Lorentz transformations for spinors The lorentz transform for spinors is not unitary, that is $S(\Lambda)^{\dagger}\neq S(\Lambda)^{-1}$. I understand that this is because it is impossible to choose a representation of the Clifford Algebra where all the $\gamma$ matrices are Hermitian.
However, does this not go against the conventional wisdom of Wigner's theorem which says that symmetry operations are needed to be either unitary of anti-unitary for the invariance of observable properties across frames? How can one reconcile this with the transformation above?
 A: The most famous theorem by Wigner states that, in a complex Hilbert space $H$, every bijective map sending rays into rays (a ray is a unit vector up to a phase) and preserving the transition probabilities is represented (up to a phase) by a unitary or antiunitary (depending on the initial map if $\dim H>1$) map in $H$.  
Dealing with spinors $\Psi \in \mathbb C^4$, $H= \mathbb C^4$ and there is no Hilbert space product (positive sesquilinear form) such that the transition probabilities are preserved under the action of $S(\Lambda)$, so Wigner theorem does not enter the game. 
Furthermore $S$ deals with a finite dimensional Hilbert space $\mathbb C^4$ and it is possible to prove that in finite-dimensional Hilbert spaces no non-trivial unitary representation exists for a  non-compact connected semisimple Lie group that does not include proper non-trivial closed normal subgroups. The orthochronous proper Lorentz group has this property. An easy argument extends the negative result to its universal covering $SL(2, \mathbb C)$.  
Non-trivial unitary representations of $SL(2,\mathbb C)$  are necessarily infinite dimensional. One of the most elementary case is described by the Hilbert space $L^2(\mathbb R^3, dk)\otimes \mathbb C^4$ where the infinite-dimensional factor $L^2(\mathbb R^3, dk)$ shows up. 
This representation is the building block for constructing other representations and in particular the Fock space of Dirac quantum field.
A: This is a common misconception.
Lorentz group $SO(3,1)$ (or its double-cover $SL(2,\mathbb{C})$ if you want to follow Wigner's analysis of symmetries in QM) is non-compact. This means that it has no finite-dimensional unitary representations (the inability to choose hermitian Clifford basis elements is just a consequence of this).
When you're constructing classical Dirac spinors, you don't need unitarity. Indeed, there's no need for $S(\Lambda)$ to be a unitary representation. We are dealing with a classical field here, and unitarity is not required in classical physics.
In QFT we are dealing with a quantum Dirac field. The state space (fermionic Fock space) of the free QFT is infinite-dimensional and unitary. This doesn't contradict the original claim precisely because of infinite dimensionality.
Classification of the infinite-dimensional unitary irreps of the Poincare group (inhomogenous Lorentz group if you wish) was done by Wigner. It uses finite-dimensional nonunitary representation theory of $SL(2,\mathbb{C})$ (which is equivalent to the finite-dimensional representation theory of the complexified Lie algebra $\mathfrak{so}(4)\sim\mathfrak{D}_2$) heavily.
Summarizing: the difference is that Poincare generators of the QFT act unitary on the infinite-dimensional Fock space. The classical transformation of the spinor field is respected by this action, but does not have to be and is not unitary.
