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I am struggling to work out correct Lorentz transformation for a boost in the 3-direction on a Dirac spinor, $u(p)$. According to Peskin & Schroeder pg. 46, I need to use the equations:

$$S^{0i} = -\frac{i}{2}\pmatrix{\sigma ^i &0 \\ 0 & -\sigma ^i} \hspace{10mm} (3.26)$$ $$\Lambda _{\frac{1}{2}} = exp\left(-\frac{i}{2}\omega _{\mu \nu} S^{\mu \nu} \right) \hspace{6mm} (3.30) $$

Where (3.26) gives me the generator for the boost and (3.30) gives the transformation.

My Attempt

Using the above, $$S^{03} = -\frac{i}{2}\pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3}$$
$$ \Lambda _{\frac{1}{2}} = exp \left (-\frac{1}{4}\omega_{03}\pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3} \right) $$

Then the transformation would be: $$ u(p) = exp \left [-\frac{1}{4}\omega_{03}\pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3} \right] u(p) $$

The Issue

The book states, on the first line of (3.49), that the transformation is instead:

$$ u(p) = exp \left[-\frac{1}{2}\eta \pmatrix{\sigma ^3 & 0 \\ 0 &-\sigma ^3} \right] u(p)$$

Where $\eta$ is the rapidity (whatever that is). I can't see where the $\omega_{\mu \nu}$ has gone, do they group it in with the $\eta$? Thanks!

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  • $\begingroup$ Leaving spinors aside, have you compared (3.21) to (3.48) for the hyperbolic rotations of the simplest boost? What is the infinitesimal angle of this rotation? WP. $\endgroup$ Jun 1, 2020 at 16:02
  • $\begingroup$ @G.Smith Ah, sorry, I meant $\omega_{03}$ to match the $S^{03}$ generator - will edit $\endgroup$ Jun 1, 2020 at 16:11
  • $\begingroup$ @CosmasZachos I had not, I think this would make the boost, setting $\omega_{03} = -\omega_{30}= \beta$: $\pmatrix{1 & 0 & 0 & \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \beta & 0 & 0 & 1} $. $\endgroup$ Jun 1, 2020 at 16:17
  • $\begingroup$ @CosmasZachos Taking my result for the transformation, $$ u(p) = exp \left[-\frac{1}{4}\omega_{03} \pmatrix{\sigma^3 & 0 \\ 0 & -\sigma ^3} \right]u(p)$$, And setting $\eta = \frac{\omega_{03}}{2}$, I recover the stated result: $$u(p) = exp\left[-\frac{1}{2}\eta \pmatrix{\sigma^3 & 0 \\ 0 & -\sigma ^3}\right]u(p) $$ I think $\omega$ is a real number. This is the only way I can see it working out - why they don't absorb the other $-\frac{1}{2}$ into $\eta$ escapes me, though. $\endgroup$ Jun 1, 2020 at 20:45
  • $\begingroup$ The correct way, which hasn't been mentioned yet, is to notice that the sum in the exponential yields $\omega_{\mu\nu}S^{\mu\nu}=\omega_{03}S^{03}+\omega_{30}S^{30}=2\omega_{03}S^{03}$ due to the antisymmetric nature of both $S^{\mu\nu}$ and $\omega_{\mu\nu}$. Setting $\omega_{03}=\eta$ yields the correct result. $\endgroup$
    – pjHart1000
    Apr 16, 2021 at 2:43

