Spin Angular Momentum and the Stern-Gerlach Experiment

Electron spin, first exposed by the crude two-spot pattern on a Frankfurt screen, is now one of the central organizing concepts of modern chemistry and technology. In electron paramagnetic resonance (EPR/ESR) spectroscopy, chemists directly measure the Zeeman splitting $\Delta E = g_s\mu_B B_z$ derived above to probe radicals, transition-metal complexes, and defect centers in materials — the technique is a routine diagnostic in mechanistic organic and inorganic chemistry for detecting unpaired electrons. Nuclear magnetic resonance (NMR), ubiquitous in structure determination for organic and biological molecules, rests on the same $s=\tfrac12$ formalism applied to nuclear spins (protons, $^{13}$C, $^{31}$P), with spinors and the Pauli algebra reappearing verbatim in the density-matrix treatment of pulse sequences. Spintronics exploits the electron's spin degree of freedom, rather than only its charge, for information storage and processing — giant magnetoresistance (GMR) read-heads in hard drives, based on spin-dependent electron scattering, earned Fert and Grünberg the 2007 Nobel Prize in Physics and now underpin essentially all magnetic data storage. In quantum computing, single electron or nuclear spins trapped in quantum dots, nitrogen-vacancy centers in diamond, or superconducting circuits function as literal physical qubits, with $\alpha$ and $\beta$ playing the role of $|0\rangle$ and $|1\rangle$; sequential Stern–Gerlach-type measurements are the conceptual ancestor of modern single-spin readout protocols. Finally, spin underlies the entire structure of the periodic table through the Pauli exclusion principle, which requires that the total wavefunction (spatial × spin) of electrons be antisymmetric — the reason atomic subshells hold at most two electrons of opposite spin, a foundational fact for building up electron configurations and rationalizing chemical bonding. The coupling of this intrinsic spin with orbital angular momentum, needed to explain fine structure and the anomalous Zeeman effect quantitatively, is the natural next step captured by Coupling of Angular Momenta and Total J.

Ask the tutor: How does adding spin–orbit coupling change the energy-level splitting pattern predicted here once the electron also has nonzero orbital angular momentum ($\ell \neq 0$)?

Everything you know about orbital angular momentum came from solving $\hat L^2$ and $\hat L_z$ as differential operators acting on functions of angle, $Y_\ell^m(\theta,\phi)$, and finding that $\ell$ must be a non-negative integer because the wavefunction must be single-valued as $\phi \to \phi + 2\pi$. Spin cannot be derived that way — there is no "spin coordinate" analogous to $\theta,\phi$ for a structureless point particle. Instead, spin is postulated: we assert that the electron possesses an additional angular momentum operator $\hat{\mathbf S} = (\hat S_x, \hat S_y, \hat S_z)$ obeying exactly the commutation relations

which are the same relations satisfied by $\hat L_x,\hat L_y,\hat L_z$. Because these commutation relations alone (via the ladder-operator algebra) force the general angular-momentum eigenvalue pattern $j(j+1)\hbar^2$ for the squared operator and $m\hbar$ for the projection, with $j$ allowed to be either an integer or a half-integer, spin is permitted to take half-integer values where orbital angular momentum was not. Nature chose $s = \tfrac12$ for the electron (also for the proton and neutron), and — unlike $\ell$, which can be $0,1,2,\ldots$ for different states of the same particle — $s$ is a fixed, immutable property of the electron itself. Every electron in the universe has $s=\tfrac12$; there is no way to put an electron into an $s=1$ state. The two allowed eigenvalues of $\hat S_z$, namely $m_s = +\tfrac12$ and $m_s = -\tfrac12$, correspond to the two spin eigenstates conventionally labeled $\alpha$ (spin up, "$\uparrow$") and $\beta$ (spin down, "$\downarrow$"). Because there are only two of them and no spatial ($\theta,\phi$) dependence, they can be represented most economically as abstract two-component column vectors (spinors),

and the operators $\hat S_x,\hat S_y,\hat S_z,\hat S^2$ become $2\times 2$ matrices acting on this two-dimensional space — the smallest possible nontrivial representation of the angular-momentum algebra. Stern and Gerlach did not know any of this in 1922; they were testing Sommerfeld's space quantization hypothesis for orbital angular momentum, which predicted that an atom's magnetic moment could only point in a discrete set of directions relative to an external field, rather than any direction (as classical physics would suggest, giving a continuous smear on a screen). The experimental design is itself worth understanding in detail, because why an inhomogeneous field is required and why the atoms must be neutral are both physically essential, not incidental. A magnetic dipole $\boldsymbol{\mu}$ in a uniform field $\mathbf B$ feels only a torque, $\boldsymbol{\tau} = \boldsymbol{\mu}\times\mathbf B$, which makes it precess but exerts zero net force — the energy $U = -\boldsymbol\mu\cdot\mathbf B$ is position-independent in a uniform field, so $\mathbf F = -\nabla U = 0$. To get a measurable spatial deflection, the field must vary in space. With the field predominantly along $z$ and varying along $z$,

so atoms with different $\mu_z$ (equivalently different $m_s$, since $\mu_z \propto -g_s m_s$) are pushed by different amounts, splitting the beam. Neutral atoms are essential because a charged particle moving through such a field experiences a much larger, velocity-dependent Lorentz force ($q\mathbf v\times \mathbf B$) that would curve the whole beam and swamp the tiny spin-dependent splitting; silver's neutral atoms, with a single unpaired $5s$ electron and all other angular momenta paired off, are the cleanest possible probe of a lone electron's intrinsic magnetic moment.

