Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does every observable commute with the hamiltonian?

  1. Mar 16, 2008 #1
    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?
  2. jcsd
  3. Mar 17, 2008 #2
    Usually operators q_i and p_i don't commute with the H.
  4. Mar 17, 2008 #3
    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.
  5. Mar 17, 2008 #4
    I think only conserved quantity/observable commute with hamiltonian.
  6. Mar 17, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.

  7. Mar 17, 2008 #6
    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.
  8. Mar 17, 2008 #7


    User Avatar

    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.
  9. Mar 17, 2008 #8
    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?
  10. Mar 17, 2008 #9
    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!!!

  11. Mar 17, 2008 #10
    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.
  12. Mar 17, 2008 #11
    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 :)

  13. Mar 17, 2008 #12


    User Avatar

    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.
  14. Mar 17, 2008 #13
    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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Does every observable commute with the hamiltonian?