Lecture 2.5: Generators, Pauli Matrices, and SU(2)

What Is a Generator?¶

Infinitesimal Rotations¶

We’ve been working with rotation matrices — $R(\theta)$ rotates a vector by angle $\theta$. But rotations can be understood more deeply by asking: what happens when the angle is very, very small?

Start with the 2D rotation matrix:

For a tiny angle $\delta\theta$, Taylor expand: $\cos\delta\theta \approx 1$ and $\sin\delta\theta \approx \delta\theta$:

where:

$G$ is called the generator of rotations. It tells you which direction you move when you begin to rotate. The identity $I$ says “stay where you are,” and $\delta\theta\, G$ is the tiny nudge that starts the rotation.

From Generator to Finite Rotation¶

How do you build a large rotation from the generator? The same way you walk a mile — one step at a time.

To rotate by a finite angle $\theta$, break it into $N$ tiny steps, each of size $\theta/N$:

For large $N$, each step is infinitesimal, so $R(\theta/N) \approx I + \frac{\theta}{N}G$:

This limit is the definition of the matrix exponential:

A finite rotation is the exponential of the generator. This is one of the most important ideas in physics: the generator encodes all rotations. You don’t need to memorize the full rotation matrix — just the generator, and exponentiation does the rest.

Evaluating the Exponential¶

Let’s verify this actually works. We need to compute $e^{\theta G}$ using the Taylor series:

First, what is $G^2$?

So $G^2 = -I$. This is the key property of the generator. The higher powers follow immediately:

The pattern repeats with period 4: $I, G, -I, -G, I, G, \ldots$

Substituting into the Taylor series:

Group the terms with $I$ and the terms with $G$:

These are the Taylor series for cosine and sine:

Let’s check:

It works. The generator $G$ contains everything we need.

The Generator and the Imaginary Unit¶

$G^2 = -I$ and $i^2 = -1$¶

We just found that the rotation generator satisfies $G^2 = -I$. Look at this property carefully — it’s the matrix version of $i^2 = -1$.

This is not a coincidence. Recall Euler’s formula for complex number rotations:

Now compare our matrix rotation formula:

The structures are identical. Everywhere that $i$ appears in Euler’s formula, $G$ appears in the matrix formula. Everywhere that 1 appears, $I$ appears.

The generator $G$ plays the role of $i$. Both satisfy the same algebraic property ($G^2 = -I$ and $i^2 = -1$), and both generate rotations through exponentiation.

Two Representations of the Same Rotation¶

We saw last lecture that complex multiplication by $e^{i\theta}$ and matrix multiplication by $R(\theta)$ do the same thing to a vector. Now we see why: they are two representations of the same object, built from generators that obey the same algebra.

The complex number $i$ and the matrix $G$ are different representations of the same mathematical structure — the generator of 2D rotations.

Generators of SO(3)¶

From 2D to 3D: Things Get Complicated¶

In 2D there is one rotation axis and one generator. In 3D there are three rotation axes — $x$, $y$, and $z$ — so we expect three generators.

The full $3\times 3$ rotation matrices are:

These are already getting unwieldy. Multiplying two of them together — say to check if they commute — would be a mess of trig identities. And we already established last lecture that 3D rotations don’t commute.

The generator approach cuts through this complexity. Instead of working with the full rotation matrices, we extract the generators — much simpler objects that encode all the essential information.

Extracting the Generators¶

Taylor expand each rotation matrix for small $\delta\theta$:

The three generators are:

Notice the pattern: each generator has the familiar SO(2) block $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ embedded in the $2\times 2$ subspace corresponding to its plane of rotation, with zeros along the rotation axis. For example, $J_z$ rotates in the $x$-$y$ plane, so the block sits in the upper-left corner. $J_x$ rotates in the $y$-$z$ plane, so the block sits in the lower-right corner.

Finite rotations are exponentials:

SO(3) Commutation Relations¶

The non-commutativity of 3D rotations — which was messy to show with the full rotation matrices — becomes clean and elegant at the generator level. Let’s compute the commutator $[J_x, J_y] = J_x J_y - J_y J_x$:

The commutator of two generators gives the third! The full set of commutation relations:

Compactly: $[J_i, J_j] = \epsilon_{ijk}J_k$, where $\epsilon_{ijk} = +1$ for cyclic permutations (xyz, yzx, zxy), -1 for anti-cyclic, and 0 if any indices repeat.

