Lecture 2.4: Complex Numbers, Rotations, and Group Theory

Why Is Quantum Mechanics Complex?¶

Review: The Classical Wave Equation¶

The wave equation for light is:

\frac{\partial^2 E}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}

(1)

This is a real equation with real solutions:

E(x,t) = A\cos(kx - \omega t), \qquad E(x,t) = B\sin(kx - \omega t)

(2)

We can also write solutions using complex exponentials:

E(x,t) = e^{\pm i(kx \pm \omega t)}

(3)

All four sign combinations — $e^{i(kx - \omega t)}$ , $e^{-i(kx - \omega t)}$ , $e^{i(kx + \omega t)}$ , $e^{-i(kx + \omega t)}$ — are valid solutions. They correspond to right- and left-traveling waves. When we use these complex exponentials for light, it’s a trick. The physical field is real; we take the real part at the end.

The dispersion relation is linear:

\omega = ck

(4)

The Schrödinger Equation¶

Now consider the equation governing matter waves — the Schrödinger equation for a free particle:

i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}

(5)

Notice the $i$ on the left side. Let’s see what it does.

Try a plane wave solution $\psi(x,t) = e^{i(kx - \omega t)}$ :

Left side:

i\hbar\frac{\partial}{\partial t}e^{i(kx - \omega t)} = i\hbar(-i\omega)e^{i(kx-\omega t)} = \hbar\omega\, e^{i(kx - \omega t)}

(6)

Right side:

-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}e^{i(kx-\omega t)} = -\frac{\hbar^2}{2m}(ik)^2 e^{i(kx-\omega t)} = \frac{\hbar^2 k^2}{2m}e^{i(kx-\omega t)}

(7)

Setting them equal gives the quantum dispersion relation:

\hbar\omega = \frac{\hbar^2 k^2}{2m} \qquad \Longrightarrow \qquad \omega = \frac{\hbar k^2}{2m}

(8)

This is quadratic in $k$ , unlike the classical $\omega = ck$ .

iClicker: Which Sign Combinations Solve the Schrödinger Equation?¶

For the classical wave equation, all four sign combinations $e^{\pm i(kx \pm \omega t)}$ are valid solutions. Do all four also solve the Schrödinger equation?

(A) Yes

(B) No ✓

Let’s check. Write a general plane wave as $\psi = e^{i(\alpha kx - \beta\omega t)}$ where $\alpha, \beta = \pm 1$ . Substituting into the Schrödinger equation:

i\hbar \frac{\partial\psi}{\partial t} = i\hbar(-i\beta\omega)\psi = \beta\hbar\omega\,\psi

(9)

-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} = -\frac{\hbar^2}{2m}(i\alpha k)^2\psi = \frac{\hbar^2 k^2}{2m}\psi

(10)

For these to be equal we need $\beta\hbar\omega = \frac{\hbar^2 k^2}{2m}$ . Since $\omega$ and $k$ are positive, we need $\beta = +1$ . The sign of $\alpha$ doesn’t matter (it only appears as $\alpha^2 = 1$ ).

So the solutions are:

\psi(x,t) = e^{-i\omega t} \cdot e^{\pm ikx}

(11)

The time dependence must be $e^{-i\omega t}$ . The Schrödinger equation forces a specific sign on the time evolution. This is fundamentally different from the classical wave equation, where both $e^{+i\omega t}$ and $e^{-i\omega t}$ are valid.

Why Complex Numbers Are Required¶

Now consider: can we use a real function? Try $\psi(x,t) = \cos(kx - \omega t)$ :

\frac{\partial\psi}{\partial t} = \omega\sin(kx - \omega t)

(12)

\frac{\partial^2\psi}{\partial x^2} = -k^2\cos(kx - \omega t)

(13)

The left side of the Schrödinger equation gives something proportional to $\sin(kx - \omega t)$ , while the right side gives something proportional to $\cos(kx - \omega t)$ . These are different functions — they cannot be equal for all $x$ and $t$ .

A real cosine does not solve the Schrödinger equation. Neither does a real sine. The complex exponential $e^{-i(\omega t \pm kx)}$ is the only option.

For classical waves, complex numbers are a convenient bookkeeping device. For quantum mechanics, complex numbers are mandatory. The wavefunction is irreducibly complex.

