← All modules 01 Qubits, superposition and measurement

SESAME · Day 1 · Module 01

Qubits, superposition
and measurement

Quantum computing manipulates qubit states and uses measurement to obtain classical results.

01 / Qubit02 / Superposition03 / Measurement

01 / Qubit · 1.1

A qubit is the unit of information we manipulate in quantum computing.

Classical bit

0 or 1

Qubit

Two reference states, 0 and 1,
and their superpositions.

Prepare

Choose a state.

Manipulate

Apply operations.

Measure

Record a result.

For example: prepare 0 → change the qubit’s state to 1.

01 / Qubit · 1.2

A quantum state describes the prepared physical system.

Physical carrier

One atom.

Qubit

Two selected internal states
and their superpositions.

Same physical carrier.
Different prepared states.

Energy 0 1 One selected state The other selected state

01 / Qubit · 1.3

Two reference vectors give us coordinates for a qubit’s state.

Ket zero

\[|0\rangle=\begin{pmatrix}1\\0\end{pmatrix}\]

Ket one

\[|1\rangle=\begin{pmatrix}0\\1\end{pmatrix}\]

These two independent vectors form a basis.

Their linear combinations describe additional states we can prepare.

02 / Superposition · 2.1

Superposition allows a qubit to have states beyond zero and one.

\[|\psi\rangle=\alpha|0\rangle+\beta|1\rangle=\begin{pmatrix}\alpha\\\beta\end{pmatrix}\]

\(\alpha,\beta\): complex amplitudes. Both are nonzero in a superposition of 0 and 1.

For example: plus

\[|+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}=\begin{pmatrix}1/\sqrt2\\1/\sqrt2\end{pmatrix}\]

One state of one physical qubit.

02 / Superposition · 2.2

Normalization fixes the total squared magnitude at one.

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

Unit-length state vectors.

Start with

\[\begin{pmatrix}1\\1\end{pmatrix}\]

Length: \(\sqrt2\)

Divide by the length

\[\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\]

The plus state.

Check

\[\frac12+\frac12=1\]

Squared magnitudes.

The amplitudes themselves need not add to one.

02 / Superposition · 2.3

Relative phase is part of the state we manipulate.

Plus

\[|+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}\]
Re Im Phase of α: 0 Re Im Phase of β: 0

Relative phase: \(0\)

Minus

\[|-\rangle=\frac{|0\rangle-|1\rangle}{\sqrt2}\]
Re Im Phase of α: 0 Re Im Phase of β: π

Relative phase: \(\pi\)

Same magnitudes. Different relative phase.

02 / Superposition · 2.3

A common phase factor leaves the physical state unchanged.

Plus

\[\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\]

Multiply both entries by \(-1\).

Same physical state

\[\frac1{\sqrt2}\begin{pmatrix}-1\\-1\end{pmatrix}\]

Both phases change by \(\pi\).

The relative phase stays \(0\).

This is a change of global phase.

02 / Superposition · 2.3

The Bloch sphere places these states on its surface.

0 and 1
Opposite poles.

Plus and minus
Opposite points on the equator.

Changing only global phase
leaves the point unchanged.

The point describes a state,
not the atom’s position in space.

|0⟩|1⟩|−⟩|+⟩

02 / Superposition · 2.3

Have we changed the physical state?

Multiply both amplitudes of plus by \(i\).

\[\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\]
\[\frac1{\sqrt2}\begin{pmatrix}i\\i\end{pmatrix}\]

No. Both phases increase by \(\pi/2\).

Their difference stays zero. The physical state is unchanged.

03 / Measurement · 3.1

Measurement gives a classical result and can change the qubit’s state.

We need information from the physical qubit that we can use as a result.

Physical system

Qubit

Interaction

Measurement

Classical record

0 or 1

No measurement can completely determine an arbitrary unknown state from a single qubit, even in principle.

Computational-basis readout \(|0\rangle\) → record 0 \(|1\rangle\) → record 1

03 / Measurement · 3.2

The Born rule connects amplitudes to outcome probabilities.

\[P(0)=|\alpha|^2\qquad P(1)=|\beta|^2\]

For plus

\[|+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}\]
\[\left|\frac1{\sqrt2}\right|^2=\frac12\]

Both outcomes have probability \(1/2\).

