Uncertainty, Symmetry, and Commutators

samalkhaiat

Any body who has taken a course on QFT should be able to show the following
[tex][ i \mathcal{ H } ( x ) , Q ] = \partial_{ \mu } J^{ \mu } =0 ,[/tex]
where
[tex]H = \int d^{ 3 } x \ \mathcal{ H } ( x ) ,[/tex]
is the Hamiltonian,
[tex]Q = \int d^{ 3 } x \ J^{ 0 } ( x ),[/tex]
is the time-independent Noether charge of a symmetry, and [itex]J^{ \mu }[/itex] is the conserved Noether current of that symmetry. Integration followed by exponentiation yield the invariance of the Hamiltonian under the symmetry group in question:
[tex]H = U^{ \dagger } H U ,[/tex]
where
[tex]U = \exp ( - i \alpha Q ) .[/tex]

Sam

samalkhaiat

No, only those quantities that generate CONTINUEOUS (Lie) group. Parity can be conserved (i.e., commute with H) but it is not a Noether charge, i.e., there is NO parity current.

Again, No. Not all classical symmetries stay intact in the quantum theory.

atyy

@Sam, thanks! What happens to Noether's theorem where the classical theory has a symmetry that leads to a conserved quantity, but the corresponding quantum theory doesn't?

friend

Could it be a matter of being an "observable", whatever that means? For example there's a conserved charge for an invariance wrt phase. But I don't think that phase is observable. Still I wonder if there is a commutator between charge and phase that might prove useful in calculations.

atyy

In this article http://arxiv.org/abs/hep-th/9907162 about the chiral anomaly, they give an example in which the classical Noether current is not conserved in the quantum theory. So naively it seems that Noether's theorem fails, but I'm not sure if that is so, and whether there is a generalization.

jfy4

I tried to mention this at the beginning, but the phase is not an operator, it is a number, like [itex]\pi[/itex], like [itex]e^{i\pi}[/itex]. It has a vanishing commutator with everything.

friend

So is it true then, that the uncertainty principle relates only observables that are represented as operators (no scalars) in the non-zero commutator? The question then is how do we know when to use operators in the Lagrangian that we then put into the commutator? Do we first recognize, by other means, that something is an observable and represented as an operator, and then put it into the lagrangian? Or can we recognize observables in the lagrangian, based on first principles such as symmetries, before puting them into the commutator?

jfy4

Observable quantities are represented in quantum mechanics as hermitian operators. For something to be observable, it must be just that, measurable. For instance, the phase of a state vector, [itex]e^{i\theta}|\psi \rangle[/itex] is not an observable (try and devise an experiment where you can measure it). But [itex]|\psi|^2= \langle \psi |e^{-i\theta}e^{i\theta}|\psi \rangle [/itex] is an observable, but now the phase is gone... Does this help?

friend

Yes, I know all that. But what about, say, the color charge of quarks. Is that observable? It seems we will never be able to isolate a quark to measure its properties.

But what I'm really trying to get at is a generalization of what an observable is. Is an observable something we've already measured and then backward incorporate it into the lagrangian? Or are there more fundamental things lurking in the lagrangian that we should be able, at least in principle to make a measurement of?