Spin 1/2 and the Pauli matrices

The previous section left two loose threads: the algebra allows half-integer angular momentum but wavefunctions on the sphere cannot house it; and in one experiment a beam of silver atoms split into exactly two. This section ties the threads together.

First, the experimental evidence that forced spin into existence. By 1925, spectroscopy had piled up a stack of debts that “orbits + integer $l$” could not pay:

• Spectral lines come in pairs. Under a spectrometer, sodium’s yellow D line is actually two lines, at wavelengths 589.0 nm and 589.6 nm, split by a mere 0.002 eV. Hydrogen’s lines show the same “doublet” fine structure. It was as if something unknown was gently snapping every level in two.
• The anomalous Zeeman effect. Lines in a magnetic field often split into an even number of components. But we just counted in the previous section: orbital angular momentum gives $2l+1$ lines with integer $l$, and that number is always odd. An even count means some angular momentum has an even $2j+1$ — and $j$ must be a half-integer.

Goudsmit and Uhlenbeck (both graduate students at the time) proposed that the electron itself carries an angular momentum $s=\tfrac12$, along with a magnetic moment to match. They named it spin.

First, defuse the name: the electron is not a little top

“Spin” is a historically misleading name. It suggests the electron is a little ball rotating about its own axis. That picture is wrong, and wrong in a way you can demonstrate quantitatively:

Then what is spin? This is exactly where the reversal at the end of section 5.1 earns its keep: angular momentum is defined by the commutation relations, not by $\vec r\times\vec p$. Spin is three Hermitian operators $\hat S_x,\hat S_y,\hat S_z$ satisfying

$$[\hat S_x,\hat S_y]=\ii\hbar\hat S_z\ \text{and its cyclic partners}\tag{5.3.2}$$

Nothing more. It is attached to no position wavefunction, so the single-valuedness argument of section 5.2 — “one full turn must return the original value” — has no grip on it: the half-integer seats stand open. The electron occupies the smallest non-trivial one: $s=\tfrac12$, with eigenvalues $\hat S^2=\tfrac12\cdot\tfrac32\hbar^2=\tfrac34\hbar^2$ and $\hat S_z=\pm\tfrac\hbar2$ — just $2s+1=2$ states.

Spin is an intrinsic property of the electron, on the same footing as charge and mass: you cannot “slow down” or “speed up” the electron’s spin. $s=\tfrac12$ is a factory setting; all that can change is how its components along various directions come out.

A two-dimensional space and the Pauli matrices

A system with only two basis vectors is the smallest stage quantum mechanics can offer. Write

$$\ket{\uparrow}=\ket{s=\tfrac12,m=+\tfrac12} =\begin{pmatrix}1\\0\end{pmatrix},\qquad \ket{\downarrow}=\ket{s=\tfrac12,m=-\tfrac12} =\begin{pmatrix}0\\1\end{pmatrix}\tag{5.3.3}$$

A general state is a two-component complex vector $\ket{\chi}=a\ket{\uparrow}+b\ket{\downarrow}$ with $|a|^2+|b|^2=1$. Chapter 3’s abstract language — “states are vectors, observables are matrices” — here becomes for the first time concrete enough to compute by hand: every operator is a $2\times2$ matrix.

Building the Pauli matrices from the ladder-operator matrix elementsbasic~8 minexpand

Step 1: $\hat S_z$ comes for free. The basis vectors are its eigenstates; put the eigenvalues on the diagonal:

$$\hat S_z=\frac\hbar2\begin{pmatrix}1&0\\0&-1\end{pmatrix}\tag{5.3.4}$$

Step 2: write $\hat S_\pm$ with last section’s general formula. The formula $\hat S_\pm\ket{s,m}=\hbar\sqrt{s(s+1)-m(m\pm1)}\ket{s,m\pm1}$ has only one non-zero matrix element when $s=\tfrac12$. For $\hat S_+$: acting on $\ket{\downarrow}$ ($m=-\tfrac12$) gives the coefficient $\hbar\sqrt{\tfrac34-(-\tfrac12)(\tfrac12)}=\hbar\sqrt{\tfrac34+\tfrac14}=\hbar$; acting on $\ket{\uparrow}$ hits the top of the ladder and annihilates. So

$$\hat S_+=\hbar\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad \hat S_-=\hat S_+^\dagger=\hbar\begin{pmatrix}0&0\\1&0\end{pmatrix}\tag{5.3.5}$$

Step 3: solve back for $\hat S_x,\hat S_y$. Inverting the definition $\hat S_\pm=\hat S_x\pm\ii\hat S_y$:

$$\hat S_x=\frac{\hat S_++\hat S_-}{2}=\frac\hbar2\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad \hat S_y=\frac{\hat S_+-\hat S_-}{2\ii}=\frac\hbar2\begin{pmatrix}0&-\ii\\ \ii&0\end{pmatrix}\tag{5.3.6}$$

Step 4: factor out the common piece. All three matrices carry $\hbar/2$; give the remaining skeletons their own names:

$$\hat{\vec S}=\frac\hbar2\vec\sigma,\qquad \sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad \sigma_y=\begin{pmatrix}0&-\ii\\ \ii&0\end{pmatrix},\quad \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\tag{5.3.7}$$

These are the Pauli matrices.