Probabilities for one readout.

The state determines the probabilities. An individual result can be random.

03 / Measurement · 3.3

Measurement can change the qubit’s state.

\[|+\rangle=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\]

Ideal computational-basis measurement.

Probability ½

Classical record

0

State afterwards

\[|0\rangle=\begin{pmatrix}1\\0\end{pmatrix}\]

Probability ½

Classical record

1

State afterwards

\[|1\rangle=\begin{pmatrix}0\\1\end{pmatrix}\]

The qubit remains available. Its state is now the reference state matching the record.

03 / Measurement · 3.3

The next calculation uses the state left by measurement.

Before

\[\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\]

Suppose we record 1.

After

\[|1\rangle=\begin{pmatrix}0\\1\end{pmatrix}\]
Measure againRecord 1 with certainty.

Assume no operation or other change between the ideal measurements.

To repeat the experiment with plus, prepare plus again.

Worked close / Qubit → Superposition → Measurement

Follow one prepared state through the calculation.

\[|\psi\rangle=\frac35|0\rangle+\frac{4i}{5}|1\rangle=\begin{pmatrix}3/5\\4i/5\end{pmatrix}\]

One physical qubit. Two nonzero amplitudes.

Does this vector satisfy the normalization condition?

\[\left|\frac35\right|^2+\left|\frac{4i}{5}\right|^2=\frac9{25}+\frac{16}{25}=1\]

Amplitude phases: \(0\) and \(\pi/2\). Relative phase: \(\pi/2\).

Worked close / Qubit → Superposition → Measurement

The Born rule reuses the squared magnitudes.

\[|\psi\rangle=\frac35|0\rangle+\frac{4i}{5}|1\rangle\]

Computational-basis measurement

\[P(0)=\left|\frac35\right|^2=\frac9{25}=0.36\]
\[P(1)=\left|\frac{4i}{5}\right|^2=\frac{16}{25}=0.64\]

The two probabilities add to one.

Worked close / Qubit → Superposition → Measurement

The recorded result determines the state afterwards.

Before measurement

\[\begin{pmatrix}3/5\\4i/5\end{pmatrix}\]

\(P(1)=0.64\)

Suppose we record 1.

After measurement

\[\begin{pmatrix}0\\1\end{pmatrix}\]

The state is \(|1\rangle\).

Measure again immediately\(P(1)=1\)

Assume no change between the ideal measurements.

Worked close / Qubit → Superposition → Measurement

Can repeated readouts give the original distribution?

The first measurement recorded 1. The qubit is now in \(|1\rangle\).

If we keep measuring this same qubit 100 times,
should we expect about 64 ones?

Keep measuring the same qubit

No. Every further result is 1.

Its state remains \(|1\rangle\), assuming no change between measurements.

Prepare the original state each time

\[P(1)=0.64\quad\text{on each shot}\]

100 independent shots: 64 ones in expectation. The actual count can differ.

Qubit / Superposition / Measurement

Quantum computing manipulates qubit states and uses measurement to obtain classical results.

01 A qubit is the unit of information we manipulate in quantum computing.

02 Superposition allows a qubit to have states beyond zero and one.

03 Measurement gives a classical result and can change the qubit’s state.

Exercise / Worked card

What is the probability of measuring 1?

Use a computational-basis measurement.

\[|+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}\]

Find the amplitude attached to \(|1\rangle\).

Take its squared magnitude.

\[P(1)=\left|\frac1{\sqrt2}\right|^2=\frac12=\boxed{0.50}\]

Exercise / Your turn

Find the probability of measuring 1 in each state.

Use a computational-basis measurement.

01
\[|0\rangle\]
02
\[\frac{\sqrt3}{2}|0\rangle+\frac12|1\rangle\]
03
\[\frac{\sqrt3}{2}|0\rangle+\frac{e^{i\phi}}2|1\rangle,\qquad\phi\in\mathbb{R}\]

Next: Gates and interference →