A Ladder of Worked Examples: Spin Operators in Positronium

Example 0.1 (An eigenstate). Compute $S_z|\uparrow ⟩$.

Solution. $|\uparrow ⟩$ is defined as the $S_z$ eigenstate with eigenvalue $+\hslash /2$:

$$S_z|\uparrow ⟩=+\frac{\hslash }{2}|\uparrow ⟩.$$

The state comes back unchanged, scaled by its eigenvalue.

What it tests. The meaning of "eigenstate": the operator returns the same state times a number. Nothing happened to the physics; the state was already "definite" with respect to this measurement.

Example 0.2 (Not an eigenstate). Compute $S_x|\uparrow ⟩$, using $S_x=\frac{1}{2}(S_{+}+S_{-})$.

Solution.

$$S_x|\uparrow ⟩=\frac{1}{2}(S_{+}|\uparrow ⟩+S_{-}|\uparrow ⟩)=\frac{1}{2}(0+\hslash |\downarrow ⟩)=\frac{\hslash }{2}|\downarrow ⟩.$$

The output points along a different basis vector.

What it tests. What "not an eigenstate" looks like at the level of kets: the operator rotates the state out of itself. $|\uparrow ⟩$ has no definite $S_x$ value.

The physical point. Definiteness is basis-relative. A state that is certain for one measurement is a superposition for another. This single idea, iterated, is the entire positronium problem: states definite for the hyperfine interaction are superpositions for the Zeeman interaction.

Example 0.3 (Eigenvalue zero is not nothing). True or false: if $A|\psi ⟩=0$, the state $|\psi ⟩$ is destroyed.

Solution. False. $A|\psi ⟩=0=0\cdot |\psi ⟩$ means $|\psi ⟩$ is an eigenstate with eigenvalue zero — a perfectly healthy state whose measured value of $A$ is $0$. The state still exists and is normalized; only the measurement outcome is zero.

What it tests. A misconception that will otherwise detonate at rung 3, where $S_z|1,0⟩=0$ and $S^2|0,0⟩=0$ both matter and mean different things.

Example 1.1 (Operators act only on their own particle). Compute $S_{1z}|\uparrow \downarrow ⟩$ and $S_{2z}|\uparrow \downarrow ⟩$.

Solution. $|\uparrow \downarrow ⟩=|\uparrow {⟩}_1\otimes |\downarrow {⟩}_2$. The operator $S_{1z}$ is really $S_z\otimes 1$: it reads slot 1 and passes through slot 2.

$$S_{1z}|\uparrow \downarrow ⟩=+\frac{\hslash }{2}|\uparrow \downarrow ⟩S_{2z}|\uparrow \downarrow ⟩=-\frac{\hslash }{2}|\uparrow \downarrow ⟩.$$

What it tests. The tensor-product rule. Every two-spin computation in this ladder reduces to this move plus linearity.

Example 1.2 (Total spin projection). Compute $(S_{1z}+S_{2z})$ on all four uncoupled states.

Solution. Add the arrows:

$$|\uparrow \uparrow ⟩\to +\hslash ,|\uparrow \downarrow ⟩\to 0,|\downarrow \uparrow ⟩\to 0,|\downarrow \downarrow ⟩\to -\hslash .$$

All four are eigenstates of total $S_z$. Two of them share the eigenvalue $0$.

What it tests. That product states have definite total $m$, and — crucially — that the $m=0$ eigenvalue is degenerate. Circle that fact. The entire structure of the coupled basis, and the entire Zeeman mixing phenomenon, lives inside that two-dimensional degenerate subspace.

Example 1.3 (A physical question: the magnetic moment of $|\uparrow \uparrow ⟩$). In positronium, what is the net magnetic moment of the state $|\uparrow \uparrow ⟩$?

