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Commuting operators => Common eigenfunctions?

  1. Jun 21, 2009 #1

    JK423

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    My book on quantum physics says that if two Hermitian operators commute then it emerges that they have common eigenfunctions.

    Is that true?

    If A,B hermitian commuting operators and Ψ a random wavefunction then:
    [A,B]Ψ=0 => ABΨ=BAΨ
    If we assume that Ψ is B`s eigenfunction:
    b*AΨ=BAΨ
    From this equation, how does it emerge for A to have common eigenfunctions with B?
    I agree that if Ψ was A`s eigenfunction then the equation would be satisfied, but in the more general case of a random Ψ it doesnt emerge that Ψ is also A`s eigenfunction!
    Another argument to support this is the following.
    If those operators commute, then their uncertainties satisfy the following inequality:
    ΔΑ*ΔΒ >=0
    Ψ is B`s eigenfunction so: ΔB=0.
    Now the inequality is satisfied for ANY ΔΑ (including zero).

    So is the statement << if two hermitian operators commute then it emerges that they have common eigenfunctions. >> true or false?
     
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  3. Jun 21, 2009 #2

    malawi_glenn

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    Sakurai have the proof on page 30-32 in "Modern Quantum Mechanics"

    <a2|[A,B]|a1> = (a2-a1)<a2|B|a1> = 0 (assuming nondegeneracy), the matrix elements <a2|B|a1> are then diagonal, since <a2|B|a1> must vanish when a2 is not equal to a1, i.e. <a''|B|a'> = delta(a'',a')<a'|B|a'>

    Thus, A and B can be represented by diagonal matrices with the same set of base kets |ai>, i = 1,2, ...

    Thus:

    B = sum (over a'') |a''><a''|B|a''><a''|

    we have used two identity operators, since the matrix representation of B is diagonal in |a'> basis

    now act with B on |a'> :

    B|a'> = exercise for you if you don't own Sakurai = (<a'|B|a'>) |a'>

    thus |a'> is simultaneous eigenket to both A and B operator.

    The statemant is true, and in the case of nondegeneracy, is same spirit but a bit more involved.
     
  4. Jun 21, 2009 #3

    JK423

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    So you`ve shown that if two opperators commute then they can be both represented by diagonal matrices in the same set of base kets. You`ve also shown that the elements on the diagonal are the operator`s eigenvalues.

    One question:
    <a|B|a'>=delta(a,a')
    That means that the random eigenvalue <a|B|a> is infinite? (Because delta(a,a') goes to infinity for a=a')
     
  5. Jun 21, 2009 #4

    malawi_glenn

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    no, by delta I mean Kronecker delta. I am sorry for confusion, I thought it was obvious from the context that I worked in the discrete set =)
     
    Last edited: Jun 21, 2009
  6. Jun 21, 2009 #5

    JK423

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    Ah ok i didn`t get it :)
    Everything is so much easier and obvious when using state vectors in the particular occasion.
    Thank you!
    I`ll come back on the subject if i have any more unanswered questions! :)
     
  7. Jun 21, 2009 #6

    malawi_glenn

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    it's ok, try to get a good classic book on QM like sakurai or ballentine =)
     
  8. Jun 21, 2009 #7
    The degenerate case used to bother me somewhat, so I'll tell you how I like to think about it. Since the commutation relation AB=BA implies that A(B|a>)=a(B|a>), B|a> is a obviously a member of the eigenspace associated with the eigenvalue a. But that just means that the eigenspace associated with the eigenvalue a is closed under the B operation, so if you view the eigenspace as a vector space in its own right, it has all of the same properties as the full space. That is, B is a Hermitian operator on the a eigenspace, so there exists a complete basis of states that span it, states which are still eigenstates of A.
     
