Time-reversal symmetry is an anti-unitary operator.

I understand the mathematical definition of this, but what are the implications? What should/would one expect from anti-unitary operators?
Are they interesting for some fundamental reason?


The effect of anti-unitary operator $A$ on the length of a Hilbert state is the same as the effect of an unitary operator $U$:

$$ <Ax, Ay> = \overline{<x,y>} $$ therefore $$|<Ax, Ay> |= |<x,y>|$$

as well as

$$ <Ux, Uy> = <x,y> $$ therefore $$|<Ux, Uy> |= |<x,y>|$$

This is one of the main assertions of Wigner's theorem.

Compressed the effect of an anti-unitary transformation could be written $A i A^{-1}=-i$.

So the class of transformations which keep Hilbert state's length invariant is much larger than if only unitary transformations were considered. But apart from this there are on equal footing with the unitary operators.

As time-reversal is the most well-known and important anti-unitary transformation, its purpose is to check by its application if a theory is T-invariant. Most should indeed be T-invariant, only CP-violation should also lead to T-invariance violation (Only a very few theories actually do that) as CPT is always valid for a Lorentz-invariant Quantum field theory. Even that (CPT) is worth while to be checked for new theories (if they are in agreement with the experiment) in order to see if QFT "axioms" are always fulfilled.

For instance applied on a Dirac-operator $\hat{\psi}$ it yields: $$T\hat{\psi}(t,x) T = \gamma^1 \gamma^3 \hat{\psi}(-t,x)$$.

With this it can for instance be studied the effect of the T-operator on V-A interaction.

  • $\begingroup$ Ok but then what’s the difference between unitary and anti-unitary in terms of physics? $\endgroup$ – SuperCiocia Aug 30 '18 at 21:46
  • $\begingroup$ is there a differene in trrms of physics between unitary amd anti unitary? $\endgroup$ – SuperCiocia Oct 6 '18 at 15:26
  • $\begingroup$ I guess, every transformation which is anti-unitary contains a time-reversal (among other possible non-T-reversal stuff). $\endgroup$ – Frederic Thomas Oct 6 '18 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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