Lorentz boost of Dirac spinor

Question

Let $\psi_\vec{0}^+$ be a Dirac wavefunction describing a motionless particle,

$$\psi_\vec{0}^+(x) = \sqrt{2m} \begin{pmatrix} \chi \\ 0 \end{pmatrix} e^{ip \cdot x}$$

where $p = (m, \vec{0})$. Acted by a Lorentz boost, say, in the $\vec{x}$ direction, I want to show that

$$\psi_\vec{0}^+(x) \to u(\vec{q})e^{iq\cdot x}$$

where $q = (E_\vec{q},\vec{q})$.

My attempt

I used the usual procedure to transform Dirac spinors:

$$\psi_\vec{0}^{'+}(x) = U_{\delta_x} \psi_\vec{0}^+(x) = S(\delta_x) \psi_\vec{0}^+(\Lambda^{-1}x)$$

where $\delta_x$ is the rapidity. Since $p \cdot \Lambda^{-1} x = \Lambda^{-1}p \cdot x$,

$$\Lambda^{-1}p = \begin{pmatrix} \cosh(\delta_x) & \sinh(\delta_x) \\ \sinh(\delta_x) & \cosh(\delta_x) \end{pmatrix} \begin{pmatrix} m \\ \vec{0} \end{pmatrix}$$

At last, it remains to transform the Dirac spinor, $u(\vec{p})$. Using $S(\delta_x) = e^{\frac{i}{4}\omega_{\mu \nu} \sigma^{\mu \nu}}$ we get

$$ S(\delta_x) = e^{\frac{\delta_x}{2} \begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}}$$

My problem is in simplifying this exponential. According to Peskin & Schroeder, this should yield something with $\cosh(\delta_x)$ and $\sinh(\delta_x)$, but I can't see how!

Edit: There was a 4 factor which was wrong, as @G. Smith stated.

Edit 2: There was vector column missing in $\Lambda^{-1}p$ .

Answer 1

Thanks to @G. Smith and @mike stone, I've come to a solution.

Expanding in Taylor Series,

$$ S(\delta_x) = e^{\frac{\delta_x}{2} \begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}} = \sum_{n=0}^{\infty} \frac{(\delta_x/2)^n}{n!}\begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}^n$$

Grouping odd and even terms,

$$\sum_{n=0}^{\infty} \frac{(\delta_x/2)^n}{n!}\begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}^n = \sum_{n=0}^{\infty}\frac{(\delta_x/2)^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \sum_{n=0}^{\infty}\frac{(\delta_x/2)^{2n+1}}{(2n+1)!}\begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}$$

Recalling the Taylor Series Expansion for cosh($x$) and sinh($x$) we get

$$ S(\delta_x) = \text{cosh}(\delta_x/2) + \text{sinh}(\delta_x/2) \begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}$$

We can now apply S($\delta_x$) to u($\vec{p}$):

$$ S(\delta_x)u(\vec{p}) = \sqrt{2m} \begin{pmatrix} \cosh(\delta_x/2) & \sigma_x \sinh(\delta_x/2)\\ \sigma_x \sinh(\delta_x/2) & \cosh(\delta_x/2) \end{pmatrix} \begin{pmatrix} \chi \\ 0 \end{pmatrix} = \sqrt{2m}\begin{pmatrix} \cosh(\delta_x/2)\chi \\ \sigma_x \sinh(\delta_x/2) \chi \end{pmatrix}$$

Since $E_\vec{q} = m \cosh(\delta_x)$ and $\vec{q} = m \sinh(\delta_x) \vec{e_i}$ and using

$$\cosh(\delta_x/2) = \sqrt{\frac{\cosh(\delta_x)+1}{2}}$$

$$\tanh(\delta_x/2) = \frac{\sinh(\delta_x)}{\cosh(\delta_x)+1}$$

we get

$$u(\vec{q}) = \sqrt{E_\vec{q}+m} \begin{pmatrix} \chi \\ \frac{\vec{\sigma} \cdot \vec{q}}{E_\vec{q}+m} \chi \end{pmatrix}$$

as we wanted!

Edit: There was a $\sigma_x$ missing, as @G. Smith kindly pointed.