Case Study 1: The Spin-1/2 System — A Complete Worked Example in Dirac Notation

Contributors to Quantum Mechanics: From Wavefunctions to Qubits

Case Study 1: The Spin-1/2 System — A Complete Worked Example in Dirac Notation

Introduction

The spin-1/2 system is the single most important example in modern quantum mechanics. It is the simplest non-trivial quantum system (two-dimensional Hilbert space), it has no classical analogue (there is no wave function $\psi(x)$ for spin), and it is the foundation of everything from the Pauli exclusion principle to quantum computing. This case study works through the entire spin-1/2 system using Dirac notation, applying every technique from Chapter 8.

This case study serves as a complete template. If you can follow every step here, you are ready for the rest of the book.

Part 1: The Hilbert Space and Basis

The spin-1/2 Hilbert space $\mathcal{H}$ is two-dimensional. We choose the eigenstates of $\hat{S}_z$ as our standard basis:

$$\hat{S}_z|\uparrow\rangle = +\frac{\hbar}{2}|\uparrow\rangle, \qquad \hat{S}_z|\downarrow\rangle = -\frac{\hbar}{2}|\downarrow\rangle$$

These kets satisfy:

Orthonormality: $\langle\uparrow|\uparrow\rangle = \langle\downarrow|\downarrow\rangle = 1$, $\langle\uparrow|\downarrow\rangle = \langle\downarrow|\uparrow\rangle = 0$
Completeness: $|\uparrow\rangle\langle\uparrow| + |\downarrow\rangle\langle\downarrow| = \hat{I}$

An arbitrary spin state is:

In column-vector representation:

$$|\uparrow\rangle \leftrightarrow \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad |\downarrow\rangle \leftrightarrow \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \qquad |\chi\rangle \leftrightarrow \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$$

Why spin-1/2 has no wave function: The spin degree of freedom is purely internal — it does not depend on position. There is no function $\psi(x)$ that encodes spin. The Hilbert space is two-dimensional (not infinite-dimensional like $L^2(\mathbb{R})$), and the state is completely specified by two complex numbers $\alpha$ and $\beta$ (subject to normalization). This is why we need Dirac notation: it handles this system naturally, while wave mechanics cannot.

Part 2: Operators and Their Matrix Representations

The spin operators

The three spin operators are $\hat{S}_x$, $\hat{S}_y$, and $\hat{S}_z$. In the $S_z$ eigenbasis:

$$[S_z] = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad [S_x] = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad [S_y] = \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$$

In Dirac notation, these are:

$$\hat{S}_z = \frac{\hbar}{2}\left(|\uparrow\rangle\langle\uparrow| - |\downarrow\rangle\langle\downarrow|\right)$$

$$\hat{S}_x = \frac{\hbar}{2}\left(|\uparrow\rangle\langle\downarrow| + |\downarrow\rangle\langle\uparrow|\right)$$

$$\hat{S}_y = \frac{\hbar}{2}\left(-i|\uparrow\rangle\langle\downarrow| + i|\downarrow\rangle\langle\uparrow|\right)$$

Verifying $\hat{S}_x$ from its Dirac form: Apply $\hat{S}_x$ to $|\uparrow\rangle$:

$$\hat{S}_x|\uparrow\rangle = \frac{\hbar}{2}\left(|\uparrow\rangle\underbrace{\langle\downarrow|\uparrow\rangle}_{0} + |\downarrow\rangle\underbrace{\langle\uparrow|\uparrow\rangle}_{1}\right) = \frac{\hbar}{2}|\downarrow\rangle$$

And to $|\downarrow\rangle$:

$$\hat{S}_x|\downarrow\rangle = \frac{\hbar}{2}\left(|\uparrow\rangle\underbrace{\langle\downarrow|\downarrow\rangle}_{1} + |\downarrow\rangle\underbrace{\langle\uparrow|\downarrow\rangle}_{0}\right) = \frac{\hbar}{2}|\uparrow\rangle$$

So $\hat{S}_x$ flips the spin: it takes $|\uparrow\rangle$ to $|\downarrow\rangle$ and vice versa (up to a factor of $\hbar/2$). This is exactly what the matrix representation says.

