Spin-1/2 and the Pauli Matrices

Derivation

Spin-1/2 and the Pauli Matrices

D-219 Home PU-301 Threads symmetry · matter Depends on Angular Momentum Spectrum from Commutators

Statement

For the angular-momentum quantum number \( j = \tfrac{1}{2} \), the abstract algebra \( [\hat{J}_i, \hat{J}_j] = i\hbar\,\varepsilon_{ijk}\hat{J}_k \) admits a unique (up to unitary equivalence) two-dimensional irreducible representation. In it the spin operators are \( \hat{S}_i = \tfrac{\hbar}{2}\sigma_i \), where the three Pauli matrices \( \sigma_i \) satisfy \( \sigma_i\sigma_j = \delta_{ij}\mathbf{1} + i\,\varepsilon_{ijk}\sigma_k \), and any rotation by angle \( \theta \) about the unit axis \( \hat{\mathbf{n}} \) is represented on the two-component spinor by \( \hat{U}(\theta,\hat{\mathbf{n}}) = \exp\!\left(-\tfrac{i}{2}\theta\,\hat{\mathbf{n}}\cdot\vec{\sigma}\right) = \cos\tfrac{\theta}{2}\,\mathbf{1} - i\sin\tfrac{\theta}{2}\,\hat{\mathbf{n}}\cdot\vec{\sigma} \).

Why it matters

Spin-\( \tfrac{1}{2} \) is the smallest non-trivial carrier of the rotation group and the template for every two-level quantum system: qubits, ammonia inversion, neutrino flavour, isospin. The Pauli algebra is the concrete face of \( \mathfrak{su}(2) \), and reading it back into physics gives the electron its magnetic moment, fine structure, and the Zeeman effect.

The rotation operator exposes the signature feature of half-integer spin: a full \( 2\pi \) turn returns the spinor to \( -1 \) times itself, not \( +1 \). This double-valuedness under \( SO(3) \) — resolved only by passing to its double cover \( SU(2) \) — is not a mathematical curiosity but an experimentally confirmed phase, and it underlies the Pauli exclusion principle and the spin-statistics connection.

Assumptions

The three spin operators close under the angular-momentum commutator Without \( [\hat{S}_i,\hat{S}_j]=i\hbar\varepsilon_{ijk}\hat{S}_k \) there is no ladder structure, no quantised spectrum, and the matrices below are unconstrained. The representation is the \( j=\tfrac{1}{2} \) irreducible block, so \( \hat{S}^2 = \tfrac{3}{4}\hbar^2\,\mathbf{1} \) is a multiple of the identity If \( \hat{S}^2 \) were not proportional to \( \mathbf{1} \) the space would be reducible and the two states would not share a single \( j \); the closed \( \sigma_i \) algebra would fail. The carrier space is exactly two-dimensional and the representation is irreducible Drop irreducibility and Schur's lemma no longer forces \( \hat{S}^2 \propto \mathbf{1} \); the operators could be block-diagonal sums, and the Pauli identity \( \sigma_i\sigma_j=\delta_{ij}\mathbf{1}+i\varepsilon_{ijk}\sigma_k \) — which uses \( d=2 \) crucially — would not hold. Rotations act unitarily and are generated by \( \hat{\mathbf{J}}=\hat{\mathbf{S}} \) If the generator of rotations were not the spin itself (e.g. an added orbital part), the exponential would represent a different physical operation and the half-angle would not appear.

