Does every observable commute with the hamiltonian?

[CPL.Luke]

I would guess that they would as every observable is a function of the q's and p's and as those commute with the hamiltonian I couldn't imagine an observable that wouldn't commute, however are there any other cases where an observable won't commute with the hamiltonian?

 

[jostpuur]

Usually operators q_i and p_i don't commute with the H.

 

[kanato]

q and p don't commute with each other, so if the hamiltonian is composed of both q's and p's then neither q or p will commute with H.

 

[Barmecides]

I would guess that they would as every observable is a function of the q's and p's and as those commute with the hamiltonian I couldn't imagine an observable that wouldn't commute, however are there any other cases where an observable won't commute with the hamiltonian?

I think only conserved quantity/observable commute with hamiltonian.

 

[ZapperZ]

I would guess that they would as every observable is a function of the q's and p's and as those commute with the hamiltonian I couldn't imagine an observable that wouldn't commute, however are there any other cases where an observable won't commute with the hamiltonian?

If you write down an Hamiltonian that as a q dependence, i.e. a particle in a central potential, let's say, you can already see that it is never going to commute with an operator such as p. As has been mentioned already, H can consist of both p and q. Unless you have an operator that does nothing, then an operator most usually do not commute with the Hamiltonian. Only in special cases, such a a free particle, would H commute with p.

Zz.

 

[CPL.Luke]

hmm thtat poses a problem as it would necessarily imply that there existed an uncertainty relation between H and q,p which I have not heard of, and it would mean that you could not have simultaneous eigenvalues of the energy and momentum or position.

 

[kdv]

hmm thtat poses a problem as it would necessarily imply that there existed an uncertainty relation between H and q,p which I have not heard of, and it would mean that you could not have simultaneous eigenvalues of the energy and momentum or position.

If you look at the one-dimensional infinite square well, to take a simple example, it's clear that the energy eigenstates are not states of definite position or definite momentum! (the particle in an energy eigenstate is not located at a specific position nor does it have a specific momentum). In other words, the energy eigenstates are not Dirac delta functions in real space nor in momentum space.

If an observable does not commute with H it means that if you measure that observable at one time and then wait a while and measure it again, you are not guaranteed to obtain again the same value. The time evolution of the system (governed by H) will mix in states of different eigenvalues for that observable.

By contrast, L_z commutes with the Hamiltonian of a spherically symmetric potential. So if you measure the L_z of a a hydrogen atom, say, you are assured that it will stay it the eigenstate corresponding to the eigenvalue you obtained.

 

[CPL.Luke]

yeah I just got my pencil and paper out and computed the commutator, your right, the commutator of H,P is i (partial v, x) (in the position representation)

so if the partial of v with respect to x is zero then you can have a simultaneous eignevalue of p and H but this wouldn't hold generally.

I wonder is there any observable then that could be constructed such that it would commute with all others?

 

[Marco_84]

Usually in physics the interesting observables that commute with H are the casimirs of some group G (Galileo/Poincarè, etc).
If some observable commute with H, u get that it is a constant of motion.
Certainly there are many obs. that are not so, i.e. q_i and p_i of every singular particule if we are dealing with many body problems.
Lz is not always a constant of motion, it depends on the potential.
If we explicity broke translational invariance also Ptot is not anymore a const!

regards
marco

 

[CPL.Luke]

interesting, do you know of any books that cover this in more detail? I'm taking a course in group theory and quantum mechnics now, however the only thing we really covered for finite groups were degeneracies in various symmetry enviroments.

 

[Marco_84]

interesting, do you know of any books that cover this in more detail? I'm taking a course in group theory and quantum mechnics now, however the only thing we really covered for finite groups were degeneracies in various symmetry enviroments.

I think that a "bible" for application of group theory in physics is Cornwell... and also tha Hamermesh is a good one...

i did like to have a look on the oldest ones also, weyl and friends :)

regards
marco

 

[kdv]

interesting, do you know of any books that cover this in more detail? I'm taking a course in group theory and quantum mechnics now, however the only thing we really covered for finite groups were degeneracies in various symmetry enviroments.

Wigner showed in 1939 that in relativistic physics, physical states were classified according to their mass and spin (or helicity for massless particles). It was an amazing application of group theory!

Buit this is getting a bit far from your initial question on nonrelativistic qm. There is no observable that will necessarily commute with all th eother observables or with the Hamiltonian, in general.

 

[Marco_84]

Wigner showed in 1939 that in relativistic physics, physical states were classified according to their mass and spin (or helicity for massless particles). It was an amazing application of group theory!

Buit this is getting a bit far from your initial question on nonrelativistic qm. There is no observable that will necessarily commute with all th eother observables or with the Hamiltonian, in general.

I agree with u :

On Unitary Representations of the Inhomogeneous Lorentz Group

E. Wigner

The Annals of Mathematics, 2nd Ser., Vol. 40, No. 1. (Jan., 1939), pp. 149-204.

regards
marco