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Pauli gates: X, Y, Z
The three flips every circuit is built from, as matrices and as Bloch rotations.
- 01Write the X, Y and Z matrices and apply them to |0⟩ and |+⟩.
- 02Explain why Z does nothing visible to |0⟩ but flips |+⟩ to |−⟩.
- 03Use the gate cards in the composer to check your answers.
X is the quantum NOT: it swaps the amplitudes of |0⟩ and |1⟩. Z leaves |0⟩ alone and multiplies |1⟩ by −1, so it changes only phase; on |+⟩ that phase flip turns it into |−⟩. Y does both at once, with a factor i you will rarely need to track by hand.
Z is invisible until you look sideways
Simulate: the histogram shows only 0, exactly as without the Z. Now insert an H before the measurement (palette → H on q0, after Z) and simulate again: Z turned |+⟩ into |−⟩, so H brings it to |1⟩.
OPENQASM 3.0;
include "stdgates.inc";
qubit[1] q;
bit[1] c;
h q[0];
z q[0];
c[0] = measure q[0];
2 questions · one try each counts
What is Z|+⟩?
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Applying X twice to any state gives…
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