Nature Is Fundamentally Complex¶

This has deep consequences for how we think about quantum states. When we write:

|\psi\rangle = c_0|0\rangle + c_1|1\rangle

(14)

the coefficients $c_0$ and $c_1$ are complex numbers. Each carries two pieces of information: an amplitude and a phase. Both are physical — the amplitude determines probabilities, and the relative phase determines interference. We’ve seen this on the Bloch sphere: the six cardinal states all have the same amplitudes ( $1/\sqrt{2}$ each) for some basis, but differ by their relative phases, and they represent completely different physical states.

Why Group Theory?¶

Before we dive into the mathematics, let’s ask: why should physicists care about group theory?

A group describes a collection of actions — things you can do to something. Rotations are the classic example: you can rotate a sphere around the x-axis, then the z-axis, or do them in the opposite order. The collection of all possible rotations, together with the rule for combining them, forms a group.

We’ve already been using groups without naming them. Every time we apply a Hadamard gate or a phase gate to a qubit, we’re rotating a state on the Bloch sphere. The set of all single-qubit gates forms a group — specifically, the group SU(2).

Nature seems to know group theory deeply. The mathematical structure of rotations is built into the fabric of physics. Here’s a preview of where we’re headed:

SO(3) is the group of rotations in 3D space — the kind you do when you rotate a ball in your hands. A full $2\pi$ rotation brings everything back to where it started: $R(2\pi) = +1$ .

SU(2) is the group of qubit transformations — rotations on the Bloch sphere. It is almost the same as SO(3), but with one crucial difference: a full $2\pi$ rotation gives $R(2\pi) = -1$ . You need to rotate by $4\pi$ to get back to the identity.

Nature cares about this distinction. Particles called fermions (electrons, quarks — the matter particles) transform under SU(2) and pick up a minus sign under $2\pi$ rotation. Particles called bosons (photons, gluons — the force carriers) transform under SO(3) and return to +1. This is connected to the spin-statistics theorem, one of the deepest results in physics. We’ll say more about this later.

For now, let’s build the mathematics from the ground up.

Complex Numbers as Rotations¶

A complex number can be written in Cartesian or polar form:

z = x + iy = re^{i\theta}

(15)

What happens when we multiply two complex numbers in polar form?

e^{i\theta_1} \cdot e^{i\theta_2} = e^{i(\theta_1 + \theta_2)}

(16)

Geometrically, multiplying by $e^{i\theta}$ rotates a point in the complex plane by angle $\theta$ . The magnitudes multiply and the angles add. For numbers on the unit circle ( $r = 1$ ), multiplication is pure rotation.

What Is a Group?¶

A group is a set together with an operation that satisfies four rules.

Take our example: the set is $\{e^{i\theta} : \theta \in (-\pi, \pi]\}$ (all complex numbers on the unit circle), and the operation is multiplication.

Rule 1: Closure¶

Combining two elements gives another element in the set.

e^{i\theta_1} \cdot e^{i\theta_2} = e^{i(\theta_1 + \theta_2)}

(17)

The product of two unit-magnitude complex numbers is another unit-magnitude complex number. The result stays in the set. ✓

Rule 2: Associativity¶

(a \cdot b) \cdot c = a \cdot (b \cdot c)

(18)

Multiplication of complex numbers is associative. ✓

Rule 3: Identity¶

There exists an element $\mathbb{1}$ such that $\mathbb{1} \cdot a = a \cdot \mathbb{1} = a$ for all $a$ .

The identity is $e^{i \cdot 0} = 1$ . Multiplying any element by 1 leaves it unchanged. ✓

Rule 4: Inverses¶

Every element $a$ has an inverse $a^{-1}$ such that $a \cdot a^{-1} = \mathbb{1}$ .

(e^{i\theta})^{-1} = e^{-i\theta} = \overline{e^{i\theta}}

(19)

The inverse is the complex conjugate. Multiplying gives $e^{i\theta} \cdot e^{-i\theta} = e^{0} = 1$ . ✓

All four rules are satisfied. The unit complex numbers under multiplication form a group. This group is called U(1) — the unitary group in 1 dimension.

