Back to the on-screen lesson ·
Spin-½ spinors and Pauli matrices, probabilities along any axis, magnetic energy and Larmor precession, and magnetic resonance.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to work with spin-½ states, find spin probabilities and averages along any axis, and predict the splitting and precession of spins in a magnetic field.
You know that the angular momentum algebra allows half-integer values of $l$ that orbital motion cannot use, that two-state systems are described by $2 \times 2$ Hermitian matrices, and that stationary states evolve by phases. From modern physics you may recall the Stern–Gerlach experiment. This lesson puts those together: spin is the half-integer angular momentum, and its states form the most important two-state system of all.
| Term | What it means |
|---|---|
| Spin | An intrinsic angular momentum of a particle, not due to motion through space. |
| Spin-½ | Spin with $s = \tfrac{1}{2}$: $S^2 = \tfrac{3}{4}\hbar^2$ and $S_z = \pm\hbar/2$; electrons, protons, neutrons. |
| Spinor | A two-component column $\chi = (a, b)$ describing a spin-½ state in the $z$ basis. |
| Pauli matrices | $\sigma_x, \sigma_y, \sigma_z$, with $\hat{\vec{S}} = \tfrac{\hbar}{2}\vec{\sigma}$. |
| Gyromagnetic ratio | $\gamma$, the ratio of a particle's magnetic moment to its spin: $\vec{\mu} = \gamma\vec{S}$. |
| Larmor precession | The rotation of a spin about a magnetic field at angular frequency $\omega = \gamma B$. |
| Bohr magneton | $\mu_B = e\hbar/2m_e = 5.788 \times 10^{-5}$ eV/T, the natural unit of electron magnetism. |
In 1922 Otto Stern and Walther Gerlach sent silver atoms through a nonuniform magnetic field and saw the beam split in two. Orbital angular momentum gives $2l + 1$ beams, always an odd number, so something else was at work: the electron carries an intrinsic angular momentum, spin, with $s = \tfrac{1}{2}$ and two values of any component, $\pm\hbar/2$.
A spin-½ state is a spinor, $\chi = a\chi_\uparrow + b\chi_\downarrow = (a, b)$, with $|a|^2 + |b|^2 = 1$. The spin operators are $\hat{\vec{S}} = \tfrac{\hbar}{2}\vec{\sigma}$ with the Pauli matrices
$$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$$
which obey the angular momentum commutators and give $S^2 = \tfrac{3}{4}\hbar^2$. Spin up along an axis tilted by $\theta$ from $z$ in the $xz$ plane is $(\cos\tfrac{\theta}{2}, \sin\tfrac{\theta}{2})$ — a half-angle, a signature of spin-½.
A spin carries a magnetic moment $\vec{\mu} = \gamma\vec{S}$. In a field along $z$, $\hat{H} = -\gamma B\hat{S}_z$: the up and down states split by $\hbar\gamma B$, and any other state precesses around the field at the Larmor frequency $\omega = \gamma B$, its average spin vector turning like a gyroscope. For the electron $|\gamma| = g\mu_B/\hbar$ with $g \approx 2$, giving $28.0$ GHz per tesla; for the proton $\gamma/2\pi = 42.58$ MHz per tesla.
Another way: picture
Picture the spin as an arrow on a sphere, the Bloch sphere: spin up along $z$ at the north pole, down at the south, and every other spinor a point on the surface whose direction is the spin's average. A measurement along any axis gives up with probability $\cos^2$ of half the angle between the arrow and the axis. A magnetic field along $z$ makes the arrow circle the pole at the Larmor frequency, like a tilted top in gravity.
Another way: steps
Checks. Every component's eigenvalues are $\pm\hbar/2$. The probabilities along any axis add to one. The average spin vector of a pure state has length $\hbar/2$ exactly. And rotating a spinor by $360°$ multiplies it by $-1$: the half-angles mean it takes $720°$ to return to itself, a sign change neutron interferometry has observed directly.
If an electron were a tiny sphere spinning on its axis, then carrying $\hbar/2$ of angular momentum with a radius as small as experiments allow, less than $10^{-18}$ m, would need a surface speed far above the speed of light. And a spinning object could have any orientation and any amount of spin, while an electron always has exactly $\hbar/2$ and any measurement finds $\pm\hbar/2$.
Spin is better thought of as an intrinsic property, like charge or mass, that happens to behave under rotations the way angular momentum does. Relativistic quantum mechanics makes this precise: Dirac's 1928 equation for the electron requires four-component wave functions and automatically includes spin-½ with $g = 2$. Measurements find $g = 2.00231930436$, the small excess explained by quantum electrodynamics to about one part in a trillion, one of the most precise agreements between theory and experiment in all of science.
