I'm a bit confused about Lorentz group representations. I see that Lorentz group is non compact and therefore there is no faithful irreducibile unitary finite dimensional representation. In fact I can see that rotation group generators are antihermitian, while boost generators are hermitian and because of this the representation is not unitary. The same thing is in Dirac representation.

When I introduce Poincaré group, I see that translation operator acts upon $C_{\infty}$ functions, so I also need generators for boosts and rotations that may give a infinite representation of lorentz group; I therefore use generic angular momentum operator.

My question is, why don't I have infinite dimensional representation also for spinors? Is it because spinorial representation of Lorentz group act on spin degrees of freedom only?

Moreover, in "An Introduction to Quantum Field Theory" by Peskin and Schroeder at page 41 it says: "In fact the Lorentz group, being non compact, has no faithful, finite-dimensional representations that are unitary. But that does not matter to us, since $\psi$ is not a wavefunction; it is a classical field." What does this mean?! I'm confused!

  • $\begingroup$ Why would you say that there aren't infinite-dimensional unitary spinor representations? Dirac field $\psi(x)$ is just one example of such a representation. $\endgroup$ Commented Jul 11, 2017 at 11:45
  • 5
    $\begingroup$ I think these answers of mine should answer your question: physics.stackexchange.com/a/174908/50583, physics.stackexchange.com/a/322837/50583, physics.stackexchange.com/a/286085/50583 $\endgroup$
    – ACuriousMind
    Commented Jul 11, 2017 at 12:05
  • $\begingroup$ @SolenodonParadoxus Could you explain to me why the spinor is a representation? $\endgroup$ Commented Jul 11, 2017 at 13:31
  • $\begingroup$ @user129511 Just the $\psi(x)$ field with the usual transformation properties... Details are given in Frederic Thomas's answer. $\endgroup$ Commented Jul 12, 2017 at 3:27
  • $\begingroup$ "But that does not matter to us, since 𝜓 is not a wavefunction; it is a classical field." This statement is false, I think 𝜓 is a wavefunction in the Hilbert space; if you write EOM of 𝜓 as $$ i \partial_t \psi = H \psi$$ and $H =- i \gamma^0 \gamma^j \partial_j + m \gamma^0.$ $\endgroup$ Commented Apr 27, 2021 at 23:23

1 Answer 1


Dirac-Spinors in relativistic theory are finite-dimensional representations of the Lorentz-group with peculiarity of being non-unitary.

$\psi^\alpha = \Lambda^\alpha_\beta \psi^\beta$

with finite-dimensional $\Lambda \approx 1+\frac{1}{2}\omega_{\mu \nu}M^{\mu\nu}$ where $M^{\mu\nu}$ is a matrix belonging to the generators of the Lorentz group and $\omega_{\mu \nu}$ an antisymmetric matrix containing the 3-dim. angle $\bf{\alpha}$ parameter and the 3-dim.velocity $\bf{v}/c$ parameter.

In non-relativistic theory you need the modulus of the wavefunction $|\psi'|^2= (U\psi)^{\dagger} U\psi = \psi^{\dagger}\psi= |\psi|^2$ being invariant, indeed that can only be achieved with unitary representations, for instance with those of the $SO(3)$-group. However, the corresponding modulus $|\psi|^2=\psi^+ \psi= \bar{\psi}\gamma^0\psi$ in the relativistic theory actually transforms like a 0-component of a 4-vector, so it is not invariant under Lorentz-transformations. So non-unitarity of the Lorentz-transformations is not a problem.

However, if you consider the Poincare'-group, one obtains also infinite-dimensional representations with spinors which transform like this:

$\psi'^{\alpha}(x')= U(\Lambda)\psi^\alpha(x) U(\Lambda^{-1})=D(\Lambda^{-1})^\alpha_\beta\psi^\beta(\Lambda x)$

And they are infinite-dimensional and unitary.

For relativistic Weyl-spinors it is rather similar, the formal expressions in this answer are the same apart from the detail that the size of the matrices and their matrix elements then used has to be adapted. Above all a tentative expression like $|\psi|^2\sim \psi^{\dagger \dot{A}}\psi_A$ (if such an expression made any sense I am not sure of) formed by Weyl-spinors neither does not need to be invariant, therefore unitarity of the finite Lorentz transformation representation is not required.

For more details the literature has to be consulted. A choice in this respect is for instance the book of Srednicki.


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