In Weinberg's QFT textbook (volume 1), he proved the symmetry representation theorem in Appendix A, chapter 2, which states

Any symmetry transformation can be represented on the Hilbert space of physical states by an operator that is linear and unitary or antilinear and antiunitary.

But in section 5.4 (on page 218), he showed that the representation of the homogeneous Lorentz group is not unitary.

Do these two statements contradict with each other? Is the catch here the fact that the representation of the Lorentz group don't really act on Hilbert space?

  • $\begingroup$ I don't see why this should be a contradiction.... He said that every symmetry transformation CAN be represented by an (anti) unitary operator, but this does no mean that every representation has to be (anti) unitary... $\endgroup$ Commented Oct 27, 2021 at 21:16
  • $\begingroup$ @G.Blaickner not sure if I understood this correctly, but I thought in the context of section 5.4 the representation of the Lorentz group would be a symmetry transformation on the Dirac spinor? $\endgroup$
    – haha
    Commented Oct 27, 2021 at 22:30
  • 2
    $\begingroup$ Related? physics.stackexchange.com/q/99051/36793 and physics.stackexchange.com/q/293539/36793 $\endgroup$
    – SRS
    Commented Oct 28, 2021 at 8:38

1 Answer 1


The point is that the notion of symmetry is not the same in the 2 situations:

  1. For Wigner's theorem, a symmetry transformation refers to a property of a transformation on a ray space.

  2. But a symmetry could also refer to invariance of an object (typically an action functional) under a group action. In particular, if a group $G$ acts on a vector space $V$, the corresponding group representation need not be unitary.

TL;DR: The no-go theorem (that Weinberg implicitly alludes to) does not affect Wigner's theorem in any way.

  • $\begingroup$ I'm trying to think of an example in nonrelativistic quantum mechanics. If we consider acting a rotation on a wave function (element in a Hilbert space) $U(R) \psi(x)$ and the same rotation on a ket (an abstract object) $U(R)|\psi>$, wouldn't the symmetry operator $U(R)$ be the same for both? $\endgroup$
    – haha
    Commented Nov 2, 2021 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.