# How to find the 3-direction Lorentz boost transformation on Dirac spinor?

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!

• 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. Jun 1, 2020 at 16:02
• @G.Smith Ah, sorry, I meant $\omega_{03}$ to match the $S^{03}$ generator - will edit Jun 1, 2020 at 16:11
• @CosmasZachos I had not, I think this would make the boost, setting $\omega_{03} = -\omega_{30}= \beta$: $\pmatrix{1 & 0 & 0 & \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \beta & 0 & 0 & 1}$. Jun 1, 2020 at 16:17
• @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. Jun 1, 2020 at 20:45
• 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. Apr 16, 2021 at 2:43

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