Commutation relations

Using the Dirac notation expressions, we can verify the fundamental spin commutation relation $[\hat{S}_x, \hat{S}_y] = i\hbar\hat{S}_z$.

Compute $\hat{S}_x\hat{S}_y$:

$$\hat{S}_x\hat{S}_y = \frac{\hbar^2}{4}\left(|\uparrow\rangle\langle\downarrow| + |\downarrow\rangle\langle\uparrow|\right)\left(-i|\uparrow\rangle\langle\downarrow| + i|\downarrow\rangle\langle\uparrow|\right)$$

Expanding (using orthonormality to evaluate the middle overlaps):

$$= \frac{\hbar^2}{4}\left(-i|\uparrow\rangle\underbrace{\langle\downarrow|\uparrow\rangle}_{0}\langle\downarrow| + i|\uparrow\rangle\underbrace{\langle\downarrow|\downarrow\rangle}_{1}\langle\uparrow| - i|\downarrow\rangle\underbrace{\langle\uparrow|\uparrow\rangle}_{1}\langle\downarrow| + i|\downarrow\rangle\underbrace{\langle\uparrow|\downarrow\rangle}_{0}\langle\uparrow|\right)$$

$$= \frac{\hbar^2}{4}\left(i|\uparrow\rangle\langle\uparrow| - i|\downarrow\rangle\langle\downarrow|\right) = \frac{i\hbar^2}{4}\left(|\uparrow\rangle\langle\uparrow| - |\downarrow\rangle\langle\downarrow|\right)$$

Similarly, $\hat{S}_y\hat{S}_x = -\frac{i\hbar^2}{4}(|\uparrow\rangle\langle\uparrow| - |\downarrow\rangle\langle\downarrow|)$.

Therefore:

$$[\hat{S}_x, \hat{S}_y] = \hat{S}_x\hat{S}_y - \hat{S}_y\hat{S}_x = \frac{i\hbar^2}{2}\left(|\uparrow\rangle\langle\uparrow| - |\downarrow\rangle\langle\downarrow|\right) = i\hbar\hat{S}_z \quad \checkmark$$

Part 3: Eigenstates of $\hat{S}_x$ and $\hat{S}_y$

Finding the $S_x$ eigenstates

We seek states $|+x\rangle$ and $|-x\rangle$ such that $\hat{S}_x|\pm x\rangle = \pm\frac{\hbar}{2}|\pm x\rangle$.

Writing $|+x\rangle = a|\uparrow\rangle + b|\downarrow\rangle$ and applying $\hat{S}_x$:

$$\hat{S}_x(a|\uparrow\rangle + b|\downarrow\rangle) = \frac{\hbar}{2}(b|\uparrow\rangle + a|\downarrow\rangle) = \frac{\hbar}{2}(a|\uparrow\rangle + b|\downarrow\rangle)$$

This gives $b = a$. Normalization: $|a|^2 + |b|^2 = 2|a|^2 = 1$, so $a = 1/\sqrt{2}$:

$$|+x\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle + |\downarrow\rangle)$$

For eigenvalue $-\hbar/2$: $b = -a$, giving:

$$|-x\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle - |\downarrow\rangle)$$

The $S_y$ eigenstates

By the same method:

$$|+y\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle + i|\downarrow\rangle), \qquad |-y\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle - i|\downarrow\rangle)$$

Verification of completeness

$$|+x\rangle\langle +x| + |-x\rangle\langle -x| = \frac{1}{2}(|\uparrow\rangle + |\downarrow\rangle)(\langle\uparrow| + \langle\downarrow|) + \frac{1}{2}(|\uparrow\rangle - |\downarrow\rangle)(\langle\uparrow| - \langle\downarrow|)$$

$$= \frac{1}{2}(|\uparrow\rangle\langle\uparrow| + |\uparrow\rangle\langle\downarrow| + |\downarrow\rangle\langle\uparrow| + |\downarrow\rangle\langle\downarrow|) + \frac{1}{2}(|\uparrow\rangle\langle\uparrow| - |\uparrow\rangle\langle\downarrow| - |\downarrow\rangle\langle\uparrow| + |\downarrow\rangle\langle\downarrow|)$$

$$= |\uparrow\rangle\langle\uparrow| + |\downarrow\rangle\langle\downarrow| = \hat{I} \quad \checkmark$$

Every orthonormal basis gives the same identity operator — this is the basis-independence of the completeness relation.

