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Identity operator

  1. Sep 28, 2011 #1
    I am trying to follow a derivation in a book which is written without bra-ket notation, and presumably without the concept of state vectors. I can easily follow it if I may use the fact that [itex]\sum_{n}|\varphi_{n}\rangle\langle\varphi_{n}|[/itex] is the identity operator.

    Analogously to the way I would prove that the above expression is the identity operator: I write [itex]\psi=\sum_{n}c_{n}\varphi_{n}[/itex] as the expansion of a wavefunction on the complete basis set [itex]\varphi_{n}[/itex]. If I use the fact that [itex]c_{m}=\int\varphi^{*}_{m}\psi d^{3}\textbf{r}[/itex], I can write [itex]\psi=\sum_{n}\varphi_{n}\int\varphi^{*}_{n}\psi d^{3}\textbf{r}[/itex]. The statement which I would like to prove is identical to this, If I replace [itex]\psi[/itex] on both sides by another expression [itex]\hat{w}\varphi_{n}[/itex]. Thus I would like to extract [itex]\psi[/itex] from the RHS and equate the rest of the expression to identity. How can I do this? The operator [itex]\hat{w}[/itex] has no special relation to the basis functions [itex]\varphi_{n}[/itex].
     
  2. jcsd
  3. Sep 28, 2011 #2
    Hello,

    To translate from braket notation to your notation:
    [tex]\mid \phi_n \rangle \leftrightarrow \phi_n[/tex]
    [tex]\langle x \mid y \rangle \leftrightarrow \int x^* y \mathrm d^3 r[/tex]

    Note that in the braket notation one formally rewrites [itex]\langle x \mid y \rangle[/itex] as [itex]\left( \langle x \mid \right) \left( \mid y \rangle \right)[/itex] where the x-thing is seen as an operator working on the y-thing.

    In this formalism, and translating, we get

    [tex]\int \phi_m^* \psi \mathrm d^3 r \leftrightarrow \langle \phi_m \mid \psi \rangle = \left( \langle \phi_m \mid \right) \left( \mid \psi \rangle \right)[/tex]

    and thus:

    [itex]\boxed{ \mid \psi \rangle} \leftrightarrow \psi=\sum_{n}\varphi_{n}\int\varphi^{*}_{n}\psi d^{3}\textbf{r} \leftrightarrow \boxed{ \sum_n \mid \phi_n \rangle \langle \phi_m \mid \left( \mid \psi \rangle \right) }[/itex]
     
  4. Sep 28, 2011 #3
    Thanks for your answer. Actually though, I was more wondering whether it was possible to show that without introducing the concept of state vectors at all. The book doesn't use them so it seems there must be a way.
     
  5. Sep 29, 2011 #4
    Oh, I see, you didn't want to prove the bra-ket identity theorem, but rather wanted an analogous expression without the Dirac notation (such that my whole post was redundant), correct?

    If I understand you correctly, I think this might be an answer:
    yes you can,
    just define a set of http://en.wikipedia.org/wiki/Linear_functional" [Broken] as [itex]\omega_n: H \to \mathbb C: \psi \mapsto \int \phi_n(\textbf r)^* \psi(\textbf r) \mathrm d^3 \textbf r[/itex] where H is the hilbert space that we're working in.

    With this entity, you can see (using your calculations) that the operator [itex]A : H \to H: \psi \mapsto \sum_n \phi_n \omega_n(\psi)[/itex] is equal to the identity operator.

    In shorthand, you can write [itex]I = \sum_n \phi_n \omega_n[/itex], where the definition of the operator is implicit. This is actually the same as what happens in the Dirac-notation, but less ambiguous.
     
    Last edited by a moderator: May 5, 2017
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