Derivation

\[ \hat{S}^2\,|s,m\rangle = \hbar^2 s(s+1)\,|s,m\rangle, \qquad \hat{S}_z\,|s,m\rangle = \hbar m\,|s,m\rangle, \qquad s=\tfrac{1}{2},\; m=\pm\tfrac{1}{2}. \]

Specialise the ladder spectrum (prior result) to the smallest half-integer \( s \). The dimension is \( 2s+1 = 2 \), giving the orthonormal basis \( |{\uparrow}\rangle\equiv|\tfrac12,\tfrac12\rangle \), \( |{\downarrow}\rangle\equiv|\tfrac12,-\tfrac12\rangle \). A

\[ \hat{S}_z \doteq \frac{\hbar}{2}\begin{pmatrix}1&0\\[2pt]0&-1\end{pmatrix}. \]

Write \( \hat{S}_z \) as a matrix in the ordered basis \( (|{\uparrow}\rangle,|{\downarrow}\rangle) \). It is diagonal because both basis kets are \( \hat{S}_z \)-eigenstates with eigenvalues \( +\tfrac{\hbar}{2}, -\tfrac{\hbar}{2} \). A

\[ \hat{S}_\pm = \hat{S}_x \pm i\hat{S}_y, \qquad \hat{S}_\pm\,|s,m\rangle = \hbar\sqrt{s(s+1)-m(m\pm1)}\;|s,m\pm1\rangle. \]

Recall the ladder operators and their matrix elements from the prior spectrum result; symbols first, evaluation next. A

\[ \hat{S}_+\,|{\downarrow}\rangle = \hbar\sqrt{\tfrac{3}{4}-\left(-\tfrac12\right)\left(\tfrac12\right)}\;|{\uparrow}\rangle = \hbar\,|{\uparrow}\rangle, \qquad \hat{S}_+\,|{\uparrow}\rangle = 0, \] \[ \hat{S}_-\,|{\uparrow}\rangle = \hbar\,|{\downarrow}\rangle, \qquad \hat{S}_-\,|{\downarrow}\rangle = 0. \]

Insert the numbers \( s=\tfrac12 \), \( m=\mp\tfrac12 \). The surd equals \( \sqrt{3/4+1/4}=1 \); raising the top state and lowering the bottom state annihilate them. A

\[ \hat{S}_+ \doteq \hbar\begin{pmatrix}0&1\\[2pt]0&0\end{pmatrix}, \qquad \hat{S}_- \doteq \hbar\begin{pmatrix}0&0\\[2pt]1&0\end{pmatrix}. \]

Read off matrix elements \( (\hat{S}_\pm)_{ab}=\langle a|\hat{S}_\pm|b\rangle \) from step 4. Only the off-diagonal coupling of \( |{\uparrow}\rangle \) and \( |{\downarrow}\rangle \) survives. A

\[ \hat{S}_x = \tfrac{1}{2}(\hat{S}_+ + \hat{S}_-) \doteq \frac{\hbar}{2}\begin{pmatrix}0&1\\[2pt]1&0\end{pmatrix}, \qquad \hat{S}_y = \tfrac{1}{2i}(\hat{S}_+ - \hat{S}_-) \doteq \frac{\hbar}{2}\begin{pmatrix}0&-i\\[2pt]i&0\end{pmatrix}. \]

Invert the definitions of the ladder operators to recover the Cartesian components, then substitute the matrices of step 5. A

\[ \hat{S}_i = \frac{\hbar}{2}\,\sigma_i, \quad\text{with}\quad \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\; \sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\; \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. \]

Factor out the common scale \( \hbar/2 \) from steps 2 and 6 and define the dimensionless Pauli matrices \( \sigma_i \). This is a definition, legal because each \( \hat{S}_i \) is real-linear in the extracted factor. A

\[ \sigma_x^2=\sigma_y^2=\sigma_z^2=\mathbf{1}, \qquad \operatorname{Tr}\sigma_i = 0, \qquad \det\sigma_i=-1. \]

Direct matrix multiplication. Squaring each Pauli matrix returns the identity; the traces vanish because the diagonal entries are opposite (or zero). A

\[ \sigma_x\sigma_y=\begin{pmatrix}i&0\\0&-i\end{pmatrix}=i\sigma_z, \qquad \sigma_y\sigma_x = -i\sigma_z, \] \[ \Rightarrow\quad \{\sigma_x,\sigma_y\}=0, \qquad [\sigma_x,\sigma_y]=2i\sigma_z. \]