2 Answers 2

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Let us first find the boost of the momentum four-vector ($p^\mu$). In the particle rest frame, $p^\mu$ is given by, $p^\mu = (m, \textbf{0)}$. Now we go to a boosted frame in the 3-direction. In the boosted frame, we define, $p^\mu = (E, \textbf{p})$. Then we have, % \begin{align} \begin{pmatrix} E\\ p^3 \end{pmatrix} = \mathrm{exp}\left[\eta \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \right] \begin{pmatrix} m\\ 0 \end{pmatrix} \end{align} Let us define, $A = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$. % Then we have, % \begin{align} A^2 &= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} = I_{2}\\ % \implies A^3 &= A^2 \cdot A = I_{2} A = A\\ % \implies A^{2n} &= I_{2}\\ % \implies A^{2n +1} &= A \end{align} % \begin{align} \mathrm{exp}\left[\eta \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \right] &= e^{\eta A} = I_{2} + \eta A + \frac{\eta^2 A^2}{2!} + \frac{\eta^3 A^3}{3!} + \frac{\eta^4 A^4}{4!} + \frac{\eta^5 A^5}{5!} + \cdots\nonumber\\ % &= I_{2} + \eta A + \frac{\eta^2}{2!} I_{2} + \frac{\eta^3 A}{3!} + \frac{\eta^4}{4!} I_{2} + \frac{\eta^5 A}{5!} + \cdots\nonumber\\ % &= I_{2} \left(\frac{\eta^2}{2!} + \frac{\eta^4}{4!} + \cdots \right) + A \left(\eta + \frac{\eta^3}{3!} + \frac{\eta^5}{5!} + \cdots \right)\nonumber\\ % &= I_{2} \cosh{\eta} + A \sinh{\eta}\nonumber\\ % &= \cos{\eta} \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} + \sin{\eta} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{align} Then we have, \begin{align} \begin{pmatrix} E\\ p^3 \end{pmatrix} &= \left[\cos{\eta} \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} + \sinh{\eta} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \right] \begin{pmatrix} m\\ 0 \end{pmatrix} = \begin{pmatrix} m \cosh{\eta}\\ m\sinh{\eta} \end{pmatrix} \end{align} where, $\eta$ is the rapidity. So we have, $E = m \cosh{\eta}$ and $p^3 = m \sinh{\eta}$. Now we look at boost of the column vector, $u(p) = \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix}$. While considering the boost along 3-direction, we have the boosting matrix given by, \begin{align} \Lambda_{\frac{1}{2}} &= \mathrm{exp}\left(-\frac{i}{2} \omega_{\mu\nu} S^{\mu\nu}\right) = \mathrm{exp}\left(-\frac{i}{2} \omega_{03} S^{03}\right)\nonumber\\ % &= \mathrm{exp}\left[-\frac{1}{4} \omega_{03} \begin{pmatrix} \sigma^3 & 0\\ 0 & - \sigma^3 \end{pmatrix}\right] = \mathrm{exp}\left[-\frac{\eta}{2} \begin{pmatrix} \sigma^3 & 0\\ 0 & - \sigma^3 \end{pmatrix}\right] \end{align} % where, we defined, $\eta = \omega_{03}/2$. So, the boosted version of $u(p)$ is, % \begin{align} u(p) &= \Lambda_{\frac{1}{2}} \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix} = \mathrm{exp}\left[-\frac{\eta}{2} \begin{pmatrix} \sigma^3 & 0\\ 0 & - \sigma^3 \end{pmatrix}\right] \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix} \end{align} Now, we note that, $\sigma^3 = \begin{pmatrix} 1 & 0\\ 0 & - 1 \end{pmatrix} \implies (\sigma^3)^2 = \begin{pmatrix} 1 & 0\\ 0 & - 1 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & - 1 \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} =I_{2} \implies (\sigma^3)^{2n} = I_{2}$ and $(\sigma^{3})^{2n + 1} = \sigma^{3}$. Proceeding as before, by defining $B = \begin{pmatrix} \sigma^3 & 0\\ 0 & - \sigma^3 \end{pmatrix}$, we can prove, \begin{align} \mathrm{exp}\left[-\frac{\eta}{2} \begin{pmatrix} \sigma^3 & 0\\ 0 & - \sigma^3 \end{pmatrix}\right] = \cosh{(\eta/2)} \begin{pmatrix} I_{2} & 0 \\ 0 & I_{2} \end{pmatrix} - \sinh{(\eta/2)} \begin{pmatrix} \sigma^3 & 0 \\ 0 & - \sigma^3 \end{pmatrix} \end{align} % Hence we find, % \begin{align} u(p) &= \mathrm{exp}\left[-\frac{\eta}{2} \begin{pmatrix} \sigma^3 & 0\\ 0 & - \sigma^3 \end{pmatrix}\right] \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix}\nonumber\\ &= \left[\cosh{(\eta/2)} \begin{pmatrix} I_{2} & 0 \\ 0 & I_{2} \end{pmatrix} - \sinh{(\eta/2)} \begin{pmatrix} \sigma^3 & 0 \\ 0 & - \sigma^3 \end{pmatrix}\right] \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix}\nonumber\\ &= \left[ \begin{matrix} I_{2}\cosh{(\eta/2)} - \sigma^3 \sinh{(\eta/2)} & 0\\ 0 & I_{2} \cosh{(\eta/2)} + \sigma^3 \sinh{(\eta/2)}\\ \end{matrix}\right] \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix}\nonumber\\ &= \left[ \begin{matrix} \left(\frac{e^{\eta/2} + e^{-\eta/2}}{2}\right) I_{2} - \left(\frac{e^{\eta/2} - e^{-\eta/2}}{2}\right) \sigma^3 & 0\\ 0 & \left(\frac{e^{\eta/2} + e^{-\eta/2}}{2}\right) I_{2} + \left(\frac{e^{\eta/2} - e^{-\eta/2}}{2}\right) \sigma^3\\ \end{matrix}\right] \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix}\nonumber\\ &= \left[ \begin{matrix} e^{\eta/2} \left( \frac{I_{2} - \sigma^3}{2}\right) + e^{-\eta/2} \left( \frac{I_{2} + \sigma^3}{2}\right) & 0\\ 0 & e^{\eta/2} \left( \frac{I_{2} + \sigma^3}{2}\right) + e^{-\eta/2} \left( \frac{I_{2} - \sigma^3}{2}\right)\\ \end{matrix}\right] \sqrt{m} \begin{pmatrix} \xi\\ \xi \end{pmatrix}\nonumber\\ &= \left[ \begin{matrix} \left(\sqrt{m} e^{\eta/2}\right) \left( \frac{I_{2} - \sigma^3}{2}\right) + \left(\sqrt{m} e^{-\eta/2}\right) \left( \frac{I_{2} + \sigma^3}{2}\right)\\ \left(\sqrt{m} e^{\eta/2}\right) \left( \frac{I_{2} + \sigma^3}{2}\right) + \left(\sqrt{m} e^{-\eta/2}\right) \left( \frac{I_{2} - \sigma^3}{2}\right)\\ \end{matrix}\right] \begin{pmatrix} \xi\\ \xi \end{pmatrix}\nonumber\\ &= \left( \begin{matrix} \left[\sqrt{E + p^3} \left( \frac{I_{2} - \sigma^3}{2}\right) + \sqrt{E - p^3} \left( \frac{I_{2} + \sigma^3}{2}\right)\right]\xi\\ \\ \left[\sqrt{E + p^3} \left( \frac{I_{2} + \sigma^3}{2}\right) + \sqrt{E - p^3} \left( \frac{I_{2} - \sigma^3}{2}\right)\right]\xi\\ \end{matrix}\right) \end{align}

