Chapter 1

Qubit

State vectors, amplitudes, and computational-basis measurement

Build the single-qubit pure-state model and learn how complex amplitudes become measurement probabilities.

Establish the notation and probability model used by every later chapter.

Available24 minIntroductory

Qubit Basis state Computational basis Probability amplitude Normalization

Learning objectives

Distinguish a classical bit value from a qubit state vector.
Use Dirac notation for computational-basis states and pure qubit states.
Compute computational-basis probabilities from complex amplitudes.

Alice In Quantumland editorial illustration

In this lesson

Learning objectives

Distinguish a classical bit value from a qubit state vector.
Use the computational basis states and Dirac notation without treating the basis as hidden classical reality.
Compute measurement probabilities from complex amplitudes and the normalization condition.
Explain why the slogan 'both 0 and 1' is incomplete.

A restrained scene

Alice enters a quiet observatory where a brass instrument has only two marks: 0 and 1. A classical device would already point to one mark before anyone reads it. This instrument is different, but the guide is careful: it is not indecisive, magical, or secretly storing two answers. It is described by a state vector, and the marks are outcomes of a particular measurement.

The point of the scene is limited. It gives us a reason to ask what is being represented. The formal answer is not a mood or a metaphor; it is a vector model.

Classical bit versus qubit state

Classical bit

Has value 0 or 1 in the model.
A read operation can reveal that value without changing the ideal bit.
Probabilities describe ignorance about the value.

Pure qubit

Is represented by a normalized state vector.
Measurement returns one classical outcome in a chosen basis.
Amplitudes and relative phase can affect later operations.

A qubit measurement produces a classical result, but the pre-measurement state is not itself a classical result.

Classical bit versus qubit

A classical bit is modeled as a variable whose value is either 0 or 1. If we are unsure which value it has, we can assign classical probabilities. That uncertainty is about our information.

A single pure qubit state is different. In the standard introductory model, it is a vector in a two-dimensional complex vector space. We often choose two reference vectors, written |0⟩ and |1⟩, and call them the computational basis. The measurement outcomes are labeled 0 and 1 because the computational basis is designed to interface with digital information, but the state before measurement is not merely an unknown classical bit.

Checkpoint

Why is 'a qubit is both 0 and 1' incomplete?

Because a qubit always has a hidden classical value.Because a qubit state is described by amplitudes in a chosen basis, not by two simultaneous classical outcomes.Because measurement always returns both outcomes.

The two-dimensional complex state space

Dirac notation writes state vectors as kets. The computational basis states are

|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad |1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.

Any pure single-qubit state can be written as a linear combination:

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

The coefficients $\alpha$ and $\beta$ are complex probability amplitudes. They are not probabilities themselves. A complex number has magnitude and phase, and both features matter in quantum mechanics. Probabilities appear only after applying the Born rule to a selected measurement basis.

Normalization

|\alpha|^2 + |\beta|^2 = 1

The total probability over a complete basis must be one.

Normalization is not an arbitrary convention. It ensures that a measurement in the computational basis returns either 0 or 1 with total probability one.

State form

|ψ⟩ = α|0⟩ + β|1⟩

The symbols |0⟩ and |1⟩ name basis vectors; α and β are complex coordinates of the state in that basis.

Qubit state explorer

State: cos(45°/2)|0⟩ + e^(i·0°) sin(45°/2)|1⟩

Angle: 45°Relative phase: 0°

Computational-basis probabilities

085%

115%

Interference preview: after applying H

085%

115%

The computational-basis probabilities above never change with phase, but the outcome after applying H does — this is the interference the phase was hiding.

Measurement in the computational basis

When the state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ is measured in the computational basis, the Born rule gives

P(0) = |\alpha|^2, \qquad P(1) = |\beta|^2.

The outcome is a classical result. In the ideal projective model, the post-measurement state for that trial becomes the basis state corresponding to the observed outcome. If outcome 0 occurs, the post-measurement state is $|0\rangle$ ; if outcome 1 occurs, it is $|1\rangle$ .

Example measurement probabilities

0NaN%

1NaN%

A state with amplitudes sqrt(0.8) and i sqrt(0.2) gives 80% and 20% computational-basis probabilities. The phase i is invisible in this one measurement but can matter later.

Worked examples

Example 1: a basis state. For $|\psi\rangle = |0\rangle$ , we have $\alpha=1$ and $\beta=0$ . Therefore

P(0)=|1|^2=1,\qquad P(1)=|0|^2=0.

This is still a quantum state, but this particular measurement is deterministic.

Example 2: an equal superposition. For

|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle,

both amplitudes have squared magnitude $1/2$ , so computational-basis measurement gives 0 and 1 with equal probability. This does not mean the qubit is a classical coin flip. The relative phase of the two amplitudes can affect future operations, as Chapter 2 will show.

Example 3: a complex amplitude. For

|\psi\rangle = \sqrt{0.8}|0\rangle + i\sqrt{0.2}|1\rangle,

normalization holds because $0.8 + 0.2 = 1$ . The computational-basis probabilities are $P(0)=0.8$ and $P(1)=0.2$ . The factor $i$ has magnitude one, so it does not change $P(1)$ . It is still part of the state and may influence later interference.

Common misconceptions

Check your understanding

For |ψ⟩ = √0.7|0⟩ + i√0.3|1⟩, what is P(1) in the computational basis?

i√0.30.3-0.3The state is invalid because one amplitude is imaginary.

Check your understanding

Which pair can have the same computational-basis probabilities but differ in relative phase?

|0⟩ and |1⟩(|0⟩ + |1⟩)/√2 and (|0⟩ - |1⟩)/√2|0⟩ and -|0⟩ only

Summary

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.
IBM Quantum Learning, “Quantum information”.
IBM Quantum Learning, “Limitations on quantum information”.

