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Maximal set of commuting observables

  1. Jun 18, 2011 #1
    Hi,

    I am having a little trouble with the concept of finding out the maximal set of commuting observables. Suppose I have n commuting operators. Then the wavefunction I use must have n parameters also. For instance, [itex] L_{3}, L^{2}[/itex] and [itex] H[/itex] where H is the Hamiltonian and L is the angular momentum operator all commute. So the common eigenstates have three variables inside i.e [itex]|n l m>[/itex].

    What I can't figure out is how to ensure that these operators are "distinct" from each other. For example, if I take a one of the above operators and multiply it by a scalar I will get a different operator, say something like [itex]2H[/itex]. Now I have four commuting operators instead of three. Obviously, this is not going to give me another independent variable inside the wavefunction. This was a trivial example but in general, how should one ensure that the commuting operators that one has are all distinct?

    My best guess so far is to check that the matrix representations form a linearly independent set but is there a) a weaker condition and b) a way to do this without resorting to matrix representation?

    Thank you.
     
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  3. Jun 18, 2011 #2

    SpectraCat

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    The answer is that, in order completely specify an eigenstate, you must specify its eigenvalues (quantum numbers will typically suffice) with respect to all commuting observables of which it is an eigenfunction. Thus, if you find degenerate eigenstates, it suggests (perhaps guarantees) that you do not have a complete specification of your quantum state yet.

    In the example of the one-electron atomic states you mentioned, if you specify only the energy (eigenvalue of H, quantum number [itex]n[/itex]), then you have an n^2-fold degeneracy to deal with. Specifying the angular momentum (eigenvalue of L2, quantum number [itex]l[/itex]), improves that, but now each state for which you know [itex]n[/itex] and [itex]l[/itex] still has a degeneracy of [itex]2l+1[/itex], which you rectify by specifying the z-projection of angular momentum (eigenvalue of Lz, quantum number [itex]m_l[/itex]). Now you have a unique designation completely specifying a particular atomic orbital.

    All of these wavefunctions are orthogonal and uniquely specified by their set of quantum numbers. As you point out, you can create arbitrary operators such as 2H for which they are also eigenfunctions, but doing so gives you no further information about the quantum states.
     
  4. Jun 19, 2011 #3
    SpectraCat, my problem is like this: Let's say I have three commuting operators. Now, when I specify the common eigenstate, I should put in three quantum numbers. But how can I be sure that my three operators are all distinct and not trivial variations of one another like H and 2H are?
     
  5. Jun 19, 2011 #4
    It's probably easier to check the eigenvalues over the simultaneous eigenvectors, let's say if [itex]L_3[/itex] is a linear combination of [itex]L^2[/itex] and [itex]H[/itex], then their eigenvalues must satisfy the same linear combination, that means if you can find a definite linear relation among those eigenvalues, then you only have 2 effective commuting observables;if you can't, then you have 3.
     
  6. Jun 19, 2011 #5

    SpectraCat

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    Well, to me that problem seems contrived. If you know what the operators are well-enough to know that they commute, then you should be able to figure out if they have trivial relationships to each other. Operators correspond to physical observables (at least the ones we typically care about do), so just saying the name of the physical observable corresponding to the operator should tell you what you want to know. Mathematically, the answer that kof9595995 gave seems reasonable to me. Additionally, if you know the mathematical form of the eigenfunctions, and there is an unresolved degeneracy, then it may mean that there is an additional commuting observable that has not yet been specified.

    I guess my point is that I don't think this issue comes up in practice ... can you give an example of what you are talking about? I mean one that doesn't rely on contrived operators like 2H, but rather on operators corresponding to physical observables that have a linear dependency that is not immediately obvious.
     
  7. Jun 19, 2011 #6
    I agree, it hasn't come up in practice yet but I'm only a couple of semesters into QM so I thought perhaps there might be situations where it would be necessary to verify that the operators were all independent.

    Just one more thing: Suppose we had a set of commuting operators where one of the elements was not a linear combination but had some other relationship with the other elements in the set. For instance let's pretend [itex]L_{1}[/itex], [itex]L_{2}[/itex], [itex]L_{3}[/itex] and [itex]L^{2}[/itex] did commute. Now [itex]L^{2}=L_{1}^{2}+L_{2}^{2}+L_{3}^{2}[/itex]. How many quantum numbers should the common eigenstate have?

    So my question is, in general, when do we know that every one of the commuting operators gives us a new quantum number? Not just for this example but more generally.

    Sorry if this sounds very contrived. I admit I've never (yet) faced a problem but I just want to know exactly how to deal with it if I ever need to. Thank you
     
  8. Jun 19, 2011 #7

    SpectraCat

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    Well, I don't want to address that specific example, because it is non-physical .. the components of angular momentum DON'T commute with one another. However, I can give another physically correct example that may be helpful, namely the 3-D harmonic oscillator. In that problem, you can separate the full 3-D Hamiltonian into 3 separate 1-D Hamiltonians, which commute with the full Hamiltonian and with each other. In that case you end up requiring 3 quantum numbers in order to uniquely label the energy eigenstates. Notice that it is 3 quantum numbers, and not 4, even though we have 4 commuting operators .. that is because the total energy (i.e. the eigenvalue of the full Hamiltonian) is already uniquely specified once each of the 3 1-D components has been specified.
     
  9. Jun 19, 2011 #8
    SpectraCat, in your example, isn't it just a linear relation? The sum of the 1-D Hamiltonians is the 3-D Hamiltonian. If one operator is a linear combination of the others, then we already know by kof95's example that the operator in question will not result in a new quantum number.

    I know my example is non-physical but I can't think of a physically correct one where there is a non linear relationship between the commuting operators. Could you perhaps think of one? Even otherwise, what would you say about the angular momentum example?

    Thank you for taking the time to reply to my questions. I appreciate it!
     
  10. Jun 19, 2011 #9

    kith

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    You can consider the case of a free particle. The Hamiltonian reads [itex]H \propto P_1^2 + P_2^2 + P_3^2[/itex], so you have 4 commuting observables. Of the four quantum numbers [itex]E, p_1, p_2, p_3[/itex] you can chose three to characterize your state.

    There is no difference between linear and non-linear combinations of observables, since the relation [itex] f(A_1, A_2, ...) |eigenstate \rangle = f(a_1, a_2,...) |eigenstate \rangle [/itex] (with eigenvalues [itex]a_i[/itex]) always holds.
     
    Last edited: Jun 19, 2011
  11. Jun 19, 2011 #10
    So as long as I can express one operator as some function of the others, it will not result in an additional quantum number. That makes sense.

    Thank you very much for all the replies.
     
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