# Lorentz boost of Dirac spinor

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 $$\Lambda^{-1}p$$ .

• My problem is in simplifying this exponential. A matrix exponential is defined by its Taylor series, which you need to compute. Mar 2, 2021 at 21:12
• Have you computed the square of the matrix with the $\sigma_x$ matrices? Once you have done this the computation should be easy. Mar 2, 2021 at 21:34
• I would double-check that 4 in the denominator. I think it may be 2 instead. Mar 2, 2021 at 21:44
• If you are using the chiral representation of the $\gamma$-matrices (as Peskin and Schroeder do), then your initial expression does not describe a stationary particle in the first place.
– Buzz
Mar 3, 2021 at 2:18

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.

• You seem to be missing a $\sigma_x$ on the bottom of the right of the 4th line of math. Mar 3, 2021 at 18:17