2
$\begingroup$

$L_{x}$ 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 .)

$\endgroup$
4
  • 1
    $\begingroup$ In the doublet representation, L² is proportional to the identity, so any and all 2-vectors (spinors) are eigenstates of it. $x_1$ is its eigenvector and that of $L_x$, but why should this imply it has to be an eigenvector of $L_z$? Which it is not. Explain your logic. $\endgroup$ Oct 9, 2021 at 0:19
  • 1
    $\begingroup$ Possible duplicate. $\endgroup$
    – march
    Oct 9, 2021 at 2:51
  • $\begingroup$ Cosmas Zachos, sir, I was of the misconception that the eigenfunctions of commutatingoperators are all common to each other. I didnt know that in the case of degeneracy the eigenstates of two commutatingoperators may be different and thus had trouble digesting the fact that calculation shows eigenstates of Lx and Lz are different( I believed all eigenstates of commutatingoperators are common)So my logic was if Lx and L*2 commute and Lz and L*2 also, and all eigenstates of L*2 are eigenstates of Lz andall eigenstates of L*2 are also eigenstates of Lx,whyLz and Lx cant have common eigenstate. $\endgroup$ Oct 9, 2021 at 6:33
  • $\begingroup$ Cosmas Zachos, Sir, Also in our Linear Algebra course it was told that if AB= BA, then the eigenvector matrix of A and B (consisting of all the eigenvectors in its columns)will be the same and hence two committing matrices have all Eigenstates in common. The fact that Degeneracy can lead to uncommon eigenfunctions were not discussed back then and neither did I think about it. $\endgroup$ Oct 9, 2021 at 6:53

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Sir, that's exactly what I wanted!! I have had seen that on computation, Lz and Lx have different Eigenstates , I just wanted to know the qualitative reason, and the answer that you provided contained the qualitative reason, that even if 2 operators commute they can have different eigenstates because of degeneracy. $\endgroup$ Oct 9, 2021 at 5:30
2
$\begingroup$

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 $\begin{pmatrix} 1\\-1 \end{pmatrix}$).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.