Checkpoint: Engineer an Entangled State

You have met all four Bell states and the template circuit that generates them. This checkpoint asks you to go further: given a target state, figure out the right setup for yourself.

The target

Prepare the Bell state

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

Like all Bell states, $|\Phi^-\rangle$ is maximally entangled — each qubit measured alone is a 50/50 mixture, yet the two outcomes are perfectly correlated. What makes $|\Phi^-\rangle$ distinct from $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ is the relative phase: the $|11\rangle$ amplitude carries a factor of $-1$ .

Why the phase matters

Two quantum states that differ only by a global phase are physically indistinguishable — there is no measurement that can tell $|\psi\rangle$ from $e^{i\theta}|\psi\rangle$ . A relative phase between two terms inside a superposition is a different story. It affects interference and is therefore observable. In particular, $|\Phi^+\rangle$ and $|\Phi^-\rangle$ are orthogonal:

\langle\Phi^+|\Phi^-\rangle = \tfrac{1}{2}(\langle 00| + \langle 11|)(|00\rangle - |11\rangle) = \tfrac{1}{2}(1 - 1) = 0.

A Bell-state measurement — used in teleportation and superdense coding — can distinguish these two states reliably, so engineering the correct relative phase is a meaningful task.

Anatomy of the template circuit

The standard two-qubit entangling circuit is: apply a Hadamard to qubit 0, then a CNOT from qubit 0 to qubit 1. Feeding this circuit a pre-rotated input state controls which Bell state emerges. If the initial state of the two-qubit register is $|xy\rangle$ , then after $H\otimes I$ followed by $\text{CNOT}_{0\to 1}$ :

$x = 0$ gives a positive relative phase; $x = 1$ gives a negative relative phase.
$y = 0$ keeps both terms in the $\{|00\rangle,|11\rangle\}$ subspace; $y = 1$ sends them to $\{|01\rangle,|10\rangle\}$ .

To reach $|\Phi^-\rangle$ you need $x = 1$ (negative phase) and $y = 0$ (the $\Phi$ family), so the desired input is $|10\rangle$ . Starting from $|00\rangle$ that means flipping qubit 0 first.

Try it

Build the circuit that maps $|00\rangle$ to $|\Phi^-\rangle$ . The grader checks the full state vector: amplitude $\tfrac{1}{\sqrt{2}}$ on $|00\rangle$ , amplitude $-\tfrac{1}{\sqrt{2}}$ on $|11\rangle$ , and exactly zero on $|01\rangle$ and $|10\rangle$ .

Run your code to see the quantum state.

Once you pass, examine the Bloch sphere: both qubits sit at the origin (maximally mixed individually), confirming the entangled nature of the joint state. The $-1$ relative phase is invisible on the individual Bloch spheres but was decisive in determining which Bell state you built.