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The Single-Qubit Toolbox

Most early QCoder problems are really asking one question:

“Can you control one qubit well?”

That sounds small, but one-qubit fluency carries a lot of the book.

The first gates to master

  • x: swaps |0> and |1>
  • z: flips the sign of |1>
  • h: changes between the computational basis and the plus/minus basis
  • rx, ry, rz: continuous rotations
  • p: adds a phase to |1>

You do not need every gate in the library yet. You do need to know what these do without hesitation.

ry as a state-preparation tool

For real amplitudes, ry(theta) is the cleanest gate to learn first:

\[ R_y(\theta)|0\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle \]

This equation matters because it lets you design a target probability instead of guessing.

For example, if you want probability 3/4 on |1>, then you want:

\[ \sin^2(\theta/2) = 3/4 \]

so one valid choice is theta = 2 * asin(sqrt(3) / 2) = 2pi/3.

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

qc = QuantumCircuit(1)
qc.ry(2 * pi / 3, 0)
print(Statevector.from_instruction(qc))

Why z feels useless until it does not

If your qubit is definitely |0>, then z appears to do nothing.

But after a Hadamard, z changes relative phase:

\[ Z\frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{|0\rangle - |1\rangle}{\sqrt{2}} \]

That is enough to change later interference.

The plus/minus basis is not optional

The states

\[ |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, \qquad |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}} \]

show up constantly:

  • |+> is what h creates from |0>
  • |-> is what h creates from |1>
  • |-> is the standard ancilla state for phase kickback tricks

You do not need to worship this notation. You do need to recognize it instantly.

Learn one tiny identity well

\[ H Z H = X \]

This is not just algebra. It is the smallest example of a powerful idea:

  1. phase information exists in one basis
  2. change basis
  3. the same action looks like a bit flip

You can check it directly:

from qiskit import QuantumCircuit
from qiskit.quantum_info import Operator

lhs_circuit = QuantumCircuit(1)
lhs_circuit.h(0)
lhs_circuit.z(0)
lhs_circuit.h(0)

rhs_circuit = QuantumCircuit(1)
rhs_circuit.x(0)

lhs = Operator(lhs_circuit)
rhs = Operator(rhs_circuit)
print(lhs == rhs)

If you prefer not to rely on operator equality, test both circuits on |0> and |1> with Statevector.

Relative phase versus global phase

This distinction becomes important very early.

  • multiplying the entire state by -1 does not change physical behavior
  • changing only one branch by -1 does change interference

So -|1> and |1> are physically equivalent, but (|0> + |1>)/sqrt(2) and (|0> - |1>)/sqrt(2) are not.

Checkpoint Exercises

  1. Prepare the minus state.
  2. Prepare a state with P(1)=3/4.
  3. Verify H Z H = X on |0> and |1>.
  4. Build three different circuits for -|1> and confirm they differ only by global phase.

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