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In an excercise regarding two particles with the same angular momentum $j_1=j_2$, I was given an operator $P$ (I don't think it is relevent here what $P$ actually is).

I was asked to show that $P$ commutes with $J^z$ and $J^\pm$ in the space spanned by the simultanuous eigenstates of the commuting operators ${J_1^2, J_2^2, J_1^z, J_2^z}$, denoted by $|j_1,j_1,m_1,m_2\rangle$. Then it is claimed without explanation, as if it was immediate, that every eigenstate $|j_1,j_1,j,m\rangle$ is also an eigenstate of $P$.

I fail to understand why this is true.

The eigenstates $|j_1,j_1,j,m\rangle$ are simultaneous eigenstates of the commuting operators ${J_1^2, J_2^2, J^2, J^z}$. The claim that every eigenstate $|j_1,j_1,j,m\rangle$ is also an eigenstate of $P$, is equivalent to the claim that we can add $P$ to the set of commuting operators ${J_1^2, J_2^2, J^2, J^z}$, so $P$ commutes with every single one of them (commuting isn't transitive).

In the exercise I showed that $P$ commutes with $J^z$, and now I'm trying to understand why commuting with $J^\pm$ and $J^z$ means commuting with $J_1^2, J_2^2,$ and $ J^2$.

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$J^x$ and $J^y$ are just linear combinations of $J^{\pm}$. So $P$ commutes with all components of the vector $J$ and so it commutes with $J^2$.

And then the particles have fixed angular momentum $j_1=j_2$ so $J_1^2=J_2^2$ are just proportional to the identity operator. That's why the book is taking it as immediate.

What if we represent this on an expanded Hilbert space where $J_1^2, J_2^2$ are not trivial, i.e. the two component particles can change their spin? Then it is not necessarily true that $P$ commutes with $J_1^2$,$J_2^2$.

As a simple counter-example is let's define an operator $P$ that has eigenstates $$|\pm\rangle_{j,m}\equiv|j,\,0,\,j,\,m\rangle\pm|0,\,j,\,j,\,m\rangle$$ $$P|\pm\rangle_{j,m}=\pm|\pm\rangle_{j,m}$$ and it annihilates every other state that's not in the linear subspace spanned by these. Clearly it does commute with $J^2,J^z$ but does not commute with $J_1^2,J_2^2$

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