These commutation relations are profound. They capture the entire structure of 3D rotations in a simple algebraic statement. Any set of three objects satisfying these relations generates a rotation group — even if those objects are $2\times 2$ matrices instead of $3\times 3$ matrices. This is the key insight that will connect SO(3) to SU(2).

The physical consequence of non-commuting generators is the uncertainty principle: if two physical quantities are described by non-commuting operators, they cannot both be known simultaneously. We’ll make this precise shortly.

From SO(3) to SU(2)¶

A Different Kind of Space¶

SO(3) describes rotations of real 3D vectors — positions, velocities, electric field directions. These are vectors in $\mathbb{R}^3$.

But throughout this course, we’ve been working with a different kind of object: a qubit.

This is a two-configuration system with complex amplitudes. The state is a vector in $\mathbb{C}^2$, and we’ve been visualizing it on the Bloch sphere. When we apply gates — Hadamard, phase gates, Pauli matrices — we rotate the state on the Bloch sphere.

We need a group that describes these rotations: transformations of $\mathbb{C}^2$ that preserve the normalization $|c_0|^2 + |c_1|^2 = 1$.

Defining SU(2)¶

This group is SU(2): the group of $2\times 2$ unitary matrices with determinant 1.

• S = Special: $\det(U) = 1$

• U = Unitary: $U^\dagger U = I$

• 2 = $2\times 2$ complex matrices

Unitarity is the complex generalization of orthogonality:

Both preserve lengths. Orthogonal matrices preserve $\vec{v}^T\vec{v}$. Unitary matrices preserve $\vec{v}^\dagger\vec{v} = |c_0|^2 + |c_1|^2$ — exactly the normalization condition we need for quantum states.

What We Expect¶

SU(2) should behave like SO(3) — after all, both describe rotations on a sphere. The Bloch sphere has three axes, so SU(2) should have three generators, and those generators should satisfy the same commutation relations as the SO(3) generators $J_x, J_y, J_z$.

But SU(2) acts on complex space, so the generators will be $2\times 2$ complex matrices instead of $3\times 3$ real matrices.

The Pauli Matrices: Generators of SU(2)¶

The three generators of SU(2) are the Pauli matrices:

The spin angular momentum operators are related by $S_i = \frac{\hbar}{2}\sigma_i$.

Key Properties¶

Hermitian: $\sigma_i^\dagger = \sigma_i$. Hermitian matrices represent observables — quantities we can measure. Their eigenvalues are guaranteed to be real.

Traceless: $\text{Tr}(\sigma_x) = \text{Tr}(\sigma_y) = \text{Tr}(\sigma_z) = 0$.

Square to identity: $\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = I$. Let’s verify one:

Since $\sigma_i^2 = I$, the eigenvalues satisfy $\lambda^2 = 1$, giving $\lambda = \pm 1$ for every Pauli matrix.

Eigenstates: The Six Cardinal States¶

Each Pauli matrix has eigenvalues +1 and -1. The eigenstates are:

$\sigma_z$:

$\sigma_x$:

where $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$.

$\sigma_y$:

where $|{+i}\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ and $|{-i}\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)$.

These are exactly the six cardinal points on the Bloch sphere:

Pauli Commutation Relations¶

Now the crucial check: do the Pauli matrices satisfy the same commutation relations as the SO(3) generators?

Compute $\sigma_x\sigma_y$:

Compute $\sigma_y\sigma_x$:

iClicker: Pauli Products¶

What is $\sigma_x\sigma_y$?

(A) $\sigma_z$

(B) $-\sigma_z$

(C) $i\sigma_z$ ✓

(D) $-i\sigma_z$

Commutators¶

The full set:

Compactly: $[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k$.

Compare to SO(3): $[J_i, J_j] = \epsilon_{ijk}J_k$. The same structure — just with a factor of $2i$. This factor is a convention: the Pauli matrices are Hermitian (so they can be observables), while the SO(3) generators are antisymmetric. The underlying pattern — which commutator gives which generator — is identical.

SU(2) and SO(3) share the same algebraic DNA.

Anticommutators¶

The anticommutator $\{A, B\} = AB + BA$ captures the symmetric part:

Different Pauli matrices anticommute to zero. Same Pauli matrices give $\{\sigma_i, \sigma_i\} = 2I$. In general:

The Master Formula¶

Adding the commutator and anticommutator:

If $i = j$: you get $I$ (squaring gives the identity). If $i \neq j$: you get $\pm i$ times the third Pauli matrix, with the sign given by cyclic ordering.

