T$. Suppose \begin{equation}p^1=|\mathbf{p}|\sin\theta\cos\phi\,,\quad p^2=|\mathbf{p}|\sin\theta\sin\phi\,,\quad p^3=|\mathbf{p}|\cos\theta\,,\end{equation}one can verify that the spatial rotation \begin{equation}S(\hat{\mathbf{p}})\equiv \mathbf{R}_z(\phi)\mathbf{R}_y(\theta)\tag{2}\end{equation} rotates $\hat{z}$ to $\hat{\mathbf{p}}\equiv[\sin\theta\cos\phi\,,\,\sin\theta\sin\phi\,,\,\cos\theta]^T$ in which \begin{eqnarray} &&\mathbf{R}_x(\theta) = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\0 & \sin\theta & \cos\theta\end{array}\right] = e^{-i \theta \mathbf{L}_1}\,,\quad \mathbf{L}_1\equiv\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & -i \\0 & i & 0\end{array}\right]\,;\nonumber\\ &&\mathbf{R}_y(\theta) = \left[\begin{array}{ccc}\cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta\end{array}\right] = e^{-i \theta \mathbf{L}_2}\,,\quad \mathbf{L}_2\equiv\left[\begin{array}{ccc}0 & 0 & i \\ 0 & 0 & 0 \\-i & 0 & 0\end{array}\right]\,;\tag{3}\\ &&\mathbf{R}_z(\theta) = \left[\begin{array}{ccc}\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\0 & 0 & 1\end{array}\right] =e^{-i \theta \mathbf{L}_3}\,, \quad \mathbf{L}_3\equiv\left[\begin{array}{ccc}0 & -i & 0 \\ i & 0 & 0 \\0 & 0 & 0\end{array}\right]\,.\nonumber \end{eqnarray}

The final solution is \begin{equation}L(\mathbf{p}) = S(\hat{\mathbf{p}})\,B(|\mathbf{p}|/\kappa)\,.\tag{4}\end{equation}