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Having trouble showing hermitian-ness.

  1. May 19, 2013 #1
    (Is that a word? I dunno.)

    Anyway,

    I'm going through Griffiths QM and I'm also supplementing it with Lifschitz QM. I can't seem to show whether or not an operator is hermitian or not.

    For instance, Lifschitz shows the hermitian-ness of the Hamiltonian,
    [tex]
    \frac{d}{dt}\int \psi \psi^* dq\,=\,\int\psi\frac{\partial \psi^*}{\partial t}dq\,=\,0
    [/tex]
    Substituting...
    [tex]
    \frac{\partial \psi}{\partial t}=\,-i\hat{H}\psi
    [/tex]
    [tex]
    \frac{\partial \psi^*}{\partial t}=\,i\hat{H}^*\psi^*
    [/tex]

    [tex]
    \int\psi\left(i\hat{H}^*\right)\psi^* dq\,-\,\int\psi^*\left(-i\hat{H}\right)\psi dq
    [/tex]

    In the next step he does away with the [itex]i[/itex] and I'm not sure how he pulls that off because, say, you're checking if the deriviative [itex]\frac{d}{dx}[/itex] is hermitian or not, it ends up being crucial to the hermitian-ness that it be multiplied by [itex]i[/itex]. Moving on with Lifschitz...
    [tex]
    \int\psi^*\hat{H}^*\psi dq \,-\,\int\psi^*\hat{H}\psi dq
    [/tex]
    [tex]
    \int\psi^*\left(\hat{H}^*\,-\,\hat{H}\right)\psi dq=\,0
    [/tex]
    Which shows that (due to the constancy of the norm'd [itex]\psi[/itex]'s) [itex]\hat{H^*}-\hat{H}=0[/itex].

    Except for the part I mentioned above, I understand how this works. I just don't know how to show it for other operators. Is the method pretty much the same?
     
  2. jcsd
  3. May 19, 2013 #2
    By definition the adjoint operator is:

    [tex]<Ax|y>=<x|A^{*}y>[/tex]

    So if it is self adjoint : [tex]A=A^{*}[/tex] then [tex]<Ax|y>=<x|Ay>[/tex]
     
  4. May 19, 2013 #3
    Yeah, I understand the definition...its the demonstration I'm having issues with. For instance if I have an operator such as the parity operator ([itex]P[/itex]), such that,
    [tex]
    P f(x) = f(-x)
    [/tex]
    and I use the definition
    [tex]
    <Pf(x)|g(x)>\,=\,<f(x)|P^*g(x)>
    [/tex]
    [tex]
    <f(-x)|g(x)>\,=\,<f(x)|g(-x)>
    [/tex]
    I don't know which step to take after this. I can't justify pulling the negative outside of the functions because that depends on whether or not they are even or odd. So confused.
     
  5. May 19, 2013 #4
    Expand f(x) and g(x') in an orthonormal basis (this exists as we have an inner product defined on the space ...).
     
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