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A few operator questions

  1. Mar 27, 2005 #1
    I think these questions may be so simple that I don't know how to properly prove them, but I am really stumped on how to show especially the first one:

    Show that [tex](A_{op}^t)^t=A_{op}[/tex] Where [tex]A_{op}[/tex] is an operator and 't' is my way of saying hermitian conjugate

    Secondly, I'm not too sure at all how to show this:

    Show that the anti-hermitian operator, [tex]I_{op}^t=-I_{op}[/tex] has at most one real eigenvalue

    Any hints would be greatly appreciated.

    Edit: oops! Hidden in the annals of the internet I found the appropriate definition for [tex]A_{op}[/tex], but I'm still having trouble with the second one.
     
    Last edited by a moderator: Mar 27, 2005
  2. jcsd
  3. Mar 28, 2005 #2

    dextercioby

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    It's straightforward to see that for an eigenvector of [itex]\hat{A}[/itex] corresponding to a spectral value [itex]\lambda[/itex]

    [tex] \lambda=\langle \psi|\hat{A}|\psi\rangle =\left(\langle\psi|\hat{A}^{\dagger}|\psi\rangle\right)^{*}=(-\lambda)^{*} [/tex]

    Therefore the spectral value [itex] \lambda [/itex] is either 0 or purely imaginary...

    Daniel.
     
  4. Mar 28, 2005 #3

    HallsofIvy

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    To show that the hermitian conjugate of the hermitian conjugate of A is A itself, use the definition of "hermitian conjugate"!
     
  5. Mar 28, 2005 #4

    dextercioby

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    For this operatorial equality

    [tex] \hat{A}=\left(\hat{A}^{\dagger}\right)^{\dagger} [/tex] (1)

    to hold,for a densly defined linear operator A,then the 2 conditions from the definition of operatorial equality on a separable Hilbert space must be met

    [tex] \mathcal{D}_{\hat{A}}=\mathcal{D}_{\left(\hat{A}^{\dagger}\right)^{\dagger}} [/tex] (2)

    and

    [tex] \hat{A}|\psi\rangle=\left(\hat{A}^{\dagger}\right)^{\dagger}|\psi\rangle,\forall |\psi\rangle \in \mathcal{D}_{\hat{A}} [/tex] (3)


    Unfortunately,there's no densly defined linear operator on a separable Hilbert space for which (1) (and implicitely (2) & (3)) to hold.

    It can be proved that for an arbitrary densly defined linear operator [itex] \hat{A} [/itex] this operatorial inclusion holds true
    [tex] \hat{A}\subset\left(\hat{A}^{\dagger}\right)^{\dagger} [/tex] (4)

    Since [itex] \hat{A}^{\dagger} [/itex] is a closed operator,we use (4) to assert that [tex] \left(\hat{A}^{\dagger}\right)^{\dagger} [/tex] is a closed extension of the operator [itex] \hat{A} [/itex].

    Ergo,equality (1) doesn't hold true for any densly defined linear operator on a separable Hilbert space...

    Daniel.

    --------------------------------------------
    NOTE:Red=Incorrect statements.Relation (4) doesn't hold for essentially self adjoint densly defined linear operators.See disclaimer in post #6.
     
    Last edited: Mar 28, 2005
  6. Mar 28, 2005 #5

    dextercioby

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    What i'm saying is that u can't prove something which is incorrect...:wink:

    Daniel.
     
  7. Mar 28, 2005 #6

    dextercioby

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    DISCLAIMER!!

    :redface: My professor of QM was wrong (and it's my fault i didn't check other sources on functional analysis)...

    The correct relation is

    [tex] \hat{A}\subseteq \left(\hat{A}^{\dagger}\right)^{\dagger} [/tex] (1)

    Among the operators for which the equality holds,one finds operators which are called ESSENTIALLY SELFADJOINT densly defined linear operators.They form a subset of the set of selfadjoint (not hermitian/symmetric) linear operators acting on a separable Hilbert space for which their adjoint is selfadjoint.They satisfty the relations

    [tex] \hat{A}=\hat{A}^{\dagger} [/tex] (the operator is selfadjoint) (2)
    [tex] \hat{A}^{\dagger}=\left(\hat{A}^{\dagger}\right)^{\dagger} [/tex] (its adjoint is selfadjoint) (3)

    The relation (1) coupled with the relations (2) & (3) prove that for most of the selfadjoint operators,their adjoint is not selfadjoint,but hermitean/symmetric.

    See [1] for a proof that H-atom Hamiltonian (Dirac formulation-Schrödinger picture-coordinate representation) is essentially selfadjoint...

    As a comment to post #5,u need to prove the relation (1) and show that the equality limit holds true,only under certain conditions (u need to find these conditions);for example,if the operator is essentially selfadjoint,then it verifies the equality limit.

    Daniel.

    --------------------------------------------------------------
    [1]J.Prugoveçki:"Quantum Mechanics in Hilbert Space".
     
    Last edited: Mar 28, 2005
  8. Mar 28, 2005 #7

    dextercioby

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    I addition to post #6,i would reccomend you to read Akhiezer & Glazman's "The Theory of Linear Operators in Hilbert Space",2nd.ed,Dover,1993.

    In section #22,he proves that for a bounded linear ooperator defined on the whole Hilbert space the equality limit holds.On page #80 he mentions the strict inclusion,though he mentions that there are operators for which the equality relation holds...:eek:

    Daniel.
     
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