Consider the orbital angular momentum in QM, labeled by $L$ ($\mathbf{L}=\mathbf{r}\times\mathbf{p}$). In spherical coordinate, the operator can be expressed as: \begin{equation*} \left\{\begin{aligned} L_x&=\frac{\hbar}{\mathrm{i}}\left(-\sin\phi\frac{\partial}{\partial \theta}-\cos\phi\cot\theta\frac{\partial}{\partial \phi}\right)\\ L_y&=\frac{\hbar}{\mathrm{i}}\left(\cos\phi\frac{\partial}{\partial \theta}-\sin\phi\cot\theta\frac{\partial}{\partial \phi}\right)\\ L_z&=\frac{\hbar}{\mathrm{i}}\frac{\partial}{\partial \phi} \end{aligned}\right. ,L^{2}=-\hbar^{2}\left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right] \end{equation*}
We know that in general, $[L_z,L_x]=\mathrm{i}\hbar L_y\neq0$, so they do not share the same basis. By solving the eigenquation: \begin{equation*} \left\{\begin{aligned} L^2\psi&=\hbar^2l(l+1)\psi\\ L_z\psi&=\hbar m\psi \end{aligned}\right. \end{equation*} We find the common eigenfunction of $L^2$ and $L_z$ is spherical harmonic funcion $Y_l^m$. But what about $L_x$ and $L_y$? Can we get the general eigenfunction of $L_x$ for state $|l\; m\rangle$(which means $L_x\psi=\hbar m\psi$) using the same method?
I know for certain cases(or each case), the eigenfunction of $L_x$ can be expressed as the linear combination of $Y_l^m$, just use ladder operator to expand the operator in $Y_l^m$ basis. For example, suppose $l=1$, we have $Y_1^{-1},Y_1^0,Y_1^1$ as the basis, so we set: \begin{equation*} \begin{aligned} Y_1^1= \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} , Y_1^0= \begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix} Y_1^{-1}= \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \end{aligned} \end{equation*} Then, same as what we did for spin, we find the matrix for $L_x$ looks like: \begin{equation*} \begin{aligned} L_{x}=\frac{1}{2}\left(L_{+}+L_{-}\right)=\frac{\sqrt{2} \hbar}{2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \end{aligned} \end{equation*}(Here I omit the calculation), and the eigenstates for $L_x$ are: \begin{equation*} \begin{aligned} \varphi_{\hbar}=\frac{1}{2}\left(\begin{array}{c} 1 \\ \sqrt{2} \\ 1 \end{array}\right) \varphi_{0}=\frac{\sqrt{2}}{2}\left(\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right) \varphi_{-\hbar}=\frac{1}{2}\left(\begin{array}{c} 1 \\ -\sqrt{2} \\ 1 \end{array}\right) \end{aligned} \end{equation*} Where $\varphi_{\hbar}$ means the eigenstate with eigenvalue $\hbar$. So in this case, the eigenstate can be expressed as the linear combination of $Y_1^m$, namely: \begin{equation*} \begin{aligned} \varphi_{\hbar}=\frac{1}{2}\left(1\; \sqrt{2}\; 1\right) \begin{pmatrix} Y_1^1\\ Y_1^0\\ Y_1^{-1} \end{pmatrix} =\frac{1}{2}\left(Y_1^1+\sqrt{2}Y_1^0+Y_1^{-1}\right) \end{aligned} \end{equation*} But we cannot do it in general cases. For example, if the particle is in state $\phi=Y_1^0+Y_2^1+Y_4^2$(without normalization), if we want to measure $L_x$, what's the probability of each value we will get?
I tried to solve the equation like what we did for $Y_l^m$, but I failed. Consider: \begin{equation*} \begin{aligned} L_x f_l^m(\theta,\phi)=\hbar mf_l^m(\theta,\phi)\Rightarrow \frac{\hbar}{\mathrm{i}}\left(-\sin\phi\frac{\partial}{\partial \theta}-\cos\phi\cot\theta\frac{\partial}{\partial \phi}\right)f_l^m(\theta,\phi)=\hbar mf_l^m \end{aligned} \end{equation*}
For PDE, the only way I know is to separate variables: set $f_l^m(\theta,\phi)=\Theta(\theta)\Phi(\phi)$, after plugging in, I found it cannot be solved like usual: \begin{equation} -\frac{1}{\Theta}\tan\theta\frac{\mathrm{d}\Theta}{\mathrm{d}\theta}-\frac{1}{\Phi}\cot\phi\frac{\mathrm{d}\Phi}{\mathrm{d}\phi}=\mathrm{i}m\tan\theta\frac{1}{\sin\phi} \end{equation} It is not a constant on the right hand side, and I cannot separate it into the sum of two functions, so I don't know what to do next.
I'v also tried to diagonalize the matrix of $L_x$ in the basis of $L_z$ directly. I find that the matrix of $L_x$ generally looks like: \begin{equation*} L_{x}=\frac{\hbar}{2}\left(\begin{array}{ccccccc} 0 & b_{s} & 0 & 0 & \cdots & 0 & 0 \\ b_{s} & 0 & b_{s-1} & 0 & \cdots & 0 & 0 \\ 0 & b_{s-1} & 0 & b_{s-2} & \cdots & 0 & 0 \\ 0 & 0 & b_{s-2} & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & b_{-s+1} \\ 0 & 0 & 0 & 0 & \cdots & b_{-s+1} & 0 \end{array}\right) \end{equation*}
Where $b_{j} \equiv \sqrt{(s+j)(s+1-j)}$. Which means the matrix element can be expressed as:$(L_x)_{jk}=\frac{\hbar}{2}b_{k+1}\delta_{j\;k+1}+\frac{\hbar}{2}b_k\delta_{j\; k-1}$. Then I write the eigenequation: $L_x|l\; m\rangle=\hbar m|l\; m\rangle$ as: \begin{equation*} \begin{aligned} &\sum_{k}\left(\frac{\hbar}{2}b_{k+1}\delta_{j\;k+1}+\frac{\hbar}{2}b_k\delta_{j\; k-1}\right)|l\; m\rangle_k=\hbar m|l\; m\rangle_j\\ &\Rightarrow \frac{1}{2}b_j|l\; m\rangle_{j-1}+\frac{1}{2}b_{j+1}|l\; m\rangle_{j+1}=m|l\; m\rangle_j\\ &\Rightarrow |l\; m\rangle_{j+1}=\frac{2m}{b_{j+1}}|l\; m\rangle_j-\frac{b_j}{b_{j+1}}|l\; m\rangle_{j-1} \end{aligned} \end{equation*} If $j>2m+1$ or $j<0$, then $|l\; m\rangle_j=0$. For the initial condition, we can just set $|l\; m\rangle_1=1$, after computing every element in the vector, we can then normalize it.
This is a recurrence relation about element of eigenstate of $L_x$, but I cannot solve it. But at least I can comput it.
So my question:
- Can the PDE above be solved to get the general solution of the eigenfunction of $L_x$?(Something looks like $Y_l^m$, maybe some ugly special functions)
- Is there other ways to figure out the general eigenfunction of $L_x$? Or for any $|l\; m\rangle$, can I calculate the coefficient of $Y_l^m$ quickly? Or solve the series by the recurrence relation I found before?
- Which book can I refer to? I have searched this question on Google, but I found nothing.
Thx for reading my long question. I'm a undergraduate physics student, and I have only learnt QM for a few weeks by myself, so there may be mistakes in the question or misunderstandings for QM. Please point them out if you find, thx!