Checkpoint: Controlled-Gate Puzzles

Module 11 introduced three families of two-qubit controlled gates — CNOT, CZ, and SWAP — and the identities that connect them. This checkpoint asks you to combine those tools to realize a specific target operation.

The target: a Bell-basis encoder

A natural and important two-qubit unitary is the Bell-basis encoder $U_\text{Bell}$ . It maps each computational basis state to a distinct Bell state:

|00\rangle \;\mapsto\; |\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr),

|01\rangle \;\mapsto\; |\Psi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|01\rangle + |10\rangle\bigr),

|10\rangle \;\mapsto\; |\Phi^-\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle - |11\rangle\bigr),

|11\rangle \;\mapsto\; |\Psi^-\rangle = \tfrac{1}{\sqrt{2}}\bigl(|01\rangle - |10\rangle\bigr).

The four Bell states form an orthonormal basis for the two-qubit space — they span the same space as the computational basis, just rotated. Because $U_\text{Bell}$ maps one orthonormal basis to another, it is unitary: $U_\text{Bell}^\dagger U_\text{Bell} = I$ .

Reading off the circuit

To figure out what gates to use, look at the first row of the map. Starting from $|00\rangle$ , the output $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ is the Bell state built in the "CNOT on Superpositions" lesson. The recipe is:

Apply $H$ to qubit 0, turning $|0\rangle$ into $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ .
Apply $\text{CNOT}(0 \to 1)$ , entangling the qubits.

Let us verify that the same two-gate sequence maps the remaining basis states correctly.

Input $|01\rangle$ :

|01\rangle \;\xrightarrow{H \otimes I}\; \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)|1\rangle = \tfrac{1}{\sqrt{2}}\bigl(|01\rangle + |11\rangle\bigr).

CNOT flips qubit 1 when qubit 0 is $|1\rangle$ , so $|11\rangle \to |10\rangle$ while $|01\rangle$ stays:

\tfrac{1}{\sqrt{2}}\bigl(|01\rangle + |10\rangle\bigr) = |\Psi^+\rangle. \checkmark

Input $|10\rangle$ :

|10\rangle \;\xrightarrow{H \otimes I}\; \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle)|0\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle - |10\rangle\bigr),

and after CNOT ( $|10\rangle \to |11\rangle$ ):

\tfrac{1}{\sqrt{2}}\bigl(|00\rangle - |11\rangle\bigr) = |\Phi^-\rangle. \checkmark

Input $|11\rangle$ :

|11\rangle \;\xrightarrow{H \otimes I}\; \tfrac{1}{\sqrt{2}}\bigl(|01\rangle - |11\rangle\bigr) \;\xrightarrow{\text{CNOT}}\; \tfrac{1}{\sqrt{2}}\bigl(|01\rangle - |10\rangle\bigr) = |\Psi^-\rangle. \checkmark

All four inputs produce the correct Bell state, confirming that $U_\text{Bell} = \text{CNOT}(0 \to 1) \cdot (H \otimes I)$ .

The unitary grader

This exercise uses the unitary grader: it extracts the full $4 \times 4$ matrix of your circuit and compares it to the reference. Every column of the matrix is the image of one computational basis state under your circuit, so each column must match one of the four Bell states shown above. An empty circuit, a circuit with only $H$ , or a circuit with only CNOT will all fail — both gates are required.

Try it

Build $U_\text{Bell}$ : apply $H$ to qubit 0, then CNOT with control 0 and target 1. The grader checks the whole unitary matrix.

Run your code to see the quantum state.

After running, inspect the Unitary panel: the four columns should each show two amplitudes of magnitude $\tfrac{1}{\sqrt{2}} \approx 0.707$ at positions $\{|00\rangle, |11\rangle\}$ , $\{|01\rangle, |10\rangle\}$ , $\{|00\rangle, |11\rangle\}$ , and $\{|01\rangle, |10\rangle\}$ (with appropriate signs), confirming the Bell-state structure of the encoder.