Debugging a Circuit

Even a one-gate mistake in a quantum circuit produces a completely wrong state. Learning to read the statevector output and trace it back to a gate error is one of the core debugging skills in quantum programming.

The target: the Bell state

The simplest entangled two-qubit state is the Bell state

The standard circuit that produces it is:

1. Apply $H$ to qubit 0 — this puts qubit 0 into the equal superposition $\tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$.
2. Apply a CNOT with qubit 0 as control and qubit 1 as target.

After step 1 the two-qubit state is $\tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\otimes|0\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |10\rangle)$. The CNOT flips qubit 1 whenever qubit 0 is $|1\rangle$, turning $|10\rangle \to |11\rangle$ while leaving $|00\rangle$ unchanged:

Both amplitudes are $\tfrac{1}{\sqrt{2}}$, and the probabilities are each $\tfrac{1}{2}$.

What a gate bug looks like

Suppose someone replaces $H$ with $X$ — the bit-flip gate. Then step 1 gives $X|0\rangle = |1\rangle$ (a definite $|1\rangle$, not a superposition), so the two-qubit state before the CNOT is $|10\rangle$. The CNOT flips qubit 1 because the control is $|1\rangle$, leaving $|11\rangle$. The result is a product state, not a Bell state:

Reading the statevector immediately reveals this: a single amplitude of 1 on the $|11\rangle$ component, instead of two equal amplitudes on $|00\rangle$ and $|11\rangle$. The fix is obvious once you see it — the wrong gate is on qubit 0.

Try it

The starter code has the bug described above. Run it, read the statevector, identify the wrong gate, and replace it so the circuit produces $|\Phi^+\rangle$.

After your fix, the Statevector panel should show two equal amplitudes $\approx 0.707 = \tfrac{1}{\sqrt{2}}$ on $|00\rangle$ and $|11\rangle$, with all other components zero. That is the signature of the Bell state.

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