The Hadamard Gate

A qubit prepared in $|0\rangle$ is perfectly predictable: measure it and you always get 0. The Hadamard gate $H$ trades that certainty for a balanced superposition — it is the single most important gate for turning a quantum computer "on."

What H does

Acting on a basis state, the Hadamard gate produces

$H|0\rangle = \tfrac{1}{\sqrt{2}}\big(|0\rangle + |1\rangle\big), \qquad H|1\rangle = \tfrac{1}{\sqrt{2}}\big(|0\rangle - |1\rangle\big).$

Its matrix is

$H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$

You can read the result straight off the matrix: applying $H$ to the column vector $\binom{1}{0}$ gives $\tfrac{1}{\sqrt{2}}\binom{1}{1}$, so both amplitudes equal $\tfrac{1}{\sqrt{2}}$. By the Born rule the probability of each outcome is $\left|\tfrac{1}{\sqrt{2}}\right|^{2} = \tfrac{1}{2}$ — a perfect 50/50 split. On the Bloch sphere, $H$ rotates the north pole $|0\rangle$ onto the $+x$ axis, the state we call $|+\rangle$.

Try it

Complete the circuit so the single qubit ends in an equal superposition. Press Run to watch the amplitudes update, then Check to grade your answer against the simulator.

Look at the output: the two amplitudes are each $\approx 0.707 = \tfrac{1}{\sqrt{2}}$, the probabilities are both $0.5$, and the Bloch vector points along $+x$. That is exactly $|+\rangle$.

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