Classical Stern-Gerlach Experiment

You cannot be shocked by two spots on a photographic plate until you know exactly what classical physics predicted instead — and earning that prediction honestly is harder than it looks. This lesson builds the 1922 apparatus from the ground up: dipoles, gradients, torques, and the one subtlety (precession) that makes the whole thing work as a measuring instrument at all.

Learning Objectives

After this lesson you will be able to:

1. Compute the force and torque on a magnetic dipole in an arbitrary field, and explain why a uniform field can never separate a beam.
2. Show that a spinning rotor's magnetic moment satisfies $\boldsymbol{\mu} = \gamma\mathbf{L}$, and derive the Larmor precession frequency from the torque equation.
3. Contrast the motion of a dipole without angular momentum (libration) against one with it (precession), and explain why only the latter yields a constant deflecting force.
4. Derive the classical prediction for the detector pattern from an isotropic source — a uniform band with sharp edges, not a Gaussian blur.
5. Estimate the deflection of a silver-atom beam with realistic 1922 apparatus numbers.
6. State precisely which classical assumptions the experiment was designed to test, so that the quantum result in the next lesson lands as a genuine contradiction.

Intuition

Everything we can learn about an atom, we learn by letting it interact with something we can see. The Stern-Gerlach experiment is the archetype: it converts an invisible internal property of an atom — its magnetism — into a visible position on a screen. The strategy is worth stating in the abstract, because the whole program uses it: to measure a quantity, arrange for it to control a force, then let the force act long enough that the resulting displacement is macroscopic.

Why magnetism? Because an atom containing moving charge is a tiny magnet, and magnets respond to magnetic fields in a way we can compute exactly. But there is a catch that turns out to be the entire engineering problem. A magnet in a uniform field feels a twist and no push: a compass needle turns toward north, it does not fly north. To get a push you need the field to be different at the two ends of the magnet — an inhomogeneous field, a gradient. That is why the Stern-Gerlach magnet has one sharp pole piece and one grooved one: it is a machine for making the field change rapidly with height.

There is a second catch, subtler and more interesting, and it is the reason this lesson is longer than "put a dipole in a gradient." A dipole that is just a little bar magnet will swing back and forth in a field like a pendulum, so the component of its moment along the field — the part that sets the force — keeps changing sign as it swings. Such an object gets pushed up, then down, then up, and the beam smears in a complicated way that encodes nothing clean. But a dipole that carries angular momentum behaves completely differently: instead of swinging, it precesses, like a gyroscope, and the component along the field stays exactly constant. Only then is the apparatus a faithful meter. Atoms, fortunately, are the second kind [Fre, §1.1].

Theory

Force and torque on a magnetic dipole

A small current loop or bar magnet is characterized by a single vector, its magnetic dipole moment $\boldsymbol{\mu}$, which points from the south pole to the north pole of the equivalent needle and whose magnitude measures the strength of the magnet. For a planar loop of current $I$ enclosing area $A$ with unit normal $\hat{n}$ given by the right-hand rule, $\boldsymbol{\mu} = IA\,\hat{n}$. Everything about how the object responds to a magnetic field $\mathbf{B}$ follows from one scalar, the interaction energy

The energy is lowest when the moment is aligned with the field, which is why compass needles point north. From $U$ we get both of the mechanical responses at once. The torque is the derivative of $U$ with respect to orientation, and the force is minus its gradient with respect to position:

Read these two formulas carefully, because they say different things. The torque depends only on the field at the dipole; it is nonzero even in a perfectly uniform field. The force depends on how the field changes from place to place; it vanishes identically when $\mathbf{B}$ is uniform, no matter how strong the field is. A uniform field reorients magnets; only a non-uniform field moves them.

