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While studying Total Angular Momentum operator, $J$ in Quantum Mechanics,
we define an operator $J^2$ in the following manner: $$J^2 = J_xJ_x + J_yJ_y + J_zJ_z$$ This operator commutes with any $J_k$ (where $k$ can be $x, y, z$) (We can verify this algebraically).
This leads us to : $[J^2, J_k] = 0$

We know that $[J_i, J_j] = i\hbar\epsilon_{ijk}J_k$ which leads us to the conculusion that $Jx, Jy, Jz$ do not commute with each other.

We know that commutator of $J^2$ with all of $J_x, J_y, J_z = 0$ ($J^2$ commutes with all of them).
We know that if two operators commute, they have simultaneous eigenkets.
By this logic, $J^2$ should have simultaneous eigenkets with all $J_x, J_y, J_z$.
But in our lectures, we were told to assume that $J^2$ commutes with exactly one of $J_x, J_y, J_z$ only (and by convention, we say that $J^2$ only commutes with $J_z$ out of the three), otherwise it would mean that $J_x, J_y, J_z$ have a set of simultaneous eigenkets, implying they all commute with each other.

Why is this so? Isn't the commutator of two operators being equal to zero sufficient to guarantee simultaneous eigenkets? What am I missing?

Qmechanic
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Ippo
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2 Answers2

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It's certainly true that $J^2$ commutes with each of $J_x$, $J_y$ and $J_z$. Additionally, it is true that for each of $J_x$, $J_y$ and $J_z$, a set of simultaneous eigenstates between that operator and $J^2$ can be found.

However, that does not imply that these are the same eigenstates, which is the assumption your question requires.

Here's a more general example, which illustrates the same point. Let $A$ and $B$ be any two arbitrary non-commuting operators, so $[A,B]\neq 0$. Then, let $I$ be the identity operator. It is certainly true that $A$ and $B$ both commute with $I$. This means we can find a set of simultaneous eigenstates $\{ |a\rangle \}$ for $A$ and $I$. (Indeed, any state is an eigenstate of $I$.) We can find a corresponding set $\{ |b\rangle \}$ for operator $B$.

But of course, these are different sets, so $A$ and $B$ don't have simultaneous eigenstates.

Jonomyster
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$\vec{J}^2$ actually commutes with all three $J_x$, $J_y$ and $J_z$. So $\vec{J}^2$ have common eigenstates with $J_x$ and have common eigenstates with $J_z$. How is it possible? Those are different eigenstates!

The reason is that for each eigenvalue of $\vec{J}^2$ there exist several eigenstates, \begin{equation} \vec{J}^2|l, m_z\rangle = l(l+1) |l, m_z\rangle, \quad J_z|l, m_z\rangle=m_z|l, m_z\rangle \end{equation} Where $m_z=-l, l+1, \ldots l$.

Similar eigenbasis exists for $J_x$, but its eigenstates $|l, m_x\rangle$ are superpositions of several different $|l, m_z\rangle$.

OON
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