A planet orbiting the sun has two different angular momenta: orbital angular momentum due to its orbit around the sun (a 365-day period for the earth) and rotational angular momentum as it rotates on its axis (a 24-hour period for the earth). We introduced angular momentum in Chapter 12, and a brief review of 《 SECTION 12.11 is highly recommended.

A classical model of the hydrogen atom would be similar. Although a circular orbit is possible, it’s more likely that the electron would follow an elliptical orbit with the proton at one focus of the ellipse. Further, the orbit need not lie in the xy-plane. FIGURE 41.1 shows a classical orbit tilted at angle $\theta$ below the xy-plane. The electron, like a planet, has orbital angular momentum, and Figure 41.1 reminds you that the orbital angular momentum vector $\vec{L}$ is perpendicular to the plane of the orbit. (The electron also has a quantum version of rotational angular momentum, called spin, that we’ll introduce in Section 41.3.) The orbital angular momentum vector has component $L_z=L\text{cos}\theta$ along the $z$-axis.

FIGURE 41.1 The angular momentum of an elliptical orbit.

Classically, $L$ and $L_z$ can have any values. Not so in quantum mechanics. Quantum conditions 2 and 3 tell us that the electron’s orbital angular momentum is quantized. The magnitude of the orbital angular momentum must be one of the discrete values

where $l$ is an integer. Simultaneously, the $z$-component $L_z$ must have one of the values $L_z=m\hslash$, where $m$ is an integer between $-l$ and $l$. No other values of $L$ or $L_z$ allow the wave function to satisfy the boundary conditions.

The quantization of angular momentum places restrictions on the shape and orientation of the electron’s orbit. To see this, consider a hydrogen atom with orbital quantum number $l=2$. In this state, the magnitude of the electron’s angular momentum must be $L=\sqrt{6}\hslash =2.45\hslash$. Furthermore, the angular momentum vector must point in a direction such that $L_z=m\hslash$, where $m$ is one of only five integers in the range $-2\le m\le 2$.

The combination of these two requirements allows $\vec{L}$ to point only in certain directions in space, as shown in FIGURE 41.2. This is a rather unusual figure that requires a little thought to understand. Suppose $m=0$ and thus $L_z=0$. With no $z$-component, the angular momentum vector $\vec{L}$ must lie somewhere in the xy-plane. Furthermore, because the length of $\vec{L}$ is constrained to be $2.45\hslash$, the tip of $\vec{L}$ must lie somewhere on the circle labeled $m=0$. These values of $\vec{L}$ correspond to classical orbits tipped into a vertical plane.

FIGURE 41.2 The five possible orientations of the angular momentum vector for $l=2$.

Similarly, $m=2$ requires $\vec{L}$ to lie along the surface of the cone whose height is $2\hslash$ and whose side has length $2.45\hslash$. These values of $\vec{L}$ correspond to classical orbits tilted slightly out of the xy-plane. Notice that $\vec{L}$ cannot point directly along the $z$-axis. The maximum possible value of $L_z$, when $m=l$, is ${\left(L_z\right)}_{\text{max}}=l\hslash$. But $l<\sqrt{l(l+1)}$, so ${\left(L_z\right)}_{\text{max}}<L$. The angular momentum vector must have either an $x$- or a $y$-component (or both). In other words, the corresponding classical orbit cannot lie in the xy-plane.

An angular momentum vector $\vec{L}$ tilted at angle $\theta$ from the $z$-axis corresponds to an orbit tilted at angle $\theta$ out of the xy-plane. The quantization of angular momentum restricts the orbital planes to only a few discrete angles. For quantum state $\left(n\text{,}l\text{,}m\right)$, the angle of the angular momentum vector is

Angles ${\theta }_{22}\text{,}{\theta }_{21}$, and ${\theta }_{20}$ are labeled in Figure 41.2. Orbital planes at other angles are not allowed because they don’t satisfy the quantization conditions for angular momentum.