Angular Momentum

Classically, angular momentum is defined as $$\vec{L} = \vec{r}\times\vec{p}$$ Replacing with operators, angular momentum in quantum mechanics is defined as $$\hat{L} = \left( \begin{matrix} \hat{y}\hat{p_z} - \hat{z}\hat{p_y} \\ \hat{z}\hat{p_x} - \hat{x}\hat{p_z} \\ \hat{x}\hat{p_y} - \hat{y}\hat{p_x} \end{matrix} \right)$$ The square of angular momentum is simply $$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$$

Commutator of Angular Momentum Component

$$\begin{align} [\hat{L}_y, \hat{L}_z] &= [\hat{z}\hat{p_x} - \hat{x}\hat{p_z}, \hat{x}\hat{p_y} - \hat{y}\hat{p_x}] \\ &= [\hat{z}\hat{p}_x, \hat{x}\hat{p}_y] - [\hat{x}\hat{p}_z, \hat{x}\hat{p}_y] - [\hat{z}\hat{p}_x, \hat{y}\hat{p}_x] + [\hat{x}\hat{p}_z, \hat{y}\hat{p}_x] \\ &= -i\hbar\hat{z}\hat{p}_y + i\hbar\hat{y}\hat{p}_z\\ &= i\hbar L_x \end{align}$$ Similarly for $L_x, L_y$. So, we have, $$\begin{align} [\hat{L}_y, \hat{L}_z] &= i\hbar L_x \\ [\hat{L}_z, \hat{L}_x] &= i\hbar L_y \\ [\hat{L}_x, \hat{L}_y] &= i\hbar L_z \\ \end{align}$$ Commutator of $L_z$ and $L^2$ is $$\begin{align} [\hat{L}_z, \hat{L}_x^2] &= [\hat{L}_z, \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2] \\ &= [\hat{L}_z, \hat{L}_x^2] + [\hat{L}_z, \hat{L}_y^2] + [\hat{L}_z, \hat{L}_z^2] \\ &= [\hat{L}_z, \hat{L}_x^2] + [\hat{L}_z, \hat{L}_y^2] \\ \end{align}$$ As $L_z$ must commute to itself. By $$\begin{align} [\hat{L}_z,\hat{L}_x] &= i\hbar\hat{L}_y \\ \hat{L}_x\hat{L}_z &= \hat{L}_z\hat{L}_x - i\hbar{L}_y \\ [\hat{L}_z,\hat{L}^2_x] &= \hat{L}_z\hat{L}_x\hat{L}_x - \hat{L}_x\hat{L}_x\hat{L}_z \\ &= (i\hbar \hat{L}_y + \hat{L}_x\hat{L}_z)\hat{L}_x \\ &-\hat{L}_x(\hat{L}_z\hat{L}_x - i\hbar\hat{L}_y) \\ &= i\hbar(\hat{L}_y\hat{L}_x - \hat{L}_x\hat{L}_y) \\ &= -i\hbar[\hat{L}_x, \hat{L}_y] \end{align}$$ Similarly, $$[\hat{L}_z, \hat{L}_y^2] = -i\hbar[\hat{L}_y, \hat{L}_x] = i\hbar[\hat{L}_x, \hat{L}_y]$$ Thus, $$[\hat{L}_z, \hat{L}_x^2] = 0$$ As $z$ direction can be chosen to be $y$ or $x$ axis as well, if $[\hat{L}_z, \hat{L}_x^2] = 0$, same must be true for $x$ or $y$ $$\begin{align} [L^2, \hat{L}_x] &= 0 \\ [L^2, \hat{L}_y] &= 0 \\ [L^2, \hat{L}_z] &= 0 \\ \end{align}$$

Ladder Operators

The ladder operators are defined as $$L_{\pm} \equiv L_x \pm iL_y$$ Let $\phi_{l,m}$ be the eigenstate of $L^2$ and $L_z$ such that $$\begin{align} \hat{L}^2 \phi_{l,m} &= \hbar^2Q_l\phi_{l,m} \\ \hat{L}_z \phi_{l,m} &= \hbar m\phi_{l,m} \\ \end{align}$$ $\hbar$ and $\hbar^2$ are put to handle the dimension. Then $$\begin{align} \hat{L}_z\hat{L}_\pm\phi_{l,m} &= \hat{L}_\pm\hat{L}_z\phi_{l,m} + [\hat{L}_z,\hat{L}_\pm]\phi_{l,m} \\ &= \hat{L}_\pm\hbar m\phi_{l,m} + \hbar\hat{L}_\pm\phi_{l,m} \\ &= (\hat{L}_\pm \hbar m \phi_{l,m})(\hbar(m\pm1)) \\ \end{align}$$ So, the eigenvalue was stepped up or down after applying the ladder operator.

Upper & Lower Bound

An upper bound exists or else $L_z$ will eventually larger than $L^2$. Let $Y_{l,m}$ be the upper bound eigenstate $$L_+Y_{l,m} = 0$$ So, $$\begin{align} < L_+Y_{l,m}|L_+Y_{l,m} > &= 0 \\ < Y_{l,m} | L_-L_+Y_{l,m} > &= 0 \\ < Y_{l,m}|\left(L_x^2 + L_y^2 + i[L_x,L_y]\right)Y_{l,m} > &= 0 \\ < Y_{l,m}|\left(L^2 -L_z^2 - \hbar L_z\right)Y_{l,m} > &= 0 \\ (\hbar^2Q_l - m^2\hbar^2 - m\hbar^2)< Y_{l,m} | Y_{l,m} > &= 0 \\ m(m+1) &= Q_l \end{align}$$ Let $$Q_l \equiv l(l+1)$$ Then, the maximum $m_+$ eigenvalue will be $$m_+= l$$ Similarly, $$m_- = -l$$ The number of states $m$ can have in $l$ state is $$N_l = 2 l + 1$$ So, $l = (N_l - 1)/2$ can either be an integer or half integer. $L_z$ in polar coordinate form is given by $$\hat{L_z} = \frac{\hbar}{i}\partial_\phi$$ Applying to the eigenfunction $$\begin{align} \hat{L}_zY_{l,m} &= \frac{\hbar}{i}\partial_\phi Y_{l,m} \\ imY_{l,m} &= \partial_\phi Y_{l,m} \\ Y_{l,m} &= e^{im\phi}P_{l,m}(\theta) \\ &= \begin{cases} Y_{l,m}(\theta,0), m \in \mathbb{Z} \\ -Y_{l,m}(\theta,0), m \in \frac{1}{2}\mathbb{Z} \end{cases} \end{align}$$ This means that the state when $l$ is half integer is always 0. As the states for $l\in \frac{1}{2}\mathbb{Z}$ do not represent anything, $l$ must be integer.