In 1927, with Schrödinger’s quantum theory brand new, the Stern-Gerlach technique was used to measure the magnetic moment of hydrogen atoms. The ground state of hydrogen is $1s$, with $l=0$, so the atoms should have no magnetic moment and there should be no deflection at all. Instead, the experiment produced the two-peaked distribution shown in FIGURE 41.12.

FIGURE 41.12 The outcome of the Stern-Gerlach experiment for hydrogen atoms.

Because the hydrogen atoms were deflected, they must have a magnetic moment. But where does it come from if $L=0?$ Even stranger was the deflection into two groupings, rather than an odd number. The deflection is proportional to $L_z$, and $L_z=m\hslash$ where $m$ ranges in integer steps from $-l$ to $+l$. The experimental results would make sense only if $l=\frac{1}{2}$, allowing $m$ to take the two possible values $-\frac{1}{2}$ and $+\frac{1}{2}$. But according to Schrödinger’s theory, the quantum numbers $l$ and $m$ must be integers.

An explanation for these observations was soon suggested, then confirmed: The electron has an inherent magnetic moment. After all, the electron has an inherent gravitational character, its mass $m_{\text{e}}$, and an inherent electric character, its charge $q_{\text{e}}=-e$. These are simply part of what an electron is. Thus it is plausible that an electron should also have an inherent magnetic character described by a built-in magnetic moment $\vec{\mu }{}_{\text{e}}$. A classical electron, if thought of as a little ball of charge, could spin on its axis as it orbits the nucleus. A spinning ball of charge would have a magnetic moment associated with its angular momentum. This inherent magnetic moment of the electron is what caused the unexpected deflection in the Stern-Gerlach experiment.

If the electron has an inherent magnetic moment, it must have an inherent angular momentum. This angular momentum is called the electron’s spin, which is designated $\vec{S}$. The outcome of the Stern-Gerlach experiment tells us that the $z$-component of this spin angular momentum is

The quantity $m_s$ is called the spin quantum number.

The $z$-component of the spin angular momentum vector is determined by the electron’s orientation. The $m_s=+\frac{1}{2}$ state, with $S_z=+\frac{1}{2}\hslash$, is called the spin-up state, and the $m_s=-\frac{1}{2}$ state is called the spin-down state. It is convenient to picture a little angular momentum vector that can be drawn $\uparrow$ for an $m_s=+\frac{1}{2}$ state and $\downarrow$ for an $m_s=-\frac{1}{2}$ state. We will use this notation in the next section. Because the electron must be either spin-up or spin-down, a hydrogen atom in the Stern-Gerlach experiment will be deflected either up or down. This causes the two groups of atoms seen in Figure 41.12. No atoms have $S_z=0$, so there are no undeflected atoms in the center.

The spin angular momentum $S$ is analogous to Equation 41.3 for L:

where $s$ is a quantum number with the single value $s=\frac{1}{2}\text{.}S$ is the inherent angular momentum of the electron. Because of the single value of $s$, physicists usually say that the electron has “spin one-half.” FIGURE 41.13, which should be compared to Figure 41.2, shows that the terms “spin-up” and “spin-down” refer to $S_z$, not the full spin angular momentum. As was the case with $\vec{L}$, it’s not possible for $\vec{S}$ to point along the $z$-axis.

FIGURE 41.13 The spin angular momentum has two possible orientations.

The electron’s spin has significant implications for atomic structure. The solutions to the Schrödinger equation could be described by the three quantum numbers $n,l,\text{and}m$, but the Stern-Gerlach experiment implies that this is not a complete description of an atom. Knowing that a ground-state atom has quantum numbers $n=1\text{,}l=0$, and $m=0$ is not sufficient to predict whether the atom will be deflected up or down in a nonuniform magnetic field. We need to add the spin quantum number $m_s$ to make our description complete. (Strictly speaking, we also need to add the quantum number $s$, but it provides no additional information because its value never changes.) So we really need four quantum numbers $\left(n\text{,}l\text{,}m\text{,}{m}_s\right)$ to characterize the stationary states of the atom. The spin orientation does not affect the atom’s energy, so a ground-state electron in hydrogen could be in either the $(1\text{,}0\text{,}0\text{,}+\frac{1}{2})$ spin-up state or the $(1\text{,}0\text{,}0\text{,}-\frac{1}{2})$ spin-down state.

The fact that $s$ has the single value $s=\frac{1}{2}$ has other interesting implications. The correspondence principle tells us that a quantum particle begins to “act classical” in the limit of large quantum numbers. But $s$ cannot become large! The electron’s spin is an intrinsic quantum property of the electron that has no classical counterpart.