A spin in a static field $B_0$ precesses at $\omega_0 = \gamma B_0$ but does not change its energy. To flip it, apply a weak field $B_1$ rotating at $\omega_0$ perpendicular to $B_0$. In a frame rotating with $B_1$, the static field disappears and the spin precesses about $B_1$ instead, slowly tipping from up to down. That is the two-state Rabi problem of lesson 12, with coupling $\hbar\gamma B_1/2$ and detuning set by how far the drive frequency is from $\omega_0$.
This is magnetic resonance. A pulse long enough to tip the spins by $90°$ leaves them precessing in the plane perpendicular to $B_0$, where their rotating magnetization induces a voltage in a pickup coil. The signal decays as the spins dephase and relax, at rates that depend on their chemical surroundings — the contrast that MRI images and that nuclear magnetic resonance spectroscopy uses to identify molecules.
Spin doubles the number of states in every orbital level: hydrogen's level $n$ holds $2n^2$ states, and the shells of the periodic table hold $2$, $8$, $18$ electrons. Combined with the Pauli exclusion principle, the subject of lesson 21, that doubling fixes the periodic table's structure.
Spin also couples to orbital motion. In the electron's rest frame the nucleus circles it, producing a magnetic field that acts on the spin. This spin–orbit coupling splits hydrogen's $2p$ level by $4.5 \times 10^{-5}$ eV and sodium's $3p$ level by $2.1$ meV, producing the famous yellow doublet at $589.0$ and $589.6$ nm seen in sodium street lamps. Combining spin and orbital angular momentum to describe such states is the subject of the next lesson.
Protons and neutrons are spin-½ as well, with magnetic moments about $660$ times smaller than the electron's, because their masses are larger. In hydrogen's ground state the electron's and proton's spins can be parallel or antiparallel, and their magnetic interaction splits the two arrangements by $5.874 \times 10^{-6}$ eV, the hyperfine splitting.
A flip from parallel to antiparallel emits a photon of $1420.4$ MHz, a wavelength of $21.1$ cm. The flip is extraordinarily slow, with a lifetime of about ten million years for an isolated atom, but there is so much hydrogen in space that the line is bright. Radio astronomers use it to map the Milky Way's spiral arms, measure the rotation of galaxies — the first evidence for dark matter came partly from such curves — and, with arrays such as the Hydrogen Epoch of Reionization Array, search for the hydrogen of the universe's first billion years.
Rotating a spin-½ state by an angle $\alpha$ about the $z$ axis multiplies its components by $e^{\mp i\alpha/2}$. The half-angle has a startling consequence: a full turn, $\alpha = 2\pi$, gives $e^{\mp i\pi} = -1$, so the spinor changes sign. Since an overall sign is unobservable for a single spin, nothing seems to change. But if half of a beam of neutrons is rotated by $360°$ and then recombined with the other half, the sign shows up as destructive interference. Experiments at the Institut Laue–Langevin and the University of Missouri in 1975 saw exactly this.
The same half-angle appears in the probability $\cos^2(\theta/2)$ for finding spin up along an axis tilted by $\theta$. It is also why the two spin states along opposite directions, $\theta = 180°$ apart on the Bloch sphere, are orthogonal vectors in Hilbert space: $\cos(90°) = 0$.
About $40$ million MRI scans are performed each year in the United States. The scanner's superconducting magnet, usually $1.5$ or $3$ T, aligns a tiny excess of the body's hydrogen nuclei: at body temperature and $3$ T, only about ten protons in a million more point with the field than against it, but a cubic millimeter of tissue holds $10^{19}$ protons. A radio pulse at the Larmor frequency, $127.7$ MHz at $3$ T, tips their spins, and as they precess back they broadcast a signal at the same frequency.
To make an image, gradient coils make the field vary slightly across the body, so that each location precesses at a slightly different frequency; a Fourier transform of the signal then reveals where it came from. Contrast comes from how quickly the spins relax in different tissues — fat, water, gray matter and tumors differ — and the 2003 Nobel Prize in Physiology or Medicine went to Paul Lauterbur and Peter Mansfield for the imaging method. Every MRI image is a map of spin precession.
A hard disk stores bits as tiny magnetized regions, and its read head senses them using the electron's spin. In a giant magnetoresistance sensor, discovered by Albert Fert and Peter Grünberg in 1988 and awarded the 2007 Nobel Prize, current flows through two magnetic layers separated by a nonmagnetic spacer a few atoms thick. Electrons whose spin matches a layer's magnetization pass through easily; those with the opposite spin scatter. When the two layers are parallel, one spin channel conducts well; when antiparallel, both are impeded, and the resistance jumps by tens of percent.
