§8.2Magnetic Moments and the Stern–Gerlach Experiment

Part V Phillips pp. 158–163 · ~17 min read

  • magnetic moment
  • Landé g-factor
  • Bohr magneton
  • nuclear magneton
  • Zeeman effect
  • Stern–Gerlach experiment

A magnetic moment is what turns an angular momentum into something a laboratory can grip. The field does not change the levels’ existence; it makes them separate enough to see.

§8.1 asserted that a component of angular momentum comes only in integer or half-integer multiples of \hbar, and admitted the claim was surprising. This section makes it visible. The route is magnetism: a moving charge is a magnet, a magnet in a field has an orientation energy, and if the angular momentum is quantized then so is that energy — which you can see, because a non-uniform field will sort the atoms in space according to it.

A classical magnet is an orbiting charge

The simplest magnetic moment in classical physics is a charged particle going round in a circle. Its moment turns out to be proportional to its angular momentum:

μ=q2mL(8.7)\boldsymbol\mu = \frac{q}{2m}\mathbf L\tag{8.7}

The derivation is three lines, and worth doing because every quantum formula in this section is a modification of it.

Quantum magnets: the same formula, three corrections

In quantum physics magnetic moments are still proportional to angular momenta, but the angular momenta are now fuzzy vectors — so only μz\mu_z has a definite value. Three cases matter, and each modifies Eq. (8.7) differently.

Electron spin carries an extra factor of 2:

μz(Spin)=2e2meSz=2e2mems(8.8)\mu_z^{(\mathrm{Spin})} = -2\frac{e}{2m_e}S_z = -2\frac{e\hbar}{2m_e}m_s\tag{8.8}

Electron orbital motion does not:

μz(Orbital)=e2meLz=e2meml(8.9)\mu_z^{(\mathrm{Orbital})} = -\frac{e}{2m_e}L_z = -\frac{e\hbar}{2m_e}m_l\tag{8.9}

A whole atom gets a factor gg that interpolates between them:

μz(Atom)=ge2meJz=ge2memj(8.10)\mu_z^{(\mathrm{Atom})} = -g\frac{e}{2m_e}J_z = -g\frac{e\hbar}{2m_e}m_j\tag{8.10}

The factor gg in Eq. (8.10) is the Landé g-factor , and for an atomic state with quantum numbers jj, ll and ss it is

g=1+j(j+1)l(l+1)+s(s+1)2j(j+1)g = 1 + \frac{j(j+1) - l(l+1) + s(s+1)}{2j(j+1)}

which takes the value g=2g = 2 when the magnetism is all spin and g=1g = 1 when it is all orbital.

Two natural units

Equations (8.8)–(8.10) all carry the same combination, so it gets a name — the Bohr magneton :

μB=e2me=9.274×1024 JT1(8.11)\mu_B = \frac{e\hbar}{2m_e} = 9.274\times10^{-24}\ \mathrm{J\,T^{-1}}\tag{8.11}

Protons and neutrons are composite — they contain quarks and gluons — so their moments are not given by any clean formula. They are measured:

μz(Proton)=2.79e2mpmjandμz(Neutron)=1.95e2mpmj(8.12)\mu_z^{(\mathrm{Proton})} = 2.79\frac{e\hbar}{2m_p}m_j \qquad\text{and}\qquad \mu_z^{(\mathrm{Neutron})} = -1.95\frac{e\hbar}{2m_p}m_j\tag{8.12}

which names a second unit, the nuclear magneton :

μN=e2mp=5.05×1027 JT1(8.13)\mu_N = \frac{e\hbar}{2m_p} = 5.05\times10^{-27}\ \mathrm{J\,T^{-1}}\tag{8.13}

Magnetic energy: a continuum, or a ladder

A classical moment in a field has orientation energy

Emag=μB(8.14)E_{\mathrm{mag}} = -\boldsymbol\mu\cdot\mathbf B\tag{8.14}

and taking B\mathbf B along zz,

Emag=μzB(8.15)E_{\mathrm{mag}} = -\mu_z B\tag{8.15}

Classically μz\mu_z can be anything between +μ+\mu and μ-\mu, so the energy is a continuum between μB-\mu B and +μB+\mu B. Quantum mechanically μz\mu_z is quantized, so the energy is too. Using Eq. (8.10), an atomic state with quantum numbers jj and mjm_j has Emag=mjgμBBE_{\mathrm{mag}} = m_j\,g\mu_B B, giving 2j+12j+1 levels:

Emag=+jgμBB, +(j1)gμBB, , (j1)gμBB, jgμBB(8.16)E_{\mathrm{mag}} = +j\,g\mu_B B,\ +(j-1)g\mu_B B,\ \dots,\ -(j-1)g\mu_B B,\ -j\,g\mu_B B\tag{8.16}

Equation (8.16) — an evenly spaced ladder whose rungs you can tune

symbol
is
the only thing that varies between levels: it takes 2j+1 values from +j to −j. So a level with quantum number j does not shift in a field — it SPLITS into 2j+1.
units
dimensionless
type
integer or half-integer

Click any symbol to see what it is, what units it carries, and what kind of object it is once you put it in an array.