The projection that sets the force

Orient the apparatus so the field points mainly along $\hat{z}$ and varies mainly along $z$: $\mathbf{B} \approx B_z(z)\hat{z}$. Then $\boldsymbol{\mu}\cdot\mathbf{B} = \mu_z B_z(z)$, and

The single number that controls the deflection is $\mu_z = \mu\cos\alpha$, the projection of the moment onto the field axis, where $\alpha$ is the angle between $\boldsymbol{\mu}$ and $\mathbf{B}$. A dipole tipped by $\alpha$ is pushed as though it were a weaker dipole pointing exactly along the axis. Since $\cos\alpha$ runs continuously over $[-1, +1]$ as the dipole is tilted from parallel to antiparallel, classical physics says the deflecting force runs continuously over $[-\mu\,\partial_z B_z,\ +\mu\,\partial_z B_z]$. Hold on to that word continuously — it is the prediction the experiment was built to confirm, and the one it destroyed.

Careful. The direction of $\mathbf{B}$ and the direction in which $|\mathbf{B}|$ increases are independent choices. Below the north pole of a magnet the field points up and gets stronger as you go up; below its south pole the field points down and still gets stronger as you go up. The sign of $F_z$ depends on the gradient, so you must track both. Freericks makes this the point of a dedicated graphical construction [Fre, §1.1]; algebraically it is just the statement that $\mu_z$ and $\partial_z B_z$ each carry their own sign.

Two classical rotors: the needle and the loop

Now the physics that decides whether the apparatus works. Consider a dipole free to rotate, and ask how $\mu_z$ evolves in time. There are two very different classical answers.

Case 1 — a needle with no angular momentum. Model the magnet as a rigid body of moment of inertia $I_{\text{rot}}$ that is not spinning. The torque $\tau = -\mu B\sin\alpha$ drives the orientation angle according to

which is exactly the pendulum equation. The needle swings toward alignment, overshoots (nothing stops instantly), swings back, and — with no friction, as in vacuum — repeats the cycle forever in a motion called libration. Its projection $\mu_z = \mu\cos\alpha(t)$ oscillates, so $F_z$ oscillates too, in general changing sign. An atom of this kind entering the magnet would be pushed up and down repeatedly, and where it landed would depend on the phase of its swing when it entered. The screen pattern would be a mess that encodes the initial phase as much as the initial tilt.

Case 2 — a rotor that carries angular momentum. Now suppose the magnetism comes from something actually circulating, so that the moment is tied to an angular momentum $\mathbf{L}$ by a constant of proportionality. For any rigid body whose charge and mass are distributed alike, the current and the mass flow are the same motion, and the two vectors are strictly parallel:

where the constant $\gamma$ is called the gyromagnetic ratio. (Quick derivation for a circular orbit: charge $q$ going around a loop of radius $r$ at speed $v$ has period $T = 2\pi r/v$, hence current $I = q/T = qv/2\pi r$ and moment $\mu = I\pi r^2 = qvr/2 = (q/2m)(mvr) = \gamma L$.)

Precession: why the loop is the honest model

With $\boldsymbol{\mu} = \gamma\mathbf{L}$, the rotational equation of motion becomes a closed equation for $\mathbf{L}$ alone:

Two conservation laws fall straight out. Dotting with $\mathbf{L}$ gives $\tfrac{d}{dt}|\mathbf{L}|^2 = 2\gamma\,\mathbf{L}\cdot(\mathbf{L}\times\mathbf{B}) = 0$: the length is fixed. Dotting with $\hat{z}$ for $\mathbf{B} = B\hat{z}$ gives

because $\mathbf{L}\times\mathbf{B}$ has no $z$-component when $\mathbf{B}\parallel\hat{z}$. The projection along the field is a constant of the motion. The vector neither aligns nor swings: with fixed length and fixed $z$-component, its tip is confined to a circle, and it sweeps that circle at a steady rate — this motion is precession, the same motion you see in a spinning top or a gyroscope whose axle traces a cone instead of falling over. Writing $\mathbf{L}$ in components shows the sweep rate is independent of the tilt angle:

the Larmor frequency. Physically, precession is angular momentum conservation asserting itself: the torque cannot change $|\mathbf{L}|$ or $L_z$, so all it can do is rotate $\mathbf{L}$ about the field axis.

