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Self Adjoint Operator

  1. Oct 12, 2010 #1
    1. The problem statement, all variables and given/known data
    I am given that an operator is defined by the following:

    [tex]\hat\pi\psi(x,y)=\psi(y,x)[/tex]

    in a 2 variable Hilbert space of functions [tex]\psi(x,y)[/tex]

    I have to show that the operator is linear and Hermitian

    2. Relevant equations
    An operator is self adjoint if:

    [tex]\left\langle\psi\right|\hat{A}^{\dagger}\left|\right\phi\rangle=\left\langle\phi\right|\hat{A}\left|\right\psi\rangle^{*}[/tex]

    3. The attempt at a solution

    With the information given, how can I apply the above condition? Should I put x one side and y on the other? Or should I use psi(x,y) on each side. If so, does the fact that one of the psi's is a bra and the other is a ket change anything?

    To show that it is linear, I need to show that it is distributive. Is it as simple as saying:

    [tex]\hat{\pi}(A\psi(x,y)+B\psi(x,y))=\hat{\pi}(A\psi(x,y))+\hat{\pi}(B\psi(x,y))[/tex]

    If I remember correctly, my professor said that since the space is linear then the operators are distributive. That doesn't make sense to me. If an operator can be anything, then couldn't it be something that isn't linear as well?
     
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  3. Oct 12, 2010 #2

    Dick

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    What's your definition of the inner product? It's an integral, right? Showing it's linear isn't at all hard. And showing it's hermitian I think is just a changes of dummy variables, isn't it? An operator can be anything, but you are asked to show this one is linear. I'm not sure what your last question is about.
     
  4. Oct 12, 2010 #3
    Hi. Thanks for responding (and for the help in a different thread).
    Inner product:

    [tex]\left\langle\psi\right|\hat{A}\left|\right\psi\rangle=\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dx}[/tex]

    Hmm, dummy variables. Would that mean that the above integral changes from integrating over dx to over dy?

    [tex]\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dy}[/tex]
     
  5. Oct 12, 2010 #4

    Dick

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    They are functions of two variables, so it's probably a double integral dx*dy, isn't it? Applying pi just flips x and y, doesn't it? Does that affect the integral?
     
  6. Oct 12, 2010 #5
    Yes, a double integral does make sense. I think that flipping x & y does not change the integral because it doesn't matter in which order we integrate because the limits are the same for both x and y.

    But, does it change the complex conjugation? I say no because I do not see any evidence based off of the information given that complex numbers are affected by this permutation.
     
  7. Oct 12, 2010 #6

    Dick

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    Right. The limits for x and y are the same. Write out the integral expressions for <psi,pi(phi)> and <pi(psi),phi> and try to convince yourself interchanging the labels 'x' and 'y' makes no difference.
     
  8. Oct 12, 2010 #7
    If I say:

    [tex]\left\langle\psi\right|\hat{\pi}\left|\right\phi\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\phi{dx}{dy}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}{dx}{dy}\hat{\pi}\phi=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\phi{dx}{dy}[/tex]

    and likewise:


    [tex]\left\langle\hat{\pi}\psi\right|\left\right\phi\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{\pi}\psi^{*}\phi{dx}{dy}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{\pi}\psi^{*}{dx}{dy}\phi=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi\hat{\pi}\psi^{*}{dx}{dy}[/tex]

    Isn't that assuming the operator is linear? I must be missing something here. I was thinking that showing that moving around the integrand does not change the integration proves they're hermitian, but now I think that is based off the idea that (pi*psi*phi)=(psi*pi*phi).
     
  9. Oct 12, 2010 #8

    Dick

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    integral conjugate(psi(y,x))*phi(x,y)*dx*dy=integral conjugate(psi(x,y))*phi(y,x)*dx*dy. True or false? They only differ by an interchange of x and y, don't they? Do you see what I'm saying?
     
  10. Oct 12, 2010 #9
    Ahh. You are good. Does that mean I get a two-fer? Is the fact that the integral is unchanged enough (I think sufficient is the word) to show the operator is linear as well. Or is there a condition for linearity that I am not considering?
     
  11. Oct 13, 2010 #10

    Dick

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    No, linearity is separate. You want to show stuff like pi(A(x,y)+B(x,y))=pi(A(x,y))+pi(B(x,y)). It's really pretty direct if you think about what pi is doing. It's just switching x and y around.
     
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