1
$\begingroup$

Is there a rigorous definition of the parity operator?

I see parity come up in the context of angular momentum, magnetic fields, quantum spin/particles. It is also related to the Levi-Civita symbol vs tensor discussion. Likewise it's related to tensor densities. It's related to cross products and bivectors.

It seems like you "just have to know" whether a certain quantity has been deemed to be a psuedo- or non-pseudo tensor, and furthermore, it seems like the rules for the parity of different types of physical objects are assigned in a way which is somewhat arbitrary. It's not entirely arbitrary, the parity rules do have some sort of self-consistency. But, it seems like the only thing we gain from the rules for parity is self-consistency of the parity of different objects. It doesn't seem like parity adds any intuition otherwise and the entire concept could be dispensed with.

That said, I might change my mind about this if someone could present me with a rigorous mathematical definition of the parity operator (perhaps in the context of differentiable manifolds and coordinates) that is general and can help clarify its utility.

I'm seeking a mathematically motivated answer rather than stuff like "what happens when you look in a mirror"

$\endgroup$

1 Answer 1

2
$\begingroup$

Parity is a discrete spacetime symmetry under which the spatial coordinates $x^i$ transform as $x^i \rightarrow -x^i$ and the time coordinate transforms as $t \rightarrow t$.

Various fields also transform under parity. Ultimately the origin of these transformations is that, in a theory with parity invariance, the Lagrangian should be invariant under parity. For example, if a spin-0 field $\phi$ couples to a fermion $\psi$ via the coupling $\phi \bar\psi \psi$, then $\phi$ must be a scalar for the Lagrangian to be invariant. On the other hand, if $\phi$ couples via $\phi \bar\psi \gamma^5 \psi$, then $\phi$ must be a pseudoscalar. The transformation laws of the fields are not written on stone tablets, but come from finding transformations that leave the Lagrangian invariant.

One way to frame failure of parity to be a symmetry in the Standard Model is that there is no consistent transformation of the fields such that inverting the spatial coordinates leaves the Lagrangian invariant.

$\endgroup$
1

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.