The Angular Momentum Eigenfunctions

The angular momentum eigenstates are eigenstates of two operators.

$$L_z Y_{\ell m}(\theta,\phi)=m\hbar Y_{\ell m}(\theta,\phi)$$

$$L^2 Y_{\ell m}(\theta,\phi)=\ell(\ell+1)\hbar^2 Y_{\ell m}(\theta,\phi)$$

All we know about the states are the two quantum numbers $\ell$ and $m$. We have no additional knowledge about $L_x$ and $L_y$ since these operators don't commute with $L_z$. The raising and lowering operators $L_\pm=L_x\pm iL_y$ raise or lower $m$, leaving $\ell$ unchanged.

$$L_\pm Y_{\ell m}=\hbar\sqrt{\ell(\ell+1)-m(m\pm 1)}Y_{\ell(m\pm 1)}$$

The differential operators take some work to derive.

$$L_z={\hbar\over i}{\partial\over\partial\phi}$$

$$L_\pm=\hbar e^{\pm i\phi}\left(\pm{\partial\over\partial\theta}+i\cot\theta{\partial\over\partial\phi}\right)$$

Its easy to find functions that give the eigenvalue of $L_z$.

$$\begin{eqnarray*} Y_{\ell m}(\theta,\phi)=\Theta(\theta)\Phi(\phi)=\Theta(\thet... ... i}im\Theta(\theta)e^{im\phi}=m\hbar Y_{\ell m}(\theta,\phi)\\ \end{eqnarray*}$$

To find the $\theta$ dependence, we will use the fact that there are limits on $m$. The state with maximum $m$ must give zero when raised.

$$\bgroup\color{black} L_+ Y_{\ell\ell}=\hbar e^{i\phi}\left({\... ...r\partial\phi}\right)\Theta_\ell(\theta)e^{i\ell\phi}=0 \egroup$$

This gives us a differential equation for that state.

$$\begin{eqnarray*} {d\Theta(\theta)\over d\theta}+i\Theta(\theta)\cot\theta(i\el... ...{d\Theta(\theta)\over d\theta}=\ell\cot\theta\Theta(\theta) \\ \end{eqnarray*}$$

The solution is

$$\bgroup\color{black} \Theta(\theta)=C\sin^\ell\theta .\egroup$$

Check the solution.

$$\bgroup\color{black} {d\Theta\over d\theta}=C\ell\cos\theta\sin^{\ell-1}\theta=\ell\cot\theta\Theta \egroup$$

Its correct.

Here we should note that only the integer value of $\ell$ work for these solutions. If we were to use half-integers, the wave functions would not be single valued, for example at $\phi=0$ and $\phi=2\pi$. Even though the probability may be single valued, discontinuities in the amplitude would lead to infinities in the Schrödinger equation. We will find later that the half-integer angular momentum states are used for internal angular momentum (spin), for which no $\theta$ or $\phi$ coordinates exist.

Therefore, the eigenstate $Y_{\ell\ell}$ is.

$$Y_{\ell\ell}=C\sin^\ell(\theta)e^{i\ell\phi}$$

We can compute the next state down by operating with $L_-$.

$$\bgroup\color{black} Y_{\ell(\ell-1)}=C L_-Y_{\ell\ell} \egroup$$

We can continue to lower $m$ to get all of the eigenfunctions.

We call these eigenstates the Spherical Harmonics. The spherical harmonics are normalized.

$$\begin{eqnarray*} \int\limits_{-1}^{1}d(\cos\theta)\int\limits_{0}^{2\pi}d\phi\... ...m}Y_{\ell m}=1 \\ \int d\Omega\; Y^*_{\ell m}Y_{\ell m}=1 \\ \end{eqnarray*}$$

Since they are eigenfunctions of Hermitian operators, they are orthogonal.

$$\int d\Omega\; Y^*_{\ell m}Y_{\ell' m'}=\delta_{\ell\ell'}\delta_{mm'}$$

We will use the actual function in some problems.

$$\begin{eqnarray*} Y_{00} &=& {1\over \sqrt{4\pi}}\\ Y_{11} &=& -\sqrt{3\over ... ...heta\\ Y_{20} &=& \sqrt{5\over 16\pi}\; (3 \cos^2\theta-1)\\ \end{eqnarray*}$$

The spherical harmonics with negative $m$ can be easily compute from those with positive $m$.

$$Y_{\ell(-m)} = (-1)^m Y_{\ell m}^*$$

Any function of $\theta$ and $\phi$ can be expanded in the spherical harmonics.

$$\bgroup\color{black} f(\theta,\phi)=\sum\limits_{\ell=0}^\inf... ...mits_{m=-\ell}^\ell C_{\ell m}Y_{\ell m}(\theta,\phi) \egroup$$

The spherical harmonics form a complete set.

$$\bgroup\color{black} \sum\limits_{\ell=0}^{\infty}\sum\limits... ...{m=-\ell}^\ell \vert\ell m\rangle \langle\ell m\vert=1 \egroup$$

When using bra-ket notation, $\vert\ell m\rangle$ is sufficient to identify the state.

The spherical harmonics are related to the Legendre polynomials which are functions of $\theta$.

$$\begin{eqnarray*} Y_{\ell 0}(\theta,\phi)&=&\left({2\ell+1\over 4\pi}\right)^{1... ...ell+m)!}\right]^{1\over 2} P_\ell^m(\cos\theta)e^{im\phi} \\ \end{eqnarray*}$$