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A simple proof involving degeneracy and commutators

  1. Jan 17, 2015 #1
    1. The problem statement, all variables and given/known data
    In the absence of degeneracy, prove that a sufficient condition for the equation below (1), where [itex]\left|a'\right>[/itex] is an eigenket of [itex]A[/itex], et al., is (2) or (3).

    2. Relevant equations
    [tex]\sum_{b'} \left<c'|b'\right>\left<b'|a'\right>\left<a'|b'\right>\left<b'|c'\right> = \sum_{b',b''} \left<c'|b'\right>\left<b'|a'\right>\left<a'|b''\right>\left<b''|c'\right>\qquad (1)[/tex]
    [tex]\left[A,B\right] = 0\qquad (2)[/tex]
    [tex]\left[B,C\right] = 0\qquad (3)[/tex]

    3. The attempt at a solution
    I know degeneracy is about one eigenvalue being associated with more than one eigevector (or eigenket). The summation looks intuitive to me. On the right side, a second completeness relation is used, where the completeness relation is
    [tex]\mathbb{1} = \sum_{b''} \left|b''\right>\left<b''\right|[/tex]
    I know the commutation relations of (2) and (3) can be written as
    [tex]\left[X,Y\right] = XY - YX[/tex]
    for any two operators [itex]X[/itex] and [itex]Y[/itex]. I know that since [itex]\left|a'\right>[/itex] is an eigenket of [itex]A[/itex], [itex]\left|b'\right>[/itex] is an eigenket of [itex]B[/itex] and [tex]\left|c'\right>[/itex] is an eigenket of [itex]C[/itex].

    Half the solution is understanding the problem, and the problem seems to be saying that no eigenvalue has the same eigenvector (eigenket). Knowing that prove that the sum is valid if either pair of operators commute. I would greatly appreciate some help moving forward. Thanks!
     
  2. jcsd
  3. Jan 19, 2015 #2
    [iteI am still not sure what the commutation relation has to do with the summation. The problem seems to be asking me if

    [tex]\left[A,B\right]=AB-BC=0[/tex]
    or
    [tex]\left[B,C\right]=BC-CB=0[/tex]

    then

    [tex]\sum_{b'} \left<c'|b'\right>\left<b'|a'\right>\left<a'|b'\right>\left<b'|c'\right>=\sum_{b'}\sum_{b''} \left<c'|b'\right>\left<b'|a'\right>\left<a'|b''\right>\left<b''|c'\right>[/tex]

    I know that an operator [itex]A[/itex] has a matrix representation of

    [tex]A = \sum_{a''}\sum_{a'} \left|a''\right>\left<a''|A|a'\right>\left<a'\right|[/tex]

    Note that [itex]a'[/itex] and [itex]a''[/itex] are just dummy variables that are iterating through the rows and columns respectively. The term [itex]\left<a''|X|a'\right>[/itex] is just a number since [itex]X[/itex], the operator, operates on a ket to produce another ket. This ket forms an inner product with the bra [itex]\left<a''\right|[/itex] and the result of an inner product is just a complex number.

    Furthermore, if the operator has eigenkets [itex]\left|a'\right>[/itex], then the matrix representation is even simpler (given in terms of the projection operator).

    [tex]
    A = \sum_{a''}\sum_{a'} \left|a''\right>\left<a''|A|a'\right>\left<a'\right|
    [/tex]

    Sakurai says that the number [itex]\left<a''|A|a'\right>[/itex] is [itex]\left<a'|A|a'\right>\delta_{a'a''}=a'\delta_{a'a''}[/itex], though I'm not sure how (perhaps if the operator is Hermitian?)

    Delta functions reduce the number of sums needed, so the matrix representation of the operator becomes

    [tex]
    A = \sum_{a'} a'\left|a'\right>\left<a'\right| = \sum_{a'} a' \Lambda_{a'}
    [/tex]

    I'm still collecting thoughts, but I really hope someone else can lend me a hand. Thanks!
     
  4. Jan 19, 2015 #3

    DrClaude

    User Avatar

    Staff: Mentor

    The fact that two operators commute tells you something very fundamental about the eigenstates of these operators. Do you know what that is?
     
  5. Jan 19, 2015 #4
    I have heard it means that their eigenstates are the same, but that doesn't really make sense to me intuitively. I haven't formally gotten to that section in my book, yet, but I"m working towards understanding everything until I get there (a few more pages).

    I was trying to convince myself of this using the simplest case I could think of... position.

    Of course the x and y operators would commute, but why would they have the same basis? One discusses x and the other covers y. Their bases should all be orthogonal? But they commute because I have read that [x_i,x_j] = 0.
     
  6. Jan 19, 2015 #5

    DrClaude

    User Avatar

    Staff: Mentor

    To be more precise: If two operators commute, then there exists a complete state of states which are eigenstates of both operators at the same time.

    Do you mean to say that the above problem should be solved without the knowledge of them having common eigenstates? If so, I'll have to think of another way of approaching the problem.

    It means you can define both x and y at the same time. I think it is easier to understand if you take quantized values, like energy and angular momentum for an electron in the hydrogen atom.
     
  7. Jan 19, 2015 #6
    What does it mean intuitively for them to have a complete set of states which are eigenstates of both operators at the same time? I haven't done quantum mechanics before, so this is all drinking from the fire hose.

    And, no, I'm sorry. I didn't mean to mislead. I just meant that it was going to take me some time before I got to that theorem, which I have now gotten to, but I feel stupid because I still don't get what is necessary to prove the equality of these two sums, whose physical meaning I'm not even sure I understand. I see a bunch of outer products / completeness relations, and that's about it.

    So far the only quantized values I'm familiar with are energy and wavenumber in a one-dimensional potential well. Do you mind elaborating on energy and angular momentum for an electron in a hydrogen atom? Thanks.
     
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