Creating |+⟩ and |−⟩ with H

Two of the most useful states in quantum computing are the X-basis states $|+\rangle$ and $|-\rangle$. Both are equal superpositions of $|0\rangle$ and $|1\rangle$, differing only in the relative sign between the two amplitudes:

On the Bloch sphere, $|+\rangle$ sits on the $+x$ axis and $|-\rangle$ on the $-x$ axis — both halfway between the two poles.

Producing them with H

The Hadamard gate $H$ is defined by its action on the computational-basis states:

You can verify this directly from the matrix:

Multiply $H$ by the column vector for $|0\rangle = \binom{1}{0}$:

Multiply by $|1\rangle = \binom{0}{1}$:

Both results are normalized: each amplitude has magnitude $\tfrac{1}{\sqrt{2}}$, so the squared magnitudes sum to $\left(\tfrac{1}{\sqrt{2}}\right)^2 + \left(\tfrac{1}{\sqrt{2}}\right)^2 = 1$.

Why these states matter

The states $|+\rangle$ and $|-\rangle$ appear throughout quantum algorithms. Interference is most powerful when a qubit is in a superposition, and $H$ is the standard way to create one from a definite input. Grover's algorithm, quantum Fourier transforms, and many error-correction gadgets all start by producing $|+\rangle$ with a Hadamard on $|0\rangle$.

Try it

The circuit starts in $|0\rangle$. Apply $H$ to produce $|+\rangle$ and return the circuit. The grader compares the full state vector, so the amplitudes must match $\tfrac{1}{\sqrt{2}} \approx 0.707$ for both $|0\rangle$ and $|1\rangle$ with no relative phase difference.

After running, check the State vector panel: you should see two equal real amplitudes of $+\tfrac{1}{\sqrt{2}}$ and the Bloch vector pointing along $+x$.

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