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I know about Noether's theorem, but I don't actually know how to use it myself. Suppose our universe were symmetric with respect to reflections about planes. What conserved quantity would then exist by Noether's theorem?

Qmechanic
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Alex Grounds
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1 Answers1

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Noether's theorem states that there exists a conservation law for every continuous (in fact, differentiable) symmetry. Reflection is a discrete symmetry, so the theorem is not applicable here.

But, in quantum mechanics, you have the parity operator $P$, that reflects the coordinates $$P\psi(\vec{r}) = \psi(-\vec{r})$$ Since $P^2 = I$, the operator $P$ has eigenvalues $\pm 1$. If the hamiltonian is invariant under reflection, i.e. $[H,P]=0$, every eigenfunction of the hamiltonian is also a eigenfunction of the parity operator with eigenvalue $p$ (called intrinsic parity). Then, the intrinsic parity is conserved.

Bosoneando
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  • If we had reflectional symmetry, then we would have $O(3)$ symmetry, which is a Lie group. – Alex Grounds Jun 20 '15 at 22:03
  • No, reflections are a discrete subgroup of $O(3)$, namely $O(3) = SO(3) \times {I, -I}$ is the direct product of the group of rotations (which is the Lie group) and the reflection group. In fact, is quite easy to find examples with reflection symmetry but not the whole $O(3)$ symmetry: a cube or an octahedron with the origin at their centers, for example. – Bosoneando Jun 20 '15 at 22:47
  • I meant that our universe is already symmetric under $SO(3)$, so adding in rotational symmetry yields $O(3)$ symmetry. – Alex Grounds Jun 20 '15 at 22:56
  • Right, but the conserved quantities via Noether's theorem for $SO(3)$ and $O(3)$ are the same (angular moment). Adding the reflections doesn't add any new conserved quantity because they aren't a Lie group but a discrete group. – Bosoneando Jun 20 '15 at 23:06
  • That seems weird to me...does the conserved quantity only depend on the connected component of the identity or something? Why do different Lie groups give rise to the same conserved quantity? – Alex Grounds Jun 20 '15 at 23:08
  • Yes, the derivation of Noether's theorem starts considering infinitesimal transformations $x \to x + \delta x$, which means that only the connected component of the identity really matters. See, for example, https://en.wikipedia.org/wiki/Noether%27s_theorem#Field-theoretic_derivation – Bosoneando Jun 20 '15 at 23:20