Multiply the explicit matrices in both orders. The symmetric part (anticommutator) vanishes; the antisymmetric part (commutator) reproduces \( 2i\sigma_z \). Cyclic permutation gives the other pairs. B

\[ \boxed{\;\sigma_i\sigma_j = \delta_{ij}\,\mathbf{1} + i\,\varepsilon_{ijk}\,\sigma_k\;} \]

Combine step 8 (\( i=j \) gives \( \mathbf{1} \)) with step 9 (\( i\neq j \) gives \( i\varepsilon_{ijk}\sigma_k \)) into one identity valid for all \( i,j \). This single relation encodes both \( \{\sigma_i,\sigma_j\}=2\delta_{ij}\mathbf{1} \) and \( [\sigma_i,\sigma_j]=2i\varepsilon_{ijk}\sigma_k \). B

\[ [\hat{S}_i,\hat{S}_j] = \frac{\hbar^2}{4}[\sigma_i,\sigma_j] = \frac{\hbar^2}{4}\,(2i\varepsilon_{ijk}\sigma_k) = i\hbar\,\varepsilon_{ijk}\,\frac{\hbar}{2}\sigma_k = i\hbar\,\varepsilon_{ijk}\hat{S}_k. \]

Consistency check: reinsert the scale. The constructed matrices satisfy exactly the angular-momentum algebra we started from, confirming the representation is faithful. B

\[ (\hat{\mathbf{n}}\cdot\vec{\sigma})^2 = n_i n_j\,\sigma_i\sigma_j = n_i n_j\big(\delta_{ij}\mathbf{1}+i\varepsilon_{ijk}\sigma_k\big) = |\hat{\mathbf{n}}|^2\,\mathbf{1} = \mathbf{1}. \]

Contract the master identity (step 10) with the symmetric tensor \( n_i n_j \). The \( \varepsilon_{ijk} \) term dies against the symmetric contraction, leaving \( \mathbf{1} \) since \( \hat{\mathbf{n}} \) is a unit vector. B

\[ \hat{U}(\theta,\hat{\mathbf{n}}) = \exp\!\left(-\frac{i}{\hbar}\,\theta\,\hat{\mathbf{n}}\cdot\hat{\mathbf{S}}\right) = \exp\!\left(-\frac{i\theta}{2}\,\hat{\mathbf{n}}\cdot\vec{\sigma}\right). \]

A rotation by \( \theta \) about \( \hat{\mathbf{n}} \) is generated by the component of angular momentum along that axis, \( \hat{\mathbf{n}}\cdot\hat{\mathbf{S}} \); substitute \( \hat{\mathbf{S}}=\tfrac{\hbar}{2}\vec{\sigma} \). The \( \hbar \) cancels, leaving the half-angle generator. C

\[ e^{-i\frac{\theta}{2}\hat{\mathbf{n}}\cdot\vec\sigma} = \sum_{k=0}^{\infty}\frac{1}{k!}\left(-\frac{i\theta}{2}\right)^{k}(\hat{\mathbf{n}}\cdot\vec\sigma)^{k} = \sum_{\text{even}}\frac{(-1)^{k/2}}{k!}\!\left(\tfrac{\theta}{2}\right)^{k}\mathbf{1} \;-\; i\sum_{\text{odd}}\frac{(-1)^{(k-1)/2}}{k!}\!\left(\tfrac{\theta}{2}\right)^{k}(\hat{\mathbf{n}}\cdot\vec\sigma). \]

Expand the exponential and use \( (\hat{\mathbf{n}}\cdot\vec\sigma)^2=\mathbf{1} \) (step 12), so even powers give \( \mathbf{1} \) and odd powers give \( \hat{\mathbf{n}}\cdot\vec\sigma \). The two subsums are the Taylor series of cosine and sine. C

\[ \boxed{\;\hat{U}(\theta,\hat{\mathbf{n}}) = \cos\frac{\theta}{2}\,\mathbf{1} - i\sin\frac{\theta}{2}\;\hat{\mathbf{n}}\cdot\vec{\sigma}\;} \]

