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.(adsbygoogle = window.adsbygoogle || []).push({});

#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.

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# Transformation of observables by permutation

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