Chapter 2

Superposition

Coherence, relative phase, and interference

Explain superposition as a coherent linear combination and show why phase affects later operations.

Show how phase makes a quantum state more than a list of outcome probabilities.

Available26 minIntroductory

Superposition Relative phase |+> and |->Hadamard gate Interference

Learning objectives

Separate coherent superposition from ordinary classical uncertainty.
Explain why |+> and |-> have the same Z-basis probabilities but different future behavior.
Track a simple Hadamard evolution to see constructive and destructive interference.

In this lesson

Learning objectives

Define superposition as a coherent linear combination of basis states.
Separate coherent superposition from classical uncertainty.
Use |+⟩, |−⟩, and the Hadamard gate to track phase-sensitive evolution.
Explain interference as addition and cancellation of amplitudes.

A scene about paths

Alice returns to the observatory and finds two corridors marked 0 and 1. The guide refuses the easy story that she walks down both corridors as a tiny classical person. Instead, the corridor labels are basis labels. The quantum state can assign amplitudes to both labels, and later operations can make those amplitudes reinforce or cancel.

The analogy ends there. A superposition is not a split traveler. It is a linear combination in a vector space.

Superposition as a linear combination

In Chapter 1, we wrote a pure qubit as

|\psi\rangle = \alpha|0\rangle + \beta|1\rangle.

When both amplitudes are nonzero in the chosen basis, we often say the state is in a superposition of those basis states. The word "superposition" does not add a new kind of object; it names the linear structure already present in the state vector model.

Two important equal superpositions are

|+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, \qquad |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}}.

Both give (P(0)=P(1)=1/2) when measured in the computational basis. Yet they are not the same state. The minus sign is a relative phase, and later operations can convert that phase difference into different measurement outcomes.

Computational-basis probabilities for |+⟩ and |−⟩

0NaN%

1NaN%

The identical bars do not mean the states are physically identical; this measurement basis does not reveal their relative phase.

Why superposition is not ordinary uncertainty

A classical mixture could describe a preparation that produces ( |0\rangle ) half the time and ( |1\rangle ) half the time, with us not knowing which was prepared. That mixture also gives 50/50 computational-basis results. But it lacks a coherent relative phase between basis components.

A coherent superposition such as ( |+\rangle ) has phase relations that gates can use. If we only look at computational-basis probabilities, the distinction is hidden. If we apply another operation first, the distinction can become visible.

Equal probabilities can hide different structure

Classical mixture

Preparation is |0⟩ or |1⟩ with classical randomness.
No stable relative phase connects the alternatives.
Later interference is absent in the single-qubit model.

Coherent superposition

One pure state has amplitudes on both basis vectors.
Relative phase is part of the state.
Later gates can reveal the phase through interference.

Density matrices give the formal distinction, but the operational difference already appears when we insert gates before measurement.

Checkpoint

Why can |+⟩ and |−⟩ have the same computational-basis probabilities but still be different states?

Because probabilities ignore the relative sign between amplitudes.Because one state is actually a classical mixture.Because measurement always makes both outcomes happen.

Relative phase

Relative phase is the phase difference between components of a state. In ( |+\rangle ), the coefficients of ( |0\rangle ) and ( |1\rangle ) have the same phase. In ( |-\rangle ), the ( |1\rangle ) component differs by a phase of ( \pi ), represented here by a minus sign.

Global phase is different. Multiplying every amplitude by (-1) gives (-|+\rangle), which has the same physical predictions as ( |+\rangle ). Changing only one component, as in ( |+\rangle ) versus ( |-\rangle ), changes the relative phase and can matter.

The Hadamard gate

The Hadamard gate connects computational-basis states with the ( |+\rangle, |-\rangle ) basis:

H|0\rangle = |+\rangle, \qquad H|1\rangle = |-\rangle.

It also reverses those relationships:

H|+\rangle = |0\rangle, \qquad H|-\rangle = |1\rangle.

These equations show why phase is not decorative. The two states ( |+\rangle ) and ( |-\rangle ) have the same computational-basis probabilities before the final H, but after H one becomes ( |0\rangle ) and the other becomes ( |1\rangle ).

Hadamard followed by Hadamard

1
Start in |0⟩.
2
Apply H to obtain |+⟩.
3
Apply H again; amplitudes interfere to recover |0⟩.

The second H does not randomly choose an outcome. It recombines amplitudes so that the |1⟩ amplitude cancels.

Relative phase and Hadamard explorer

Computational-basis probabilities

050%

150%

After applying H

0100%

10%

Interference

Interference is the addition of amplitudes before probabilities are computed. Consider starting in ( |0\rangle ), applying H, and then applying H again:

|0\rangle \xrightarrow{H} \frac{|0\rangle + |1\rangle}{\sqrt{2}} \xrightarrow{H} \frac{|+\rangle + |-\rangle}{\sqrt{2}} = |0\rangle.

The ( |1\rangle ) paths cancel because one contribution is positive and the other is negative. This is destructive interference. The ( |0\rangle ) contributions add constructively.

If the middle state were ( |-\rangle ), the second H would produce ( |1\rangle ). The same visible probabilities before the second H lead to different final outcomes because the amplitudes carry phase.

Check your understanding

What does H|−⟩ equal?

|0⟩|1⟩(|0⟩ + |1⟩)/√2A 50/50 classical mixture

Check your understanding

Which statement best describes interference?