Step 5: inspection. Direct multiplication verifies $[\hat S_x,\hat S_y]=\ii\hbar\hat S_z$ (i.e. $\sigma_x\sigma_y=\ii\sigma_z$), and

$$\hat S^2=\frac{\hbar^2}{4}(\sigma_x^2+\sigma_y^2+\sigma_z^2) =\frac{3\hbar^2}{4}\begin{pmatrix}1&0\\0&1\end{pmatrix}\tag{5.3.8}$$

— the identity matrix times $\tfrac34\hbar^2$: every two-component state is an eigenstate of $\hat S^2$, with eigenvalue exactly $s(s+1)\hbar^2$ at $s=\tfrac12$. The algebra is fully self-consistent.

The Pauli algebra is useful enough to memorise outright. Each matrix squares to the identity, any two distinct ones anticommute, and both facts fold into one master product formula:

$$\sigma_i\sigma_j=\delta_{ij}\,\mathbb{1}+\ii\sum_k\epsilon_{ijk}\,\sigma_k\tag{5.3.9}$$

($\epsilon_{ijk}$ is the cyclic symbol: $+1$ for the order $xyz$ and its cycles, $-1$ for the reverse, 0 with any repeated index.) An immediate corollary: the spin along any direction $\hat n$, $\hat n\cdot\vec\sigma$, satisfies $(\hat n\cdot\vec\sigma)^2=\mathbb{1}$, so its eigenvalues can only be $\pm1$, i.e. $\hat S_n=\pm\tfrac\hbar2$. No matter which direction you measure along, the answer is always just “up” or “down”, always of size $\hbar/2$. The next section will watch this play out in the laboratory again and again.

Spin states along an arbitrary direction

The state “up” along $\hat n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ (the eigenstate of $\hat n\cdot\hat{\vec S}$ with eigenvalue $+\tfrac\hbar2$) works out to be:

$$\ket{+\hat n}=\cos\frac\theta2\,\ket{\uparrow}+\ee^{\ii\phi}\sin\frac\theta2\,\ket{\downarrow}\tag{5.3.10}$$

∎Key formulas

• Spin-1/2 spectrum

  $$\hat S^2=\tfrac34\hbar^2\,\mathbb{1},\qquad \hat S_z\ket{\uparrow}=\tfrac\hbar2\ket{\uparrow},\ \ \hat S_z\ket{\downarrow}=-\tfrac\hbar2\ket{\downarrow}$$

  Intrinsic property: s = 1/2 is unchangeable; only the orientation can vary

• Pauli matrices

  $$\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\ \sigma_y=\begin{pmatrix}0&-\ii\\ \ii&0\end{pmatrix},\ \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

  S = (ħ/2)σ; Hermitian, traceless, eigenvalues ±1

• Pauli algebra

  $$\sigma_i\sigma_j=\delta_{ij}\mathbb{1}+\ii\,\epsilon_{ijk}\sigma_k$$

  Squares to 1, mutual anticommutation, cyclic products — all in one line

• Spin state along any direction

  $$\ket{+\hat n}=\cos\tfrac\theta2\ket{\uparrow}+\ee^{\ii\phi}\sin\tfrac\theta2\ket{\downarrow}$$

  Half-angle θ/2: a 360° turn in space flips the sign; 720° restores the state

?Self-check4 questions

1. 1.

   What is wrong with the claim "the electron's spin comes from the electron rotating about its own axis"?

   • Using the classical electron radius, the required surface speed is about 170 times the speed of light — and the electron is in fact a point particle
   • The electron is too light to rotate
   • A rotating charge would radiate energy and the electron would collapse
   • The claim is entirely correct; the rotation is just very fast

   Submit
2. 2.

   Which statements about the Pauli matrices are correct? (Select all that apply.)

   Select all that apply

   • Each Pauli matrix squares to the identity matrix
   • σ_x σ_y = iσ_z
   • The three Pauli matrices commute pairwise
   • For any direction, n̂·σ has eigenvalues ±1

   Submit
3. 3.

   Rotate a spin state through 360° in space about some axis. What happens to the state vector?

   • It becomes −1 times itself; a 720° rotation is needed to return to the original state
   • It returns to itself, like any vector
   • It becomes an orthogonal state
   • The change is unpredictable

   Submit
4. 4.

   In the S_z = +ħ/2 state, what is the angle in degrees between the "spin vector" and the z axis? (Use cos θ = S_z/|S|; give one decimal place.)

   °30% relative tolerance

   Submit

What comes next

We now hold all the mathematics of spin: a two-dimensional space, three matrices, and only “up” or “down” as answers along any direction.

But these conclusions sound frankly bizarre — exactly $\pm\hbar/2$ from any direction, with not a single value in between? Time to go back to that Frankfurt laboratory in 1922. One magnet, one beam of silver atoms, one glass plate: the apparatus not only demonstrates every claim above directly, it turns chapter 3’s measurement postulate into something you can see. The next section is devoted entirely to this one experiment.