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Hermitian conjugate of a Hermitian Conjugate

  1. Feb 8, 2015 #1
    I know that [itex](A\mp )\mp =A[/itex] . Where A is an Hermitian operator How does one go about proving this through the standard integral to find Hermitian adjoint operators?

    I should mention, I don't want anyone to just flat out show me step by step how to do it. I'd just like a solid starting place.

    [itex]\int (A\mp \psi*) \varphi dx = \int (\psi*)A\varphi dx[/itex]
     
    Last edited: Feb 8, 2015
  2. jcsd
  3. Feb 8, 2015 #2
    Edited with the question. Forums are difficult sometimes :)
     
  4. Feb 8, 2015 #3

    stevendaryl

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    I prefer [itex]\dagger[/itex] for Hermitian conjugate. Anyway...

    Writing [itex]\langle \psi|\phi \rangle[/itex] for [itex]\int \psi^* \phi dx[/itex], we have the definition of [itex]A^\dagger[/itex]:

    Equation 1: [itex]\langle A^\dagger \psi | \phi \rangle = \langle \psi | A \phi \rangle[/itex]

    But we also have a fact about inner products:

    Equation 2: [itex]\langle X | Y \rangle^* = \langle Y | X \rangle[/itex].

    Now, apply equation 2 to both sides of equation 1 to get another fact about Hermitian conjugates.
     
  5. Feb 9, 2015 #4

    DrDu

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    That's not true. An operator is hermitian or synonymously symetric, if ##\langle A^\dagger \phi|\psi\rangle =\langle \phi| A\psi \rangle ## for ## \phi,\; \psi \in \mathcal{D}(A)##. However, the domain of ##A^\dagger ## may be larger than that of A, ##\mathcal{D}(A)\subset \mathcal{D}(A^\dagger)## and then also ##\mathcal{D}(A)\subset \mathcal{D}(A^{\dagger \dagger})##. However if A is self adjoint, i.e. if additionally ##\mathcal{D}(A)=\mathcal{D}(A^\dagger)## then also ##\mathcal{D}(A)=\mathcal{D}(A^{\dagger \dagger})##.
     
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