Matrix Representation of SO(2)¶

There’s another way to describe 2D rotations: as matrices acting on vectors.

A rotation by angle $\theta$ in the $x$ - $y$ plane is:

R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

(20)

It acts on a position vector:

\begin{pmatrix} x' \\ y' \end{pmatrix} = R(\theta)\begin{pmatrix} x \\ y \end{pmatrix}

(21)

turning it by angle $\theta$ counterclockwise.

Connection to Complex Multiplication¶

The matrix rotation and complex multiplication are doing the same thing. A complex number $z = x + iy$ corresponds to the vector $(x, y)^T$ . Multiplying by $e^{i\theta} = \cos\theta + i\sin\theta$ :

e^{i\theta}z = (\cos\theta + i\sin\theta)(x + iy) = (\cos\theta\cdot x - \sin\theta\cdot y) + i(\sin\theta\cdot x + \cos\theta\cdot y)

(22)

Reading off the real and imaginary parts:

x' = \cos\theta\cdot x - \sin\theta\cdot y, \qquad y' = \sin\theta\cdot x + \cos\theta\cdot y

(23)

This is exactly:

\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}

(24)

Complex multiplication by $e^{i\theta}$ and matrix multiplication by $R(\theta)$ are two representations of the same rotation.

Verifying the Group Axioms for $R(\theta)$ ¶

The set of all rotation matrices $\{R(\theta)\}$ also forms a group:

Closure: $R(\theta_1 + \theta_2) = R(\theta_1)R(\theta_2)$ . The product of two rotations is another rotation. ✓

Associativity: Matrix multiplication is associative. ✓

Identity: $R(0) = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . ✓

Inverses: $R(\theta)^{-1} = R(-\theta) = R(\theta)^T$ . The inverse is rotation by the opposite angle, which is the transpose. You can verify: $R(\theta)^T R(\theta) = I$ . ✓

Naming: SO(2)¶

This group is called SO(2):

S = Special: $\det R(\theta) = \cos^2\theta + \sin^2\theta = 1$
O = Orthogonal: $R^T R = I$ (the inverse is the transpose; this preserves lengths)
2 = $2 \times 2$ real matrices

iClicker: Does Order Matter in SO(2)?¶

Is $R(\theta_2)R(\theta_1) = R(\theta_1)R(\theta_2)$ ? Does the order of two rotations matter?

(A) Yes, order matters

(B) No, order doesn't matter ✓

Since $R(\theta_1)R(\theta_2) = R(\theta_1 + \theta_2) = R(\theta_2 + \theta_1) = R(\theta_2)R(\theta_1)$ , the order doesn’t matter. SO(2) is a commutative (abelian) group. Rotating by 30° then 45° is the same as 45° then 30° — it’s all rotations in a single plane, and angles just add.

From 2D to 3D: Non-Commutativity¶

Now let’s move to three dimensions. In 3D, we can rotate around three independent axes: $x$ , $y$ , and $z$ .

The Key Question¶

Consider two sequences of rotations applied to an object:

Rotate 90° around the x-axis, then 90° around the z-axis
Rotate 90° around the z-axis, then 90° around the x-axis

R_x(90°)\,R_z(90°) \qquad \text{vs.} \qquad R_z(90°)\,R_x(90°)

(25)

iClicker: Do 3D Rotations Commute?¶

Do these two sequences give the same result?

(A) Yes, they commute

(B) No, they don't commute ✓

Try it with a book! Hold a book in front of you:

Sequence 1: Rotate 90° around the x-axis (tip the top edge away from you), then rotate 90° around the z-axis (turn the book counterclockwise as seen from above).

Sequence 2: Rotate 90° around the z-axis first, then 90° around the x-axis.

The book ends up in a different orientation. 3D rotations do not commute.

R_x(\theta_1)\,R_z(\theta_2) \neq R_z(\theta_2)\,R_x(\theta_1)

(26)

This is the essential new feature of 3D. In 2D, all rotations are around the same axis (perpendicular to the plane), and angles simply add — order doesn’t matter. In 3D, there are multiple rotation axes, and the order in which you compose rotations changes the outcome.

The group of 3D rotations is called SO(3): special orthogonal $3 \times 3$ real matrices. Unlike SO(2), it is non-commutative (non-abelian).

