# Transformation of observables by permutation

1. May 25, 2012

### buttolo

Hi. Consider two isomorphic state spaces $\mathcal{E}(1)$ and $\mathcal{E}(2)$. The first belongs to a proton, the second to an electron and they both have the same spin.

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

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

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

What I understood is: By putting $B(1)$ 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 $(1)$. By applying $B(2)$ to the same ket I obtain the eigenvalue corresponding to the j-th element of the element of the basis belonging to the index $(2)$, but I don't get the point :uhh:

Im am studying that on Coehn-Tannoudji.

2. May 25, 2012

### Morgoth

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.

3. May 26, 2012

### buttolo

B(1) acts on the vectors of the first space E(1)
B(2) acts on the vectors of the second space E(2)

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

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

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
4. May 26, 2012

### Morgoth

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.