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Hermitian operator

  1. Nov 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Since the momentum operator is Hermitian why is this wrong:

    <psi| (p-hat)^2 |psi> = <psi| p-hat p-hat |psi> = <p-hat psi| p-hat |psi> = (p)^2 where p is the expectation value of the momentum.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 28, 2007 #2

    malawi_glenn

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    <p-hat psi| p-hat |psi> this is not a allowed thing.

    p-operator on a state gives you the state back + its eigenvalue, so you dont change the state.

    now to your question:
    since psi is a general state here, we dont know if psi is a eigenstate to p-hat, so one must insert some completness relations to take care of the the operator. Now there are many ways to do this, start with the simplets case and see if you can make progress.
     
  4. Nov 28, 2007 #3
    I see. But when psi is an eigenstate of p-hat, then my equations hold, right?
     
  5. Nov 28, 2007 #4

    malawi_glenn

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    yes, but only then. And then you usally label them as: |p>

    Do you use any perticular book in this course?
     
  6. Nov 28, 2007 #5
    Robinett (2nd Edition)

    I think this quote from my book is downright wrong then:

    [tex] <Y_{l,m}|(L_{x}^2+L_{y}^2+L_{z}^2)|Y_{l,m}> = <L_x Y_{l,m}|L_x Y_{l,m}> + <L_y Y_{l,m}|L_y Y_{l,m}> + <L_z Y_{l,m}|L_z Y_{l,m}>[/tex]

    where the L_i is the ith component of the angular momentum operator. The angular momenta components definitely do not commute so we know that Y_{l,m} is an eigenfunction of L_{z} only.

    This is equation 16.49 on Robinett page 456.
     
    Last edited: Nov 28, 2007
  7. Nov 28, 2007 #6

    malawi_glenn

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    I think the notation is very bad, <| is a state! You never take in the operator inside the bracket.

    Then Y_lm is not the state!

    [tex] Y_{lm} = <\vec{n}| lm> [/tex]

    Y_lm is a wavefunction in direction space (n is direction vector). It is a difference of state and wavefunction.

    I have never seen this strange notation anyware in scientific litterature, it is very confusing.

    Rather write:
    [tex] <l,m| L_x^2 |l,m> = <l,m| L_xL_x |l,m> [/tex]
    Now L_x is not hermitian, so it MUST operate to the right, on the ket (if you operate with a non hermitian operator to the left, you get the comlex conjugate of its eigenvalue, see dual correspondence). So the book of yours do a fatal misstake there also. L_z is hermitian so it can operate both to left and to the right. In the case of L_x, you can rewrite it as a superposition of the ladder operators and see for yourself.

    Best book about this is Sakurai, modern quantum mechanics. A classic.
     
  8. Nov 28, 2007 #7
    But L_x is Hermitian. It corresponds to an observable.
     
    Last edited: Nov 28, 2007
  9. Nov 28, 2007 #8

    malawi_glenn

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    oh yes, sorry, my mistake :) Was confusing it with the ladder operators.

    But my point was that the notation your book uses is quite confusing.
     
  10. Nov 28, 2007 #9
    But the equation in post #5 is still wrong, isn't it?
     
  11. Nov 28, 2007 #10
    Actually, my book does justify that:

    [tex] <\psi_1|O|\psi_2> = <O ^\dagger \psi_1|\psi_2> [/tex]

    So, that leaves me still confused since isn't this true:

    [tex] <(p-hat)^2> = <\psi|(p-hat)^2|\psi> = <(p-hat) \psi|(p-hat)\psi> = <p-hat>^2 [/tex]

    which one of those equalities is wrong? Is it the last one?
     
    Last edited: Nov 28, 2007
  12. Nov 28, 2007 #11

    George Jones

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    Robinett's equation is fine,

    The last one.
     
  13. Nov 28, 2007 #12
    I think I figured out what is wrong my thinking. It should be:

    [tex] <\hat{p}^2> = <\psi|\hat{p}^2|\psi> = <\hat{p} \psi|\hat{p}\psi> [/tex]

    which is equal to
    = p^2

    only when psi is an eigenstate of of the momentum operator.
     
  14. Nov 28, 2007 #13

    George Jones

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    Common notation:

    [tex]\left< \psi |A| \phi \right> = \left< \psi |A \phi \right> = \left<A^\dagger \psi | \phi \right> .[/tex]
     
  15. Nov 28, 2007 #14
    But was my statement in post #12 right?
     
  16. Nov 28, 2007 #15

    George Jones

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    Yes, if [itex]\hat{p} \left| \psi \right> = p \left| \psi \right>[/itex], then

    [tex]\left< \psi | \hat{p}^2 | \psi \right> = p \left< \psi | \hat{p} | \psi \right> = p^2 \left< \psi | \psi \right> = p^2.[/tex]
     
  17. Nov 28, 2007 #16

    malawi_glenn

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    Well yes it is full leagal to take in the operator in the ket, etc. But as I have learned, it is not a clear notation. I have seldom seen it, but George is more experienced than me, so :)
     
  18. Nov 28, 2007 #17

    George Jones

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    Well, what you find confusing, I might find clear, and what I find clear, you might find confusing. A third person might have completely different ideas than either of us on what's confusing and what's clear. We are all individuals, and we all look at things in different ways. This is part of what makes life interesting! The important thing is to have enough common ground that communication is possible.

    For example, I find Dirac notation's handling of general operators and adjoints to be confusing. I always have to translate this into the notation I learned in math class, fool around bit until I (hopefully!) get the answer, and then translate back into Dirac notation. On the other hand, I find that Dirac notation deals with projection operators and insertions of complete sets of state in a manner that is much more transparent than standard math notation.

    Cheers!
     
  19. Dec 20, 2007 #18
    OK. I now think that it is me that was wrong in writing this post, not my QM book. Can someone verify that? I mean verify that [tex] <\psi O^{\dagger}|\chi> = <\psi|O\chi> [/tex] is true regardless of whether psi and chi are eigenfunctions of O.
     
    Last edited: Dec 20, 2007
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