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?
Answer
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.
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