Qubit

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

• 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.

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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.

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 ).

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.

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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

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Summary

Summary

• A qubit is represented by a normalized vector in a two-dimensional complex state space.
• The computational basis ( |0\rangle, |1\rangle ) is a coordinate choice used to describe states and measurements.
• A pure qubit has the form ( |\psi\rangle = \alpha|0\rangle + \beta|1\rangle ) with ( |\alpha|^2 + |\beta|^2 = 1 ).
• Measurement probabilities in the computational basis are (P(0)=|\alpha|^2) and (P(1)=|\beta|^2).
• Phase information can be hidden from one measurement distribution but become important after later operations.

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, single-qubit states and measurement chapters.