Modern drives use tunnel magnetoresistance, where electrons tunnel through a thin insulator at a rate that depends on their spin, giving resistance changes of over $100$ percent. The same devices, in magnetic random-access memory, store data without power, and the field built on them — spintronics — uses the electron's two spin states, not just its charge, to carry information.
The name suggests a charged sphere rotating on its axis, and that picture explains why spin carries a magnetic moment. It fails everywhere else. An electron is pointlike to the best of our measurements, and a sphere small enough would have to rotate faster than light to carry $\hbar/2$. A classical spinning top can point any way with any amount of spin; an electron always has $S^2 = \tfrac{3}{4}\hbar^2$, and any component measures $\pm\hbar/2$, never anything in between.
A related misconception is that spin up along $z$ means the spin points exactly along $+z$. As with orbital angular momentum, $S_z = \hbar/2$ is less than $|S| = \tfrac{\sqrt{3}}{2}\hbar$, and $S_x$ and $S_y$ are completely uncertain: measuring $S_x$ on a spin-up state gives $\pm\hbar/2$ with equal probability.
A spin is in $\chi = \tfrac{1}{5}(3, 4i)$. Check its normalization.
$\chi^\dagger\chi = \tfrac{1}{25}(9 + 16) = 1$
$|4i|^2 = 16$.
Write the $x$ eigenvectors.
$\chi_\pm^{(x)} = \tfrac{1}{\sqrt{2}}(1, \pm 1)$
The eigenvectors of $\sigma_x$.
Project onto spin up along $x$.
$\langle\chi_+^{(x)}|\chi\rangle = \tfrac{1}{5\sqrt{2}}(3 + 4i)$
Real bra, so no conjugation is needed.
Square the magnitude.
$P_+^{(x)} = \dfrac{|3 + 4i|^2}{50} = \dfrac{25}{50} = 0.5$
$|3 + 4i|^2 = 9 + 16$.
Find the probability of down.
$P_-^{(x)} = \dfrac{|3 - 4i|^2}{50} = 0.5$
A state with a purely imaginary relative phase lies in the $yz$ plane, so $x$ is a coin toss.
Take the state $\chi = \tfrac{1}{5}(3, 4)$ and compute $\langle S_z\rangle$.
$\langle S_z\rangle = \dfrac{\hbar}{2}\left(\dfrac{9}{25} - \dfrac{16}{25}\right) = -0.14\hbar$
Up and down weighted by their probabilities.
Compute $\langle S_x\rangle$.
$\langle S_x\rangle = \dfrac{\hbar}{2}\chi^\dagger\sigma_x\chi = \dfrac{\hbar}{2} \cdot \dfrac{2 \times 3 \times 4}{25} = 0.48\hbar$
$\sigma_x$ swaps the components.
Compute $\langle S_y\rangle$.
$\langle S_y\rangle = \dfrac{\hbar}{2}\chi^\dagger\sigma_y\chi = \dfrac{\hbar}{2} \cdot \dfrac{3(-4i) + 4(3i)}{25} = 0$
Real components give zero $y$ spin.
Find the length of the average vector.
$|\langle\vec{S}\rangle| = \hbar\sqrt{0.48^2 + 0^2 + 0.14^2} = \hbar\sqrt{0.25} = 0.5\hbar$
Exactly $\hbar/2$ for any pure spin state.
Find its angle from the $z$ axis.
$\cos\theta = \dfrac{-0.14}{0.5} = -0.28 \quad\Rightarrow\quad \theta = 106°$
The spin points just below the equator of the Bloch sphere.
Check with the half-angle formula.
$\cos^2\dfrac{106°}{2} = \cos^2 53° = 0.36 = \dfrac{9}{25}$
The probability of up along $z$, as it should be.
Write the electron's spin Hamiltonian in a field $B$ along $z$.
$\hat{H} = g\mu_BB\,\dfrac{\hat{S}_z}{\hbar}$
The electron's moment points opposite to its spin.
Find the energies of the two states.
$E_\pm = \pm\tfrac{1}{2}g\mu_BB$
Spin up is higher for the electron.
Find the splitting in a $0.35$ T field.
$\Delta E = 2 \times 5.788 \times 10^{-5} \times 0.35 = 4.05 \times 10^{-5}\ \text{eV}$
With $g = 2$.
Convert the splitting to a frequency.