The outcome is unambiguous: two discrete spots, symmetric about the undeflected position, with no beam surviving in between and none further out. A classical magnetic dipole with a continuously variable orientation angle $\theta$ relative to $\mathbf B$ would give $\mu_z = \mu\cos\theta$, smoothly distributed between $-\mu$ and $+\mu$, producing a smear, not two lines. Sommerfeld's orbital space-quantization picture, applied to the silver valence electron with $\ell=0$, predicts a single undeflected spot (there is no orbital angular momentum to couple to the field at all). Neither classical physics nor $\ell$-based quantum theory anticipated two spots from an $s$-electron; only the later hypothesis of a $\tfrac12$-valued intrinsic angular momentum with $2s+1 = 2$ projections explains it.

The eigenvalue equations for spin are written directly by analogy with orbital angular momentum, replacing $L$ by $S$ and $\ell$ by $s$:

where: $\hat S^2 = \hat S_x^2+\hat S_y^2+\hat S_z^2$ is the total-spin-squared operator, $s$ is the spin quantum number, and $\hbar$ is the reduced Planck constant. Reading: unlike $\ell$, which ranges over $0,1,2,\dots$ for different orbitals of the same atom, $s$ never changes for the electron — it is a fixed label, like the electron's charge or mass.

where: $m_s$ is the spin magnetic quantum number, restricted to $2s+1 = 2$ values for $s=\tfrac12$. Reading: this is the reason the Stern–Gerlach beam splits into exactly two components — there are exactly two eigenvalues of $\hat S_z$, full stop. The magnetic moment operator associated with spin is

where: $g_s \approx 2.0023$ is the electron spin g-factor (the anomalous doubling relative to the orbital value $g_\ell = 1$, a hallmark that spin is not ordinary orbital motion), $\mu_B$ is the Bohr magneton, $e$ is the elementary charge, and $m_e$ is the electron mass. Reading: the minus sign reflects the electron's negative charge, so $\hat{\boldsymbol\mu}_S$ points opposite to $\hat{\mathbf S}$; this is why the $m_s=+\tfrac12$ state (spin "up") is the higher-energy state in a field along $+z$ for an electron, a frequently-missed sign subtlety. The interaction energy in an external field $\mathbf B = B_z\hat z$ is the spin-Zeeman Hamiltonian:

where: the eigenvalues are $E_{m_s} = g_s\mu_B B_z\, m_s$, giving $E_{+1/2} = +\tfrac12 g_s\mu_B B_z$ and $E_{-1/2} = -\tfrac12 g_s\mu_B B_z$. Reading: this is exactly the two-level splitting drawn in the figure above, and it is the energy difference $\Delta E = g_s\mu_B B_z$ that magnetic resonance techniques (EPR/ESR) measure directly. Finally, the explicit matrix representations in the $\{\alpha,\beta\}$ basis (the Pauli spin matrices scaled by $\hbar/2$) are

where: each matrix acts on column vectors $\alpha=\binom10,\ \beta=\binom01$. Reading: $\hat S^2$ is proportional to the identity — every state in this 2D space has the same total spin magnitude $\sqrt{s(s+1)}\hbar = \frac{\sqrt3}{2}\hbar$; only the orientation (encoded in $S_z$, or a superposition of $\alpha,\beta$) varies.

Example 1 — Foundational: eigenvalues and the magnitude of the spin vector. Compute the eigenvalue of $\hat S^2$ for the electron and compare $|\mathbf S| = \sqrt{\langle \hat S^2\rangle}$ to the maximum observable $z$-component $\hbar/2$. With $s = \tfrac12$:

so

Comparing to the maximum measurable projection, $S_z^{\max} = \tfrac12\hbar = 0.500\,\hbar$, we see $|\mathbf S| > S_z^{\max}$: the total spin vector can never point exactly along $z$, since $0.866\hbar > 0.500\hbar$ — there is always a nonzero "leftover" transverse component, consistent with the fact that $S_x$ and $S_y$ have completely undetermined values ($\langle S_x\rangle=\langle S_y\rangle=0$, but $\langle S_x^2\rangle,\langle S_y^2\rangle\neq0$) whenever $S_z$ is sharp. Numerically, $\langle S_x^2\rangle + \langle S_y^2\rangle = \tfrac34\hbar^2 - \tfrac14\hbar^2 = \tfrac12\hbar^2$, split equally, $\langle S_x^2\rangle = \langle S_y^2\rangle = \tfrac14\hbar^2$, each contributing $\sqrt{1/4}\hbar=0.5\hbar$ of root-mean-square transverse spin. Limiting-case check: if one (incorrectly) imagined a classical limit where the vector could align exactly with $z$, you would need $|\mathbf S| = S_z^{\max}$; the fact that this never happens for any finite $s$ (the ratio $S_z^{\max}/|\mathbf S| = s/\sqrt{s(s+1)} \to 1$ only as $s\to\infty$) is precisely the statement that quantum angular momentum vectors are never fully "along" an axis except in the classical large-$s$ limit.

