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Hermitian conjugate of an operator

  1. May 6, 2012 #1
    Hey guys, I'm doing a third year course called 'Foundations of Quantum Mechanics' and there's this thing in my notes I don't quite get. I was hoping to get your help on this, if you don't mind. It's about Hermitian conjugate operators. The sentences go

    (v, Au) = (A†v|u)

    <v|A|u> = <v|(A|u>)
    <v|A|u> = (<v|A)|u>
    (<v|A)† = A†|v>
    <v|A|u>*=<u|A†|v>

    I am wondering how the fourth line might follow from the third and if the fifth line belongs to this chunk or if it's a separate piece of info on its own.
    I would appreciate any form of help!
     
  2. jcsd
  3. May 6, 2012 #2

    Fredrik

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    The right-hand side of the first line looks weird. Did you mean (A†v,u)?

    I don't think it's very useful to define the adjoint of a bra ket, but if we define it, it's natural to define it as the bra that's assigned to it by the isomorphism from the Hilbert space into its dual space that's guaranteed to exist by the Riesz representation theorem for Hilbert spaces. (The adjoint of a bra is defined similarly, using the inverse of the same isomorphism, to ensure that |u>††=|u>).

    So the fourth line says that ket that corresponds to the bra <v|A is A†|v>. This follows from the previous lines and the properties of the inner product.

    If we denote the ket that corresponds to <v|A by |w>, and use the notation ##\big(|\alpha\rangle,|\beta\rangle\big)## for the inner product of ##|\alpha\rangle## and ##|\beta\rangle##, we have
    $$\big(|w\rangle,|u\rangle\big) =\langle w|u\rangle =\big(\langle v|A\big)|u\rangle =\langle v|\big(A|u\rangle\big) =\big(|v\rangle,A|u\rangle\big) =\big(A^\dagger|v\rangle,|u\rangle\big).$$
     
    Last edited: May 6, 2012
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