Preparing Uniform Superpositions

A quantum register of $n$ qubits has $2^n$ computational-basis states. Feeding every algorithm with a uniform superposition over all of them is the starting point of Deutsch–Jozsa, Grover, the quantum Fourier transform, and most other quantum algorithms. Building that superposition costs exactly $n$ gates.

One Hadamard per qubit

Recall what the Hadamard gate does to a single-qubit basis state:

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

When every qubit starts in $|0\rangle$ and we apply one $H$ to each, the $n$ -qubit state factors into a tensor product:

H^{\otimes n}|0\rangle^{\otimes n} = \bigotimes_{k=0}^{n-1} \frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{1}{\sqrt{2^n}} \sum_{x=0}^{2^n - 1} |x\rangle.

Each amplitude equals $1/\sqrt{2^n}$ , so by the Born rule every basis state is equally likely with probability

\left(\frac{1}{\sqrt{2^n}}\right)^2 = \frac{1}{2^n}.

For $n = 3$ that is $1/8 \approx 0.125$ for each of the eight strings $|000\rangle$ through $|111\rangle$ .

Counting amplitudes

It is worth building intuition with small $n$ :

$n = 1$ : $H|0\rangle$ has 2 amplitudes, each $1/\sqrt{2}$ .
$n = 2$ : $H^{\otimes 2}|00\rangle = \tfrac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$ , each amplitude $1/2$ .
$n = 3$ : 8 amplitudes, each $\tfrac{1}{2\sqrt{2}} \approx 0.354$ .
$n = 10$ : 1024 amplitudes, each $1/32$ .

The preparation cost grows only as $O(n)$ gates, while the number of states encoded grows as $O(2^n)$ — this exponential encoding is the resource that quantum algorithms exploit.

Try it

Prepare the uniform superposition on 3 qubits and measure. The grader checks that all 8 outcomes appear with equal probability $1/8$ .

Run your code to see the quantum state.

After running, open the Probabilities tab: every bar should reach exactly $0.125$ .