# Are there simultaneous eigenstates of $L_x$ and $L_y$?

$$L_x$$ and $$L_y$$ don't commute since $$[L_x,L_y] = i \hbar L_z$$. From the uncertainty principle, we know that $$\Delta L_x \cdot \Delta L_y \geq \frac{1}{2} \cdot |\langle i \hbar L_z \rangle|$$. There would be no simultaneous eigenstates of $$L_x$$ and $$L_y$$ if $$\langle L_z \rangle$$ was zero for all eigenstates of $$L_x$$ and $$L_y$$. Is it possible to show that ?

Otherwise I would be happy with an other explanation of why or why not there would be simultaneous eigenstates of $$L_x$$ and $$L_y$$.

• Related Jan 3 at 17:03

According to the closure relationship we have for any two components of the angular momentum: $$[\hat{J}_i,\hat{J}_j]=i\hbar\epsilon_{ijk}L_k\tag{1}$$ where $$\epsilon_{ijk}$$ is Levi-Civita symbol and the sum is understood over the repeated indexes. A non-zero commutator implies that there is no common basis of eigenvectors.

That said, common eigenstates can still exist: in this case the state vector such that all components of angular momentum are zero. Please notice this happens only because the commutator is an operator: if it were a number (I mean a multiple of the identity operator), not a single common eigenstate would be possible. This is the case of position and momentum. Consider the 1D case: $$[\hat{x},\hat{p}]=i\hbar\mathbb{1}\tag{2}$$ Suppose there exists a common eigenstate $$|\alpha\rangle$$ such that: $$\hat{x}|\alpha\rangle=x|\alpha\rangle \\ \hat{p}|\alpha\rangle=p|\alpha\rangle$$ Then $$[\hat{x},\hat{p}]|\alpha\rangle=xp|\alpha\rangle-px|\alpha\rangle=0$$ that is contradiction with $$(2)$$.

On the other hand if the commutator were an operator as in $$(1)$$ it could possibly be zero for that particular state, as it happens for angular momentum.

• The first part of your answer is good but I don't think it's wise in the second part to use $\hat x$ and $\hat p$ as an example given it is plagued by subtleties with normalization and unboundedness. Jan 3 at 17:45
• Do you mean because their eigeinbases are non-normalizable states? Jan 3 at 17:48
• In part yes. Also because in finite dimensional spaces as there is the contradiction of $\hbox{Tr}[\hat x,\hat p]=\hbox{Tr}(\hat x \hat p)-\hbox{Tr}(\hat p\hat x)=i\hbar \hbox{Tr}(\mathbb{1})$, whereas in infinite dimensional spaces of course this identity must be handled carefully because of the unbounded nature of the operators. Jan 3 at 18:38

It is true that $$\langle \alpha\vert L_z\vert\alpha\rangle=0$$ if $$\vert\alpha\rangle$$ is an eigenstate of $$L_x$$ OR an eigenstate of $$L_y$$. Indeed, if it’s an eigenstate of $$L_x$$ then $$\Delta L_x=0$$ so the uncertainty relation yields $$0\times \Delta L_y=0\ge \frac{1}{2}\vert\langle \alpha\vert L_z\vert\alpha\rangle\vert \tag{1}$$ which implies the right hand side (which is necessarily real and non-negative) must be $$0$$. The same argument can be trivially modified to show the same result if $$\vert\alpha\rangle$$ is an eigenstate of $$L_x$$ instead.

I don’t know how this proves there are no simultaneous eigenstates of $$L_x$$ and $$L_y$$ and nothing in (1) prevents $$\Delta L_y=0$$.

The converse is not true: there are states with $$\langle L_z\rangle=0$$ which are not eigenstates of $$L_x$$ or $$L_y$$.

There is one state which is a simultaneous eigenstate of $$L_x$$ AND $$L_y$$: it is the $$L=0$$ state.

Now if you have a basis where $$L_x$$ is diagonal, then you can construct ladder operators $$\tilde L_\pm$$ in terms of $$L_y$$ and $$L_z$$, and in particular express $$L_y$$ as a combo of $$\tilde L_\pm$$, which do not act diagonally on the eigenstates of $$L_y$$. The exception is (again) the $$L=0$$ state, which is killed by $$\tilde L_\pm$$.

One tool for finding your way towards a proof is to construct some representation matrices for a few different $$L_{x,y,z}$$ and see how you feel about them. Here’s a representation for spin two, which gives you a sense of how it goes. The size of the matrix is $$2\ell+1$$. The $$z$$-axis matrix has the eigenvalues on the diagonal already,

$$L_z = \left(\begin{array}{ccccc} \ell \\ &\ell-1 \\ && \ddots \\ &&& -\ell+1 \\ &&&& - \ell \end{array}\right),$$

and the raising and lowering matrices are one-off-the-diagonal,

$$L_+ = \left(\begin{array}{ccccc} 0& \sqrt{2\ell} & \\ &0& \sqrt{\cdots} & \\ &&\ddots \\ && \cdots & 0& \sqrt{2\ell} \\ &&&&0 \end{array}\right) \qquad L_- = (L_+)^T \renewcommand{\ket}{\left|{#1}\right>}$$

so that $$L_\pm\ket{\ell,m} = \sqrt{\ell(\ell+1) - m(m\pm1)} \ket{\ell, m\pm1}$$. From these raising and lowering matrices $$L_\pm$$ you construct

\begin{align} 2 L_x &= L_+ + L_- \\ 2i L_y &= L_+ - L_- \end{align}

and then just find the eigenvectors. For spin-half (which isn’t allowed for orbital angular momentum $$L$$, but the algebra is similar), the $$x$$-axis eigenvectors are $${1}\choose{\pm1}$$, with eigenvalues $$\pm1$$, and the $$y$$-axis eigenvectors are $$1\choose{\pm i}$$, with eigenvalues $$\pm i$$. Those basis vectors are orthogonal to each other. However, it is the case that the eigenstates of $$\sigma_x$$ have expectation value $$\left<\sigma_y\right> = \left<\sigma_z\right> = 0$$.