Definitions

In classical mechanics, the angular momentum of an object with position ${\bf x}$ and momentum ${\bf p}$ is defined as

$${\bf L}= {\bf x}\times {\bf p},$$ (295)

that is the cross product (vector product) of the position and the momentum. In components, this is

$${\bf L}= \left\vert \begin{array}{lll} .&.&.\\ x_1&x_2&x_3\\... ...array}\right\vert =(x_2p_3-x_3p_2,x_3p_1-x_1p_3,x_1p_2-x_2p_1).$$ (296)

The corresponding quantum mechanical operator is obtained from the correspondence principle that replaces ${\bf x}\to {\bf\hat{x}}$, i.e. the position operator, and ${\bf p} \to -i\hbar \nabla$, i.e. the momentum operator. Because we are in three dimensions, the operators are vectors:

$${\bf\hat{x}}=(\hat{x}_1,\hat{x}_2,\hat{x}_3),\quad {\bf p}=-i\hba... ...\partial_{x_1},\partial_{x_2},\partial_{x_3}) =(\hat{p}_1,\hat{p}_2,\hat{p}_3).$$ (297)

The angular momentum operator therefore becomes

$$\hat{\bf {L}}=-i\hbar {\bf\hat{x}} \times \nabla =(\hat{x}_2\hat{... ...2,\hat{x}_3\hat{p}_1-\hat{x}_1\hat{p}_3,\hat{x}_1\hat{p}_2-\hat{x}_2\hat{p}_1).$$ (298)

In polar coordinates, one has to use the spherical polar expression for the Nabla operator that you learned in vector analysis. The corresponding expression for the components of the angular momentum operator become

$$\hat{L}_x = -i\hbar \left( -\sin \varphi \frac{\partial}{\partial \theta} -\cos \varphi \cot \theta \frac{\partial}{\partial \varphi} \right)$$

$$\hat{L}_y = -i\hbar \left( \cos \varphi \frac{\partial}{\partial \theta} -\sin \varphi \cot \theta \frac{\partial}{\partial \varphi} \right)$$

$$\hat{L}_z = -i\hbar\frac{\partial}{\partial \varphi}.$$ (299)

Notice that in spherical polar coordinates, the $z$-axis is the central axis of the coordinate system with the angle $\varphi$ revolving around it. The angular momentum $L_z$ corresponds to revolutions around the $z$-axis, its quantum mechanical expression becomes quite simple: basically, just a differentiation with respect to the angle $\varphi$ `around the $z$ axis'.

Another important observation is that the angular momentum square, that is the operator ${\bf\hat{L}}^2$, becomes

$${\bf\hat{L}}^2= -\hbar^2 \left[\frac{1}{\sin \theta}\frac{\partia... ...a}\right) +\frac{1}{\sin^2 \theta}\frac{\partial^2}{\partial \varphi^2}\right].$$ (300)

We notice that this is just $-\hbar^2$ times the expression $\hat{\Omega}$, eq. (4.61), for the angular part of the Laplacian $\Delta$! In particular, the eigenfunctions $Y_{lm}$ of $\hat{\Omega}$ are the eigenfunctions of the angular momentum square, cf. eq.(4.63) with $c=-l(l+1)$,

$${\bf\hat{L}}^2 Y_{lm}(\theta,\varphi) = \hbar^2 l(l+1)Y_{lm}(\theta,\varphi),\quad l=0,1,2,3,...$$ (301)

Furthermore, the dependence of $Y_{lm}(\theta,\varphi)$ on the angle $\varphi$ is only through the exponential $e^{im\varphi}$. We thus have

$$\hat{L}_z Y_{lm}(\theta,\varphi) = \hbar m Y_{lm}(\theta,\varphi),$$ (302)

which means that $Y_{lm}(\theta,\varphi)$ are eigenfunctions of the $z$-component of the angular momentum, too.

The eigenfunctions $\Psi_{nlm}(r,\theta,\varphi)$ of the Hamiltonian $\hat{H}$ for the hydrogen atom therefore are also eigenfunctions of ${\bf\hat{L}}^2$ and $\hat{L}_z$. We summarize this in three equations

$$\hat{H} \Psi_{nlm}(r,\theta,\varphi) = E_n \Psi_{nlm}(r,\theta,\varphi)$$

$${\bf\hat{L}}^2 \Psi_{nlm}(r,\theta,\varphi) = \hbar^2l(l+1) \Psi_{nlm}(r,\theta,\varphi)$$

$$\hat{L}_z \Psi_{nlm}(r,\theta,\varphi) = \hbar m \Psi_{nlm}(r,\theta,\varphi).$$ (303)