Physical Consequence: The Uncertainty Principle¶

Non-commuting generators mean that the corresponding physical quantities cannot be simultaneously known.

Since $[\sigma_x, \sigma_z] = -2i\sigma_y \neq 0$:

• No state can be an eigenstate of both $\sigma_x$ and $\sigma_z$

• If $S_z$ is definite ($|0\rangle$ or $|1\rangle$), then $S_x$ is completely uncertain

• If $S_x$ is definite ($|+\rangle$ or $|-\rangle$), then $S_z$ is completely uncertain

You cannot simultaneously know the spin along two perpendicular axes. This is fundamentally different from classical physics, where all components of angular momentum can be known at once. The non-commutativity of the Pauli matrices — the same non-commutativity we saw for 3D rotations — is the mathematical origin of quantum uncertainty.

The Double Cover: What Does $R(2\pi) = -I$ Mean?¶

SU(2) $\neq$ SO(3)¶

SU(2) and SO(3) have the same commutation relations, but they are not the same group. They differ in a fundamental way.

In SO(3), a $2\pi$ rotation around any axis brings you back to the identity:

This makes intuitive sense — spin a ball $360°$ and it’s back where it started.

In SU(2), a $2\pi$ rotation gives minus the identity:

You need to rotate by $4\pi$ — two full turns — to get back to $+I$. SU(2) is the double cover of SO(3): it wraps around twice for every single winding of SO(3).

What Does the -1 Mean Physically?¶

In quantum mechanics, the state of a system is described by a complex amplitude:

If $|\psi\rangle \to -|\psi\rangle$, every amplitude picks up a minus sign: $c_0 \to -c_0$, $c_1 \to -c_1$. That -1 is a phase — specifically, it’s the phase $e^{i\pi} = -1$.

For a single isolated state, this phase is unobservable. The probabilities $|c_0|^2$ and $|c_1|^2$ don’t change when you multiply by -1.

But in a superposition, the phase matters. Consider a state that is a superposition of a rotated part and an unrotated part. The rotated part picks up -1, the unrotated part doesn’t. Where the two parts used to add constructively (interference maximum), they now add destructively (interference minimum). Where they used to cancel (minimum), they now reinforce (maximum).

The interference pattern shifts. The -1 is physically real.

This has been confirmed experimentally. In neutron interferometry experiments, a neutron beam is split, one path is rotated by $2\pi$ using a magnetic field, and the beams are recombined. The interference pattern shifts — exactly as predicted by the -1 phase from SU(2). The neutron must be rotated by $4\pi$ to restore the original interference pattern.

Fermions vs. Bosons¶

Nature has two kinds of particles:

Fermions (electrons, quarks, neutrinos — the matter particles) have half-integer spin and transform under SU(2): $R(2\pi) = -1$.

Bosons (photons, gluons, the Higgs — the force carriers) have integer spin and transform under SO(3): $R(2\pi) = +1$.

This distinction is connected to the spin-statistics theorem: fermions obey the Pauli exclusion principle (no two can occupy the same state), while bosons don’t. The -1 under $2\pi$ rotation and the exclusion principle are two manifestations of the same deep mathematical structure. More on this later in the course.

Rotations on the Bloch Sphere¶

The General Rotation¶

A rotation by angle $\theta$ around a unit axis $\hat{n} = (n_x, n_y, n_z)$ on the Bloch sphere is:

where $\hat{n}\cdot\vec{\sigma} = n_x\sigma_x + n_y\sigma_y + n_z\sigma_z$.

The factor of $1/2$ is because of the double cover — SU(2) matrices make half-angle rotations on the Bloch sphere.

Deriving the Closed Form¶

We can evaluate this exponential using the same Taylor series technique that worked for SO(2). Define $A = \hat{n}\cdot\vec{\sigma}$.

Step 1: Compute $A^2$ using the master formula $\sigma_i\sigma_j = \delta_{ij}I + i\epsilon_{ijk}\sigma_k$:

The symmetric piece gives $\sum_i n_i^2\,I = |\hat{n}|^2\,I = I$. The antisymmetric piece $\sum_{ij}n_in_j\epsilon_{ijk}$ vanishes because $n_in_j$ is symmetric while $\epsilon_{ijk}$ is antisymmetric.