Resum the even and odd series into \( \cos(\theta/2) \) and \( \sin(\theta/2) \). This closed form holds because the algebra truncates — the same reason \( e^{i\phi}=\cos\phi+i\sin\phi \) does, with \( (\hat{\mathbf{n}}\cdot\vec\sigma) \) playing the role of \( i \). C

Result

\[ \hat{S}_i=\frac{\hbar}{2}\sigma_i, \qquad \sigma_i\sigma_j=\delta_{ij}\mathbf{1}+i\varepsilon_{ijk}\sigma_k, \qquad \hat{U}(\theta,\hat{\mathbf{n}})=\cos\tfrac{\theta}{2}\,\mathbf{1}-i\sin\tfrac{\theta}{2}\,\hat{\mathbf{n}}\cdot\vec{\sigma}. \]

Reading. The three Pauli matrices are the generators of \( SU(2) \) in its fundamental representation; scaled by \( \hbar/2 \) they are the physical spin components of a spin-\( \tfrac12 \) particle. Their product rule packages orthonormality (the \( \delta_{ij}\mathbf{1} \) piece, from squaring) and the rotation algebra (the \( i\varepsilon_{ijk}\sigma_k \) piece, from commuting) into one line. The rotation operator rotates a spinor by turning the "spin vector" \( \langle\vec\sigma\rangle \) through the ordinary angle \( \theta \) on the Bloch sphere, while the state itself only advances by \( \theta/2 \) — so a \( 2\pi \) physical rotation yields \( \hat{U}=-\mathbf{1} \).

Units check. The \( \sigma_i \) are pure numbers (dimensionless), so \( \hat{S}_i=\tfrac{\hbar}{2}\sigma_i \) carries units of \( \hbar \), i.e. \( \mathrm{J\,s} \) — the correct dimension of angular momentum. In \( \hat{U}=\exp(-i\theta\,\hat{\mathbf{n}}\cdot\hat{\mathbf{S}}/\hbar) \) the exponent is \( [\text{rad}]\times[\mathrm{J\,s}]/[\mathrm{J\,s}] \), dimensionless as an exponent must be; the eigenvalues of \( \hat{U} \) have unit modulus, confirming unitarity.

Limiting cases

\( \theta\to0 \): \( \hat{U}\to\mathbf{1} \) — no rotation, identity on the spinor.
\( \theta=\pi \) about \( \hat{\mathbf{z}} \): \( \hat{U}=-i\sigma_z=\operatorname{diag}(-i,\,i) \); the two spin states pick up opposite phases, a half-turn on the Bloch sphere.
\( \theta=2\pi \): \( \hat{U}=-\mathbf{1} \) for any axis — the celebrated spinor sign change.
\( \theta=4\pi \): \( \hat{U}=+\mathbf{1} \); \( SU(2) \) elements are single-valued only after two full turns.
Large \( s \) (heavy top): for \( s\gg1 \) the \( (2s+1) \)-dimensional representation approaches classical rotation and the half-angle anomaly is invisible in expectation values.

Breaks when

The system is not a single spin-\( \tfrac12 \). For \( s\neq\tfrac12 \) the space is not two-dimensional, \( \sigma_i^2=\mathbf{1} \) fails, and the master product identity \( \sigma_i\sigma_j=\delta_{ij}\mathbf{1}+i\varepsilon_{ijk}\sigma_k \) is simply false (e.g. spin-1 uses the \( 3\times3 \) generators, whose squares are not the identity).
Relativistic / Lorentz regime. Under boosts the rotation group is enlarged to the Lorentz group; the two-component Pauli spinor no longer transforms consistently and must be doubled into a four-component Dirac spinor. The non-relativistic \( SU(2) \) rotation operator is then only the small-velocity block.
Coupling to other angular momenta. When spin is not conserved alone — spin–orbit coupling, or a total \( \hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}} \) — rotations are generated by \( \hat{\mathbf{J}} \), and \( \exp(-i\theta\,\hat{\mathbf{n}}\cdot\hat{\mathbf{S}}/\hbar) \) is no longer the physical rotation of the full state.