This is the result that rescues the experiment. Because $\mu_z$ is constant during precession, the force $F_z = \mu_z\,\partial_z B_z$ is constant throughout the flight through the magnet. Each atom receives a steady transverse impulse proportional to its own $\mu_z$, and the landing position on the screen is a faithful, monotonic readout of that one number. A librating needle would give no such clean map; a precessing rotor does. An inhomogeneous magnet is therefore a $\mu_z$-meter, and only for objects that carry angular momentum.

Why an atom should behave like a current loop

In 1922, the reigning picture of the atom was Rutherford's nucleus orbited by electrons, quantized after Bohr and Sommerfeld. Whatever else was wrong with it, it made one prediction that is still right: the orbiting electron is a current loop, so an atom should carry a magnetic moment tied to its orbital angular momentum by $\gamma = -e/2m_e$ (negative, because the electron's charge is negative, so $\boldsymbol{\mu}$ is antiparallel to $\mathbf{L}$). With angular momentum of order $\hbar$, the natural unit of atomic magnetism is the Bohr magneton

So the expected moment of a silver atom is "of order one Bohr magneton," and the expected motion in the magnet is precession at the Larmor rate with a constant deflecting force. Everything up to here is classical mechanics and classical electromagnetism, with one quantum input (the scale $\hbar$) and no quantum weirdness at all.

The apparatus

flowchart LR
    OVEN["Oven<br/>silver vapor, ~1300 K"] --> S1["Slit 1"]
    S1 --> S2["Slit 2<br/>(collimation)"]
    S2 --> MAG["Inhomogeneous magnet<br/>length ~5 cm<br/>gradient ~10³ T/m"]
    MAG --> DRIFT["Field-free drift"]
    DRIFT --> PLATE["Glass plate<br/>(detector)"]

Four stages, each doing one job. A hot oven of metallic silver emits atoms through a pinhole, giving a thermal beam with mean speed of order several hundred metres per second. A second aperture downstream performs collimation: it discards all but the atoms travelling nearly parallel to the axis, so that the undeflected beam makes a narrow line on the plate. Everything the experiment can resolve is set here — a deflection smaller than the collimated beam width is invisible. Next, the beam crosses the field gradient, where each atom picks up a transverse velocity proportional to its $\mu_z$. Finally a field-free drift region lets those tiny transverse velocities accumulate into a visible separation before the atoms strike a glass plate and stick.

Silver was an excellent choice: it is heavy enough to travel in straight lines, evaporates at accessible temperatures, condenses permanently where it lands, and — the part nobody could have designed on purpose — a silver deposit far too thin to see develops into a visible dark trace when exposed to sulfur, which the experimenters supplied by smoking cheap cigars over the plates [Fre, §1.2].

The classical prediction: a uniform band with sharp edges

Now assemble the prediction. The oven has no reason to prefer any direction, so the moments emerge isotropically: the unit vector $\hat{\mu}$ is uniformly distributed over the sphere. The deflection of an atom is proportional to $\mu_z = \mu\cos\alpha$, so we need the distribution of $\cos\alpha$, not of $\alpha$. Solid angle on the sphere is

so a uniform distribution over the sphere is a uniform distribution in $\cos\alpha$ on $[-1,+1]$. (This is Archimedes' hat-box theorem: equal slices of a sphere by parallel planes have equal area.)

Therefore classical physics predicts something sharper than the "smear" of folklore. The pattern on the plate is a band of uniform intensity, symmetric about the undeflected line, extending out to $\pm z_{\max}$ where $z_{\max}$ corresponds to $|\mu_z| = \mu$, with abrupt edges and no concentration anywhere — least of all at the extremes. Real effects soften this: the thermal spread in speeds means slower atoms deflect more (the deflection scales as $1/v^2$), and the finite beam width blurs the edges. But the qualitative signature is unmistakable: one connected blob covering every intermediate deflection, densest nowhere, with nothing special about the endpoints.