Just like $\sigma_i^2 = I$ for each individual Pauli, the combination $(\hat{n}\cdot\vec\sigma)^2 = I$ for any unit vector $\hat{n}$.

Step 2: Higher powers cycle: $A^2 = I$, $A^3 = A$, $A^4 = I$, ... — exactly the same pattern as $G$ in SO(2).

Step 3: Taylor expand with $\alpha = \theta/2$:

Step 4: Group even powers ($I$) and odd powers ($A$):

Step 5: Substitute back $\alpha = \theta/2$:

Compare to our SO(2) result: $e^{\theta G} = \cos\theta\,I + \sin\theta\,G$. The same structure — the generator multiplied by sine, the identity multiplied by cosine. The $-i$ and $\theta/2$ reflect the Hermitian convention and the double cover.

Verification: At $\theta = 2\pi$: $R_{\hat{n}}(2\pi) = \cos\pi\,I - i\sin\pi\,(\hat{n}\cdot\vec\sigma) = -I$. ✓

Explicit Rotations and Gate Connections¶

Rotation Around the $z$-Axis¶

Connection to the phase gate:

Up to a global phase (which is unobservable), the phase gate IS a z-rotation. This is why applying $P(\phi)$ to $|+\rangle$ traces the equator of the Bloch sphere — it rotates around the z-axis.

Connection to the Z gate:

The Z gate is a $\pi$ rotation around the z-axis. It maps $|+\rangle \to |-\rangle$ and $|-\rangle \to |+\rangle$ — it flips the equatorial states.

Rotation Around the $x$-Axis¶

Connection to the X gate:

The X gate is a $\pi$ rotation around the x-axis. It swaps $|0\rangle \leftrightarrow |1\rangle$ — the quantum NOT gate, flipping the north and south poles.

Rotation Around the $y$-Axis¶

Note: $R_y(\theta)$ is a real matrix. This is because $-i\sigma_y = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, which is the SO(2) generator $G$. So y-rotations on the Bloch sphere look exactly like classical 2D rotations.

Connection to the Y gate:

The Y gate is a $\pi$ rotation around the y-axis.

The Hadamard Gate¶

The Hadamard doesn’t correspond to a rotation around a coordinate axis. Instead:

This is a $\pi$ rotation around the axis $\hat{n} = (\hat{x} + \hat{z})/\sqrt{2}$, halfway between $x$ and $z$ on the Bloch sphere. Geometrically, it swaps the z-axis and the x-axis:

The Fundamental Result¶

Any $U \in$ SU(2) can be written as $R_{\hat{n}}(\theta) = e^{-i\theta(\hat{n}\cdot\vec\sigma)/2}$ for some axis $\hat{n}$ and angle $\theta$. There are no other single-qubit gates. The three Pauli matrices generate all possible qubit operations through rotation.

Summary¶

Key Results¶

1. Generators encode infinitesimal rotations. Exponentiating gives finite rotations: $R(\theta) = e^{\theta G}$.

2. $G$ and $i$ are the same thing in different representations. $G^2 = -I$ is the matrix version of $i^2 = -1$. Both generate rotations.

3. SO(3) has three generators $J_x, J_y, J_z$ satisfying $[J_i, J_j] = \epsilon_{ijk}J_k$. Non-commutativity of generators → uncertainty principle.

4. SU(2) is the quantum rotation group. Its generators — the Pauli matrices — satisfy the same commutation relations as SO(3). They are Hermitian (observables), traceless, and square to the identity (eigenvalues $\pm 1$). Their eigenstates are the six Bloch sphere cardinal states.

5. The double cover: $R_{SU(2)}(2\pi) = -I$. The minus sign is a physical phase that shifts interference patterns. Fermions transform under SU(2); bosons under SO(3).

6. Every qubit gate is a Bloch sphere rotation:

Phase gate $\sim R_z$. X gate $\sim R_x(\pi)$. Hadamard $\sim R_{(\hat{x}+\hat{z})/\sqrt{2}}(\pi)$.

Next Lecture¶

The Stern-Gerlach experiment: nature hands us a real physical qubit. Electron spin, measurement in different bases, and the birth of the qubit.

Homework 2.5¶

Problem 1: A Different Kind of Generator — Lorentz Boosts¶

Not all generators produce rotations. In special relativity, a Lorentz boost along the x-direction by rapidity $\beta$ is described by:

This matrix mixes space and time coordinates, just as a rotation matrix mixes $x$ and $y$.

