Derivation of generators of Lorentz group for spinor representation I want to prove
$$S^{\mu \nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu].$$
I started from 
$$[\gamma^\mu,S^{\alpha\beta}]=(J^{\alpha\beta})^\mu_\nu \gamma^\nu$$
Putting the value of $(J^{\alpha\beta})^\mu_\nu$
$$=i(\eta^{\alpha\mu}\delta^\beta_\nu-\eta^{\beta\mu}\delta^\alpha_\nu)\gamma^\nu$$
we get
$$\gamma^\mu S^{\alpha\beta}-S^{\alpha\beta}\gamma^\mu=i(\eta^{\alpha\mu}\gamma^\beta-\eta^{\beta\mu}\gamma^\alpha)$$
Whats the next step?
Also tell me if there is any other decent method. Note I am using metric $Diag(1.-1).$
 A: We do not have to guess the structure of $S^{\mu\nu}$. You are really close, just replace
$$2\eta^{\mu\nu}= \{\gamma^\mu,\gamma^\nu\}.$$
Rearrange the gammas on left side you will automatically make structure like your desired one by comparison on both sides. 
$$\gamma^\mu S^{\alpha\beta}-S^{\alpha\beta}\gamma^\mu=i(\eta^{\alpha\mu}\gamma^\beta-\eta^{\beta\mu}\gamma^\alpha)$$
$$\gamma^\mu S^{\alpha\beta}-S^{\alpha\beta}\gamma^\mu=\frac{i}{2}( \{\gamma^\alpha,\gamma^\mu\}\gamma^\beta- \{\gamma^\beta,\gamma^\mu\}\gamma^\alpha)$$
There might be difference of some constant factor. Fix it yourself.
A: You'll probably be a little disappointed but it seems that one had to guess the answer:
Existence of a solution to your second equation: Check that your first equation does satisfy the second.
Unicity??: I first thought that it could not be shown, but as usually for linear equations, if you have a particular solution $S^{\alpha\beta}$ of a non-homogeneous equation then the others are obtained by solving the associated homogeneous ones, here
$$[\gamma^\mu, T^{\alpha\beta}]= \gamma^\mu \cdot T^{\alpha\beta}-T^{\alpha\beta}\cdot \gamma^\mu=0$$
This is equivalent to
 $$\gamma^\mu \cdot T^{\alpha\beta} = T^{\alpha\beta}\cdot \gamma^\mu\quad \Longleftrightarrow\quad  T^{\alpha\beta} = (\eta^{\mu\mu})\, \gamma^{\mu}\cdot T^{\alpha\beta}\cdot \gamma^\mu $$
I leave it to others to find possible solutions...

For those who do not know where that comes from: second equation is the linearized form of 
$$S(\Lambda)\cdot \gamma^{\mu}\cdot S^{-1}(\Lambda) = \Lambda^{\mu}{}_{\nu}\, \gamma^{\nu}$$
One wants to find a representation $S(\Lambda)$ of the Lorentz group that satisfies such a relation and the previous $S^{\alpha\beta}$ are the generators in this representation
$$S(\Lambda)= \mathrm{Id} - \frac{i}{2 i \hbar}\, \omega_{\mu\nu}\, S^{\mu\nu} + o(\omega) $$
while the $J^{\alpha\beta}$ are the generators in the defining representation.

There are constructions in books on Clifford algebras where the $S(\Lambda)$ can explcitly be constructed without exhibithing a solution from nowhere. Sketch: Lorentz transfo can always be written as a composition of particular "symmetries" $\Lambda_0$ (of the kind $x$ mapped to $-x$ and identity for orthogonal vectors). For those, one find a simple associated $S(\Lambda_0)$...
