Calculating field transformation

[Markus Kahn]

Homework Statement

Let $\psi(x)=u(p)e^{-ipx}$, where
$$
u((m,0)) = \sqrt{m}\begin{pmatrix}
\xi\\\xi
\end{pmatrix}\quad\text{where}\quad \xi = \sum_{s\in \{+,-\}}c_s\xi^s\quad \text{and}\quad \xi^+\equiv\begin{pmatrix}
1\\ 0
\end{pmatrix}\quad \xi^-\equiv\begin{pmatrix}
0\\ 1
\end{pmatrix},
$$
be a solution of the Dirac equation of a particle with momentum $p^\mu= (m,0)$. Find the solution of the Dirac equation for a particle with momentum $p^\mu=(E,0,0,p^3)$.

Homework Equations

A spinor transforms under a Lorentz transformation $\Lambda$ according to
$$\psi(x)\longmapsto \psi'(x')= \exp\left(-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}\right)\psi(x),$$
where $M^{\mu\nu}=\frac{1}{4}[\gamma^\mu,\gamma^\nu]$ for a set of $\gamma$-matrices.

The Attempt at a Solution

First of all, we will be working in the Weyl representation of the $\gamma$-matrices. Since $u(p)$ transforms like a spinor we can change coordinate frames by a transfomration of the form $p\mapsto \Lambda p$. A boost along the $z$-axis can be described by a parameter $\eta_3$ called rapidity and is given by
$$
\begin{pmatrix}
m\\0\\0\\0
\end{pmatrix}
\longmapsto \begin{pmatrix}
E\\0\\0\\p^3
\end{pmatrix} = (\mathbb{I}_4 \cosh \eta_3+ J_3 \sinh \eta_3)\begin{pmatrix}
m\\0\\0\\0
\end{pmatrix} = m\begin{pmatrix} \cosh\eta_3\\0\\0\\ \sinh \eta_3
\end{pmatrix}.
$$
With this we can now express $\eta_3$ in terms of $E$ and $p^3$. We find
$$
\frac{p}{E}=\tanh(\eta_3)\Longleftrightarrow \eta_3=\frac{1}{2}(\ln(E+p^3)-\ln(E-p^3))\Longleftrightarrow \eta_3 = \frac{1}{2} \ln \frac{E+p^{3}}{E-p^{3}}.
$$
We now need to figure out the field transformation. Since $\omega$ is zero everywhere, except for $\omega_{30}=-\omega_{03}=-\eta_3$ we get
$$
\begin{align*}
\exp\left(-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}\right) &= \exp\left(-\frac{i}{2}(\omega_{03}M^{03}+\omega_{30}M^{30})\right) = \exp\left(-\frac{i}{2}(\eta_3M^{03}-\eta_3 M^{30})\right)\\
&= \exp\left(-\frac{i}{2}(\eta_3M^{03}+\eta_3 M^{03})\right) = \exp\left(-i\eta_3M^{03}\right)\\
&=
\begin{pmatrix}
\sqrt[4]{\frac{E-p^3}{E+p^3}} &0&0&0\\
0& \sqrt[4]{\frac{E+p^3}{E-p^3}} &0&0\\
0&0&\sqrt[4]{\frac{E-p^3}{E+p^3}} &0\\
0& 0&0& \sqrt[4]{\frac{E+p^3}{E-p^3}}
\end{pmatrix}.
\end{align*}
$$
We can now apply this to the spinor in the rest frame and find
$$\begin{align*}\exp\left(-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}\right)u(p) &=
\begin{pmatrix}
\sqrt[4]{\frac{E-p^3}{E+p^3}} &0&0&0\\
0& \sqrt[4]{\frac{E+p^3}{E-p^3}} &0&0\\
0&0&\sqrt[4]{\frac{E-p^3}{E+p^3}} &0\\
0& 0&0& \sqrt[4]{\frac{E+p^3}{E-p^3}}
\end{pmatrix} \sqrt{m} \begin{pmatrix}\xi\\\xi\end{pmatrix} = \dots\\
&= \begin{pmatrix}{\sqrt{E-p^{3} \sigma_{3}} \xi^{s}} \\ {\sqrt{E+p^{3} \sigma_{3}} \xi^{s}}\end{pmatrix}.\end{align*}
$$
And this is exactly where I fail... How does one get from line one to line two in the last equation... I'm not sure where my error happens, maybe its the transformation matrix of the field, or I'm just unable how the math in this last equation works, but I just can't figure it out...

 

[Markus Kahn]

Figured out the solution and thought that maybe it would be interesting for other as well... The calculation from above is correct up until this point. Now we can make use of the fact that the energy momentum relation $m^2=p^2$ holds and we therefore have $m = \sqrt{E^2-(p^3)^2} = \sqrt{(E+p^3)(E-p^3)}$ which directly leads to
$$
\frac{E+p^3}{m} =\sqrt{\frac{(E+p^3)^2}{m^2}} = \sqrt{\frac{(E+p^3)^2}{(E+p^3)(E-p^3)}} = \sqrt{\frac{E+p^3}{E-p^3}}.
$$
With this we find
$$
\begin{pmatrix}
\sqrt[4]{\frac{E-p^3}{E+p^3}} &0&0&0\\
0& \sqrt[4]{\frac{E+p^3}{E-p^3}} &0&0\\
0&0&\sqrt[4]{\frac{E+p^3}{E-p^3}} &0\\
0& 0&0& \sqrt[4]{\frac{E-p^3}{E+p^3}}
\end{pmatrix} \sqrt{m} =
\begin{pmatrix}\sqrt{E-p^3}&0&0&0\\ 0& \sqrt{E+p^3} &0&0\\ 0&0&\sqrt{E+p^3}&0 \\ 0&0&0&\sqrt{E-p^3}\end{pmatrix},
$$
which by applying onto $u^s(p)$ and noting that $[(1\pm\sigma_3)/2]^2=(1-\sigma_3)/2$ gives the desired result.