Eigenstates of $L_x$, $L^2$, and $L_z$

Question

$L_{x}$ and $L^{2}$ commute, while $L_{x}$ and $L_{z}$ do not. However, $L_{z}$ and $L^{2}$ also commute, and hence, they also have common eigenstates. So the problem is:

If $\lvert\psi\rangle$ is an eigenstate of $L_{x}$ and $L^{2}$, won't it also be an eigenstate of $L_{z}$ (since $L^{2}$ and $L_{z}$ have common eigenstates), thus violate the uncertainty relationship? ( Edit -Actually, I knew the quantitative answer that is by computing the eigenstates of Lx and Lz, do turn out to be different, however, I wanted to know qualitatively why Lz and Lx do not have common eigenstates .)

Answer 1

The fact that $A$ and $B$ commute does not necessarily mean that every eigenstate of $A$ is also an eigenstate of $B$. It just means that there is basis of eigenstates of $A$ that are also eigenstates of $B$. However, in situations when the eigenvalues of $A$ are degenerate, it is also possible to use a different basis, in which the basis vectors are not eigenstates of $B$.

Let's see this concretely by looking at the simplest possible case, which is isomorphic to a $j=\frac{1}{2}$ angular momentum system. The angular momentum operators are $$J_{x}=\frac{\hbar}{2}\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] \\ J_{y}=\frac{\hbar}{2}\left[\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right] \\ J_{x}=\frac{\hbar}{2}\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right].$$ If you calculated the total angular momentum squared ${\bf J}^{2}=J_{x}^{2}+J_{y}^{2}+J_{z}^{2}$, you will find that ${\bf J}^{2}$ has the explicit form $${\bf J}^{2}=\frac{3\hbar^{2}}{4}\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right].$$ Note that since ${\bf J}^{2}$ is proportional to the identity matrix, any two-component state will be an eigenstate: ${\bf J}^{2}|\psi\rangle=(3\hbar^{2}/4)|\psi\rangle$.

However, only certain vectors will be eigenstates of the individual angular momentum components. For instance, the eigenstates of $J_{z}$ are $$|z+\rangle=\left[\begin{array}{c} 1 \\ 0 \end{array}\right], \quad |z-\rangle=\left[\begin{array}{c} 0 \\ 1 \end{array}\right].$$ It is straightforward to verify that these satisfy $J_{z}|z\pm\rangle=(\pm\hbar/2)|z\pm\rangle$, making them eigenstates of $J_{z}$, with eigenvalues $\pm\hbar/2$. However, they are not eigenstates of $J_{x}$. Instead, the action of $J_{x}$ on the eigenstates of $J_{z}$ is $J_{x}|z\pm\rangle=(\pm\hbar/2)|z\mp\rangle$. So the states $|z\pm\rangle$ are simultatneous eigenstates of ${\bf J}^{2}$ and $J_{z}$, but not of $J_{x}$.

On the other hand, there is another basis of states that are eigenstates of $J_{x}$. These are $$|z\pm\rangle=\frac{1}{\sqrt{2}}\left(|z+\rangle\pm|z-\rangle\right) =\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1 \\ \pm1 \end{array}\right].$$ These satisfy $J_{x}|x\pm\rangle=(\pm\hbar/2)|x\pm\rangle$. However, they cannot be eigenstates of $J_{z}$, since they are linear combinations of two states $|z+\rangle$ and $|z-\rangle$ that have different $J_{z}$ eigenvalues.

Answer 2

Consider angular momentum operators for spin 1/2: $$ L^2=\frac{3}{4}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\,L_z=\frac12\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \,L_x=\frac12\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. $$ Eigenstates for $L_z$ are vectors $\begin{pmatrix} 1\\0 \end{pmatrix}$ and $\begin{pmatrix} 0\\1 \end{pmatrix}$, but they are not eigenstates for $L_x$ (which has eigenstates $\begin{pmatrix} 1\\1 \end{pmatrix}$ and $\begin{pmatrix} 1\\-1 \end{pmatrix}$).