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About the significance of commutivity

  1. Jul 14, 2010 #1
    This pertains to the quantum mechanics of angular momentum, so I'll ask it here:

    If operators A and B commute, does it follow that

    (i.) all of A's eigenvectors are eigenvectors of B and vice versa,

    or that

    (ii.) merely some of them are?

    According to Gillespie (see below) the commutivity of A and B implies that they "possess a common eigenbasis" (and the converse) and his proof seems to imply (i.), but it assumes that the operators are non-degenerate, and I don't know if the angular momentum operators are. Also, if (i.) and its converse are true, then it would seem to me that if A and B commute and A and C commute, then B and C would also commute, and I know that that's not supposed to be true. Nevertheless, so far I haven't found anyone explicitly say that (ii.) is the case.

    The book is "A Quantum Mechanics Primer" by Daniel T. Gillespie, 1970.
    Last edited: Jul 14, 2010
  2. jcsd
  3. Jul 14, 2010 #2
    If A and B are [itex]n\times n[/itex] matrices so that they commute, then following claims are true:

    If A and B are non-degenerate (no eigenvalue appears more than once in diagonal form), then the eigenbasis of A and B are the same. All eigenvectors of A are also eigenvectors of B.

    If A is degenerate (at least some eigenvalues appear more than once in diagonal form), then it is possible to choose a common eigenbasis for the both A and B. Not all eigenvectors of A are necessarily eigenvectors of B.
  4. Jul 14, 2010 #3


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    I. All of A's eigenvectors are eigenvectors of B.

    If A has degenerate eigenvalues, then the B matrix will be block-diagonal, meaning it can be diagonalized by a unitary operator. So by a suitable change of basis, it will be diagonal in both, and you will have the same eigenvectors.

    Edit: Oh well, Jostpuur beat me to it, and said it better too ;)
  5. Jul 14, 2010 #4
    So then, the angular momentum operators [itex] L_x , L_y , L_z , L^2 [/itex] are all non-degenerate?
  6. Jul 14, 2010 #5
    I'm also looking at Richard L. Liboff's "Introductory Quantum Mechanics", and in section 9.1 he indicates that every eigenvector of the [itex] L_z [/itex] operator is also an eigenvector of the [itex] L^2 [/itex] operator, but the reverse is not true. The same is true for the [itex] L_x [/itex] and [itex] L_y [/itex] operators (in their relation to [itex] L^2 [/itex]). I take it then that [itex] L^2 [/itex] is degenerate, but I don't yet know about the other three.
  7. Jul 14, 2010 #6
    They are all degenerate.

    If two operators a and b commute, and a has some eigenspace L for an eigenvalue l then b cannot connect two eigenspaces of a.

    If v1 is an eigenvector in the eigenspace L, then there are two possibilities when b is acting on v1: Either v1 is also an eigenvector of b or b transforms v1 into another vector that is also in L.
  8. Jul 14, 2010 #7
    I'm sorry but I don't know what this means. What does it mean for an operator to connect two eigenspaces of another operator? That both eigenspaces of a are also eigenspaces of b?
  9. Jul 15, 2010 #8
    I used the word "connect" for convenience without a formal definition. I thought the next sentence would explain it. If you apply b to a vector in L the resulting vector must stay in L. b cannot produce vectors outside of L but in another eigenspace say M from vectors in L.
  10. Jul 15, 2010 #9
    I get it now. Thanks, 0xDEADBEEF.
    Do you mean the four operators I mentioned in entry #4, or all operators in quantum mechanics?
  11. Jul 15, 2010 #10
    I meant the four that you mentioned. The momentum operator is not degenerate for example.
  12. Jul 15, 2010 #11
    But how does one know if an operator is degenerate or not?
  13. Jul 16, 2010 #12
    You'll learn about all the eigenspaces of the important operators soon enough. There is no generic way to know if an operator is degenerate, and it is not that important.
  14. Jul 16, 2010 #13
    It does seem odd that


    \hat {p_x} = - i \hbar \frac {\partial}{\partial x}


    would be non-degenerate while


    \hat {L_z} = - i \hbar \frac {\partial}{\partial \phi}


    is degenerate.
  15. Jul 16, 2010 #14


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    Lx,Ly,Lz are not degenerate, as is L**2. To get two states with the same Lz=m, one must use two different values of l, hence these two states are not degenerate. Work out the matrices for L=2,3,4, and see directly the lack of degeneracy of the Li operators.
    Reilly Atkinson
  16. Jul 16, 2010 #15
    I confess that I suspected as much.
    Unfortunately, I don't know how to do that. Perhaps I can find out.
  17. Jul 18, 2010 #16
    l does not matter for the discussion of the degeneracy of Lz. All States with the same m number can be superimposed and form the respective eigenspace for that eigenvalue.
  18. Jul 18, 2010 #17
    It is a bit more tricky. [tex]p_x[/tex] itself is degenerate, on the three dimensional space, because it doesn't matter what the dependency of the wave function w.r.t. y and z is.
    But in three dimensional space the full momentum operator that yields a vector as an eigenvalue is not degenerate. In two dimensions Lz is not degenerate, because we can only turn in one dimension.

    What reilly describes is the fact that you can get a complete set of operators that all commute and then you can number the states in your system. So although the energy operator, [tex]L^2[/tex] and [tex]L_z[/tex] are all degenerate, because there are multiple functions with n=5, l=3 or m =2 the state that fulfills all three at the same time is unique. He confused this with degeneracy of a single operator.

    So really don't get too confused. If you can write down two different functions that give the same eigenvalue, then the operator is degenerate. Simple as that.
  19. Jul 18, 2010 #18
    can anyone put this into a practical use or just formulas for the sake of formulas? just trying to figure out peoples interest in this if not to develop new ideas or fix what is broken.
  20. Jul 18, 2010 #19
    Two dimensions? Which dimension are you leaving out? Does it matter?
    Hi ShadeTreePhys. I'm just trying to understand quantum mechanics, with no practical application in mind. I'm not trying to develop new ideas, and I don't assume that anything about QM is broken.
  21. Jul 18, 2010 #20
    totaly cool. i love seeking the know just to know. qm by it's self is not broken. the ref was to the lack of compatibility with the classical.
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