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.
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