1
$\begingroup$

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 $\Lambda^{-1}p$ .

$\endgroup$
4
  • 1
    $\begingroup$ My problem is in simplifying this exponential. A matrix exponential is defined by its Taylor series, which you need to compute. $\endgroup$
    – G. Smith
    Mar 2, 2021 at 21:12
  • 1
    $\begingroup$ Have you computed the square of the matrix with the $\sigma_x$ matrices? Once you have done this the computation should be easy. $\endgroup$
    – mike stone
    Mar 2, 2021 at 21:34
  • 1
    $\begingroup$ I would double-check that 4 in the denominator. I think it may be 2 instead. $\endgroup$
    – G. Smith
    Mar 2, 2021 at 21:44
  • $\begingroup$ 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. $\endgroup$
    – Buzz
    Mar 3, 2021 at 2:18

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ You seem to be missing a $\sigma_x$ on the bottom of the right of the 4th line of math. $\endgroup$
    – G. Smith
    Mar 3, 2021 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.