Discrete symmetries and conserved quantities

[dRic2]

TL;DR

Discrete symmetries

Hi, please correct me if I use a wrong jargon.

If I have discrete symmetries (like for example in a crystal lattice) can I find some conserved quantity ? For example crystal momentum is conserved up to a multiple of the reciprocal lattice constant and it is linked (I think) to the periodicity of the lattice. Can I have something similar for rotations for example ? Do you have any reference at the level of advanced undergrad/first-year grad student ?Thanks
Ric

 

[fresh_42]

There is no discrete version of Noether's theorem. At her times the jargon was even "theory of variations, continuous groups" which already had continuity in its names. Moreover conservation quantities are usually conserved under some kind of (continuous) transformations, not under "hopping around".

Now assumed the answer to your question would by "yes". By which means shall we distinguish between an accidental coincidence and a fundamental connection? And how would it be different from a simple fixed point? There is of course a close connection between crystals and their symmetry groups, but this is only classical geometry, not physics.

 

[vanhees71]

In some sense there is a conserved quantity associated with discrete symmetries. Take, e.g., parity, defined by the unitary transformation $\hat{P}$ that obeys
$$\hat{P} \hat{\vec{x}} \hat{P}^{\dagger}=-\hat{\vec{x}}, \quad \hat{P} \hat{\vec{p}} \hat{P}^{\dagger}=-\hat{\vec{p}}, \quad \hat{P} \hat{\vec{\sigma}} \hat{P}^{\dagger},\hat{\vec{\sigma}}$$
for the "fundamental observable operators", position, momentum, and spin of a non-relativistic particle. Since further $\hat{P}^{\dagger}=\hat{P}^{-1}=\hat{P}$, the system is symmetric under this space-reflection transformation, if
$$\hat{P} \hat{H} \hat{P}^{\dagger}=\hat{P} \hat{H} \hat{P}=\hat{H} \; \Rightarrow [\hat{H},\hat{P}]=0.$$
This implies that a wave function that is initially an eigenstate of the parity operator (with eigenvalues +1 or -1, i.e., and even or odd function $\psi_0(-\vec{x},\sigma)=\pm \psi_0(\vec{x},\sigma)$) stays an eigenstate of the parity operator with the same eigenvalue, because
$$\psi(t,\vec{x},\sigma)=\exp(-\mathrm{i} \hat{H} t/\hbar) \psi_0(\vec{x},\sigma)$$
and thus if $\psi_0$ is an eigenstate of parity, due to $[\hat{H},\hat{P}]=0$, also $\psi(t,\vec{x},\sigma)$ is a parity eigenstate with the same eigenvalue. In this sense parity is conserved, and transitions between states between different parity don't occur.

 

[dRic2]

Sorry for the late reply. Going back to the example of a crystal, I have one last question. If I have a periodic potential, I find that "crystal momentum" (ℏk) is almost a "good" quantum number: it is not enough to label a state, but it is conserved to within a lattice vector. It is not conserved in the real sense, because we can't write a conservation law for this quantity, but it has similarities with the classical momentum that we know. I find interesting that if we shrink the lattice spacing to zero (we take le limit a→0), since k is "conserved" in the interval [−π/a,π/a] it seems to me that we recover the continuous symmetry (and Noether's theorem). Is this just a coincidence, or is there a deeper connection between this limiting procedure and Noether's theorem and discrete symmetries ? Or is it me who wants to see things where there is nothing to see ?

 

[vanhees71]

If you have a periodic potential, the continuous translation invariance of free particles is broken to a corresponding discrete translation invariance of the lattice. If you make the lattice continuous again by taking $a \rightarrow 0$ you restore the full continuous translation invariance.

 

[dRic2]

Thanks for the replies.