Solution. The moments are ${\mathbf{\mu }}_1=-\frac{g{\mu }_B}{\hslash }{\mathbf{S}}_1$ (electron, charge $-e$) and ${\mathbf{\mu }}_2=+\frac{g{\mu }_B}{\hslash }{\mathbf{S}}_2$ (positron, charge $+e$). With both spins up:

$${\mu }_z=-\frac{g{\mu }_B}{\hslash }\cdot \frac{\hslash }{2}+\frac{g{\mu }_B}{\hslash }\cdot \frac{\hslash }{2}=0.$$

Aligned spins on opposite charges give canceling moments. Conversely, $|\uparrow \downarrow ⟩$ has anti-aligned spins but adding moments: ${\mu }_z=-g{\mu }_B$, a full double moment.

What it tests. Whether you are tracking physics or just arrows. Students who answer "$|\uparrow \uparrow ⟩$ has the biggest moment because both spins point up" are pattern-matching from hydrogen.

The physical point. This is the single fact from which everything at rung 5 follows: a magnetic field cannot shift the stretched states (no moment to grab) and couples maximally to the mixed-arrow states. Positronium's Zeeman physics is upside down relative to naive intuition, and the inversion happens here, in a two-line moment calculation.

Example 2.1 (Act on a stretched state). Compute ${\mathbf{S}}_1\cdot {\mathbf{S}}_2|\uparrow \uparrow ⟩$.

Solution. The zz term gives $(+\frac{\hslash }{2})(+\frac{\hslash }{2})=+\frac{{\hslash }^2}{4}$. Both ladder terms annihilate the state: $S_{1+}$ can't raise an up spin, and neither can $S_{2+}$.

$${\mathbf{S}}_1\cdot {\mathbf{S}}_2|\uparrow \uparrow ⟩=+\frac{{\hslash }^2}{4}|\uparrow \uparrow ⟩.$$

An eigenstate. The swap part has nothing to swap into — there is no other state with $m=+1$.

What it tests. Ladder-operator bookkeeping, and a structural insight: states alone in their $m$ sector are automatically eigenstates of anything that conserves $m$.

Example 2.2 (Act on a mixed state: the swap appears). Compute ${\mathbf{S}}_1\cdot {\mathbf{S}}_2|\uparrow \downarrow ⟩$.

Solution. zz term: $(+\frac{\hslash }{2})(-\frac{\hslash }{2})=-\frac{{\hslash }^2}{4}$, diagonal. Ladders: $S_{1+}{S}_{2-}|\uparrow \downarrow ⟩=0$ (can't raise spin 1), but $S_{1-}{S}_{2+}|\uparrow \downarrow ⟩={\hslash }^2|\downarrow \uparrow ⟩$ — the arrows swap.

$${\mathbf{S}}_1\cdot {\mathbf{S}}_2|\uparrow \downarrow ⟩=-\frac{{\hslash }^2}{4}|\uparrow \downarrow ⟩+\frac{{\hslash }^2}{2}|\downarrow \uparrow ⟩.$$

Not an eigenstate: the output leaks into $|\downarrow \uparrow ⟩$.

What it tests. Recognizing non-eigenstate behavior in a two-particle setting (rung 0.2 scaled up), and seeing where matrices come from: the two coefficients you just computed are a column of the matrix below.

The physical point. ${\mathbf{S}}_1\cdot {\mathbf{S}}_2$ physically exchanges the spins. "The electron is up and the positron is down" is not a stationary situation under a spin-spin interaction — the interaction relentlessly trades the orientation back and forth. Stationary states will have to be combinations that look the same under the trade. You can now guess the eigenstates before deriving them: the symmetric and antisymmetric combinations.

Example 2.3 (Assemble the matrix). Write ${\mathbf{S}}_1\cdot {\mathbf{S}}_2$ as a matrix in the ordered basis $\{|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩\}$.

Solution. From 2.1, 2.2, its mirror image on $|\downarrow \uparrow ⟩$, and the $|\downarrow \downarrow ⟩$ analog of 2.1:

$${\mathbf{S}}_1\cdot {\mathbf{S}}_2\doteq \frac{{\hslash }^2}{4}\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 2& 0\\ 0& 2& -1& 0\\ 0& 0& 0& 1\end{array}\right).$$

What it tests. The ket-computation → matrix-element → matrix pipeline. Rule to internalize: always declare the basis ordering before writing a matrix. Most sign errors in this subject are silent reorderings.