Part 4: Measurement and Probability

Stern-Gerlach measurement

A particle is prepared in state $|\chi\rangle = \cos\theta|\uparrow\rangle + e^{i\phi}\sin\theta|\downarrow\rangle$ (the most general normalized spin-1/2 state). We measure $S_z$.

Probabilities:

$$P(\uparrow) = |\langle\uparrow|\chi\rangle|^2 = \cos^2\theta$$

$$P(\downarrow) = |\langle\downarrow|\chi\rangle|^2 = \sin^2\theta$$

Check: $\cos^2\theta + \sin^2\theta = 1$. $\checkmark$

Expectation value:

$$\langle\hat{S}_z\rangle = \langle\chi|\hat{S}_z|\chi\rangle = \frac{\hbar}{2}\cos^2\theta + \left(-\frac{\hbar}{2}\right)\sin^2\theta = \frac{\hbar}{2}\cos(2\theta)$$

Now measure $S_x$ instead:

$$P(+x) = |\langle +x|\chi\rangle|^2 = \left|\frac{1}{\sqrt{2}}(\cos\theta + e^{i\phi}\sin\theta)\right|^2 = \frac{1}{2}(1 + \sin(2\theta)\cos\phi)$$

$$P(-x) = |\langle -x|\chi\rangle|^2 = \frac{1}{2}(1 - \sin(2\theta)\cos\phi)$$

Check: $P(+x) + P(-x) = 1$. $\checkmark$

Expectation value:

$$\langle\hat{S}_x\rangle = \frac{\hbar}{2}\sin(2\theta)\cos\phi$$

Part 5: Change of Basis — $S_z$ to $S_x$

The unitary matrix connecting the $S_z$ and $S_x$ bases is:

$$[U] = \begin{pmatrix} \langle +x|\uparrow\rangle & \langle +x|\downarrow\rangle \\ \langle -x|\uparrow\rangle & \langle -x|\downarrow\rangle \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$

Transforming $\hat{S}_z$ to the $S_x$ basis:

$$[S_z]_{S_x \text{ basis}} = [U]^\dagger [S_z] [U] = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \cdot \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$

(Note: $[U]^\dagger = [U]$ here since $[U]$ is real and symmetric.)

Computing step by step:

$$\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{\hbar}{2\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$$

Then:

$$[S_z]_{S_x} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \cdot \frac{\hbar}{2\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} = \frac{\hbar}{4}\begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix} = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

This is $\frac{\hbar}{2}\sigma_x$. That is: $\hat{S}_z$ in the $S_x$ basis has the same form as $\hat{S}_x$ in the $S_z$ basis. This is a consequence of the symmetry between $x$ and $z$ — they are related by a rotation.

Key check: $\text{Tr}([S_z]_{S_x}) = 0 = \text{Tr}([S_z])$. The trace is preserved under the basis change, as it must be.

Part 6: Time Evolution — Larmor Precession

A spin-1/2 particle with magnetic moment $\hat{\boldsymbol{\mu}} = \gamma\hat{\mathbf{S}}$ sits in a uniform magnetic field $\mathbf{B} = B\hat{z}$. The Hamiltonian is:

$$\hat{H} = -\hat{\boldsymbol{\mu}} \cdot \mathbf{B} = -\gamma B \hat{S}_z = \omega_0 \hat{S}_z$$

where $\omega_0 \equiv -\gamma B$ is the Larmor frequency.

