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- Thread starter CPL.Luke
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Usually operators q_i and p_i don't commute with the H.

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I think only conserved quantity/observable commute with hamiltonian.

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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.

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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.

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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?

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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

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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

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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.

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I agree with u :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.

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

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