where, we have used the following facts, $\sqrt{m}\, e^{\eta/2} = (m e^\eta)^{1/2} = \sqrt{m \cosh{\eta} + m \sinh{\eta}}$ and, $\sqrt{m}\, e^{-\eta/2} = (m e^{-\eta})^{1/2} = \sqrt{m \cosh{\eta} - m \sinh{\eta}}$.

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Let us evaluate the sum in the exponential: \begin{equation} -\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}=-\frac{i}{2}(\omega_{03}S^{03}+\omega_{30}S^{30}). \end{equation} Due to $S^{\mu\nu}$ and $\omega_{\mu\nu}$ being antisymmetric, we get: \begin{equation} -\frac{i}{2}(\omega_{03}S^{03}+\omega_{30}S^{30})=-\frac{i}{2}(\omega_{03}S^{03}+(-\omega_{03})(-S^{03})) = -i\,\omega_{03}S^{03}. \end{equation} Now set $\omega_{03}=\eta$ (on page 40 in Peskin & Schroeder, they do the same for a boost in the $x$-direction, with $\omega_{01}=\eta$ instead). Thus: \begin{equation} \exp\left(-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}\right) = \exp\left(-i\eta\left(-\frac{i}{2}\begin{pmatrix}\sigma^i&0\\0&-\sigma^i\end{pmatrix}\right)\right)=\exp\left(-\frac{1}{2}\eta\begin{pmatrix}\sigma^i&0\\0&-\sigma^i\end{pmatrix}\right). \end{equation}

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