Failure modes

Half-angle amnesia. Writing \( \hat{U}=\cos\theta\,\mathbf{1}-i\sin\theta\,\hat{\mathbf{n}}\cdot\vec\sigma \). The generator is \( \hat{\mathbf{S}}=\tfrac{\hbar}{2}\vec\sigma \), so the angle is halved; forgetting this loses the \( 2\pi\to-\mathbf{1} \) sign.
Confusing \( \hat{S}_i \) with \( \sigma_i \). Quoting eigenvalues of spin as \( \pm1 \) instead of \( \pm\tfrac{\hbar}{2} \); the \( \sigma_i \) are dimensionless and their eigenvalues are \( \pm1 \), the spins are \( \tfrac{\hbar}{2} \) times these.
Sign of \( \sigma_y \). Dropping the \( i \) or flipping its position, giving a non-Hermitian or wrong-handed \( \sigma_y \); it must read \( \begin{pmatrix}0&-i\\i&0\end{pmatrix} \) so that \( [\sigma_x,\sigma_y]=+2i\sigma_z \).
Treating \( \hat{\mathbf{n}}\cdot\vec\sigma \) as a scalar. Using \( e^{-i\theta/2\,\hat{\mathbf{n}}\cdot\vec\sigma}=e^{-i\theta/2} \) times something; the exponential is a \( 2\times2 \) matrix and only collapses because \( (\hat{\mathbf{n}}\cdot\vec\sigma)^2=\mathbf{1} \), not because it is a number.
Assuming \( \sigma_x,\sigma_y,\sigma_z \) commute. Applying rotations about different axes as if the generators added like scalars; they do not, and \( \hat{U}(\alpha,\hat{\mathbf{x}})\hat{U}(\beta,\hat{\mathbf{y}})\neq\hat{U}(\beta,\hat{\mathbf{y}})\hat{U}(\alpha,\hat{\mathbf{x}}) \).

Discussion

The deep content of this construction is that quantum mechanics represents rotations not by \( SO(3) \) but by its universal cover \( SU(2) \). The map \( SU(2)\to SO(3) \) is two-to-one: both \( +\mathbf{1} \) and \( -\mathbf{1} \) send every physical vector back to itself, yet they are distinct spinor states. A spin-\( \tfrac12 \) particle "knows" whether it has been turned once or twice, a fact made visible in neutron interferometry, where a rotating magnetic field on one beam path shifts the interference fringes with period \( 4\pi \), not \( 2\pi \). The \( -\mathbf{1} \) is not a gauge artefact; it is measurable relative phase.

The product rule \( \sigma_i\sigma_j=\delta_{ij}\mathbf{1}+i\varepsilon_{ijk}\sigma_k \) is the workhorse of two-level physics. Its symmetric part gives \( \{\sigma_i,\sigma_j\}=2\delta_{ij}\mathbf{1} \), the defining relation of a Clifford algebra in three Euclidean dimensions — the same structure that, generalised to four spacetime dimensions with a Minkowski metric, produces the Dirac gamma matrices. Its antisymmetric part reproduces the Lie algebra \( \mathfrak{su}(2)\cong\mathfrak{so}(3) \). Thus the Pauli matrices sit precisely at the meeting point of the rotation Lie algebra and the spatial Clifford algebra, which is why they recur across the whole subject.

