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Eigenvalue equation, prove that a is real.

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    You are given a self-adjoint operator [itex]\hat{A}[/itex] and the equation and [itex]\hat{A}\Phi_{i} ~=~ \Phi_{i}a_{i}[/itex]. Prove that ai are real numbers.

    2. Relevant equations

    There are instructions to guide me along with the question. The first step it says to do is write the eigenvalue equation (what I have already written above) and its complex conjugate.

    3. The attempt at a solution

    I feel like I would have a decent shot of doing this on my own if I could get this first step. I'm not sure what the complex conjugate of an eigenvalue equation looks like. The eigenfunction is [itex]\Phi[/itex]. A hat is an operator, and ai is the eigenvalue. So is the complex conjugate just replacing a with a*? I've tried to look this up online and in my book and I can't find anything.
    Last edited: Feb 8, 2012
  2. jcsd
  3. Feb 8, 2012 #2


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    It should just mean take the conjugate of everything. Not just a.
  4. Feb 8, 2012 #3
    Ok I lied, I still need help.

    So write both equations.
    1) [itex]\hat{A}\Phi_{i} ~=~ \Phi_{i}a_{i}[/itex]

    2) [itex]\hat{A*}\Phi_{i}* ~=~ \Phi_{i}*a_{i}*[/itex]

    Multiply (1) by [itex]\Phi[/itex], and (2) by [itex]\Phi*[/itex]

    1) [itex]\Phi_{i} \hat{A}\Phi_{i} ~=~ \Phi_{i}\Phi_{i}a_{i}[/itex]

    2) [itex]\Phi_{i}* \hat{A}*\Phi_{i}* ~=~ \Phi_{i}*\Phi_{i}*a_{i}*[/itex]

    Take the integral of both equations:

    1) [itex]\int \Phi_{i} \hat{A}\Phi_{i} ~=~ \int \Phi_{i}\Phi_{i}a_{i}[/itex]

    2) [itex]\int \Phi_{i}* \hat{A}*\Phi_{i}* ~=~ \int \Phi_{i}*\Phi_{i}*a_{i}*[/itex]

    I know that [itex]\left\langle \Phi A | \Phi \right\rangle ~=~ \int \Phi A \Phi[/itex]

    But I don't know where to go from here. I can right the left side of the equations in dirac notation, but I'm not sure what happens to the right side. And I don't know how this relates the complex conjugates...
    Last edited: Feb 8, 2012
  5. Feb 8, 2012 #4
    Self-adjoint means that [itex]<\Phi|\hat{A}\Phi>=<\hat{A}\Phi|\Phi>[/itex], knowing [itex]\Phi[/itex] is an eigenvector with eigenvalue a, you should easily figure out that a*=a.
  6. Feb 8, 2012 #5
    The property of it being self-adjoint tells me that:

    [itex]\int A^{*} \Phi^{*} \Phi ~=~ \int \Phi^{*} A \Phi[/itex]

    How does this help me with those integrals I wrote above (if my integrals are even correct)? The method we are given to solve it is write the eigenvalue equation and its complex conjugate. Multiply each side by phi and phi*, respectively. Take the integral, and then use the fact that it is a self adjoint operator to prove that a*=a.
  7. Feb 8, 2012 #6
    Your equation is wrong, the correct form is
    [itex]\int (A\Phi)^{*} \Phi = \int \Phi^{*} A \Phi[/itex]
    [itex]LHS=a^*\int \Phi^{*} \Phi[/itex]
    [itex]RHS=a\int \Phi^{*} \Phi[/itex]
  8. Feb 8, 2012 #7
    I guess I need to go talk to my professor because I am really lost here. I don't know why [itex] (A\Phi)^{*} \neq A^{*} \Phi^{*}[/itex].

    I also don't know how you got your second and third equations from the first one. And my textbook doesn't cover this at all so I don't know where to turn....

    EDIT: Would you be able to go through the steps I listed in order to prove it? I think my professor knows we don't know very much about this subject so he gave us a path for solving the question.

    In the first response I got, does that mean the complex conjugate is (A Phi)*= Phi*a*, not A*Phi* etc.
    Last edited: Feb 8, 2012
  9. Feb 8, 2012 #8
    I'm using X for easy typing here. If you think of A as nxn matrix and X as nx1 vector, (AX)* is 1xn, and A*X* is not even defined (nxn matrix times 1xn vector), the * represent conjugate transpose of a vector or a matrix.

    The step is X*(AX)=X*(aX)=a(X*X), and X*(AX)=X*(A*X)=(X*A*)X=(AX)*X=(aX)*X=a*(X*X), therefore a*=a since (X*X)=|X|^2≠0. The step you don't understand is probably (AX)*=X*A*, which should be familiar from your freshmen linear algebra class.
  10. Feb 8, 2012 #9


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    [itex]\int \Phi^{*} A \Phi=\int \Phi^{*} a \Phi=a \int \Phi^{*} \Phi[/itex] and since [itex]A^{*} \Phi^*=a^* \Phi^*[/itex], [itex]\int A^{*} \Phi^{*} A =a^{*} \int \Phi^{*} \Phi[/itex]. If they are equal since A is self adjoint, doesn't that show [itex]a=a^{*}[/itex]?
    Last edited: Feb 9, 2012
  11. Feb 9, 2012 #10
    In this case, are you multiplying the complex conjugate by Phi and the regular equation by Phi*? Your way makes sense and I was thinking that was how I should be doing it, I just was confused because the wording of the question made it seem like the complex conjugate should be multiplied by Phi*...

    Also, should that last part be:

    [itex]\int A^{*} \Phi^{*} \Phi =a^{*} \int \Phi^{*} \Phi[/itex]

    Your answer makes perfect sense to me. The wording had just confused me because of the way he said to multiply by Phi and Phi* "respectively". But it seems to work your way so that it good enough for me. Part of me was thinking I should have been doing it your way, but he seemed so adamant about just following his steps in order to solve it that I brushed those thoughts aside.
    Last edited: Feb 9, 2012
  12. Feb 9, 2012 #11


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    Yes, I think "Multiply (1) by Φ, and (2) by Φ∗" should have been the opposite. And I did substitute an 'A' for a 'Φ'. Sorry.
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