# Observables as a function of energy

1. Jan 25, 2005

### zitek

The question is:
Prove that, for a 1-D harmonic oscillator, every conserved observable is a function of the energy. Find, for the 3-D harmonic oscillator, some conserved observable not a function of the energy and angular momentum.

My first problem with this question is I'm not sure what a "conserved observable" is. I know that the state of a 1-D harmonic oscillator can be expressed by only one position component and one velocity component, and from that you can get a function of energy, but I don't know how to write a proof for this.

Thanks.

2. Jan 26, 2005

### vincentchan

EASY, in 1D, energy(KE or PE, not total, total energy is constant) is a function of one variable (velocity, position...etc, let's say that variable called q) ONLY, so E=E(q), or q=q(E)
however, in 3D, energy(KE or PE) is a function of 3 variable, (3 components for velocity, position...etc) therefore E=E(p,q,r)....and p=p(E,q,r), knowing E alone can't determent p...

3. Jan 29, 2005

### reilly

What else could it be? Is p conserved? Is x conserved? What else is there?
Regards,
Reilly Atkinson

4. Jan 31, 2005

### Galileo

A time-independant observable is a constant of motion if it commutes with the hamiltonian. That may be what they mean with a conserved observable.

5. Jan 31, 2005

### pmb_phy

By definition, an constant of motion is an observable (and therefore an operator) which (a) is not an explicit function of time and (b) commutes with the Hamiltonian. A conservative system is a system in which the Hamiltonian is not an explicit function of time.

The reason its called a "constant of motion" is because the expectation value of the observable is not a function of time. I.e. if A is a constant of motion then d<A>/dt = 0.

For your question to make sense you'd have to assert that all constants of motion are functions of the Hamiltonian. I assume that whomever made that assertion is thinking about the physical quantity associated with the observable. In any case I don't believe that assertion is true. For example, there are two observables I know of which have nothing to do with energy. One is spin and the other position. Each is an observable and thus there is a operator for each and corresponding physical quantity. Each has nothing to do with energy in general. The energy and spin of a particle are unrelated when, for example, the particle is an electron moving in the absence of a EM field.

Pete

Last edited: Feb 1, 2005
6. Jan 31, 2005

### pmb_phy

What is a "velocity component"?

Velocity is pretty much undefined in quantum mehanics. Do you mean the ratio of momentum to mass (some people like to define a velocity operator. But I think that they're just plain crazy. :tongue: )?

Pete

7. Feb 1, 2005

### ZapperZ

Staff Emeritus
I think the original poster of this question/thread needs to clarify if the question has any connection with QM. I suspect it does not since it was posted in the classical physics section. So this may be purely a classical mechanics problem and thus all the connection with "commutation" with the Hamiltonian may be irrelevant (but then again, people on PF have posted questions in the wrong section all the time).

If this is a classical mechanics question, then a "conserved" quantity as defined within the Lagrangian/Hamiltonian mechanics is simply a quantity in which the time derivative is zero. In that formulation, there are only two "observables", the generalized coordinate q and the generalized momentum p (and the associated derivatives).

Zz.

8. Feb 1, 2005

### pmb_phy

Oy! What was I thinking! Yep. You're right. For some reason I thought he was asking about quantum mechanics. I guess it was term "observerable" since I've never heard of that term being used outside quantum mechanics before.

Pete