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Homework Help: Sakurai, p. 59, Pr 1.6 - critique the proof

  1. Aug 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that |i> and |j> are eigenkets of some Hermitian operator. Under what condition can we conclude that |i> + |j> is also an eigenket of A? Justify your answer.


    2. Relevant equations

    It seems that all that is needed is for "A" to be a linear operator and for |i> and |j> to have the same eigenvalue. Justification:

    A(u + v) = Au + Av (that's A's linearity at work)

    Au = u[0]*u
    Av = v[0]*v

    If we have u[0] = v[0] (matching eigenvalues), then:
    A(u + v) = u[0]*(u + v) = v[0]*(u + v)

    ...and thus:
    A(u + v) = [some common scalar]*(u + v)

    ...meaning (u + v) is eigenstate of linear operator "A".

    My question: qed?
     
  2. jcsd
  3. Aug 20, 2010 #2

    kuruman

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    For the eigenkets that you are considering, what is < i | j > equal to?
     
  4. Aug 20, 2010 #3
    If they are part of the same orthonormal basis, <i|j> ought to be equal to zero. I'll permit the assumption that since |i> and |j> have the same eigenvalues, they ought to be in the same basis.
     
  5. Aug 20, 2010 #4

    kuruman

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    Good. The problem is asking you under what conditions is the sum of two orthonormal eigenstates of A also an eigenstate of A? There is no "qed" here; qed implies that you need to show something. You need to specify the conditions and of course justify them. So what are the conditions and why? Your initial answer is not clear on that.
     
    Last edited: Aug 20, 2010
  6. Aug 20, 2010 #5
    I'd have to say the necessary and possibly-sufficient reasons are 1) for those eigenstates to have matching eigenvalues and 2) for the operator A to be linear.

    I said that before, but am I missing something? Or did my initial answer not clearly tout that as The Conditions? :-|

    We seem to be in a most dangerous situation where I am convinced a possibly-wrong answer is irrefutably correct :-|
     
  7. Aug 20, 2010 #6

    kuruman

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    Change "matching" to "identical" and you are correct. If A is the Hamiltonian, |i> and |j> would be degenerate states.
     
  8. Aug 20, 2010 #7
    Oh, with the same energy eigenvalues. This question makes sense now.

    Is there some subtle difference between "matching" and "identical" that I'm not aware of? I thought I could use the word "matching" in this case without dashing myself against some subtle technicality of a distinction. Perhaps not?


    Thank you very much for your help and patience : )
     
  9. Aug 21, 2010 #8

    kuruman

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    There is no subtle difference, except that the usage that I have seen includes "identical" and (sometimes) "the same" but not "matching." Therefore, I was not sure how you understood "matching." If I and my spouse are wearing "matching" clothes, it means something different from "identical" which is different from "the same" clothes.
     
  10. Aug 21, 2010 #9
    Oh, gotcha. I thought I missed a technicality, which I almost always do. I'm so bad with little details, and that just cost me a large amount of time over summer's research.

    Thanks again : )
     
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