Independent Gates on Each Qubit

A quantum register holds multiple qubits. When no gate couples them, each qubit evolves independently — you can think of the register as two separate one-qubit systems that happen to share the same wire.

Tensor products and independent evolution

A two-qubit state that factorises is written as a tensor product:

|\psi_0\rangle \otimes |\psi_1\rangle.

Applying a gate $U$ to qubit 0 and a gate $V$ to qubit 1 simultaneously means the combined operation is $U \otimes V$ :

(U \otimes V)\big(|\psi_0\rangle \otimes |\psi_1\rangle\big) = U|\psi_0\rangle \otimes V|\psi_1\rangle.

The qubits do not need to communicate; each gate acts on its own qubit and the results are combined through the tensor product. This works only when the state is separable — not entangled.

A concrete example

Start from $|00\rangle$ (both qubits initialised to $|0\rangle$ ). Apply $H$ to qubit 0 and $X$ to qubit 1:

H|0\rangle \otimes X|0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}} \otimes |1\rangle = \frac{|01\rangle + |11\rangle}{\sqrt{2}}.

Here qubit 0 is the most significant bit, so $|01\rangle$ means qubit 0 = 0, qubit 1 = 1. Both basis strings have amplitude $\tfrac{1}{\sqrt{2}}$ , and no entanglement has been created.

Try it

Build the circuit that transforms $|00\rangle$ into $\tfrac{1}{\sqrt{2}}(|01\rangle + |11\rangle)$ by applying $H$ to qubit 0 and $X$ to qubit 1.

Run your code to see the quantum state.

Check the State Vector tab: qubits 0 and 1 are shown separately, and the amplitudes should each read $\approx 0.707$ .