Superposition Across Two Qubits

With a single qubit the Hadamard gate creates two equal possibilities; with two qubits we can reach four at once. This lesson shows how independent Hadamard gates expand a two-qubit register into a uniform superposition of all four basis states.

Starting from $|00\rangle$

A two-qubit register initialised to $|00\rangle$ has state vector $\begin{pmatrix}1 & 0 & 0 & 0\end{pmatrix}^T$ in the ordering $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ (qubit 0 is the most significant bit). Applying $H$ to qubit 0 alone gives the product state

Now apply $H$ to qubit 1:

Expanding the tensor product gives

Each of the four basis states carries amplitude $\tfrac{1}{2}$, so the Born-rule probability for each outcome is $\left|\tfrac{1}{2}\right|^2 = \tfrac{1}{4}$.

Why use two independent Hadamards?

Because the two qubits do not interact — there is no entanglement here — the operation decomposes as $H \otimes H$. In circuit notation this is simply a Hadamard on qubit 0 followed (or preceded, since they commute) by a Hadamard on qubit 1. The order does not matter because single-qubit gates on different qubits always commute — this is a consequence of the tensor-product structure of the operators, independent of the entanglement of the state.

Try it

Apply a Hadamard gate to each qubit so the register ends in the uniform superposition $\tfrac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$. The grader checks the full statevector, so every amplitude must be correct.

After running, inspect the amplitudes tab: all four entries should read $\approx 0.500$, and the probabilities tab should show four equal bars at $25\%$ each.

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