Two qubits, four amplitudes
Read a joint statevector before introducing controlled gates.
Intuition
With two bits there are four possible pairs. Two qubits use the same four labels, but a quantum state assigns an amplitude to each pair. Four basis states do not mean four simultaneous measurement results.
Ketwise follows Qiskit's |q1 q0⟩ ordering: the rightmost bit is q0. Start at |00⟩ and apply H to q0. Only |00⟩ and |01⟩ have nonzero amplitudes. Then add H to q1: all four outcomes have probability 1/4. This product state is not entangled.
The mathematics
Look a little deeper
A tensor product combines independent state descriptions. Expand the brackets to get the joint amplitudes. For this state every amplitude is 1/2, so every probability is 1/4. Entanglement appears when the joint pure state cannot be factored into individual qubit states.
Use it
Explore H on each wire. For the linked Bell challenge, reset and use H(q0), CX(0,1). Compare its two nonzero amplitudes with the four in the product state.
This exercise uses the same server-side state and circuit checks as its linked practice problem.
Build your circuit
Click a gate to append it, or drag it onto a wire. CX uses the selected control; SWAP uses it as the second operand. Basis order: |qₙ … q₀⟩.
Returned logical circuit · before transpilation · q0 on top
Connect to code
qc = QuantumCircuit(2) qc.h(0) qc.h(1)
These operations go inside a circuit-building program. Practice problems provide a complete solve() template.
Check your understanding
First complete the circuit exercise successfully.