Superposition and Interference at Scale

algorithmsCircuit workspace

9 min read

Superposition and Interference at Scale

Put every qubit into superposition at once with H, and see amplitudes cancel exactly via interference.

Suggested before this lesson

Measurement, Shots, and Running on a Simulator

Everything up to this point has used one or two qubits at a time. Quantum algorithms (starting with Deutsch-Jozsa a few lessons from now) get their power from doing this at scale: putting an entire register into superposition at once, then using interference to cancel out the amplitudes of wrong answers before you ever measure. This lesson builds both pieces.

Hadamard on Every Qubit

You already know qc.h([0, 1, 2]) broadcasts H across a list of qubits (from How Circuits and Gates Take Their Arguments). Do that to every qubit in an $n$ -qubit register, starting from $|00\ldots0\rangle$ , and you get a perfectly uniform superposition over all $2^n$ basis states:

$H^{\otimes n}|0\rangle^{\otimes n} = \frac{1}{\sqrt{2^n}}\sum_{x=0}^{2^n-1}|x\rangle$

Every one of the $2^n$ basis states gets exactly the same amplitude, $\frac{1}{\sqrt{2^n}}$ , not approximately equal, exactly equal. That exactness is the whole point: it's what lets later interference cancel some of them down to exactly zero, rather than just "small."

Uniform Superposition Over 3 Qubits

All 8 basis states, each with the same exact amplitude, 1/√8 ≈ 0.354. The Exact State table below shows the precise numbers.

Circuit

Step through the circuit: after gate 3 of 3

4 / 4

Exact State

Basis state	Re	Probability
$∣000 ⟩$	0.354	12.5%
$∣001 ⟩$	0.354	12.5%
$∣010 ⟩$	0.354	12.5%
$∣011 ⟩$	0.354	12.5%
$∣100 ⟩$	0.354	12.5%
$∣101 ⟩$	0.354	12.5%
$∣110 ⟩$	0.354	12.5%
$∣111 ⟩$	0.354	12.5%

Exercise

Build a 3-qubit circuit and apply

H

to all three qubits in a single broadcast call, so the circuit ends in a uniform superposition over all 8 basis states.

Live preview of the circuit detected in your code

Need a hint?

Interference: When Amplitudes Cancel Exactly

Superposition alone isn't enough to make an algorithm useful: you also need a way to steer probability toward the right answer. That's interference: routing two different computational "paths" so their amplitudes for the same outcome add up (constructive) or cancel out (destructive).

The simplest possible example is one qubit and two Hadamards in a row. You already know $H|0\rangle = |+\rangle$ . What does a second $H$ do?

H

, then

H

Again

Step through the two gates with the scrubber below the circuit. Watch the Exact State table after the second

H

: the amplitude on

|1\rangle

isn't just small, it's exactly 0.

Circuit

Step through the circuit: after gate 2 of 2

3 / 3

Exact State

Basis state	Re	Im	Probability
$∣0 ⟩$	1.000	0	100.0%
$∣1 ⟩$	0	0	0.0%

Each basis state after the second $H$ is reached by two different paths through the intermediate $|+\rangle$ state: one contributing a positive amplitude to $|1\rangle$ , one contributing a negative amplitude of the same size. They cancel to exactly zero, and the state lands back on $|0\rangle$ with certainty. $H$ is its own inverse ( $H^2 = I$ ), and this cancellation is why: it's not a coincidence of these particular numbers, it's interference doing exactly what it's supposed to do.

This (steering amplitude toward a "yes" answer and cancelling it away from every "no" answer) is the mechanism every algorithm in the rest of this track relies on.

Check your understanding

In the

H

-then-

H

circuit above, why does the amplitude on

|1\rangle

end up at exactly 0.000 rather than just something small?

Rounding error in the simulator happens to hit zero.

Two paths to

|1\rangle

contribute equal-and-opposite amplitudes, which cancel exactly: destructive interference, not approximation.

The second

H

gate doesn't actually do anything to a qubit already in superposition.

Measurement forces it to zero.

Hadamard on Every Qubit

You already know qc.h([0, 1, 2]) broadcasts H across a list of qubits (from How Circuits and Gates Take Their Arguments). Do that to every qubit in an

n

-qubit register, starting from

|00\ldots0\rangle

, and you get a perfectly uniform superposition over all

2^n

basis states:

H^{\otimes n}|0\rangle^{\otimes n} = \frac{1}{\sqrt{2^n}}\sum_{x=0}^{2^n-1}|x\rangle

Every one of the

2^n

basis states gets exactly the same amplitude,

\frac{1}{\sqrt{2^n}}

, not approximately equal, exactly equal. That exactness is the whole point: it's what lets later interference cancel some of them down to exactly zero, rather than just "small."

Python

Uniform Superposition Over 3 Qubits

All 8 basis states, each with the same exact amplitude, 1/√8 ≈ 0.354. The Exact State table below shows the precise numbers.

Circuit

Step through the circuit: after gate 3 of 3

4 / 4

Exact State

Basis state	Re	Probability
$∣000 ⟩$	0.354	12.5%
$∣001 ⟩$	0.354	12.5%
$∣010 ⟩$	0.354	12.5%
$∣011 ⟩$	0.354	12.5%
$∣100 ⟩$	0.354	12.5%
$∣101 ⟩$	0.354	12.5%
$∣110 ⟩$	0.354	12.5%
$∣111 ⟩$	0.354	12.5%

Exercise

Build a 3-qubit circuit and apply

H

to all three qubits in a single broadcast call, so the circuit ends in a uniform superposition over all 8 basis states.

Python

Live preview of the circuit detected in your code

Need a hint?

Interference: When Amplitudes Cancel Exactly

The simplest possible example is one qubit and two Hadamards in a row. You already know

H|0\rangle = |+\rangle

. What does a second

H

do?

Python

H

, then

H

Again

Step through the two gates with the scrubber below the circuit. Watch the Exact State table after the second

H

: the amplitude on

|1\rangle

isn't just small, it's exactly 0.

Circuit

Step through the circuit: after gate 2 of 2

3 / 3

Exact State

Basis state	Re	Im	Probability
$∣0 ⟩$	1.000	0	100.0%
$∣1 ⟩$	0	0	0.0%

Each basis state after the second

H

is reached by two different paths through the intermediate

|+\rangle

state: one contributing a positive amplitude to

|1\rangle

, one contributing a negative amplitude of the same size. They cancel to exactly zero, and the state lands back on

|0\rangle

with certainty.

H

is its own inverse (

H^2 = I

), and this cancellation is why: it's not a coincidence of these particular numbers, it's interference doing exactly what it's supposed to do.

This (steering amplitude toward a "yes" answer and cancelling it away from every "no" answer) is the mechanism every algorithm in the rest of this track relies on.

Check your understanding

In the

H

-then-

H

circuit above, why does the amplitude on

|1\rangle

end up at exactly 0.000 rather than just something small?

Rounding error in the simulator happens to hit zero.

Two paths to

|1\rangle

contribute equal-and-opposite amplitudes, which cancel exactly: destructive interference, not approximation.

The second

H

gate doesn't actually do anything to a qubit already in superposition.

Measurement forces it to zero.