Physically, coupling \( \hat{\mathbf{S}} \) to a magnetic field via \( \hat{H}=-\vec{\mu}\cdot\mathbf{B}=-\gamma\,\mathbf{B}\cdot\hat{\mathbf{S}} \) makes the time-evolution operator \( e^{-i\hat{H}t/\hbar} \) identical in form to our rotation operator with \( \theta=\gamma|\mathbf{B}|t \). Larmor precession is therefore literally a rotation in spin space, and the factor \( \tfrac12 \) in \( \hat{S}=\tfrac{\hbar}{2}\sigma \) is what sets the electron's gyromagnetic response and, with the \( g\approx2 \) from Dirac theory, its magnetic moment.

More abstractly, the two-dimensional representation is the fundamental representation of \( SU(2) \) from which all others are built by tensor products: \( \tfrac12\otimes\tfrac12=0\oplus1 \) is the singlet–triplet decomposition of two spins, and iterating this Clebsch–Gordan series generates every integer and half-integer spin. Half-integer representations are exactly the ones that are faithful on \( SU(2) \) but only projective on \( SO(3) \); the spin-statistics theorem then ties this projective sign to Fermi–Dirac statistics, so the \( -\mathbf{1} \) under \( 2\pi \) rotation and the antisymmetry of the electron wavefunction are two faces of one topological fact — that \( SO(3) \) is not simply connected, \( \pi_1(SO(3))=\mathbb{Z}_2 \).

Common misconceptions. The spinor sign change is often dismissed as "just a phase and therefore unobservable." It is unobservable as a global phase of an isolated state, but perfectly observable as a relative phase in interference — which is exactly how it is measured. Second, students often think the electron is a tiny spinning ball; the \( \hbar/2 \) is intrinsic angular momentum with no classical rotation behind it, as a literal surface would have to exceed light speed.

Worked examples

Rotate a spin-up state \( |{\uparrow}\rangle \) by \( \theta=\tfrac{\pi}{2} \) about \( \hat{\mathbf{y}} \); find the new state and the probability of measuring spin-up afterward.

Setup: apply \( \hat{U}(\tfrac{\pi}{2},\hat{\mathbf{y}}) \) to \( |{\uparrow}\rangle=\binom{1}{0} \). A

\[ \hat{U}(\tfrac{\pi}{2},\hat{\mathbf{y}}) = \cos\tfrac{\pi}{4}\,\mathbf{1} - i\sin\tfrac{\pi}{4}\,\sigma_y = \frac{1}{\sqrt2}\begin{pmatrix}1&0\\0&1\end{pmatrix} - \frac{i}{\sqrt2}\begin{pmatrix}0&-i\\i&0\end{pmatrix} = \frac{1}{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}. \]

Insert \( \theta/2=\pi/4 \), so \( \cos=\sin=1/\sqrt2 \), and \( -i\sigma_y=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \). A

\[ \hat{U}\,|{\uparrow}\rangle = \frac{1}{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \frac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix} = \frac{1}{\sqrt2}\big(|{\uparrow}\rangle+|{\downarrow}\rangle\big). \]

Matrix–vector product; the result is the \( +x \) eigenstate, as expected since rotating \( +z \) by \( 90^\circ \) about \( y \) points it along \( +x \). A

\[ P_{\uparrow} = \left|\langle{\uparrow}|\hat{U}|{\uparrow}\rangle\right|^2 = \left|\tfrac{1}{\sqrt2}\right|^2 = \tfrac12. \]

Born rule on the upper component. A

\[ \hat{U}(\tfrac{\pi}{2},\hat{\mathbf{y}})\,|{\uparrow}\rangle = \tfrac{1}{\sqrt2}(|{\uparrow}\rangle+|{\downarrow}\rangle) = |{+}x\rangle, \qquad P_{\uparrow}=\tfrac12. \]

Reading. A quarter-turn about \( y \) maps the spin from the \( +z \) to the \( +x \) direction; a subsequent \( \hat{S}_z \) measurement is now maximally uncertain, giving up and down with equal weight. Probabilities are dimensionless.

An electron (\( g=2 \), \( \gamma=-g\mu_B/\hbar \)) sits in \( \mathbf{B}=B\hat{\mathbf{z}} \) with \( B=1.00\ \mathrm{T} \). Starting in \( |{+}x\rangle \), find the Larmor precession frequency and the state after \( t=1.00\ \mathrm{ns} \).