$f = \dfrac{\Delta E}{h} = \dfrac{4.05 \times 10^{-5}}{4.1357 \times 10^{-15}} = 9.8\ \text{GHz}$
The X band used by most ESR spectrometers.
Find the precession period.
$T = \dfrac{1}{f} = 0.10\ \text{ns}$
The spin circles the field ten billion times per second.
Compare the populations at room temperature.
$\dfrac{N_\uparrow}{N_\downarrow} = e^{-\Delta E/k_BT} = e^{-4.05 \times 10^{-5}/0.0259} = 0.9984$
Only a small excess in the lower state produces the signal.
Find the excess fraction.
$\dfrac{N_\downarrow - N_\uparrow}{N} \approx \dfrac{\Delta E}{2k_BT} = 7.8 \times 10^{-4}$
Less than a tenth of a percent.
Compare with protons in the same field.
$f_p = 42.58 \times 0.35 = 14.9\ \text{MHz}$
About $660$ times lower, and with a correspondingly smaller population excess.
Write the Larmor frequency.
$f = \dfrac{\gamma}{2\pi}B$
For protons $\gamma/2\pi = 42.58$ MHz/T.
Substitute the values.
$f = 42.58 \times 3.0$
Megahertz per tesla times tesla.
Evaluate the frequency.
A beam of silver atoms, each with a single unpaired electron and no orbital angular momentum, passes through a Stern–Gerlach magnet. Into how many beams does it split?
Complete the worked solution: a spin is in the state $\chi = \tfrac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix}$. Find the probabilities of up and down along $z$ and $\langle S_z\rangle$ in units of $\hbar$.
Square the upper component.
$P_\uparrow = \dfrac{4^2}{5^2} =$ u
The probability of $+\hbar/2$.
Square the lower component.
$P_\downarrow = \dfrac{3^2}{5^2} =$ d
The probability of $-\hbar/2$.
Weight the two values by their probabilities.
$\dfrac{\langle S_z\rangle}{\hbar} = \tfrac{1}{2}P_\uparrow - \tfrac{1}{2}P_\downarrow =$ m
The average over many measurements.
Match each spin-½ quantity to its value or form.
| $\tfrac{3}{4}\hbar^2$ | $\pm\tfrac{1}{2}\hbar$ | $\tfrac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ | $\gamma B$ | |
|---|---|---|---|---|
| $S^2$ | ||||
| the possible values of $S_z$ | ||||
| the matrix of $\hat{S}_x$ | ||||
| the Larmor precession rate |
An electron is prepared spin up along $z$. Its spin is then measured along an axis tilted by $\theta$ from $z$. Fill in the probabilities of up and down along that axis.
| $P_\uparrow$ | $P_\downarrow$ | |
|---|---|---|
| $\theta = 60°$ | ||
| $\theta = 90°$ | ||
| $\theta = 120°$ |
A free electron sits in a magnetic field of $7$ T. How far apart in energy are its two spin states, in μeV? Use $g = 2$ and $\mu_B = 57.88$ μeV/T.
Answer: μeV
An electron's spin points along $+x$ at $t = 0$, in a magnetic field of $0.3$ T along $z$. What is the probability of finding it along $+x$ after $40$ ps? Use $2\mu_B/\hbar = 1.7588 \times 10^{11}$ rad s⁻¹ T⁻¹.
Answer:
An MRI scanner detects hydrogen nuclei (protons) precessing in its main magnetic field of $4.2$ T. At what frequency, in MHz, must its radio coils transmit and listen? The proton's $\gamma/2\pi = 42.58$ MHz/T.
Answer: MHz
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Complete the worked solution: a spin is in the state $\chi = \tfrac{1}{25}\begin{pmatrix} 24 \\ 7 \end{pmatrix}$. Find the probabilities of up and down along $z$ and $\langle S_z\rangle$ in units of $\hbar$.
Square the upper component.
$P_\uparrow = \dfrac{24^2}{25^2} =$ u
The probability of $+\hbar/2$.
Square the lower component.
$P_\downarrow = \dfrac{7^2}{25^2} =$ d
The probability of $-\hbar/2$.
Weight the two values by their probabilities.
$\dfrac{\langle S_z\rangle}{\hbar} = \tfrac{1}{2}P_\uparrow - \tfrac{1}{2}P_\downarrow =$ m
The average over many measurements.
You can describe and measure spin-½. Explain to someone how an MRI scanner uses the precession of protons.
17. Your turn: at what frequency do protons precess in a $3.0$ T MRI scanner?, step 3
$f = 127.7\ \text{MHz}$
Just above the FM radio band.