The |+⟩ and |−⟩ States

Two of the most important states in quantum computing are not $|0\rangle$ or $|1\rangle$ but equal superpositions of them. They appear constantly in algorithms, error correction, and teleportation, and they have their own names: $|+\rangle$ and $|-\rangle$ .

Expanding |+⟩ and |−⟩ in the computational basis

Written explicitly in the $\{|0\rangle, |1\rangle\}$ basis:

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

Both states are unit vectors: the probabilities of the two outcomes sum to one,

\left|\frac{1}{\sqrt{2}}\right|^2 + \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} + \frac{1}{2} = 1.

They differ only in the relative sign of the $|1\rangle$ amplitude. That sign is invisible to a single measurement in the computational basis — both give outcome 0 or 1 with probability $\frac{1}{2}$ each — but it becomes observable when the states are made to interfere with one another, for example by applying a second Hadamard gate.

Preparing these states with the Hadamard gate

The Hadamard gate $H$ has the matrix

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

Applying it to the column vectors for $|0\rangle$ and $|1\rangle$ immediately recovers the definitions above:

H|0\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = |+\rangle, \qquad H|1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} = |-\rangle.

So starting from the ground state $|0\rangle$ , a single Hadamard puts the qubit into $|+\rangle$ .

Try it

Prepare the qubit in the $|+\rangle$ state. The simulator will compare your circuit's state vector to the expected amplitudes $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$ , so the starting $|0\rangle$ alone will not pass.

Run your code to see the quantum state.

After running, inspect the State Vector panel: you should see two amplitudes both equal to $\approx 0.707 = \frac{1}{\sqrt{2}}$ , with identical signs — that is $|+\rangle$ .