Example 2 — Applied: quantitative Stern–Gerlach beam splitting. A beam of silver atoms (electron $g_s = 2.0023$, mass of one Ag atom $M = 1.792\times10^{-25}\,\mathrm{kg}$) travels at $v = 420\,\mathrm{m/s}$ through a magnet of length $L = 3.5\,\mathrm{cm}$ with field gradient $\partial B_z/\partial z = 15\,\mathrm{T/cm} = 1500\,\mathrm{T/m}$. Find the transverse separation of the two spots at a screen $D = 15\,\mathrm{cm}$ beyond the magnet.

Step 1 — Force on each spin state. The force along $z$ is

using $\mu_z = -g_s\mu_B m_s$ with $m_s=\pm\tfrac12$. Numerically, with $\mu_B = 9.274\times10^{-24}\,\mathrm{J/T}$:

so

(with opposite sign for the two spin states).

Step 2 — Time inside the magnet and transverse deflection there.

The transverse deflection acquired inside the magnet (starting from rest transversely):

Step 3 — Transverse velocity gained, and additional drift to the screen.

Time to cross the drift region:

Additional deflection:

Step 4 — Total one-sided deflection and full spot separation.

Since the $m_s=+\tfrac12$ and $m_s=-\tfrac12$ atoms deflect by equal and opposite amounts, the full separation between the two spots is

This millimeter-scale splitting is exactly the order of magnitude Stern and Gerlach observed (their original apparatus, with cruder gradients and a much longer flight path, achieved comparable resolvable separations) and is easily resolved against the beam's thermal width — which is why the experiment gives two clean spots rather than a blur.

Limiting-case check: if the field were homogeneous, $\partial B_z/\partial z \to 0$, then $F_z\to0$ and $\Delta z\to0$: no splitting at all, consistent with the earlier statement that a uniform field exerts torque but zero net force on a magnetic dipole.

Example 3 — Exam/competition level: three sequential Stern–Gerlach magnets at an off-axis angle. A beam of spin-$\tfrac12$ particles is first filtered by a $z$-oriented Stern–Gerlach magnet, selecting only the $m_s=+\tfrac12$ ($\alpha$) sub-beam. This beam is then sent through a second magnet oriented at angle $\theta$ to the $z$-axis in the $xz$-plane, and only the "spin-up along $\hat n(\theta)$" sub-beam is kept. That sub-beam is finally sent through a third magnet oriented back along $z$. What fraction of the original $\alpha$ beam emerges as $m_s=+\tfrac12$ from the third (final) $z$-magnet, if $\theta = 37^\circ$ (an intentionally non-special angle)?

Step 1 — Spin eigenstate along an arbitrary direction $\hat n(\theta)$ in the $xz$-plane. The general spin-up eigenstate of $\hat S_{\hat n} = \sin\theta\,\hat S_x + \cos\theta\,\hat S_z$ (eigenvalue $+\hbar/2$) is

This is a standard result obtained by diagonalizing $\hat S_{\hat n} = \frac{\hbar}{2}\begin{pmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta\end{pmatrix}$ and picking the $+\hbar/2$ eigenvector.

Step 2 — Probability that the $\alpha$ beam passes the second (tilted) magnet as spin-up along $\hat n$.

With $\theta = 37^\circ$, $\theta/2 = 18.5^\circ$, $\cos(18.5^\circ) = 0.9483$, so

Step 3 — Probability that the selected $\hat n$-beam then passes the third ($z$) magnet as $m_s=+\tfrac12$. By the symmetric formula (the angle between $\hat n$ and $\hat z$ is again $\theta$, traversed in reverse),

(identical to $P_1$ because $|\langle a|b\rangle|^2 = |\langle b|a\rangle|^2$ always).

Step 4 — Combine the two independent probabilistic filtering steps.

So 80.9% of the original $\alpha$ beam survives all three filters and re-emerges as $m_s=+\tfrac12$ along $z$ — despite having been forced through an intermediate measurement along a different axis that had a nonzero (10.1%) chance of rejecting each particle at each stage. Note this is strictly less than 100%, which is the signature that the intermediate tilted measurement genuinely disturbed the ensemble; only in the trivial limiting case $\theta\to0$ (the second magnet coincides with the first and third) does $\cos^2(\theta/2)\to1$ and $P_{\text{total}}\to1$, recovering perfect transmission with no disturbance — exactly as physically expected.