Probabilities are added after measurement.Amplitudes combine before probabilities are computed.A qubit chooses the fastest path.

Summary

A superposition is a coherent linear combination of basis states.
( |+\rangle ) and ( |-\rangle ) have identical computational-basis probabilities but different relative phase.
A classical mixture can mimic one measurement distribution without having coherent phase.
The Hadamard gate creates and resolves equal superpositions.
Interference occurs when amplitudes add or cancel before probabilities are calculated.

References and further study

Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information.
John Preskill, Lecture Notes for Physics 229: Quantum Information and Computation.
Qiskit Textbook, sections on single-qubit gates and interference.

Chapter exercises

Practice the ideas from this chapter with short interactive exercises.

Check your understanding

Coherent superposition

Distinguish superposition from classical uncertainty.

Not started

Two corridors, one state

The forest presents two paths labeled 0 and 1. The guide refuses the picture of a tiny traveler walking both routes at once.

"The labels are basis labels," she says. "The quantum state assigns amplitudes to both. That is superposition — a linear combination in a vector space, not classical indecision."

Superposition as a linear combination

Any pure qubit can be written:

|\psi\rangle = \alpha|0\rangle + \beta|1\rangle.

When both amplitudes are nonzero, the state is in a superposition of $|0\rangle$ and $|1\rangle$ . The word names the linear structure already present in the state-vector model — it does not add a new kind of object.

Equal superpositions include:

|+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, \qquad |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}}.

Which situation best describes coherent superposition (not classical uncertainty)?

Equal probabilities, different structure

Both $|+\rangle$ and $|-\rangle$ give $P(0)=P(1)=1/2$ in the computational basis. Yet they are different states because the relative sign between amplitudes differs. This distinction is hidden until an operation sensitive to phase — such as a Hadamard — is applied.

Paths align in phase

The forest paths shimmer together — not as two classical routes you forgot, but as one coherent state whose amplitudes may reinforce or cancel later.

Interactive example

Relative phase

Explore how relative phase affects interference.

Not started

Twin reflections

The Phase Mirror shows two states with identical computational-basis probabilities — fifty-fifty for 0 and 1 — yet different futures after a Hadamard gate.

"Probabilities alone do not tell the whole story," the guide whispers. "Look at the relative phase between amplitudes."

|+⟩ versus |−⟩

|+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, \qquad |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}}.

Both satisfy $P(0)=P(1)=1/2$ when measured in the computational basis. The minus sign in $|-\rangle$ is a relative phase of $\pi$ between the $|1\rangle$ component and the $|0\rangle$ component.

The Hadamard gate reveals this difference:

H|+\rangle = |0\rangle, \qquad H|-\rangle = |1\rangle.

Why do |+⟩ and |−⟩ produce different outcomes after H even though Z-basis probabilities match?

The mirror clears

You adjusted the explorer until $|-\rangle$ was prepared and confirmed $H|-\rangle = |1\rangle$ . The mirror no longer treats the two reflections as interchangeable.

Guided exercise

Hadamard transformation

Connect computational and X-basis states via H.

Not started

The crossing of amplitudes

At Hadamard Crossing, paths recombine. The guide writes the gate action on a mossy stone:

H|0\rangle = |+\rangle, \qquad H|1\rangle = |-\rangle.

"Phase decides whether amplitudes add or cancel when they meet again," she says.

Hadamard action and reverse mapping

The Hadamard gate is its own inverse (up to global phase) and maps between computational and X-basis states:

H|+\rangle = |0\rangle, \qquad H|-\rangle = |1\rangle.

|0\rangle \xrightarrow{H} |+\rangle \xrightarrow{H} |0\rangle.

What does H|−⟩ equal?

Interference in one line

Interference is amplitude addition before probabilities are computed. Constructive interference increases a path's amplitude; destructive interference can reduce it to zero — as when $H$ twice on $|0\rangle$ cancels the $|1\rangle$ component.

Which statement best describes quantum interference?

You cross with control

You predict Hadamard outcomes without treating the gate as a random coin flip. The crossing opens toward the twin-state trial ahead.

Chapter review

Chapter 2 review

Combine superposition, phase, Hadamard, and interference.

Not started

Mirrored states diverge

Twin platforms hold two preparations that look identical under computational-basis statistics — both fifty-fifty — yet evolve differently under the Hadamard gate.

"Track both states through phase and interference," the forest warden demands. "The measurement realm will not forgive sloppy superposition reasoning."

Boss synthesis — superposition forest

Review the operational chain:

Superposition: $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ with coherent relative phase.
Equal Z probabilities: $|+\rangle$ and $|-\rangle$ both give 50/50 but differ by a sign.
Hadamard: $H|+\rangle = |0\rangle$ , $H|-\rangle = |1\rangle$ .
Interference: $H(H|0\rangle) = |0\rangle$ because amplitudes recombine with phase.

Trial I — Starting from |0⟩, apply H twice. What state results (ignoring global phase)?

Trial II — Which pair shares the same computational-basis probabilities but differs in relative phase?

Trial III — What does H|1⟩ equal?

Twin symmetry broken

You traced both twins through Hadamard evolution. The forest acknowledges your phase navigation — the Born Gate of the measurement realm glows on the horizon.

Chapter completion

Exercises completed: 0/4

Next: Chapter 3