The book’s Fig. 8.2 draws the ladder for three values of jj:

Fig. 8.2 — magnetic energy levels for j = ½, 1 and 3/2

energy (gμ_B B)j = 3/2, m_j = +3/21.50j = 1, m_j = +11.00j = 1/2, m_j = +1/2 · j = 3/2, m_j = +1/20.50j = 1, m_j = 00.00j = 1/2, m_j = −1/2 · j = 3/2, m_j = −1/2-0.50j = 1, m_j = −1-1.00j = 3/2, m_j = −3/2-1.50

Click a gold level for its energy and degeneracy.

No transitions shown in this view.

Levels in units of gμ_B B, at fixed field. A state with quantum number j splits into 2j+1 evenly spaced levels — two, three and four here. Click a level for its m_j.

The Stern–Gerlach experiment

The Stern–Gerlach experiment turns that energy ladder into a picture. A uniform field would only twist the atoms; the trick is a non-uniform one: if BB varies with zz, the energy Emag=μzB(x,y,z)E_{\mathrm{mag}} = -\mu_z B(x,y,z) varies with position, and a position-dependent energy is a force,

F=Emagz=μzBzF = -\frac{\partial E_{\mathrm{mag}}}{\partial z} = \mu_z\frac{\partial B}{\partial z}

Each atom is pushed along zz by an amount proportional to its own μz\mu_z. A classical beam would smear into a continuous band, because μz\mu_z could be anything from μ-\mu to +μ+\mu. A quantum beam splits into 2j+12j+1 separate beams.

The Stern–Gerlach experiment: a beam that splits instead of smearing
ovencollimated beamNS∂B/∂z ≠ 0is what splits itmj = 1/2mj = -1/2screen2 beams = 2j + 1a classical magnet would give one continuous band, shown belowclassical

With j = 1/2 the moment has 2 allowed z components, so the beam lands in 2 places. This is the case Stern and Gerlach actually saw. Silver atoms split into two beams, which says j = ½ — and a half-integer j cannot come from orbital motion, so it is spin.

The Stern–Gerlach experiment: a beam that splits instead of smearing

Spin-½ throughout. Each magnet measures the spin along its own axis; block a beam to send only the other one onward.

magnet 1axis0°block
magnet 2axis90°block
unpolarized0°50.0%50.0%90°25.0%25.0%
surviving beam
50.0% of the original atoms
if you now measured Sz on what is left:
up 50.0% · down 50.0%
Bloch sphere — the surviving state
|↑⟩|↓⟩x

A measurement snaps the arrow onto its magnet's axis. Two magnets at 90° are perpendicular here too — and knowing one says nothing about the other.

Press the z → x → z surprise: keep only spin-up along z, keep only spin-up along x, then measure z again. Half the atoms come out down — the outcome the first magnet had already filtered away. The x-measurement did not add anything; it destroyed the z-information, becausex, Ŝz] ≠ 0.

What a Stern–Gerlach experiment tells you

you observetherefore j iswhich means
one undeflected spotno magnetic moment — nothing to push on
two beamshalf-integer, so spin — this is silver
three beamsinteger, and one beam is undeflected
a continuous bandclassical behaviour; not observed

The beam count is a direct read-out of j. Click any cell for the reasoning.

Check yourself

0 / 6 answered

  1. 1.Why does a Stern–Gerlach apparatus need a *non-uniform* magnetic field?

  2. In the widget's chained view, press “the z → x → z surprise”.

    2.Half the atoms emerge spin-down at the third magnet, even though the first magnet removed every spin-down atom. What happened?

  3. 3.Silver's beam splits into two. Why does that establish the existence of *spin* rather than just some angular momentum?

  4. 4.The nuclear magneton is about 1836 times smaller than the Bohr magneton. Why?

  5. 5.Which statement about the Landé g-factor is correct?

  6. 6.Why is the Zeeman effect described as *indirect* evidence for quantized magnetic energies, while Stern–Gerlach is *direct*?