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Transformation of observables by permutation

  1. May 25, 2012 #1
    Hi. Consider two isomorphic state spaces [itex]\mathcal{E}(1)[/itex] and [itex]\mathcal{E}(2)[/itex]. The first belongs to a proton, the second to an electron and they both have the same spin.

    #Let [itex]B(1)[/itex] be an observable defined on the first space, spanned by [itex]|1,u_{i}\rangle[/itex], eigenvectors of [itex]B(1)[/itex] with eigenvalues [itex]b_i[/itex].
    #Let [itex]B(2)[/itex] be an observable defined on the first space, spanned by [itex]|2,u_{j}\rangle[/itex], eigenvectors of [itex]B(2)[/itex] with eigenvalues [itex]b_j[/itex].

    Consider the tensor product of the two spaces: [itex]\mathcal{E}=\mathcal{E}(1)\otimes \mathcal{E}(2)[/itex] with basis [itex]|1,u_{i},2,u_{j}\rangle =|1,u_{i}\rangle \otimes 2,u_{j}\rangle[/itex]. [itex]P_{21}[/itex] is the permutation operator. What does [itex]P_{21}B(1)P^{\dagger}_{21}=B(2)[/itex] really mean? I know that:

    #[itex]P_{21}B(1)P^{\dagger}_{21}|1,u_{i},2,u_{j}\rangle=b_{j}|1,u_{i},2,u_{j}\rangle [/itex]
    #[itex]B(2)|1,u_{i},2,u_{j}\rangle=b_{j}|1,u_{i},2,u_{j} \rangle[/itex]

    What I understood is: By putting [itex]B(1)[/itex] between the permutation operators I obtain from an arbitrary element of the basis the eigenvalue corresponding to the j-th element of the element of the basis belonging the to index [itex](1)[/itex]. By applying [itex]B(2)[/itex] to the same ket I obtain the eigenvalue corresponding to the j-th element of the element of the basis belonging to the index [itex](2)[/itex], but I don't get the point :uhh:

    Im am studying that on Coehn-Tannoudji.
     
  2. jcsd
  3. May 25, 2012 #2
    I don't understand your question.
    What does that mean?
    It means what you wrote. the operator on the left is the same as the operator on the right.

    exchange and make 1 into 2, and 2 into 1 (the action of the right P*[21])
    act with B(1) according to your 1st premise above according to B(1).
    then re-exchange 2 into 1, and 1 into 2 (the action of the left P[21])
    it will give you the same result as acting immediately with B(2).

    I guess that's it, and that's what you wrote.
     
  4. May 26, 2012 #3
    B(1) acts on the vectors of the first space E(1)
    B(2) acts on the vectors of the second space E(2)

    [itex]B(1)[|1,u_{i}\rangle \otimes 2,u_{j}\rangle][/itex]
    [itex][B(1)|1,u_{i}\rangle] \otimes 2,u_{j}\rangle[/itex]

    [itex]B(2)[|1,u_{i}\rangle \otimes 2,u_{j}\rangle][/itex]
    [itex]|1,u_{i}\rangle] \otimes [B(2) |2,u_{j}\rangle[/itex]

    How can they be equal if they are different operators defined on different spaces? The only thing that they return equal is an index. I don't understand why Cohen-Tannoudji calls their eigenvalues using the same letter, b.
     
    Last edited: May 26, 2012
  5. May 26, 2012 #4
    It is because the P takes your states from E2 and brings them to E1 and vice versa.
    So in fact B1 acting on E1 sees the elements of E2 instead.
     
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