(a) Verify that the set of boost matrices $\{B(\beta)\}$ forms a group: show $B(\beta_1)B(\beta_2) = B(\beta_1 + \beta_2)$, identify the identity element and the inverse of $B(\beta)$.

Hint: You will need the hyperbolic addition formulas: $\cosh(\alpha+\beta) = \cosh\alpha\cosh\beta + \sinh\alpha\sinh\beta$ and $\sinh(\alpha+\beta) = \sinh\alpha\cosh\beta + \cosh\alpha\sinh\beta$.

(b) Find the generator $K$ by expanding $B(\delta\beta)$ for small $\delta\beta$ (use $\cosh\delta\beta \approx 1$, $\sinh\delta\beta \approx \delta\beta$):

Write out $K$.

(c) Compute $K^2$. Compare to the rotation generator where $G^2 = -I$. What is different?

(d) Using $K^2 = +I$, expand the exponential $e^{\beta K}$ as a Taylor series. Group even and odd powers (as we did in lecture for $e^{\theta G}$). What functions appear instead of cosine and sine?

(e) Verify that your result matches $B(\beta)$.

(f) Compute $\det B(\beta)$. Compare to $\det R(\theta) = 1$. Is a Lorentz boost a rotation?

(g) Reflect on the difference: $G^2 = -I$ gives oscillatory behavior ($\cos, \sin$) — rotations go around in circles. $K^2 = +I$ gives hyperbolic behavior ($\cosh, \sinh$) — boosts grow without bound. What physical difference does this correspond to? (Can you rotate forever? Can you boost forever?)

Problem 2: The Pauli Vector Identity¶

(a) Using the master formula $\sigma_i\sigma_j = \delta_{ij}I + i\epsilon_{ijk}\sigma_k$, prove the identity:

where $\vec{a}$ and $\vec{b}$ are arbitrary real 3-vectors.

Hint: Write $\vec{a}\cdot\vec\sigma = \sum_i a_i\sigma_i$ and $\vec{b}\cdot\vec\sigma = \sum_j b_j\sigma_j$, multiply, and use the master formula on $\sigma_i\sigma_j$. You will need to recognize $\sum_i a_i b_i = \vec{a}\cdot\vec{b}$ and $\sum_{ijk}\epsilon_{ijk}a_i b_j = (\vec{a}\times\vec{b})_k$.

(b) As a special case, set $\vec{a} = \vec{b} = \hat{n}$ (a unit vector). Show that $(\hat{n}\cdot\vec\sigma)^2 = I$.

(c) Set $\vec{a} = \hat{x}$ and $\vec{b} = \hat{y}$. Evaluate both sides of the identity and verify they match.

(d) Set $\vec{a} = \vec{b} = \hat{x} + \hat{z}$ (not a unit vector). What is $[(\hat{x}+\hat{z})\cdot\vec\sigma]^2$? How does this relate to the Hadamard gate?

Problem 3: Mystery Matrix¶

Consider the unitary matrix:

(a) Verify that $U$ is unitary: show $U^\dagger U = I$.

(b) Verify that $\det U = 1$, confirming $U \in$ SU(2).

(c) Every element of SU(2) can be written as:

By comparing $U$ to this formula, determine the rotation angle $\theta$ and axis $\hat{n}$.

Hint: The coefficient of $I$ gives $\cos(\theta/2)$. The coefficients of $\sigma_x$, $\sigma_y$, $\sigma_z$ in the remainder give $\sin(\theta/2)\, n_x$, $\sin(\theta/2)\, n_y$, $\sin(\theta/2)\, n_z$.

(d) Describe geometrically what this rotation does on the Bloch sphere. Where does it send $|0\rangle$? Where does it send $|+\rangle$?

(e) Compute $U^2$. What rotation is this? What about $U^4$?

Problem 4: Visualizing Rotations on the Bloch Sphere (Qiskit)¶

In this problem you’ll use Qiskit to see rotations in action.

Setup: You will need the following imports:

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector
from qiskit.visualization import plot_bloch_multivector, plot_bloch_vector
import numpy as np
import matplotlib.pyplot as plt

(a) Start with $|0\rangle$ and apply $R_y(\theta)$ for $\theta = 0, \pi/6, \pi/3, \pi/2, 2\pi/3, 5\pi/6, \pi$. For each value of $\theta$: - Create a circuit with qc.ry(theta, 0) - Extract the statevector - Visualize on the Bloch sphere (or record the Bloch coordinates)