Example 3.1 (Construct $|1,0⟩$ with the lowering operator). Starting from $|1,1⟩=|\uparrow \uparrow ⟩$, generate $|1,0⟩$.

Solution. The general lowering rule $S_{-}|s,m⟩=\hslash \sqrt{s(s+1)-m(m-1)}|s,m-1⟩$ gives $S_{-}|1,1⟩=\hslash \sqrt{2}|1,0⟩$. But also $S_{-}=S_{1-}+S_{2-}$ acting on the product state:

$$S_{-}|\uparrow \uparrow ⟩=\hslash |\downarrow \uparrow ⟩+\hslash |\uparrow \downarrow ⟩.$$

Equate and solve:

$$|1,0⟩=\frac{|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩}{\sqrt{2}}.$$

The $1/\sqrt{2}$ falls out of the ladder normalization — it is not imposed by hand. The singlet is then fixed (up to phase) as the unique $m=0$ state orthogonal to this one: $|0,0⟩=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$.

What it tests. The ladder construction — the same machine that generates every Clebsch-Gordan table. If you can do this, you never need to memorize two-spin coefficients.

Example 3.2 (Verify the quantum numbers of $|0,0⟩$, the strong way). Show that all three components of total spin annihilate the singlet.

Solution. $S_z$: each component of $|0,0⟩$ has one up and one down arrow, so the $\pm \frac{\hslash }{2}$ cancel term by term $\to 0$. Raising:

$$S_{+}|0,0⟩=(S_{1+}+S_{2+})\frac{|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩}{\sqrt{2}}=\frac{\hslash |\uparrow \uparrow ⟩-\hslash |\uparrow \uparrow ⟩}{\sqrt{2}}=0$$

— the two surviving terms cancel because of the minus sign. Same for $S_{-}$. Since $S_x,S_y$ are combinations of $S_{\pm }$: all components give zero, hence $S^2=S_x^2+S_y^2+S_z^2$ gives zero.

What it tests. Eigenvalue zero as a real statement (rung 0.3): the singlet has no spin along any axis. It is annihilated by every generator of rotations — the definition of a rotational scalar. No rotated Stern-Gerlach apparatus can find an axis in it, because there isn't one.

The physical point. Contrast to run immediately: $S_{+}|1,0⟩=\hslash \sqrt{2}|1,1⟩\ne 0$. The plus sign makes the terms add. So $|1,0⟩$ has $m=0$ but is emphatically not spinless — it is one orientation of a spin-1 object. Zero projection and zero spin are different claims, and the relative sign inside the superposition is the entire difference.

Example 3.3 (The cancellation inside $S^2|0,0⟩=0$). Compute $S^2|0,0⟩$ via $S^2=S_1^2+S_2^2+2{\mathbf{S}}_1\cdot {\mathbf{S}}_2$ and identify what cancels.

Solution. The constituent terms are fixed: $S_1^2+S_2^2=\frac{3{\hslash }^2}{4}+\frac{3{\hslash }^2}{4}=\frac{3{\hslash }^2}{2}$, always, for any two-spin state — you cannot turn off the constituents' spins. On the singlet, the swap picks up a sign (antisymmetry), giving ${\mathbf{S}}_1\cdot {\mathbf{S}}_2|0,0⟩=(-\frac{{\hslash }^2}{4}-\frac{{\hslash }^2}{2})|0,0⟩=-\frac{3{\hslash }^2}{4}|0,0⟩$. So

$$S^2|0,0⟩=\left(\frac{3{\hslash }^2}{2}-\frac{3{\hslash }^2}{2}\right)|0,0⟩=0.$$

The physical point. The zero is an exact cancellation between constituent spin and correlation. The spins don't vanish; their perfect anticorrelation cancels them. (Triplet check: correlation term $+\frac{{\hslash }^2}{2}$, total $2{\hslash }^2=1(1+1){\hslash }^2$. $✓$) The same computation, run on both states, is the hyperfine eigenvalue calculation — nothing here is wasted.

Example 3.4 (A product state is not a spin state). Expand $|\uparrow \downarrow ⟩$ in the coupled basis and interpret.

