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Qubits & Quantum States

A qubit is the fundamental unit of quantum information. Unlike a classical bit which is definitively 0 or 1, a qubit exists in a superposition of both until measured. This topic establishes the mathematical framework — Dirac notation, the Bloch sphere, and the Born rule — that underpins all of quantum computing.

Key Concepts

Qubit
The quantum analogue of the classical bit. A qubit occupies a two-dimensional complex Hilbert space spanned by basis states 0|0\rangle and 1|1\rangle (the computational basis). Physically realized as electron spin, photon polarization, superconducting circuits, or trapped ion energy levels.
Superposition
A qubit can exist in any linear combination ψ=α0+β1|\psi\rangle = \alpha|0\rangle + \beta|1\rangle where α,βC\alpha, \beta \in \mathbb{C}. The normalization constraint α2+β2=1|\alpha|^2 + |\beta|^2 = 1 ensures total probability equals one. Superposition is destroyed upon measurement.
Born Rule
When a qubit ψ=α0+β1|\psi\rangle = \alpha|0\rangle + \beta|1\rangle is measured in the computational basis, the probability of obtaining 0|0\rangle is α2|\alpha|^2 and of obtaining 1|1\rangle is β2|\beta|^2. After measurement the qubit collapses to the observed state.
Bloch Sphere
A geometric representation of a single qubit state as a point on a unit sphere: ψ=cos(θ/2)0+eiϕsin(θ/2)1|\psi\rangle = \cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle. The north pole is 0|0\rangle, south pole is 1|1\rangle, and the equator holds equal superpositions. Global phase has no physical meaning.
Global Phase
Multiplying a state by eiγe^{i\gamma} (a global phase) produces no observable difference — all measurement probabilities are unchanged. Only relative phases between 0|0\rangle and 1|1\rangle components are physically meaningful.
Multi-Qubit States
An nn-qubit system lives in a 2n2^n-dimensional Hilbert space. Two qubits have basis states 00,01,10,11|00\rangle, |01\rangle, |10\rangle, |11\rangle. The exponential growth of Hilbert space dimension is the source of quantum computing's potential power.

Key Equations

General Qubit State
ψ=α0+β1,α,βC|\psi\rangle = \alpha|0\rangle + \beta|1\rangle, \quad \alpha,\beta \in \mathbb{C}
Any single-qubit state is a complex linear combination of computational basis states.
Normalization
α2+β2=1|\alpha|^2 + |\beta|^2 = 1
The sum of measurement probabilities must equal 1.
Born Rule
P(0)=α2,P(1)=β2P(|0\rangle) = |\alpha|^2, \quad P(|1\rangle) = |\beta|^2
Measurement probability equals the squared modulus of the amplitude.
Bloch Sphere Parametrization
ψ=cos ⁣θ20+eiϕsin ⁣θ21|\psi\rangle = \cos\!\tfrac{\theta}{2}|0\rangle + e^{i\phi}\sin\!\tfrac{\theta}{2}|1\rangle
Every pure qubit state maps to a unique point on the unit sphere (θ∈[0,π], φ∈[0,2π)).
Inner Product
ϕψ=αγ+βδ for ϕ=α0+β1,  ψ=γ0+δ1\langle\phi|\psi\rangle = \alpha^*\gamma + \beta^*\delta \text{ for } |\phi\rangle=\alpha|0\rangle+\beta|1\rangle,\; |\psi\rangle=\gamma|0\rangle+\delta|1\rangle
The inner product measures overlap between quantum states.
Worked Example

Measurement Probabilities

Problem

A qubit is prepared in the state ψ=350+451|\psi\rangle = \frac{3}{5}|0\rangle + \frac{4}{5}|1\rangle. What are the probabilities of measuring 0|0\rangle and 1|1\rangle? Verify normalization.

Solution

Apply the Born rule to find each probability:

P(0)=352=925=0.36P(|0\rangle) = \left|\frac{3}{5}\right|^2 = \frac{9}{25} = 0.36
P(1)=452=1625=0.64P(|1\rangle) = \left|\frac{4}{5}\right|^2 = \frac{16}{25} = 0.64

Check normalization:

P(0)+P(1)=0.36+0.64=1.00P(|0\rangle) + P(|1\rangle) = 0.36 + 0.64 = 1.00 \checkmark
Answer P(|0⟩) = 0.36, P(|1⟩) = 0.64
Practice

Exercises

5 problems
1 of 5

For ψ=350+451|\psi\rangle = \frac{3}{5}|0\rangle + \frac{4}{5}|1\rangle, what is P(0)P(|0\rangle)?

2 of 5

For ψ=120+121|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle, what is P(1)P(|1\rangle)?

3 of 5

A normalized qubit has α=0.6\alpha = 0.6 (real). What is β2|\beta|^2?

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4 of 5

For ψ=i20+321|\psi\rangle = \frac{i}{2}|0\rangle + \frac{\sqrt{3}}{2}|1\rangle, what is P(0)P(|0\rangle)?

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5 of 5

For ψ=1200+1211|\psi\rangle = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle, what is the probability of measuring the first qubit as 0|0\rangle?

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Key Takeaways

  • A qubit state ψ=α0+β1|\psi\rangle = \alpha|0\rangle + \beta|1\rangle requires α2+β2=1|\alpha|^2 + |\beta|^2 = 1.
  • The Born rule: measurement yields 0|0\rangle with probability α2|\alpha|^2 and 1|1\rangle with probability β2|\beta|^2.
  • The Bloch sphere gives a complete geometric picture of all single-qubit pure states.
  • Global phase eiγe^{i\gamma} has no physical consequence; only relative phase matters.
  • An nn-qubit system lives in a 2n2^n-dimensional Hilbert space — the exponential foundation of quantum speedups.