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

Homework Help: Good set of commuting observables?

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data
    I keep seeing this crop up throughout my QM course but i still don't understand what a "good" set of commuting observables would be. . Surely any set of observables that commute have to be a good set? I may just be stating the obvious but the way its phrased it makes me feel as though some observables that commute may be 'better' for some reason than another set that also commute but I don't understand why thats the case. Like if you have a couple of operators such that [a, b] = 0 and [a, c] = 0 are they both as good as each other for describing a quantum system or what? To me this says that observables for both a and b, or a and c can be found simultaneously so neither is a better set?

    2. Relevant equations

    3. The attempt at a solution
    Pretty much explained my thoughts on the matter above, its not so much a homework question as it is just clarification.
  2. jcsd
  3. Dec 31, 2012 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    You often see "good" associated with quantum numbers. I did a search for "good set of commuting observables (operators)" and didn't find much. But "good quantum number" yielded some hits. Looking at the results, it appears that a "good quantum number" is often taken to mean a quantum number associated with an operator that commutes with the Hamiltonian so that the value of the quantum number remains constant in time. An example would be the angular momentum quantum number ##l## for the hydrogen atom if you neglect spin-orbit interaction. With spin-orbit interaction ##l## is no longer "good".

    So, perhaps, a good set of commuting observables would be a set of commuting operators that includes the Hamiltonian. You could possibly add a further condition that the set of observables is "complete" so that a quantum state which is an eigenstate of all of the observables in the set would be uniquely determined. Don't know if this helps much.
    Last edited: Dec 31, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook