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Hermitian Operator

  1. Apr 12, 2009 #1
    If [itex]\hat{O}[/itex] is hermitian, show that [itex]\hat{O}^2[/itex] is hermitian.

    we have [itex]<\psi|\hat{O}^2|\psi>^* = <\psi|\hat{O}\hat{O}|\phi>^*=<\phi|\hat{O}^{\dagger} \hat{O}^{\dagger}|\psi>=<\phi|\hat{O}\hat{O}|\psi>=<\phi|\hat{O}^2|\psi>[/itex]
    which works (hopefully)!

    to do this in integral notation is the following ok:

    [itex]\left(\int \psi^* \hat{O} \hat{O} \phi dx \right)^* = \int \left(\hat{O}\hat{O} \phi \right)^* \psi dx = \int \phi^* \hat{O} \hat{O} \psi dx[/itex]
    ???
     
  2. jcsd
  3. Apr 12, 2009 #2

    dextercioby

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    What does 'hermitean' mean ?

    Let [itex] \mathcal{H} [/itex] be a (complex) separable Hilbert space and [itex] \mbox{O} : \mathcal{D}(O) \longleftarrow \mathcal{H} [/itex] a densely defined linear operator. Symmetry/hermiticity of this operator means that

    [tex] \forall \varphi, \psi \in \mathcal{D}(O) [/tex]

    [tex] \langle O\varphi, \psi\rangle = \langle\varphi, O\psi\rangle [/tex].

    What does the hermiticity of the squared operator mean and what extra (restraining) assumptions do you need to make ?
     
  4. Apr 12, 2009 #3
    hey. can you explain in simpler terms....

    i haven't covered hilbert spaces or densely defined linear operators.

    im basically working with the knowledge of definition of hermitian conjugate:
    [itex]<\psi|\hat{O}|\phi>^*=<\phi|\hat{O}^{\dagger}|\psi>[/itex] and that [itex]\hat{O}[/itex] is hermitian iff [itex]\hat{O}=\hat{O}^{\dagger}[/itex]
     
  5. Apr 12, 2009 #4
    To make his words simplier, it is basically as following:
    if a operator A is Hermitean, for all [tex]\Psi[/tex], [tex]\Phi[/tex]
    < [tex]\Psi[/tex]l A [tex]\Phi[/tex] > = <A [tex]\Psi[/tex]l [tex]\Phi[/tex] >

    Proceed from here. I think this is enough to solve the proof
     
  6. Apr 12, 2009 #5
    two questions:
    why is your definition of hermitian operator different from mine (i seem to have an extra "line" between my operators and my kets?
    also is my attempt at a proof in post 1 wrong?
     
  7. Apr 12, 2009 #6
    I don't think that your statement is "wrong." It is more like you are stating it in a different way, and I didn't quite get your notation.
    For instance, what is your "dagger" sign mean here? And your * sign meant to be complex conjugate?
    I study in the US, and we, at least my school, use totally different notations from yours.

    However, the integration part looks kinda right to me. Please clarify if I am right. I can follow up to you apply to complex conjugate to OO [tex]\Phi[/tex], and of course complex conjugate apply to each individual "thing."
    And since O = O*, therefore you get the second equality.
    So I assume that you are taught that if O is hermition, then < [tex]\Phi[/tex] lO [tex]\Phi[/tex] > = < [tex]\Phi[/tex] lO [tex]\Phi[/tex] >*?
    I think this is a valid statement (at least this is consistent to what I was taught). It is just that we are not taught in this particular language.
     
  8. Apr 13, 2009 #7
    dagger is hermtian conjugate.

    i dont get why you dont have a line on either side of your operator?
     
  9. Apr 13, 2009 #8
    I belive it is only a notation difference and if my memory does not fail me... In "Dirac notation" one uses the two lines and as long as one is dealing with hermitian operators we do not need to care if they are acting to the left or right (and hence dont need to specify if they belong to the ket or the bra) which makes the dirac notation convenient.
    The notation with the operator in either the ket or bra tells us that it is acting to the left/right, i.e. on the bra or ket.

    Hope that can make things a bit clearer.
     
  10. Apr 13, 2009 #9
    ok thanks. as for my working in post 1 then, does that proof look ok?
     
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