Everything the next lesson does depends on your having internalized this prediction. It is not "we had no idea what would happen." It is a specific, quantitative, edge-to-edge continuum.

Worked example: how big is the deflection?

Take a silver atom, $M = 1.79\times10^{-25}$ kg, moving at $v = 660$ m/s through a magnet of length $\ell = 5$ cm with gradient $\partial_z B_z = 10^3$ T/m, followed by a drift of $D = 20$ cm. Consider the extreme case $\mu_z = \mu_B$. The acceleration inside the magnet is

The time in the field is $t_1 = \ell/v = 7.58\times10^{-5}$ s, giving a displacement inside the magnet of $z_1 = \tfrac12 a t_1^2 = 1.49\times10^{-4}$ m and an exit transverse velocity of $v_z = a t_1 = 3.93$ m/s. The drift takes $t_2 = D/v = 3.03\times10^{-4}$ s and adds $z_2 = v_z t_2 = 1.19\times10^{-3}$ m. The total is

so the classical band should be about $2z_{\max} \approx 2.7$ mm tall. Two lessons hide in this arithmetic. First, the drift region contributes nearly nine tenths of the deflection — the magnet supplies the impulse, the flight path supplies the magnification, exactly as a lever arm does. Second, the whole effect is sub-millimetre-per-decimetre: with a beam only tens of microns wide and a plate read under a microscope, this experiment sat right at the edge of what 1922 could do, which is why it failed repeatedly before it succeeded.

A Stern-Gerlach experiment that cannot work

Suppose a colleague proposes an improved apparatus using a beautifully uniform field, arguing that a cleaner field must give a cleaner separation. The proposal is hopeless, and it is worth being able to say exactly why [Fre, §1.1.5].

In a uniform field, $\nabla(\boldsymbol{\mu}\cdot\mathbf{B}) = 0$, so $\mathbf{F} = 0$ identically: no atom is deflected, whatever its moment. The field is not inert — each atom's moment precesses at $\omega_L = |\gamma|B$ about the field axis — but precession moves no centre of mass. Every atom flies straight through and lands in the undeflected line, and the plate shows a single narrow trace identical to the one you get with the magnet switched off. The apparatus sorts nothing, because sorting requires the energy to depend on position, and in a uniform field it does not.

The comparison also sharpens what a gradient buys you. A source that emitted only one value of $\mu_z$ would, in a real inhomogeneous magnet, produce a single displaced line; in a uniform field it would produce a single undisplaced line. The two experiments differ observably — one line moves when you reverse the gradient, the other does not — so gradient-off and single-valued-source are distinguishable hypotheses, not the same null result.

Hands-on (Python)

Simulate the classical plate and confirm the uniform-band prediction, including the thermal velocity spread that softens its edges.

import numpy as np
import matplotlib.pyplot as plt

rng = np.random.default_rng(0)
N = 200_000

mu_B, M = 9.274e-24, 1.79e-25          # J/T, kg
grad, ell, D = 1.0e3, 0.05, 0.20       # T/m, m, m
v0 = 660.0                             # m/s

# Nominal band edge: |cos(alpha)| = 1 at the mean speed (the worked example).
z_edge = (mu_B * grad / M) * ell * (ell / 2 + D) / v0**2

# (a) Isotropic moments  ->  cos(alpha) uniform on [-1, 1]  (Archimedes).
cos_alpha = rng.uniform(-1.0, 1.0, N)

# (b) Thermal speed spread (deflection scales as 1/v^2, so slow atoms
#     overshoot the nominal edge and soften it into a tail).
v = np.clip(v0 * (1.0 + 0.15 * rng.normal(size=N)), 250.0, None)

