I've been studying the Dirac equation, and one thing that I can't get my head around is that the Lorentz transformations in the Hilbert space of Dirac spinors are NOT unitary. Namely, the matrices corresponding to Lorentz boosts are not unitary. In David Tong's QFT lecture notes (http://www.damtp.cam.ac.uk/user/tong/qft/four.pdf), he says that there are no finite dimensional unitary representations of the Poincaré group. Therefore, the 4-dimensional bispinor representation can't be expected to be unitary.

How is this not a problem though? A good part of what I've read on introductory QFT was convincing me that all quantum operators (acting on Hilbert spaces of quantum states) corresponding to Lorentz transformation should be unitary. After all, if a Lorentz transformation takes the states $\left|\alpha\right>$ into $\left|\alpha'\right>$ and $\left|\beta\right>$ into $\left|\beta'\right>$, relativistic invariance requires that $$ \left|\left< \alpha | \beta \right>\right|^2 = \left|\left< \alpha' | \beta' \right>\right|^2 $$ so that both observers agree on the probabilities (after all, probabilities are experimentally measurable and two inertial frames should agree on them). The above equation can only be satisfied if the primed and unprimed states are connected by a unitary operator $$ \left|\alpha'\right> = U \left|\alpha\right> \\ \left|\beta'\right> = U \left|\beta\right> $$ with $U^\dagger = U^{-1}$.

Aren't Dirac spinors supposed to represent states in a Hilbert space? So why doesn't the above argument hold for them?

  • 2
    $\begingroup$ This is the point of quantisation, the finite-dimensional Poincaré algebra reps are necessarily not unitary. However, after we "quantize" the field, $\psi(x)$ becomes an operator and it is on this infinite-dimensional Hilbert space that the Poincaré algebra can be represented unitarily. $\endgroup$
    – Charlie
    Jun 20, 2021 at 23:11
  • $\begingroup$ Note also that in non-relativistic quantum mechanics the $\mathfrak su(2)$ representation used to attach spin to particles does not admit a unitary representation of the Poincaré algebra but the states $|\uparrow\rangle$ and $|\downarrow\rangle$ etc. are still considered "quantum states". $\endgroup$
    – Charlie
    Jun 20, 2021 at 23:12
  • $\begingroup$ („Classical”) Dirac fields (in the Lagrangian or Hamiltonian formulation) become observables through quantization, not states. $\endgroup$
    – DanielC
    Jun 21, 2021 at 0:09

1 Answer 1


Dirac spinor's non-unitary representation of the Poincaré group leads me to conclude that Dirac spinors are not "quantum states" in the usual sense

Good call.

Aren't Dirac spinors supposed to represent states in a Hilbert space?


Long before I was born, there was a brief moment in history when people tried to interpret Dirac spinors that way, but it didn't work. Didn't work then, and still doesn't work today.

For the past several decades, our understanding of most of physics has been based on quantum field theory (QFT). In QFT, the components of a Dirac spinor are operators that act on a Hilbert space. The Hilbert space on which they act does carry a unitary (but not finite-dimensional) representation of the Poincaré group — actually a projective representation, because spinors are involved.

The relationship between Dirac matrices and the (projective) unitary representation is this: if $U$ is a unitary operator that implements a Poincaré transform $|\Psi\rangle\to U|\Psi\rangle$, then the effect of the same Poincaré transform on operators is $O\to UOU^{-1}$. If the operator $O$ is a component of a Dirac spinor operator, $O=\psi_a$, then $U\psi_a U^{-1}=\sum_b R_{ab}\psi_b$ for some matrix $R_{ab}$. The matrix $R_{ab}$ can be expressed in terms of Dirac matrices. The matrix $R$ is not unitary, and it doesn't need to be. The operator $U$ is unitary, and that's what matters.

Even though this was all resolved long before I was born, we still use the overloaded notation today: the same symbol $\psi$ is still frequently used either for an element of the Hilbert space or for a Dirac field operator, depending on the context. In this post, I used an upper-case $\Psi$ for an element of the Hilbert space, reserving $\psi$ for an operator like most QFT texts do. There are only so many letters in the alphabet, far too few to represent all of the distinct things we need to represent in physics, so no notation is perfect.

  • $\begingroup$ Is this just the statement that the transformation preserves the Hilbert-space inner product but not the inner product on spinors? $\endgroup$ Aug 7 at 18:23

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