Setup: \( \hat{H}=-\gamma B\hat{S}_z=\omega_L\hat{S}_z \) with \( \omega_L=g\mu_B B/\hbar \); time evolution is a rotation about \( \hat{\mathbf{z}} \). B

\[ \omega_L = \frac{g\mu_B B}{\hbar} = \frac{(2)(9.274\times10^{-24}\ \mathrm{J\,T^{-1}})(1.00\ \mathrm{T})}{1.055\times10^{-34}\ \mathrm{J\,s}} = 1.759\times10^{11}\ \mathrm{rad\,s^{-1}}. \]

Symbols first, then numbers with SI units; \( \mu_B \) and \( \hbar \) are standard constants. B

\[ \theta = \omega_L\,t = (1.759\times10^{11}\ \mathrm{rad\,s^{-1}})(1.00\times10^{-9}\ \mathrm{s}) = 175.9\ \mathrm{rad} \equiv 175.9 - 27\!\cdot\!2\pi \approx 6.24\ \mathrm{rad}. \]

The precession angle is \( \omega_L t \); reduce modulo \( 2\pi \) (\( 27\times2\pi=169.6 \)) to \( 6.24\ \mathrm{rad}\approx357.5^\circ \), essentially one full turn less \( 2.5^\circ \). B

\[ \hat{U}(\theta,\hat{\mathbf{z}}) = \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{+i\theta/2} \end{pmatrix}, \qquad |\psi(t)\rangle = \frac{1}{\sqrt2}\begin{pmatrix} e^{-i\theta/2} \\ e^{+i\theta/2}\end{pmatrix}. \]

Apply the diagonal \( \hat{U} \) to \( |{+}x\rangle=\tfrac{1}{\sqrt2}\binom{1}{1} \); the spin still points in the \( xy \)-plane, now at azimuth \( \theta \). B

\[ \omega_L = 1.76\times10^{11}\ \mathrm{rad\,s^{-1}}\ (f_L\approx28.0\ \mathrm{GHz}), \qquad \langle\hat{S}_x\rangle,\langle\hat{S}_y\rangle \text{ precess at azimuth } \theta\approx357.5^\circ. \]

Reading. After \( 1\ \mathrm{ns} \) the in-plane spin has swept almost exactly \( 28 \) full precessions and sits \( 2.5^\circ \) short of its start. \( \omega_L \) has units \( \mathrm{rad\,s^{-1}} \); dividing by \( 2\pi \) gives \( f_L\approx28\ \mathrm{GHz} \), the electron-spin-resonance frequency at \( 1\ \mathrm{T} \).