This non-commutativity will turn out to be the mathematical origin of the uncertainty principle: if two measurements don’t commute, you can’t know both simultaneously. But that’s for next lecture.

Summary¶

Quantum mechanics is fundamentally complex. The $i$ in the Schrödinger equation is not a bookkeeping trick — it’s essential. Only complex exponentials $e^{-i(\omega t \pm kx)}$ solve the Schrödinger equation; real cosines and sines do not.
Complex numbers encode rotations. Multiplication by $e^{i\theta}$ rotates by angle $\theta$ in the complex plane. This is identical to matrix multiplication by the rotation matrix $R(\theta)$ .
Rotations form a group. A group is a set with an operation satisfying closure, associativity, identity, and inverses. The unit complex numbers (U(1)) and the $2\times 2$ rotation matrices (SO(2)) are two representations of the same group.
2D rotations commute; 3D rotations do not. SO(2) is commutative — the order of rotations doesn’t matter. SO(3) is non-commutative — order matters. This non-commutativity will be central to quantum mechanics.
Looking ahead. The qubit lives on the Bloch sphere, which is acted on by the group SU(2). SU(2) is like SO(3) — three-dimensional, non-commutative — but with the twist that $R(2\pi) = -1$ . The generators of SU(2) are the Pauli matrices, which we will meet next lecture.

Homework 2.4¶

Problem 1: Is It a Group?¶

For each of the following, determine whether the set and operation form a group. If yes, verify all four axioms. If no, identify which axiom fails and explain why.

(a) The set of all integers $\{\ldots, -2, -1, 0, 1, 2, \ldots\}$ under addition.

(b) The set $\{1, -1, i, -i\}$ under multiplication. If it is a group, construct the complete $4 \times 4$ multiplication table.

(c) The set of all $2\times 2$ real matrices under multiplication.

Hint: Consider the matrix $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ . Does it have an inverse?

Problem 2: SO(2) — Matrix Properties¶

The 2D rotation matrix is:

R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

(27)

(a) Compute $R(\theta_1) R(\theta_2)$ by explicit matrix multiplication. Show that the result equals $R(\theta_1 + \theta_2)$ .

Hint: You will need the angle addition formulas: $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$ .

(b) Show that $\det R(\theta) = 1$ for all $\theta$ . This is the “S” (Special) in SO(2).

(c) Show that $R(\theta)^T R(\theta) = I$ for all $\theta$ . This is the “O” (Orthogonal) in SO(2).

(d) Using part (c), show that rotations preserve the length of any vector. That is, if $\vec{v}' = R(\theta)\vec{v}$ , show that $|\vec{v}'| = |\vec{v}|$ .

Hint: $|\vec{v}'|^2 = \vec{v}'^T \vec{v}' = (R\vec{v})^T(R\vec{v})$ .

Problem 3: Non-Commutativity in 3D¶

The 3D rotation matrices around the $x$ , $y$ , and $z$ axes by angle $\theta$ are:

R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix}

(28)

R_y(\theta) = \begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{pmatrix}

(29)

R_z(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

(30)

(a) Write out $R_x(\pi/2)$ and $R_z(\pi/2)$ explicitly (plug in $\theta = \pi/2$ ).

(b) Compute $R_x(\pi/2)\, R_z(\pi/2)$ .

(c) Compute $R_z(\pi/2)\, R_x(\pi/2)$ .

(d) Are your answers to (b) and (c) the same? What does this tell you about SO(3)?

(e) Compute the commutator $R_x(\pi/2)\, R_z(\pi/2) - R_z(\pi/2)\, R_x(\pi/2)$ . Is it the zero matrix?

Problem 4: Discovering the Generator¶

(a) Taylor expand $\cos(\delta\theta)$ and $\sin(\delta\theta)$ to first order in $\delta\theta$ (i.e., for small $\delta\theta$ ). Write the resulting approximate form of $R(\delta\theta)$ .

(b) Show that for small $\delta\theta$ :

R(\delta\theta) \approx I + \delta\theta\, G

(31)

where:

G = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

(32)

$G$ is called the generator of SO(2).

(c) Compute $G^2$ . What familiar property does it share with $i$ ?

We will learn more about this next week.