5

As stated in the question, and I have looked at other questions on this topic here, I am still confused whether the equation $$ U(\Lambda,a)\hat\phi(x) U^{-1}(\Lambda,a) = R(\Lambda,a) \hat\phi(\Lambda^{-1}x + a) $$ is something that can be derived from the theory of Poincare representations, or whether it is an axiom of Quantum Field Theory?

So $\hat\phi(x)$ denotes a quantum field operator, $U(\Lambda,a)$ an infinite dimensional unitary irrep of Poincare on the Hilbert space of particle states, and $R(\Lambda,a)$ a finite-dim irrep of Poincare acting on the (components of) corresponding classical fields $\phi(x)$, eg spinors, vectors etc.

Frido
  • 151
  • 3
    If you assume your theory to be relativistic in flat space, then it must carry a representation of the Poincare group. This means that there is an action of the group on the objects of the theory, that mimics the action of the group on spacetime events. This is what the quoted equation is saying. – QuantumFieldMedalist Jun 13 '23 at 08:44
  • 3
    Wikipedia. – Tobias Fünke Jun 13 '23 at 08:58
  • 1
    I think it just defines the transformation property of $\phi$ under the Poincare group. A nontrivial point is that what kind of $R$s allowed depends on the underlying group. – Keyflux Jun 13 '23 at 08:59
  • @QuantumFieldMedalist Right, so as you say you'd like the quantum field to mimic the spacetime behaviour of its classical counterpart (which is encapsulated by the RHS of the equality), but there is no way to actually derive the RHS from the LHS and vice versa is there? – Frido Jun 13 '23 at 09:01
  • @TobiasFünke Just saw your comment, thank you. That I think clarifies it for me for the moment at least. So this is basically one of the Wightman axioms. – Frido Jun 13 '23 at 09:01
  • Right. This relation trivially holds as a consequence of assuming there is a representation of the Poincare group, which is why it is interpreted as an axiom. – QuantumFieldMedalist Jun 13 '23 at 11:51
  • 1
    Related answers of mine: https://physics.stackexchange.com/a/525930/50583, https://physics.stackexchange.com/a/174908/50583 – ACuriousMind Jun 13 '23 at 15:53
  • @ACuriousMind I am taking your answer in the second link provided as the truth at the moment, which as I understand it says that it is an axiom and that because of this axiom the mass and spin of the infinite dim unitary irrep is determined by the mass and spin of the finite dim irrrep. – Frido Jun 15 '23 at 17:49
  • 1
    I thought my question was pretty straightforward: axiom or theorem (can be proved/derived)... – Frido Jun 15 '23 at 17:57

1 Answers1

1

Most text books discuss this "Wightman axiom" in the context of scalar $\hat\phi(x)$.

However, when it come to the spinor $\hat\psi(x)$ as a quantum field operator, the existence of the infinite-dimensional unitary $U(\Lambda,a)$ on the LHS of $$ U(\Lambda,a)\hat\psi(x) U^{-1}(\Lambda,a) = R(\Lambda,a) \hat\psi(\Lambda^{-1}x + a) $$ is highly questionable.

So far I don't see a single concrete example in any text book or forum which shows that the alleged infinite-dimensional unitary $U(\Lambda,a)$ actually exists for the spinor $\hat\psi(x)$.

See more details here.

MadMax
  • 3,737
  • I have to admit I'm back to being confused. So you are saying that for scalar field there is a constructive proof (in which case I think the term axiom should be replaced by theorem) whilst for spinor fields the axiom really is still an axiom as there is no proof yet of a realization of a unitary infinite dim irrep U? – Frido Jun 15 '23 at 17:15
  • 1
    The (Wigner) classification of irreducible unitary representations of the Poincaré group is well-known, see e.g. Weinberg's QFT book for constructions via the method of little groups ("theory of induced representations" for mathematicians). This answer is simply wrong. There are Wightman axioms that are difficult to show rigorously, this is not one of them. – ACuriousMind Jun 15 '23 at 17:19
  • 1
    Since I have A Curious Mind, I would like to see a single concrete example, with the details of the unitary $U$ worked out for a given non-unitary Lorentz boost $R$ as an example. – MadMax Jun 15 '23 at 18:47