Chapter exercises

Practice the ideas from this chapter with short interactive exercises.

Check your understandingfoundation

Classical bit versus qubit

Distinguish classical bits from qubit state vectors.

Not started

A needle that will not settle

Your brass compass has only two marks: 0 and 1. On a classical instrument, the needle would already point to one mark before anyone reads it. This needle spins between them without settling.

The guide stops you from guessing a hidden value. "The compass does not know and forget," she says. "It reads a state vector — not a classical bit with a secret answer."

Classical bit versus qubit state

A classical bit is modeled as having value 0 or 1. If we are unsure which, we assign classical probabilities — that is ignorance about a definite value.

A pure qubit state is a normalized vector in a two-dimensional complex space. We write basis vectors as $|0\rangle$ and $|1\rangle$ :

|\psi\rangle = \alpha|0\rangle + \beta|1\rangle, \qquad |\alpha|^2 + |\beta|^2 = 1.

The coefficients are probability amplitudes, not probabilities themselves. Measurement in the computational basis returns one classical outcome with probabilities $|\alpha|^2$ and $|\beta|^2$ .

Which description fits a single pure qubit before measurement?

Valid basis-state notation

These are standard ways to name computational-basis states:

|0\rangle = \begin{bmatrix}1\\0\end{bmatrix}, \qquad |1\rangle = \begin{bmatrix}0\\1\end{bmatrix}.

A valid pure state uses complex amplitudes that satisfy normalization. Writing $|0\rangle + |1\rangle$ without the $\frac{1}{\sqrt{2}}$ factor is not normalized and therefore not a valid physical state as written.

Which expression describes a valid normalized single-qubit state?

Interactive examplefoundation

Amplitude and normalization

Shape probability amplitudes and satisfy normalization.

Not started

Raw metal, unfinished states

The forge glows with half-formed vectors. Unnormalized combinations spark and sputter — the guide explains that only normalized states can be prepared as physical qubit states.

"Temper the amplitudes," she says, "until the total probability over a complete basis is exactly one."

Amplitudes and normalization

For $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ , the Born rule gives computational-basis probabilities:

P(0) = |\alpha|^2, \qquad P(1) = |\beta|^2.

Normalization requires:

|\alpha|^2 + |\beta|^2 = 1.

On the Bloch sphere, tilting away from $|0\rangle$ increases $P(1)$ . For example, when $P(0) = 0.25$ , we have $P(1) = 0.75$ — a state leaning toward $|1\rangle$ but still retaining amplitude on $|0\rangle$ .

Amplitude versus probability

Students often confuse amplitudes with probabilities. An amplitude can be complex; its squared magnitude is a probability. Two states can share the same $P(0)$ and $P(1)$ in one basis yet differ in relative phase — a distinction this forge ignores, but the forest ahead will not.

The forge accepts your work

When $P(0) = 0.25$ , the forge stamps the vector as normalized. You have prepared a legitimate single-qubit state ready for measurement statistics — not a classical 25% guess about a hidden bit.

Guided exerciseintermediate

Computational-basis measurement

Apply the Born rule in the computational basis.

Not started

Statistics from the tower

From the measurement tower you watch repeated Z-basis (computational-basis) trials. Each flash is one outcome: 0 or 1. Over many independent preparations, frequencies settle toward the Born-rule predictions.

The guide asks you to predict the split for the normalized state you forged — the one with $P(0) = 0.25$ .

Born rule and Bloch-sphere angles

For $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ , computational-basis measurement gives:

P(0) = |\alpha|^2, \qquad P(1) = |\beta|^2.

On the Bloch sphere with polar angle $\theta$ measured from $|0\rangle$ :

P(0) = \cos^2\!\left(\frac{\theta}{2}\right), \qquad P(1) = \sin^2\!\left(\frac{\theta}{2}\right).

When $P(0) = 0.25$ , normalization forces $P(1) = 0.75$ . That corresponds to $\theta = 120°$ because $\cos^2(60°) = 0.25$ .

For the forged state with Bloch angle θ = 120° (equivalently P(0) = 0.25), what is P(0)? Enter as a decimal.

Probability between 0 and 1.

Theory meets frequency

The tower's histogram approaches your prediction. Short runs fluctuate — ten trials might show two or four zeros — but the Born rule describes the limit of many independent preparations, not every finite sequence exactly.

Independent trials

Each dot of light in the tower is a fresh preparation followed by one measurement. The 25/75 split is a statement about the ensemble, not a guarantee for every block of four trials.

Chapter reviewintermediate

Chapter 1 review

Combine basis states, normalization, and Born-rule reasoning.

Not started

The guardian blocks the path

A stone guardian bars the exit from the Qubit Frontier. "Show me you understand the single-qubit model," it rumbles, "without the shortcuts that confuse beginners."

Two trials await — basis reasoning and Born-rule calculation.

Chapter 1 synthesis

Before you leave this region, confirm these pillars:

State vector: $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$ .
Measurement: One classical outcome per trial in the chosen basis.
Born rule: $P(0) = |\alpha|^2$ , $P(1) = |\beta|^2$ for computational-basis measurement.

Review I — Which statement about |0⟩ and |1⟩ is correct?

Review II — For |ψ⟩ = (3|0⟩ + 4|1⟩)/5, what is P(0) in the computational basis?

The path opens

The guardian steps aside. "You treat the basis as a coordinate system, not a hiding place. The superposition forest lies ahead — remember that amplitudes can coexist coherently there."

Chapter completion

Exercises completed: 0/4

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