Circuit Identities & Gate Cancellation

The handful of rewrite rules an optimiser uses, and a bloated circuit you can watch shrink from level 0 to level 3.

Intermediate12 min read5 sections · 2 questions+45

By the end you will be able to

01Recognise self-inverse pairs: H·H, X·X and CNOT·CNOT.
02Use H·Z·H = X and H·X·H = Z to swap bit flips and phase flips.
03Merge rotations about one axis and predict what an optimiser leaves of a bloated circuit.

An optimiser does not understand your algorithm. It knows a list of local rules, short patterns of gates that equal something shorter, and applies them until nothing more fits. Learn the rules and you can predict its output, or write the short circuit yourself.

Identity	Why it holds	Saving
H · H = I	H is its own inverse	2 gates → 0
X · X = I	a flip done twice is no flip (same for Y and Z)	2 gates → 0
CNOT · CNOT = I	the target is XOR-ed with the control twice	2 two-qubit gates → 0
H · Z · H = X	H swaps the X and Z axes of the Bloch sphere	3 gates → 1
H · X · H = Z	the same swap, the other way round	3 gates → 1
RZ(α) · RZ(β) = RZ(α+β)	two turns about one axis add up	2 gates → 1

H Z H = X, H X H = Z, R_{z} (α) R_{z} (β) = R_{z} (α + β)

H·Z·H = X is also why a phase-flip error looks like a bit-flip error when you measure in the X basis.

01When a pair does not cancel

Two CNOTs cancel only if they have the same control and target and nothing in between blocks them: an H on the target between them, for example, does not commute with the CNOT and keeps both alive. That is why optimisers check commutation. Level 1 cancels gates that are already neighbours; levels 2 and 3 also slide gates past each other to bring pairs together, and level 3 re-synthesises whole two-qubit blocks.

Hands-on

Shrink a bloated circuit

Eleven gates that do very little. Compile at level 0, then level 1 and level 3 (or press Compare levels). Level 0 keeps every gate and translates each one, so you get two two-qubit gates and a long run of rz and sx. Level 1 already cancels the pairs, turns H·Z·H into one x and merges the two rz into rz(0.8). Level 3 goes at least as far; an rz right before a measurement cannot change the counts, so it may shorten or drop it.

OpenQASM 3

OPENQASM 3.0;
include "stdgates.inc";
qubit[2] q;
bit[2] c;
h q[0];
h q[0];
x q[1];
x q[1];
cx q[0], q[1];
cx q[0], q[1];
h q[1];
z q[1];
h q[1];
rz(0.3) q[0];
rz(0.5) q[0];
c = measure q;

Open in the IDE Open in the Playground

Knowledge check

2 questions · one try each counts

Question 1

What does H·X·H equal?

Question 2

One qubit has rz(pi/4), rz(pi/4), rz(pi/2) in a row. What can the optimiser replace them with?

Identity

Why it holds

Saving

H · H = I

H is its own inverse

2 gates → 0

X · X = I

a flip done twice is no flip (same for Y and Z)

2 gates → 0

CNOT · CNOT = I

the target is XOR-ed with the control twice

2 two-qubit gates → 0

H · Z · H = X

H swaps the X and Z axes of the Bloch sphere

3 gates → 1

H · X · H = Z

the same swap, the other way round

3 gates → 1

RZ(α) · RZ(β) = RZ(α+β)

two turns about one axis add up

2 gates → 1

01When a pair does not cancel