Quantum Computing 101

Measurement

Measurement is where a quantum state becomes classical data, and it changes the state it reads.

In this lesson you will learn to

Predict measurement probabilities and the counts you expect over many shots.
Explain collapse and why repeated measurements agree.
Measure in the X or Y basis by rotating the state first.
Sample measurement counts from a state in Qiskit.

Shots and Counts

Measuring a qubit gives one classical bit. You never see α or β directly, only a 0 or a 1.

To learn the probabilities you repeat the experiment. Each repetition is called a shot. Over many shots, the fraction of 0s approaches |α|² and the fraction of 1s approaches |β|².

Kub measures the state (√3/2)|0⟩ + (1/2)|1⟩ again and again. Every shot gives a single 0 or 1; only the counts over many shots reveal the probabilities 75% and 25%.

Worked example: probabilities and shots

Use a state with a larger amplitude for |0⟩ than for |1⟩.

|\psi\rangle=\frac{\sqrt{3}}{2}|0\rangle+\frac{1}{2}|1\rangle

P(0)=\frac{3}{4}=75\%\qquad P(1)=\frac{1}{4}=25\%

1000\text{ shots}\ \longrightarrow\ \text{about }750\text{ zeros and }250\text{ ones}

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector
import math

qc = QuantumCircuit(1)
qc.ry(math.pi / 3, 0)          # amplitudes sqrt(3)/2 and 1/2

state = Statevector.from_instruction(qc)
print(state.sample_counts(shots=1000))
# e.g. {'0': 742, '1': 258}  (different every run)

Collapse

Measurement does not just read the qubit; it changes it. After you observe 0 the state is |0⟩, and after you observe 1 it is |1⟩. This is called collapse.

If no gate acts in between, measuring again gives the same answer with certainty. The randomness happens once, at the first measurement.

Measuring |+⟩ gives 0 or 1 with equal odds, but the state then becomes the result. Measuring again, with no gates in between, repeats the same answer.

Worked example: measure twice

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

\begin{aligned}\text{First result }0&:\quad |+\rangle\longrightarrow|0\rangle\quad (50\%)\\\text{First result }1&:\quad |+\rangle\longrightarrow|1\rangle\quad (50\%)\end{aligned}

|0\rangle\xrightarrow{\text{measure again}}0\ (100\%)\qquad |1\rangle\xrightarrow{\text{measure again}}1\ (100\%)

›Check yourself: A qubit in |+⟩ is measured and gives 1. You then apply H and measure again. What are the odds now?

After the first measurement the state is |1⟩. H turns it into |−⟩, so the second measurement is 50/50. Without the first measurement, H would have turned |+⟩ into |0⟩ and you would read 0 every time: the early measurement destroyed the interference.

Measuring in Other Bases

Hardware measures in the Z basis: it asks "0 or 1?". To ask a different question, rotate the state first. Applying H before measuring turns |+⟩ into 0 and |−⟩ into 1, so H followed by a measurement is an X-basis measurement.

For the Y basis, apply S† and then H before measuring. The same trick lets you estimate how much of a state points along any axis.

Worked example: reading |−⟩ in the X basis

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

S^\dagger|{+i}\rangle=|+\rangle\ \Rightarrow\ HS^\dagger|{+i}\rangle=|0\rangle\qquad HS^\dagger|{-i}\rangle=|1\rangle

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector

qc = QuantumCircuit(1)
qc.x(0)
qc.h(0)          # prepare |->
qc.h(0)          # rotate the X basis onto the Z basis
print(Statevector.from_instruction(qc).probabilities_dict(decimals=3))
# {'1': 1.0}: |-> reads as 1

Key Takeaways

A single shot returns one bit; probabilities only show up in counts over many shots.
Counts approach |α|² and |β|², with noise shrinking like 1/√shots.
Measurement collapses the state onto the result, so repeating it gives the same answer.
Rotate before measuring to measure in another basis: H for X, S† then H for Y.

What To Practice

Practice predicting outcomes, choosing the right measurement basis, and decoding two-qubit states.