Comments on Soul Physics: Why we need superselection rules

Good. I'll give it a more careful work-through...

2010-04-05T07:10:11.566-04:00

Good. I'll give it a more careful work-through when I've got a moment. Many thanks for the helpful comments!

I don't know whether that works or not. It wou...

2010-04-04T22:27:11.738-04:00

I don't know whether that works or not. It would seem to depend on whether it makes sense to take a direct product (or sum, I have a hard time keeping them straight) of different bases in the same space. If it does, then great, I think you've got a winner there, though I think it sounds a little more detailed than what you'd want for a lay audience. If you're shooting for "someone who remembers their undergrad quantum reasonably well," it should be Ok.

BG

Sorry, that should read "the eigenstates of A...

2010-04-04T17:57:31.141-04:00

Sorry, that should read "the eigenstates of A are superpositions.

Thinking about this more, perhaps this would work?...

2010-04-04T17:55:47.477-04:00

Thinking about this more, perhaps this would work? Consider again the x and y spin observables Sx and Sy, on a 2-dimensional Hilbert space with basis set {ψ,φ}. Since [Sx, Sy] = i, we normally say you can't measure spin in both directions at the same time. But now construct a new *4-dimensional* Hilbert space, using the basis set {(ψ,ψ),(ψ,φ),(φ,ψ),(φ,φ)}, with each basis-element representing the measurement of x-spin and y-spin at the same time. Then the self-adjoint operator A having these four basis-elements as eigenstates would be an "observable," which measures both x-spin and y-spin at the same time!

Presumably, A isn't really an observable (it seems it can't be, if our interpretation of [Sx, Sy]=i is correct). So maybe here a superselection rule is appropriate. The 4D Hilbert space is a direct sum of two copies of the 2D Hilbert space. And the eigenstates of A superpositions of pure states from distinct sectors. But this isn't allowed according to Wigner's rule. So, A isn't an admissible observable, because its "eigenstates" aren't admissible pure states.

Hm, maybe not so much. It can probably be argued t...

2010-04-04T17:00:19.096-04:00

Hm, maybe not so much. It can probably be argued that the vacuum state is a superposition of particle/anti-particle pairs (see the Dirac sea).

Tough call.

BG

If you want to be concrete about it, just use the ...

2010-04-04T16:53:27.765-04:00

If you want to be concrete about it, just use the eigenstates of the charge operator. While we have observed superpositions of particle and anti-particle (K' mesons, the K_long and K_short are superpositions), we have not observed superpositions of, say, electron and anti-electron, or K+ and K- for that matter.

BG

Blackgriffin: You're right, this example may j...

2010-04-03T06:54:14.087-04:00

Blackgriffin: You're right, this example may just be too simple-minded. Obviously, both A and B can be observables in this situation (as in the example of spin, as you mention), they just don't commute. I was just trying to come up with some dirt-simple situation in which someone might try to rule out entanglements between states in distinct sectors. This example obviously isn't a *plausible* rule, so I suppose I should keep looking...

Superselection and supersymmetry are closely tied together (on some accounts, you have one iff you have the other). So at least you're consistent. :)

Ooops, I just reread the article. You're not n...

2010-04-02T23:09:39.479-04:00

Ooops, I just reread the article. You're not necessarily talking about spin... My bad.

Still not a fan of superselection or supersymmetry, for that matter.

Wait, are you arguing here that the spin along the...

2010-04-02T19:35:32.668-04:00

Wait, are you arguing here that the spin along the x-axis is not a physical observable? Or are you arguing that there is a privileged basis in QM in which we must consider operators, and that the other bases are somehow deficient? That would certainly sound like it's problematic for Lorentz symmetry...

It's certainly true that since Lx and Lz don't commute it's not possible to measure both simultaneously, but they are both operators with observable eigenstates. You can even measure them sequentially (though this is most often done with photon polarizations) and observe how it's possible to measure both and how the measurement process alters the state of the measured photon.

I haven't looked closely at superselection rules, myself, but the way you present them here makes them sound like bunk to me.

What's more, to claim that it is not possible to measure the eigenstates of all possible Hermitian operators seems to be getting a little ahead of yourself. The measurement process isn't even well defined quantum mechanically, yet, so I don't see how you can make such a categorical statement without effectively adding it to the definition of what a measurement is.