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Eigenvalues of Hermitian operators

  1. Oct 16, 2006 #1
    Hi again,

    Question: [tex]\hat{A} [/tex] is an Hermitian Operator. If [tex]\hat{A}^{2}=2[/tex], find the eigenvalues of [tex]\hat{A}[/tex]

    So We have:
    [tex] \hat{A}\left|\Psi\right\rangle=a\left|\Psi\right\rangle [/tex]

    But I actually don't know how to even begin. [tex]\hat{A}[/tex] is a general Hermitian operator, and I don't know where even [tex]\hat{A}^{2}[/tex] would fit in with the question asked.

    Any help is appreciated! Thank you!
    -Rick
     
  2. jcsd
  3. Oct 16, 2006 #2

    AKG

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    A2v = A(Av).
     
  4. Oct 16, 2006 #3
    Umm...

    [tex]\hat{A}\left(\hat{A}\left|\Psi\right\rangle\right) = a\left(a\left|\Psi\right\rangle\right) [/tex]

    [tex]\hat{A}^{2}\left|\Psi\right\rangle = a^{2}\left|\Psi\right\rangle [/tex]

    [tex]2\left|\Psi\right\rangle = a^{2}\left|\Psi\right\rangle [/tex]

    [tex]a = \sqrt{2}[/tex] ?

    I'm not sure...
     
  5. Oct 16, 2006 #4

    AKG

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    How did you get the first equation? What's [itex]|\Psi \rangle[/itex]? What's [itex]a[/itex]? How do you get a = [itex]\sqrt{2}[/itex], and not, say, [itex]-\sqrt{2}[/itex]? I mean, you have the right idea, but you haven't put that right idea into the form of a proper proof.
     
  6. Oct 16, 2006 #5
    Well, I'm using Dirac bracket notation to be consistent with the rest of my work, and ket space is just a real space, so it's same thing as \

    [tex] \hat{A}^{2}\Psi = a^{2}\Psi [/tex]

    where a is the eigenvalue of the Hermitian operator.


    You're right, a can be negative too, but is the final answer really [tex] a=\pm\sqrt{2}[/tex]? It seems too simple to be true.
     
  7. Oct 16, 2006 #6
    Well, I'm using Dirac bracket notation to be consistent with the rest of my work, and ket space is just a real space, so it's same thing as \

    [tex] \hat{A}^{2}\Psi = a^{2}\Psi [/tex]

    where a is the eigenvalue of the Hermitian operator.


    You're right, a can be negative too, but is the final answer really [tex] a=\pm\sqrt{2}[/tex]? It seems too simple to be true.
     
  8. Oct 16, 2006 #7

    AKG

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    I understand the notation, but you just introduced [itex]a[/itex] and [itex]|\Psi \rangle[/itex] without saying a thing about them. Okay, so [itex]a[/itex] is an eigenvalue of [itex]\hat{A}[/itex] and [itex]|\Psi \rangle[/itex] is a (non-zero) eigenvector corresponding to [itex]a[/itex].

    [tex]2|\Psi \rangle = \hat{A}^2|\Psi \rangle = \hat{A}(\hat{A}|\Psi \rangle ) = \hat{A}(a|\Psi \rangle ) = a(\hat{A}|\Psi \rangle) = a(a|\Psi \rangle) = a^2|\Psi \rangle[/tex]

    Therefore [itex]a = \pm \sqrt{2}[/itex]. Note there is some indeterminacy. [itex]\hat{A}[/itex] could have all it's eigenvalues positive, or all negative, or some positive and some negative.
     
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