$\mathbf{L} = \mathbf{r} \times \mathbf{p}$ is the angular momentum operator where $\mathbf{r} = (x, y, z)$ is the positional operator and $\mathbf{p} = (p_x, p_y, p_z)$ is the momentum operator
Show that:
\begin{align} L_x &= - i \hbar \left(y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \right) \\ L_y &= - i \hbar \left(z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \right) \\ L_z &= - i \hbar \left(x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) \end{align}Hint: $p_x = -i\frac{\partial}{\partial x}$
Hint 2: To do this exercise you need to do a cross product
The particle on a sphere is given by:
\begin{align} L^2 Y_{m_l}^l (\varphi, \theta) = \hbar^2 l(l+1) Y_{m_l}^l (\varphi, \theta) \\ L_z Y_{m_l}^l (\varphi, \theta) = \hbar m_l Y_{m_l}^l (\varphi, \theta) \end{align}where $l = 0, 1, 2, ...$ and $m_l = -l, -l+1,...,0,..., l-1, l$
Make a surface plot of a sphere where the particle can be located
In [ ]:
def Y_10(phi, theta):
def Y_11(phi, theta):
def Y_1m1(phi, theta):
# or more generally using sph_harm from scipy.special
from scipy.special import sph_harm
The parametrization for the probability densities are:
\begin{align} x &= \cos \theta \sin \phi |Y_{m_l}^l (\varphi, \theta)|^2 \\ y &= \sin \theta \sin \phi |Y_{m_l}^l (\varphi, \theta)|^2 \\ z &= \cos \phi |Y_{m_l}^l (\varphi, \theta)|^2 \end{align}Give an explaination to why the parametrization looks like this.
Plot the probability density of $|Y_{m_l}^l (\varphi, \theta)|^2$ for $L = 0$, $L = 1$ and $L=2$ for $m_l = -l, -l+1,...,0,..., l-1, l$.
Try to plot them with the sphere from exercise 3, here it is a good idea to add the keyword $alpha = 0.3$ for the sphere plot.
The formulas for $Y_{m_l}^l (\varphi, \theta)$ for $l=1$ and $m_l=-1,0,1$ represent the three $p$ orbitals. It is easy to identify $p_z$ as the function $Y_{0}^1$, whereas $p_x$ and $p_y$ must be obtained as combinations of $Y_{-1}^1$ and $Y_{1}^1$.
Advanced (not mandatory): repeat the points above for $d$ and $f$ orbitals.