So $\hat{L}_{x}$ and $\hat{L}_{y}$ do not commute:

$$ [ \hat{L}_{x}, \hat{L}_{y}] = i\hbar \hat{L}_{z}$$

But, what if we perform this operation on a state such that:

$$\hat{L}_{z} \phi_{l, m_{l}} = \hbar m_{l}\phi_{l, m_{l}},$$ where we require that $m_{l} = 0$, so

$$\hat{L}_{z} \phi_{l, m_{l}} =0.$$

Hence, for the case $m_{l}$ = 0, $$ [ \hat L_{x}, \hat L_{y}] \phi_{l, m_{l}} = 0,$$ and thence $\hat L_{x}$ and $\hat L_{y}$ share eigenstates! Does this work?

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    $\begingroup$ Yes, that works. They don't commute as operators on the whole space, i.e. they don't share a basis of eigenvectors. It is not forbidden that they share eigenvectors at all. $\endgroup$ – ACuriousMind Jun 4 '15 at 19:25
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    $\begingroup$ Wouldn't it really be that $[\hat{L}_x,\,\hat{L}_y]\phi_{m_l=0}=0$? Which doesn't quite mean that $\hat{L}_x,\,\&\,\hat{L}_y$ commute, just that they commute with $\phi_{m_l=0}$. $\endgroup$ – Kyle Kanos Jun 4 '15 at 19:26
  • $\begingroup$ Yup for you're right Kyle I think. And thank you both of you! $\endgroup$ – SomePhysicsStudent Jun 4 '15 at 19:27
  • $\begingroup$ $[\hat{L}_x,\,\hat{L}_y]\phi_{m_l=0}=0$ does not mean that $[\hat{L}_x,\,\hat{L}_y]=0$. So, those two operators do not commute with each other. Also, $\hat{L}_x$ and $\hat{L}_y$ DOES NOT commute with $\phi_{m_l=0}$ $\endgroup$ – Lê Dũng Jun 5 '15 at 2:50
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    $\begingroup$ To put it another way: every linear operator $\mathcal{O}$ has a subspace of vectors for which $\mathcal{O} \vec{v} = 0$, called its null space or kernel. Sometimes this subspace is trivial (i.e., only the zero vector), sometimes it's a proper subspace, and sometimes it's the whole space (when the operator is the zero operator.) What you've found is that the kernel of the operator $\mathcal{O} = [\hat{L}_x, \hat{L}_y]$ is non-trivial, but for two operators to commute, the kernel of the commutator must be the entire space. $\endgroup$ – Michael Seifert Jun 5 '15 at 17:09

Elaborating on ACruiosMind's comment, assume that the matrices $A$ and $B$ are defined the following way:

$$A=\begin{pmatrix} 1 & 2 \\ 5 & 4 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$

Notice that the eigenvectors of $A$ are

$$\begin{pmatrix} 1 \\ 5/2 \end{pmatrix} \quad \text {and} \quad \begin{pmatrix} 1 \\ -1 \end{pmatrix} $$

and the eigenvectors of $B$ are degenerate and its only eigenvector is

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

However as you can easily verify the commutator does not vanish ie

$$[A,B]= \begin{pmatrix} -7 & -7 \\ 7 & 7 \end{pmatrix} $$

This shows that though one of the eigenvectors of matrices (if you want operators) are the same, they don't commute.


Although $ [ \hat{L_{x}}, \hat{L_{y}}] \phi_{l, m_{l}} = 0$, $\phi_{l, m_{l}} $ is neither $\hat{L_{x}}$'s nor $ \hat{L_{y}}$'s eigenstate.


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