Project: A Quantum Random Number

A single Hadamard on one qubit gives one random bit. What if you need more? The answer is straightforward: use $n$ independent qubits, apply a Hadamard to each, and measure all of them. The result is an $n$ -bit quantum random number whose bits are genuinely independent and uniformly distributed.

From one random bit to many

After a Hadamard, qubit $k$ sits in the equal superposition

H|0\rangle_k = \frac{1}{\sqrt{2}}\bigl(|0\rangle_k + |1\rangle_k\bigr).

Applying this to each of the $n$ qubits independently produces the product state

H^{\otimes n}|0\rangle^{\otimes n} = \frac{1}{\sqrt{2^n}} \sum_{x=0}^{2^n - 1} |x\rangle,

where the sum runs over every $n$ -bit string $x$ . This is the uniform superposition over all $2^n$ computational basis states. Measuring it collapses the state to one of those strings, each with probability

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

For $n = 4$ qubits, there are $2^4 = 16$ possible outcomes and each carries probability $\tfrac{1}{16} = 0.0625$ .

Why the bits are independent

The product state above has no entanglement: qubit 0 does not know the result of qubit 1. Formally, the state factors as a tensor product, meaning the joint probabilities factorize too — $P(x_0 x_1 x_2 x_3) = P(x_0)\,P(x_1)\,P(x_2)\,P(x_3)$ with each marginal equal to $\tfrac{1}{2}$ . Each qubit contributes exactly one independent, unbiased random bit.

Try it

Build the 4-qubit circuit that generates a 4-bit quantum random number. The grader checks that every one of the 16 outcomes has equal probability $\tfrac{1}{16}$ .

Run your code to see the quantum state.

After running, open the Probabilities panel. You should see 16 equal bars, each at height $0.0625$ . Because the bits are independent, you can read the 4-bit result as an integer from 0 to 15, drawn uniformly at random by the laws of quantum mechanics.