    Last edited: Jun 21, 2009
  9. Jun 22, 2009 #8

    JK423

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    The reason i asked the original question was to try to answer a question i have on angular momentum (L). Unfortunately i couldnt.
    If we have a spherical symmetric potential [ V=V(r) ] then the following commutators are true:
    [H,lx]=[H,ly]=[H,lz]=0
    Where H is the hamiltonian and lx,ly,lz the L`s components.
    So, if H commutes with all li then they must have common eigenfunctions!
    That means that we should be able to measure H,lx,ly,lz with no distribution.
    Ofcourse that is not true because of the commutator: [li,lj]=ihεijklk which is responsible for the uncertainty principle between any two components of angular momentum.
    So what`s going on? According to the proof that malawi_glenn posted, if two operators commute they SHOULD have common eigenfunctions regardless of anything else.
    Which means that H,lx,ly,lz should all have common eigenfunctions.
    The fact that they dont (only H and one of the li does) shows the proof to have a flaw? Or to be incomplete?


    I have another question that has nothing to do with the above.

    For the continuous spectrum we have:
    x|x>=x|x>
    We multiply with the bra <x'| :
    <x'|x|x>=x<x'|x>=x δ(x'-x)
    For x'=x : δ(x'-x)=infinite
    This means that the elements on the diagonal of the matrix <x'|x|x> are infinite?
     
  10. Jun 22, 2009 #9

    Fredrik

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    You can find a set of common eigenfunctions of H and Lx, and a set of common eigenfunctions of H and Ly, but you can't find a set of eigenfunctions that's common to all three.

    And yes, the diagonal elements of x in the x basis are infinite, but we aren't really dealing with matrices here. If you want to know what you're really dealing with, I suggest you start with these: https://www.amazon.com/Principles-M...ational-Mathematics/dp/0070856133/ref=ed_oe_p. :smile:
     
    Last edited by a moderator: May 4, 2017
  11. Jun 22, 2009 #10

    George Jones

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    Last edited by a moderator: May 4, 2017
  12. Jun 23, 2009 #11

    Fredrik

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    It looks really good. I wish I had known about it sooner. I think I'll just continue to read Sunder (my #3), because it contains some stuff that I'm interested in that Kreyszig doesn't cover, but I'm going to have to think about getting Kreyszig too.
     
  13. Jun 23, 2009 #12

    malawi_glenn

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    It is often so that one choose Lz and L-vector squared to be the the operators which you use. L_-vector squared commutes with L_x, L_y and L_z, it is called a Casimir Operator.
     
  14. Jun 23, 2009 #13

    JK423

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    I`m going to take a look at Kreyszig, thanks!
    So [A,B]=0 isn`t sufficient to consider that A and B have common eigeinstates?
     
  15. Jun 23, 2009 #14

    malawi_glenn

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    if you have [A,B]=0 and [A,C]=0 but NOT [B,C]=0, then you can not have eigenstates common to all three at once. i.e you can not have states of the form {a,b,c>, but you can have {a,b> and {a,c>
     
  16. Jun 23, 2009 #15

    JK423

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    Yes, but its |a,b>=|a,c> , isnt it? Operator A cannot have two sets of eigenfunctions!
     
  17. Jun 23, 2009 #16

    malawi_glenn

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    Read Fredriks post again "You can find a set of common eigenfunctions of H and Lx, and a set of common eigenfunctions of H and Ly, but you can't find a set of eigenfunctions that's common to all three."
     
  18. Jun 23, 2009 #17

    Avodyne

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    No, |a,b> is a linear combination of |a,c>'s with different values of c.
     
  19. Jun 23, 2009 #18

    JK423

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    ok now everything is explained! I just had the weird impression that an operator cannot have more than one sets of eigenfunctions.
     
  20. Jun 23, 2009 #19
    Note the fact that the degeneracy of H is crucial! That means there is some (or multiple) n-dimensional subspace (n>1) where all eigenvectors have the same eigenvalue with respect to H. For H, any generic orthonormal basis for this subspace is 'legitimate'.

    However, the operator Lz still distinguishes among states within this subspace. So there is a specific eigenbasis for Lz. The same goes for Lx, but because Lx and Lz do not commute this will be a different basis.
     
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