# What determines which observables are QM?

Spin, position, and velocity are observables which are QM for quantum particles. My question is, what determines whether an observable is QM or not?

For example, why is electric charge not QM? That is, why don't (or can't) particles exist in a superposition of being positive and negative?

What is the underlying mechanism in all of this?

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There are super-selection rules that forbid some superpositions. For example, an electron state cannot be in a superposition with a proton state. – Vladimir Kalitvianski Nov 11 '12 at 10:06

Which superpositions are allowed or forbidden in principle is determined by so-called superselection sectors. http://en.wikipedia.org/wiki/Superselection These are virtually absent from the standard textbook literature as they cannot be inferred from the traditional perturbative treatment of quantum mechanics or quantum field theory.

The superselection rule easiest to understand is the one that forbids the superpositions of a boson state and a fermion state. The reason is that these states transform differently under the action of the rotation group, hence there is no consistent action of the rotation group on the superposition. (Under a 360 degree rotation The boson half would rotate back to itself, while the fermion half changes its sign.)

This shows that superselection rules are tied to inequivalent representation of Lie groups or Lie algebras of quantities defining the physics of a system. The charge superselection rule is associated to inequivalent representations of an infinite-dimensional Heisenberg group defining the canonical commutation relations (CCR) of a relativistic quantum field. [My discussion assumes the bosonic case. In case of fermions, one needs instead the CAR, and s similar reasoning applies.]

Here the reasoning is more intricate, and requires the nonperturbative setting of algebraic quantum field theory. In this setting, observables form a C^*-algebra, and states are suitable positive linear functionals of this algebra.

The analogous standard QM situation is where the C^*-algebra is the algebra of bounded linear operators on a Hilbert space, and a state is the expectation mapping $\langle A\rangle =Tr ~\rho A$ with a positive semidefinite density matrix $\rho$ of trace 1. Here the uniqueness theorem of von Neumann guarantees that the canonical commutation relations have a unique unitary representation up to unitary equivalence.

However, this theorem only holds for CCR of finitely many operators, whereas field theory deals with infinitely many of these. The CCR of field theory have infinitely many inequivalent representations, and these live in different Hilbert spaces, between the elements of which no sensible inner product is defined. As it makes no sense to consider superpositions between states of two different Hilbert spaces (not embedded in a common Hilbert space with a physical meaning), inequivalent representations imply a superselection rule.

This implies the charge superselection rule for QED, as it can be shown that in QED, states of different charge must lie in inequivaent representations.

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Thanks for the detailed response! – ThisIsNotAnId Nov 16 '12 at 5:38

What I think you're really asking is: why can an elementary particle (say an electron) be in a superposition of different momenta, or different spins, but not in a superposition of different charges? The reason is that all electrons have −1 charge. If an electron could have either a +1 or a −1 charge, then quantum mechanics would say that you could put it in a superposition of these states. But the standard model says that all electrons must have certain properties, and among these is −1 charge.

Similar things happen with spin: some elementary particles have spin ½, and others have spin 0 or spin 1, but no elementary particle can ever be in a superposition of ½ and 1 spins, because then it would have to be simultaneously a boson and a fermion. But an electron can have state either ½ spin or −½ spin, so it can be in a superposition of these two states.

Why can't an electron have either $+1$ or $-1$ charge? The properties of the electron come from the standard model, and nobody really understands where that comes from.

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Particles can exist in a superposition of being positive and negative. If you pair produce an electron and positron then as long as they remain entangled they are in a superposition.

I'm far from expert on the fundamentals of QM but as far as I know every observable is quantum mechanical. All observables are given by the expectation value of a hermitian operator, and these operators obey the rules of quantum mechanics.

As for the underlying mechanism: the simple answer is that no-one knows. All we know is that the rules of QM work in the sense that they agree with experiment. The underlying mechanism of QM has been a problem from it's creation, and arguments about it continue to this day. Most of us agree with Feynmann's approach of shut up and calculate.

Response to comment:

Let me expand on my first paragraph since Vladimir objects.

It isn't possible for a single particle to be a superposition of an electron and positron. The reason is simply because if such a particle existed observation would find an electron half the time and a positron half the time, and the total charge of the universe would decrease or increase accordingly. This would violate conservation of charge.

The situation is different for an electron positron pair. In the centre of mass frame we know the two particles move in opposite directions, but as long as they remain entangled we don't know which particle went which way. The pair of particles is in a superposed state of $e_{left} + \bar{e}_{right}$ and $e_{right} + \bar{e}_{left}$. This doesn't violate conservation of charge because observation will always result in one electron and one positron and the net charge is always zero.

But this probably isn't what you meant by a superposition of charge, so I guess I have to concede Vladimir's downvote is justified.

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Whoever downvoted me please say why. I won't be able to improve my answers if you don't tell me why you found them unhelpful. – John Rennie Nov 11 '12 at 10:20
Electron is never in a superposition with a positron. – Vladimir Kalitvianski Nov 11 '12 at 10:26
@Vladimir: Electrical charges can indeed be in superposition. How can you explain quasiparticles with the observed electric charges of $1/3$ in the fractional quantum hall effect without putting electrical charges in superposition? – Peter Shor Nov 11 '12 at 12:32
Interaction of electron with positron does not make their superposition, but a compound system. In a compound system can exist quasi-particles (collective motion modes) whose properties are expressed/calculated via particle properties. – Vladimir Kalitvianski Nov 11 '12 at 15:13
The electron-positron pair does not violate the charge superselection rule as its total charge is zero. It is a [superopistion of] product states rather than a superposition of a state of an electron state and a positron state. – Arnold Neumaier Nov 12 '12 at 17:26