# (c) Deflection: impulse in the magnet, magnified over the drift, plus the
#     finite width of the collimated beam.
z = cos_alpha * z_edge * (v0 / v) ** 2
z += 0.02 * z_edge * rng.normal(size=N)

plt.hist(z / z_edge, bins=300, density=True)
plt.xlabel("deflection / nominal band edge")
plt.ylabel("intensity")
plt.title("Classical prediction: a filled band, not two spots")
plt.show()

print(f"nominal band edge        = {z_edge * 1e3:.2f} mm")
print(f"within 10% of the centre = {np.mean(np.abs(z) < 0.1 * z_edge):.3f}")
print(f"outer 10% of the band    = {np.mean(np.abs(z) > 0.9 * z_edge):.3f}")
# Expect a band edge of ~1.34 mm and ~0.10 of the atoms near the centre --
# the flat-density prediction of E3. The outer bin runs a little higher
# (~0.16) only because the 1/v^2 tail lets slow atoms overshoot the nominal
# edge. Nothing is special about the extremes; the middle is not empty.

The number to remember from the last line: the classical centre is not empty. In the next lesson, it will be.

Exercises

E1 (easy). A bar magnet is released in a perfectly uniform magnetic field. Describe its subsequent motion, and state what happens to its centre of mass. Repeat for a spinning gyroscope-like magnet with $\boldsymbol{\mu} = \gamma\mathbf{L}$.

E2 (easy). Compute the Larmor frequency for $\gamma = -e/2m_e$ in a field of $B = 0.1$ T, and compare the precession period with the $7.6\times10^{-5}$ s an atom spends inside the magnet in the worked example. How many precessions occur during the flight?

E3 (medium). Show that for moments distributed isotropically, the deflection density on the plate is uniform between its extremes, and compute the probability that an atom lands in the outer 10% of the band on either side. Contrast with the intuitive guess that most atoms land near the edges.

E4 (medium). In the worked example the drift region contributed $z_2 = 1.19$ mm out of $z_{\max} = 1.34$ mm. Derive the general expression for the ratio $z_2/z_1$ and explain, in terms of the underlying kinematics, why lengthening the drift is a cheaper way to gain signal than strengthening the magnet.

E5 (hard). A referee objects that the atom is an extended object, so $F_z = \mu_z\,\partial_z B_z$ is only the leading term. Estimate the size of the neglected corrections for an atom of radius $r \sim 10^{-10}$ m in a gradient of $10^3$ T/m with $B \sim 0.1$ T, and separately explain why $\nabla\cdot\mathbf{B} = 0$ forbids a field with $\partial_z B_z \ne 0$ and no transverse gradients — and why the transverse forces nevertheless do not spoil the measurement.

Checkpoint

1. Why does a Stern-Gerlach magnet need an inhomogeneous field? What would the plate show if the field were perfectly uniform but very strong?
2. What is the difference between the motion of a magnetic dipole that carries angular momentum and one that does not, and why does only one of them make the apparatus a reliable meter?
3. Starting from $d\mathbf{L}/dt = \gamma\mathbf{L}\times\mathbf{B}$, show that both $|\mathbf{L}|$ and $L_z$ are conserved. What motion is left?
4. Sketch the pattern that classical physics predicts on the plate for an isotropic source. Where is the intensity highest?
5. In the worked example, which stage of the apparatus contributes most of the final deflection, and why?

Further Reading

• [Fre] J. K. Freericks, Quantum Mechanics Done Right, §1.1 — the source this lesson follows, with its graphical "fat arrow" and projection constructions worked out pictorially rather than algebraically.
• [Sak] Sakurai & Napolitano, §1.1 — the classic Stern-Gerlach opening, and the standard reference for the sequential experiments taken up in the next lesson.
• [ER] Eisberg & Resnick, Ch. 8 — the full experimental and historical account of magnetic moments and space quantization.
• [Gri] Griffiths & Schroeter, §4.4 — magnetic interactions and Larmor precession in wave-mechanics language.
• P.6.3 Magnetic Moments, Stern–Gerlach & Spin — the same experiment inside the program's Pre-Term spine, reaching spin from the angular-momentum ladder.
• P.4.4 Hamiltonian Mechanics — the energy-first formulation that makes $U = -\boldsymbol{\mu}\cdot\mathbf{B}$ the natural starting point.

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