# Transformation of spinors due to Lorentz group

Assume we have a Dirac spinor $\psi(x)$ which satisfies the Dirac equation:

$$(i\gamma^{\mu}\partial_{\mu} - m)\psi(x) = 0.$$

If we boost our spacetime coordinates to a new system with a Lorentz transformation we get

$$x^{\mu} \longrightarrow x'^{\mu} = \Lambda^{\mu}_{\,\,\nu}x^{\nu},$$ $$\psi(x) \longrightarrow \psi'(x') = S(\Lambda) \psi(x) \equiv S \psi(x)$$ where $S$ is a matrix.

If we want $\psi'(x')$ to satisfy the Dirac equation we need to impose some conditions on the way the spinor transforms, i.e. on the matrix $S$:

$$(i\gamma^{\mu}\partial'_{\mu} - m)\psi'(x') = 0 \Leftrightarrow(i\gamma^{\mu}\partial_{\rho}(\Lambda^{-1})^{\rho}_{\,\,\mu} - m)S\psi(x) = 0,$$ If the following holds true: $$S^{-1} \gamma^{\mu} (\Lambda^{-1})^\rho_{\,\,\mu} S = \gamma^{\rho}.$$

The text im trying to follow concludes that thus:

$$S^{-1} \gamma^{\mu} S = \Lambda^{\mu}_{\,\,\rho} \gamma^{\rho}.$$

But I can't seem to get the $\Lambda$ matrix onto the right side, since I see no reason that it should commute with $S$. Where'd I go wrong?

Start from your expression: $$\tag{1} S^{-1} \gamma^\mu \Lambda_\mu^{\,\,\rho} S = S^{-1} \gamma^\mu (\Lambda^{-1})_{\,\,\mu}^{\rho} S = \gamma^\rho.$$ Expliciting the spinor indices and using the orthogonality of $\Lambda$ it reads $$\tag{2} S^{-1}_{\alpha\beta} \gamma^\mu_{\beta \gamma} \Lambda^{\,\,\rho}_{\mu} S_{\gamma \delta} = \gamma^\rho_{\alpha \delta},$$ where it is important to note that this are all numbers, hence every object commutes with every other. Then write it as $$\tag{3} S^{-1}_{\alpha\beta} \gamma^\mu_{\beta \gamma} S_{\gamma \delta} \Lambda^{\,\,\rho}_{\mu} = \gamma^\rho_{\alpha \delta},$$ now multiplying by $\Lambda^\mu_{\,\,\rho}$ and using the orthogonality relation $$\Lambda^\mu_{\,\,\nu} \Lambda_\mu^{\,\,\sigma} = \delta_\nu^{\,\,\sigma},$$ you have your result.
Lambda matrices have Lorentz indices, on the other hand $S$ has spinor indices. These matrices belong to the whole different kinds of linear operators, acting on different spaces.
In your situation, $\gamma^{\mu}\Lambda^{-1}$ is just a shorthand for $\gamma^{\nu} {\Lambda^{-1}}_{\nu}^{\mu}$.
Sure they commute. If you expand all Greek and spinor (Latine) indices, it will be obvious (note that gamma matrices have actually three indices: $(\gamma^{\mu})^{a}_b$).