Spectral decomposition: $\hat{H} = \frac{\omega_0\hbar}{2}|\uparrow\rangle\langle\uparrow| - \frac{\omega_0\hbar}{2}|\downarrow\rangle\langle\downarrow|$

Time-evolution operator:

$$\hat{U}(t) = e^{-i\hat{H}t/\hbar} = e^{-i\omega_0 t/2}|\uparrow\rangle\langle\uparrow| + e^{+i\omega_0 t/2}|\downarrow\rangle\langle\downarrow|$$

Initial state: $|\chi(0)\rangle = |+x\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle + |\downarrow\rangle)$

Time evolution:

$$|\chi(t)\rangle = \hat{U}(t)|+x\rangle = \frac{1}{\sqrt{2}}\left(e^{-i\omega_0 t/2}|\uparrow\rangle + e^{+i\omega_0 t/2}|\downarrow\rangle\right)$$

Expectation values at time $t$:

$$\langle\hat{S}_x\rangle(t) = \langle\chi(t)|\hat{S}_x|\chi(t)\rangle$$

$$= \frac{\hbar}{2} \cdot \frac{1}{2}\left(e^{+i\omega_0 t/2}\langle\uparrow| + e^{-i\omega_0 t/2}\langle\downarrow|\right)(|\uparrow\rangle\langle\downarrow| + |\downarrow\rangle\langle\uparrow|)\left(e^{-i\omega_0 t/2}|\uparrow\rangle + e^{+i\omega_0 t/2}|\downarrow\rangle\right)$$

$$= \frac{\hbar}{4}\left(e^{i\omega_0 t} + e^{-i\omega_0 t}\right) = \frac{\hbar}{2}\cos(\omega_0 t)$$

Similarly:

$$\langle\hat{S}_y\rangle(t) = -\frac{\hbar}{2}\sin(\omega_0 t), \qquad \langle\hat{S}_z\rangle(t) = 0$$

The spin vector precesses about the $z$-axis at frequency $\omega_0$ — this is Larmor precession. The expectation value of the spin traces a circle in the $xy$-plane, exactly as a classical magnetic moment would precess about a magnetic field. But the quantum state is always a superposition of $|\uparrow\rangle$ and $|\downarrow\rangle$, with a time-dependent relative phase.

Part 7: The Trace as a Physical Tool

Compute $\text{Tr}(\hat{S}_x^2)$ and $\text{Tr}(\hat{S}_z^2)$:

$$\text{Tr}(\hat{S}_x^2) = \text{Tr}\left(\frac{\hbar^2}{4}\hat{I}\right) = \frac{\hbar^2}{4} \cdot 2 = \frac{\hbar^2}{2}$$

$$\text{Tr}(\hat{S}_z^2) = \langle\uparrow|\hat{S}_z^2|\uparrow\rangle + \langle\downarrow|\hat{S}_z^2|\downarrow\rangle = \frac{\hbar^2}{4} + \frac{\hbar^2}{4} = \frac{\hbar^2}{2}$$

Both traces are equal, as they must be (the trace is basis-independent, and $\hat{S}_x^2 = \hat{S}_y^2 = \hat{S}_z^2 = \frac{\hbar^2}{4}\hat{I}$ for spin-1/2).

The total spin operator is $\hat{S}^2 = \hat{S}_x^2 + \hat{S}_y^2 + \hat{S}_z^2 = \frac{3\hbar^2}{4}\hat{I}$, giving $\text{Tr}(\hat{S}^2) = \frac{3\hbar^2}{2}$.

Summary and Key Lessons

This case study has demonstrated, in one concrete system:

Kets and bras as column and row vectors in a finite-dimensional space.
Orthonormality and completeness of the $S_z$ eigenbasis.
Operators as outer products: $\hat{S}_z = \frac{\hbar}{2}(|\uparrow\rangle\langle\uparrow| - |\downarrow\rangle\langle\downarrow|)$.
Matrix representations via $A_{mn} = \langle m|\hat{A}|n\rangle$.
Eigenstate finding by solving $\hat{A}|\psi\rangle = a|\psi\rangle$.
Change of basis using unitary transformations.
Time evolution using $\hat{U}(t) = e^{-i\hat{H}t/\hbar}$ and spectral decomposition.
The trace as a basis-independent computational tool.

Every one of these techniques will be used in every subsequent chapter. The spin-1/2 system is where you practice them until they become automatic.

🔗 Connection — The spin-1/2 system will return in Chapter 11 (tensor products of two spins), Chapter 13 (full spin formalism and Bloch sphere), Chapter 24 (entangled spin pairs and Bell's theorem), and Chapter 25 (the qubit — the fundamental unit of quantum information). Master it now and the rest follows.