Problems

Verify by direct multiplication that \( \sigma_y\sigma_z=i\sigma_x \) and hence that \( [\sigma_y,\sigma_z]=2i\sigma_x \).
Solution
\( \sigma_y\sigma_z=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}0&i\\i&0\end{pmatrix}=i\begin{pmatrix}0&1\\1&0\end{pmatrix}=i\sigma_x \). Reversing, \( \sigma_z\sigma_y=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix}=\begin{pmatrix}0&-i\\-i&0\end{pmatrix}=-i\sigma_x \). Therefore \( [\sigma_y,\sigma_z]=i\sigma_x-(-i\sigma_x)=2i\sigma_x \), consistent with \( \varepsilon_{231}=+1 \).
Find the normalized eigenstates of \( \sigma_x \) and their eigenvalues, and express \( |{\uparrow}\rangle \) in that basis.
Solution
\( \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix} \) has characteristic equation \( \lambda^2-1=0 \), so \( \lambda=\pm1 \). For \( \lambda=+1 \): \( \binom{a}{b} \) with \( b=a \Rightarrow |{+}x\rangle=\tfrac{1}{\sqrt2}\binom{1}{1} \). For \( \lambda=-1 \): \( b=-a \Rightarrow |{-}x\rangle=\tfrac{1}{\sqrt2}\binom{1}{-1} \). Inverting, \( |{\uparrow}\rangle=\binom{1}{0}=\tfrac{1}{\sqrt2}\big(|{+}x\rangle+|{-}x\rangle\big) \). Both outcomes of an \( S_x \) measurement on \( |{\uparrow}\rangle \) have probability \( \tfrac12 \).
Show that \( \hat{U}(2\pi,\hat{\mathbf{n}})=-\mathbf{1} \) for any axis \( \hat{\mathbf{n}} \), and that \( \hat{U}(4\pi,\hat{\mathbf{n}})=+\mathbf{1} \). Comment on observability.
Solution
From \( \hat{U}=\cos\tfrac{\theta}{2}\mathbf{1}-i\sin\tfrac{\theta}{2}\,\hat{\mathbf{n}}\cdot\vec\sigma \): at \( \theta=2\pi \), \( \cos\pi=-1 \), \( \sin\pi=0 \), so \( \hat{U}=-\mathbf{1} \) independent of \( \hat{\mathbf{n}} \). At \( \theta=4\pi \), \( \cos2\pi=+1 \), \( \sin2\pi=0 \), so \( \hat{U}=+\mathbf{1} \). A global \( -1 \) on an isolated state is physically inert, but in an interferometer where only one arm is rotated the relative phase between arms flips the interference pattern — observed in neutron interferometry, with full recovery at \( 4\pi \).
A spin is prepared in \( |{\uparrow}\rangle \) and rotated by \( \theta \) about \( \hat{\mathbf{x}} \). Compute \( \langle\hat{S}_z\rangle \) as a function of \( \theta \).
Solution
\( \hat{U}(\theta,\hat{\mathbf{x}})=\cos\tfrac{\theta}{2}\mathbf{1}-i\sin\tfrac{\theta}{2}\sigma_x=\begin{pmatrix}\cos\frac{\theta}{2}&-i\sin\frac{\theta}{2}\\-i\sin\frac{\theta}{2}&\cos\frac{\theta}{2}\end{pmatrix} \). Then \( |\psi\rangle=\hat{U}\binom{1}{0}=\binom{\cos\frac{\theta}{2}}{-i\sin\frac{\theta}{2}} \). With \( \hat{S}_z=\tfrac{\hbar}{2}\sigma_z \), \( \langle\hat{S}_z\rangle=\tfrac{\hbar}{2}\big(|\cos\tfrac{\theta}{2}|^2-|\sin\tfrac{\theta}{2}|^2\big)=\tfrac{\hbar}{2}\cos\theta \). The expectation follows the classical vector: full \( \hbar/2 \) at \( \theta=0 \), zero at \( \theta=\pi/2 \), \( -\hbar/2 \) at \( \theta=\pi \).
A proton (\( g_p=5.586 \), \( \mu_N=5.051\times10^{-27}\ \mathrm{J\,T^{-1}} \)) is placed in \( B=1.50\ \mathrm{T} \). Find its Larmor frequency \( f_L=\omega_L/2\pi \) and compare with the electron value at the same field.
Solution
\( \omega_L=g_p\mu_N B/\hbar=(5.586)(5.051\times10^{-27})(1.50)/(1.055\times10^{-34})=4.01\times10^{8}\ \mathrm{rad\,s^{-1}} \). Then \( f_L=\omega_L/2\pi=6.39\times10^{7}\ \mathrm{Hz}\approx63.9\ \mathrm{MHz} \), the familiar NMR range. The electron at \( 1.50\ \mathrm{T} \) has \( f_L=g_e\mu_B B/(2\pi\hbar)\approx42.0\ \mathrm{GHz} \), larger by \( \sim660\times \) because \( \mu_B/\mu_N=m_p/m_e\approx1836 \) partly offset by the \( g \)-factor ratio. Units: \( \mathrm{Hz} \) throughout.