2
$\begingroup$

I've read a paragraph in Schwartz-QFT where he argues that unitarity and Lorentz invariance are incompatible due to the norms being different:

Schwartz.png

Why does he assume that the boost in this basis is $(\cosh\beta,\sinh\beta)$ though? Can't you have a normal $(\cos\beta,\sin\beta)$ in this Hilbert space? I mean for some other definition of an inner product (i.e. not $\langle\psi|\psi\rangle)$ it may be $(\cosh\beta,\sinh\beta)$ again, but it doesn't have to be the same for $\langle\psi|\psi\rangle$?

$\endgroup$

1 Answer 1

3
$\begingroup$

You are correct that Schwartz shows here only that the standard 4-vector representation of the Lorentz group cannot be made unitary.

The general statement is that a non-trivial unitary representation of the Lorentz group cannot be on any finite-dimensional Hilbert space, essentially because the Lorentz group is not compact - see this answer by Valter Moretti for a proof of the precise mathematical claim.

$\endgroup$
4
  • $\begingroup$ Thanks! It cannot be finite? I saw an expression $\exp(-i(J\cdot\theta-K\cdot\phi))$ for the Lorentz group where $J,K$ are generators for rotations and boosts. It looks finite? Is that not a Hilbert space? Or why is this expression not a finite-dimensional unitary representation of the Lorentz group? $\endgroup$
    – Gere
    Commented Sep 12, 2022 at 10:06
  • 1
    $\begingroup$ @Gerenuk: If $J$ and $K$ are finite-dimensional matrices, then you'll find that one of them (usually the boosts) is not Hermitian, so the exponential is not a unitary operator. See e.g. physics.stackexchange.com/a/669881/50583. $\endgroup$
    – ACuriousMind
    Commented Sep 12, 2022 at 10:07
  • $\begingroup$ Not sure how to ask this: Is it meaningful if $\exp()$ is a representation for some definition of inner product (related to relativistic invariants), but the unitarity required for the probabilities is shown for a modified form of the inner product. Hence $J$ and $K$ are unitary, but just with another form of the inner product (an extra term which flips the sign)? I don't see a reason why $\exp()$ and the probabilities $\langle \psi|\psi\rangle$ should use the same inner product. $\endgroup$
    – Gere
    Commented Sep 12, 2022 at 10:17
  • $\begingroup$ @Gerenuk 1. the reason you want the representation to be unitary is that we don't want symmetries to change probabilities/normalizations. If you say "this is unitary but not with the product we use for the probabilities" then what's the point of wanting unitarity in the first place? 2. The statement that finite-dimensional representations of the Lorentz group are never unitary is completely independent of any physical idea of "probabilities", i.e. there are no inner products that make any finite-dimensional rep unitary. $\endgroup$
    – ACuriousMind
    Commented Sep 12, 2022 at 11:29

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.