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Homework Help: QM eigenket

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose that |[tex]\alpha[/tex]> and |[tex]\beta[/tex]> are eigenkets(eigenfunctions) of a hermitian operator A. Under what condition can we conclude that |[tex]\alpha[/tex]> + |[tex]\beta[/tex]> is also an eigenket of A?

    2. Relevant equations
    It's quite basic, I don't think any addtional equations are needed except the definations.

    3. The attempt at a solution
    From the question we know that A| [tex]\alpha[/tex] > =a|[tex]\alpha[/tex]> , A|[tex]\beta[/tex]> =b|[tex]\beta[/tex]>. And A is a hermitian operator:
    <[tex]\alpha[/tex]|A[tex]\beta[/tex]>=<[tex]\alpha[/tex]|b[tex]\beta[/tex]>=b<[tex]\alpha[/tex]|[tex]\beta[/tex]>,
    <[tex]\alpha[/tex]|A[tex]\beta[/tex]>=<A[tex]\alpha[/tex]|[tex]\beta[/tex]>=<a[tex]\alpha[/tex]|[tex]\beta[/tex]>=a<[tex]\alpha[/tex]|[tex]\beta[/tex]>,
    Therefore a=b? or <[tex]\alpha[/tex]|[tex]\beta[/tex]>=0?
    But it's nothing to do with |[tex]\alpha[/tex]>+|[tex]\beta[/tex]>+
    It seems no addition is need to constrain on them
     
    Last edited: Oct 18, 2007
  2. jcsd
  3. Oct 18, 2007 #2

    Hurkyl

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    Of course not; you never applied the condition that [itex]|\alpha\rangle + |\beta\rangle[/itex] is to be an eigenket!
     
  4. Oct 18, 2007 #3
    But what is the condition? if my first part is right:
    A(|[tex]\alpha[/tex]>+|[tex]\beta[/tex]>)=a(|[tex]\alpha[/tex]>+|[tex]\beta[/tex]>)
    is automatic right?
     
  5. Oct 18, 2007 #4

    Hurkyl

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    Well, no, it's not automatic. Your first part said
    "a = b" or "[itex]\langle \alpha | \beta \rangle[/itex] = 0".​

    So, you have to consider both cases, not just the "a = b" case.
     
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