Solution. Invert the definitions:

$$|\uparrow \downarrow ⟩=\frac{|1,0⟩+|0,0⟩}{\sqrt{2}}.$$

A 50/50 superposition of total spin 1 and total spin 0. "The electron is up and the positron is down" is a state of definite individual projections but indefinite total spin.

What it tests. Basis inversion, and the deepest conceptual content of the rung: which questions have definite answers depends on the state, and definite individual properties do not imply definite collective ones. (The singlet itself is a Bell state — the first genuinely entangled state most students ever meet.)

Example 4.1 (Two levels from four states). For $H_{\text{hf}}=\frac{A}{{\hslash }^2}{\mathbf{S}}_1\cdot {\mathbf{S}}_2$ with $A>0$, find the spectrum and its degeneracies.

Solution. From rung 3: eigenvalue $+\frac{A}{4}$ on all three triplet states, $-\frac{3A}{4}$ on the singlet. Four states, two levels, splitting $A$. In positronium $A\approx 0.841$ meV — the 203 GHz interval.

What it tests. The states-vs-levels distinction. The triplet degeneracy is not an accident: with no field there is no preferred direction, and states differing only in orientation ($m$) cannot differ in energy. Rotational symmetry protects the degeneracy.

The physical point. The two levels are two species: para-positronium (singlet, $2\gamma$, 125 ps) and ortho-positronium (triplet, $3\gamma$, 142 ns). The thousand-fold lifetime ratio — set by charge conjugation ($C=(-1{)}^{L+S}$ vs $C=(-1{)}^n$ for $n$ photons) plus an extra factor of $\alpha$ — is the loaded gun on the wall. Rung 5 fires it.

Example 4.2 (Statistical populations). Positronium forms at room temperature with random spin orientations. What fraction is ortho?

Solution. $k_{B}T\approx 25$ meV $\gg A\approx 0.84$ meV, so all four spin states are populated equally. Three of four are triplet: 75% ortho, 25% para — the 3:1 ratio directly visible as a two-component decay curve in annihilation-lifetime experiments.

What it tests. Whether degeneracy is understood as a count with physical consequences, not a formal label. The multiplicity $2S+1$ shows up in a detector.

Example 4.3 (Dynamics: the state that won't sit still). Positronium is prepared in $|\uparrow \downarrow ⟩$ at $t=0$, zero field. Find the state at time $t$.

Solution. Expand in energy eigenstates (Example 3.4) and attach phases:

$$|\psi (t)⟩=\frac{e^{-iAt/4\hslash }|1,0⟩+e^{+i3At/4\hslash }|0,0⟩}{\sqrt{2}}.$$

Up to a global phase, the relative phase is $e^{iAt/\hslash }$. When $At/\hslash =\pi$, the relative sign flips and the state has become $|\downarrow \uparrow ⟩$. The system oscillates between $|\uparrow \downarrow ⟩$ and $|\downarrow \uparrow ⟩$ at the hyperfine frequency $A/h\approx 203$ GHz.

What it tests. The full payoff of "not an eigenstate": superpositions of split levels beat at the splitting frequency. Example 2.2 said the interaction "trades the arrows back and forth" — this computation is that sentence made quantitative. (The same mathematics is neutrino oscillation and every two-level Rabi problem you will ever meet.)

Example 5.1 (Derive the Zeeman operator and act on everything). From $\mathbf{\mu }=g\frac{q}{2m}\mathbf{S}$, build $H_Z$ and act on both bases.

Solution. Opposite charges give ${\mathbf{\mu }}_1=-\frac{g{\mu }_B}{\hslash }{\mathbf{S}}_1$, ${\mathbf{\mu }}_2=+\frac{g{\mu }_B}{\hslash }{\mathbf{S}}_2$, so with $x\equiv g{\mu }_{B}B$:

$$H_Z=-({\mathbf{\mu }}_1+{\mathbf{\mu }}_2)\cdot B\hat{z}=\frac{x}{\hslash }(S_{1z}-S_{2z})$$

— the difference, exactly as Example 1.3 predicted. Uncoupled action: eigenvalues $0,+x,-x,0$ on $\{|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩\}$ — diagonal. Coupled action: the $\pm \hslash$ eigenvalues flip the internal relative sign,

$$H_Z|1,0⟩=x|0,0⟩H_Z|0,0⟩=x|1,0⟩H_Z|1,\pm 1⟩=0$$

— purely off-diagonal in the $m=0$ block. Same antisymmetry-detects-a-swap mechanism as rung 3.2, now driven by the field.

