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Okay, I've got really desperate now. I've spend 10 hours of work (for a few days) trying to prove 'trivial' equation in Quantum Electrodynamics. To anybody who want to write an answer for my questions - treat me like an idiot who only can multiplicate and sum equations. NO HYPERLINKS (I know how to use Google and sources, really!), NO SHORTCUTS. Please, it's very important to me.

So, main question is:

Let's suppose Lorentz transformation: $\Lambda^a_b=g^a_b+\Delta\omega^a_b$, where $\Delta\omega^a_b$ is just a (small) number ($\Delta\omega^2$ can be neglected). So, we can build an equation for transformation matrix S:

$S^{-1}\gamma^iS=\Lambda^i_j\gamma^j$

where $\gamma^i$ are Dirac matrix (you know, these fundamental in Dirac Equation). It is now easy to show that S:

$S=I+iS^{ab}\Delta\omega_{ab}$

$S^{-1}=I-iS^{ab}\Delta\omega_{ab}$

Something like that. $\Delta\omega$ is still a number, but $s$ is a matrix. Now black magic happens and we get:

$S^{ab} = \frac{i}{4}\big[\gamma^a,\gamma^b\big]$

In some like 5 sources proof of last equation is described as 'trivial' or as 'exercise for reader'. If my explanation is too much complicated, here is a short one:

  1. Read a http://physics.ucsd.edu/students/courses/fall2009/physics130b/Dirac_Lorentz.pdf script
  2. Go to page 26 (in pdf, at page numerated by page 22)
  3. Prove equation (46).
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Yes, it's not absolutely trivial. –  PhysiXxx Sep 10 '13 at 19:37
    
Dear @user, the derivation is pretty much written right above the equation 46 and you omitted it here. The previous displayed equation before 46 tells you exactly which components of Sigma should commute with what components of gamma. You may write Sigmas as the most general superposition of products of gamma matrices - a basis of all 4 x 4 matrices and prove that those are the only possibilities. Alternatively, you may assume the Lorentz covariance of the formula for Sigma which implies that it is only composed of gammamugammanu, gammanugammamu, and those times gamma5. –  Luboš Motl Sep 10 '13 at 19:41
    
Your reference text is explaining correctly .You have $S^{-1}\gamma^iS=\Lambda^i_j\gamma^j$, with $S=I-\frac{i}{2}\omega_{\mu_\nu}\Sigma^{\mu\nu}$ and $\Lambda^\mu_\nu = g^\mu_\nu + \omega^\mu_\nu (44)$, you have to demonstrate, as indicated a few lines before (46) : $i(g^{\alpha \mu} \gamma^\nu-g^{\alpha \nu} \gamma^\mu) = [\gamma^\alpha, \Sigma^{\mu\nu}]$You now may suppose that $\Sigma^{\mu\nu} \sim [\gamma^\mu, \gamma^\nu]$, and make the demonstration using the properties of gamma matrices $\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu=2 g^{\mu\nu} $, –  Trimok Sep 11 '13 at 9:53
    
..... and the properties of imbricated commutators : $[A, BC] + B[A,C] +[A,B]C$. Just make the developpement of $[\gamma^\alpha, [\gamma^\mu, \gamma^\nu]]$, and rearrange the terms to find something proportionnal to $(g^{\alpha \mu} \gamma^\nu-g^{\alpha \nu} \gamma^\mu)$ –  Trimok Sep 11 '13 at 9:54

1 Answer 1

The direct sum of $\left(\frac{1}{2}, 0\right), \left(0, \frac{1}{2}\right)$ gives the Dirac bispinor $\Psi $ with components $\psi_{a}, \kappa_{\dot {a}}$. Here there is the infinitesimal transformation law for them: $$ \delta \psi_{a} = \frac{i}{2}\omega^{\mu \nu}(J_{\mu \nu})_{a}^{\quad b}\psi_{b}, \quad \delta \kappa_{\dot {a}} = \frac{i}{2}\omega^{\mu \nu}(J_{\mu \nu})_{\dot {a}}^{\quad \dot {b}}\kappa_{\dot {b}}, \qquad (.1) $$ where $(J_{\mu \nu})_{a}^{\quad b}$ is the generator of the Lorentz group in spinor representation. They are given by

$$ (J_{\mu \nu})_{a}^{\quad b} = -i(\sigma_{\mu \nu})_{a}^{\quad b}, \quad (J_{\mu \nu})_{\dot {a}}^{\quad \dot {b}} = -i(\tilde {\sigma}_{\mu \nu})_{a}^{\quad b}, \qquad (.2) $$ where $$ (\sigma^{\mu \nu})_{\alpha \beta } = -\frac{1}{4}\left( (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha \beta} - (\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha \beta } \right), \quad (\tilde {\sigma}^{\mu \nu})_{\dot {\alpha } \dot {\beta }} = -\frac{1}{4}\left( (\tilde {\sigma}^{\mu}\sigma^{\nu})_{\dot {\alpha } \dot {\beta } } - (\tilde {\sigma }^{\nu}\sigma^{\mu})_{\dot {\alpha } \dot {\beta } } \right) , $$ $$ \tilde {\sigma}^{\mu} = (\hat {\mathbf E}, -\hat {\mathbf \sigma}) . $$

This is the most untrivial part of the derivation.

So, after substitution $(.2)$ to $(.1)$, you can get $$ \delta \psi_{a} = \frac{1}{2}\omega^{\mu \nu}(\sigma_{\mu \nu})_{a}^{\quad b}\psi_{b}, \quad \delta \kappa_{\dot {a}} = \frac{1}{2}\omega^{\mu \nu}(\tilde {\sigma }_{\mu \nu})_{\dot {a}}^{\quad \dot {b}}\kappa_{\dot {b}} \Rightarrow $$ $$ \delta \Psi = \delta (\psi_{a} \quad \kappa_{\dot {a}})^{T}= \frac{\omega^{\mu \nu}}{2}\begin{pmatrix} \sigma_{\mu \nu} & 0 \\ 0 & \tilde {\sigma }_{\mu \nu} \end{pmatrix}\Psi . \qquad (.3) $$ Then you can introduce gamma-matrices and their combination:

$$ \gamma_{\mu} = \begin{pmatrix} 0 & \sigma_{\mu} \\ \tilde {\sigma}_{\mu} & 0 \end{pmatrix}, \quad S_{\mu \nu} = \frac{1}{4}\left( \gamma_{\mu}\gamma_{\nu} - \gamma_{\nu }\gamma_{\mu} \right) = \begin{pmatrix} \sigma_{\mu \nu} & 0 \\ 0 & \tilde {\sigma }_{\mu \nu} \end{pmatrix}, $$ and so, by using $(.3)$, you can straightforward get the wanted result $$ \delta \Psi = \frac{\omega^{\mu \nu}}{2}S_{\mu \nu}\Psi . $$

Of course, this matrix and transformation law is also applicable for the transformations of $\gamma$-matrices.

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