Stern–Gerlach Deflection

Equivalent forms

\Delta z = \tfrac{1}{2}\frac{g\,\mu_B\,(\partial_z B)}{m}\left(\frac{L}{v}\right)^2

\mu_z = -g_s\,\mu_B\,m_s,\;m_s = \pm\tfrac{1}{2}

The experiment that made 'quantization' visible — and the canonical measurement of a qubit.

Symbol	Name	SI unit	Dimension	Range
$F_z$	Deflecting forceoutput Force along the gradient direction	N	MLT^-2	1e-25 – 1e-18
$mu_z$	Magnetic moment z-component z-component of the atom's magnetic moment (= ∓g*mu_B/2 for spin-1/2)	J/T	L^2*A	-9.274e-24 – 9.274e-24
$dB_dz$	Field gradient Spatial gradient of B_z along beam-perpendicular direction	T/m	MT^-2A^-1*L^-1	1 – 10000
$m_atom$	Atomic mass Mass of the deflected atom (e.g., silver)	kg	M	1.67e-27 – 1e-24

Unit systems

Symbol	SI	CGS	Imperial
$F_z$	N	dyn	lbf
$mu_z$	J/T	erg/G	ft*lbf/T
$dB_dz$	T/m	G/cm	T/ft
$m_atom$	kg	g	lb

Where it holds

Classical trajectory limit valid when wavelet spreads remain smaller than splitting. Breaks down for ultracold beams (interferometric Stern-Gerlach) where wave nature dominates.

Dimensional analysis

F_z \to [M*L*T^-2]

\mu _z * dB/dz \to [L^2

*A] * [M*T^-2*A^-

1*L^-1] = [M*L*T^-2]

Both sides are force.

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Discovery

Otto Stern & Walther Gerlach · 1922

Stern and Gerlach sent silver atoms through an inhomogeneous magnet expecting a continuous classical smear. They found two spots — the first direct evidence of spatial quantization and (in hindsight) of electron spin.

Try this

Why does a beam of silver atoms split into exactly two spots — not a smear?

Pass a beam of neutral atoms through an inhomogeneous magnetic field. Classically you'd see a smear; quantum mechanics says you see exactly two dots. What sets the splitting?

Research status: stable

Real-world applications

Spin-polarized atomic beams for fundamental tests
Spin-readout in quantum computing
Magnetic resonance imaging calibration
Tests of quantum measurement and EPR-Bell setups

Common misconceptions

The atom does not 'choose' spin up or down — measurement establishes the outcome
Silver was used because Ag has one unpaired 5s electron (effective spin-1/2 atom)
Sending an output beam through a perpendicular SG re-splits it — non-commutation of S_x and S_z

Experimental verification

Stern–Gerlach (1922) two silver spots. Phipps–Taylor (1927) hydrogen confirmed half-integer spin. Modern cold-atom and trapped-ion experiments use Stern–Gerlach for spin readout. Sequential SG experiments verify quantum-measurement basis change.

Derivation

Energy of dipole in field:

U = -\mu \cdot B

.

Force:

F =

-nabla*

U = (\mu \cdot

nabla)*B.

For B mostly along z and gradient along z:

F_z = \mu _z * (dB_z/dz)

.

For spin-1/2 with

g \approx 2

:

\mu _z = \pm \mu _B = \pm 9.274e-24\,\mathrm{J/T}

.

Two values of

\mu _z \to two

opposite forces

\to two

beams.

Drift time

t = L/v

, deflection

\Delta _z = (1/2)*(F_z/m

_atom)*t^2.

Limiting cases

dB/dz = 0

⟶ No deflectionWithout a gradient, force vanishes — uniform B only causes precession.

Spin-0 atom (e.g., Hg in ground state)⟶ Single spotNo magnetic moment — beam passes undeflected.

Higher spin j⟶ 2j+1 spotsMultiplet structure — Stern-Gerlach reveals total angular-momentum quantization.

What if…

What if dB/dz were uniform B (no gradient)?

No deflection — only Larmor precession of the spin around B.

What if you chained two SGs at 90 degrees?

First splits along z, second re-splits each beam into two along x — basis of quantum measurement non-commutation demos.

What if the atom had spin-1?

Three spots $(m_s = -1$ , 0, +1) instead of two — observed for triplet helium and others.

1

Splitting of silver beam in classical Stern-Gerlach

Given ·

mu z:: 9.274e-24
dB dz:: 1000
m atom:: 1.79e-25
L:: 0.1
v:: 500

Find · Delta_z (per spin)

Steps

$F = 9.274e-24 * 1000 = 9.274e-21\,\mathrm{N}$
$a = F/m = 9.274e-21 / 1.79e-25 \approx 5.18e_{4} m/s^{2}$
$t = L/v = 0.1/500 = 2e-4\,\mathrm{s}$
$\Delta _z = (1/2)*5.18e_{4}*(2e-4)^2 \approx 1.04e-3\,\mathrm{m}$

Result ·

\Delta _z = (1/2)*F/m * (L/v)^2 \approx 1.04e-5\,\mathrm{m} \approx 10 \mu m

2

Required gradient for 1 mm splitting at v = 500 m/s, L = 10 cm

Given ·

Delta z:: 0.001
L:: 0.1
v:: 500
m atom:: 1.79e-25
mu z:: 9.274e-24

Find · dB_dz

Steps

Total Delta_z (both spots $) = 2*1e-3\,\mathrm{m}$
$2*1e-3 = (\mu _z * dB_dz / m) * (L/v)^2$
$dB_dz = (2e-3 * 1.79e-25 * 500^2) / (9.274e-24 * 0.01) \approx 965\,\mathrm{T/m}$

Result ·

dB/dz = (2*\Delta _z*m*v^2)/(\mu _z*L^2) \approx 960\,\mathrm{T/m}

Spin-1/2 Operators→Pauli Exclusion Principle→Heisenberg Uncertainty Principle→