What it tests. Every prior rung at once: moments (1.3), slot-wise action (1.1), linearity on superpositions (3.x), sign bookkeeping.

Example 5.2 (The role reversal, stated as matrices). Write $H_{\text{hf}}$ and $H_Z$ in both bases and compare.

Solution. In the uncoupled basis, $H_Z$ is diagonal, $\mathrm{diag}(0,x,-x,0)$, and $H_{\text{hf}}$ has the off-diagonal $A/2$ swap couplings (Example 2.3). In the coupled basis, $H_{\text{hf}}$ is diagonal, $\mathrm{diag}(\frac{A}{4},\frac{A}{4},\frac{A}{4},-\frac{3A}{4})$, and $H_Z$ is purely off-diagonal. The operators have exactly traded places. The bridge is one $45^{\circ }$ rotation

$$R=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$$

on the $m=0$ block; verify by explicit multiplication that $R$ diagonalizes one block while un-diagonalizing the other. No rotation fixes both, because the blocks don't commute.

The physical point. Rung 0's closing thought has returned at full scale: non-commuting operators disagree about which states are definite. Here the disagreement is between two pieces of the same Hamiltonian, and the field strength decides who wins.

Example 5.3 (Assemble, exploit symmetry, diagonalize). Solve $H=H_{\text{hf}}+H_Z$ exactly.

Solution. In the coupled basis, ordering $\{|1,1⟩,|1,-1⟩,|1,0⟩,|0,0⟩\}$:

$$H\doteq \left(\begin{array}{cccc}A/4& & & \\ & A/4& & \\ & & A/4& x\\ & & x& -3A/4\end{array}\right).$$

Before computing anything, read the zeros: total $S_z$ commutes with both terms, so $H$ cannot connect different $m$ — the block structure is a conservation law, and $|1,\pm 1⟩$ are exact eigenstates at $A/4$ for any field: protected, flat. The $m=0$ block gives

$$E_{\pm }=-\frac{A}{4}\pm \sqrt{{\left(\frac{A}{2}\right)}^2+x^2}.$$

Run all three checks: $B\to 0$ recovers $A/4,-3A/4$; $x\gg A$ gives $\pm x$ (Paschen-Back — the uncoupled basis wins at high field, the mixing angle $\tan 2\theta =2x/A$ rotating from $0^{\circ }$ to $45^{\circ }$ is the dial between the bases); trace $=-A/2$ at every field.

What it tests. Symmetry-before-computation as a working habit. Every zero was earned at a lower rung; here they organize the entire solution.

Example 5.4 (The measurable: magnetic quenching). At $B=1$ T, find the lifetime of the perturbed $m=0$ triplet. Take ${\Gamma }_{\text{para}}/{\Gamma }_{\text{ortho}}\approx 1140$.

Solution. $x/A\approx 0.138$, so $\theta =\frac{1}{2}\arctan (0.276)\approx 0.135$ rad and ${\sin }^2\theta \approx 0.018$. The perturbed triplet $|+⟩=\cos \theta |1,0⟩+\sin \theta |0,0⟩$ inherits singlet decay:

$${\Gamma }_{+}\approx {\Gamma }_{\text{ortho}}\left({\cos }^2\theta +1140{\sin }^2\theta \right)\approx 21.5{\Gamma }_{\text{ortho}}.$$

Lifetime: $142\text{ns}/21.5\approx 6.6$ ns — collapsed by a factor of twenty from a 1.8% admixture, while $|1,\pm 1⟩$ are untouched. This is how Deutsch measured $A$ in the 1950s: not by driving the 203 GHz transition, but by reading the splitting out of the field dependence of the quenching.

What it tests. The complete chain, ending in a number an experiment can check. The tiny admixture matters because it multiplies a huge rate ratio — a lesson in when "small" perturbations have large consequences: always compare the mixing to the ratio of the physics it unlocks.