All Four Bell States

The four canonical maximally entangled two-qubit states are collectively called the Bell basis. You already built the first one, $|\Phi^+\rangle$. The complete set is:

These four states are orthonormal and span the entire two-qubit space, so they form a basis — the Bell basis. Any two-qubit state can be expressed as a linear combination of them; they do not, however, exhaust all maximally entangled two-qubit states, which form a continuous family.

Building all four from scratch

A single template circuit produces any Bell state. Begin in a computational basis state $|xy\rangle$ (where $x, y \in \{0,1\}$), apply a Hadamard to qubit 0, then a CNOT from qubit 0 to qubit 1:

Working out each case:

| Input $|xy\rangle$ | After $H\otimes I$ | After CNOT | Result | |---|---|---|---| | $|00\rangle$ | $\tfrac{1}{\sqrt2}(|00\rangle+|10\rangle)$ | $\tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$ | $|\Phi^+\rangle$ | | $|10\rangle$ | $\tfrac{1}{\sqrt2}(|00\rangle-|10\rangle)$ | $\tfrac{1}{\sqrt2}(|00\rangle-|11\rangle)$ | $|\Phi^-\rangle$ | | $|01\rangle$ | $\tfrac{1}{\sqrt2}(|01\rangle+|11\rangle)$ | $\tfrac{1}{\sqrt2}(|01\rangle+|10\rangle)$ | $|\Psi^+\rangle$ | | $|11\rangle$ | $\tfrac{1}{\sqrt2}(|01\rangle-|11\rangle)$ | $\tfrac{1}{\sqrt2}(|01\rangle-|10\rangle)$ | $|\Psi^-\rangle$ |

The $x$ bit controls the relative phase (0 gives $+$, 1 gives $-$), while the $y$ bit controls which pair of basis states appears (0 gives the $\Phi$ family, 1 gives the $\Psi$ family). Starting from $|00\rangle$, you can reach any of the four by pre-applying $X$ gates before the Hadamard and CNOT.

Try it

Prepare $|\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$. You need to engineer the right input to the template circuit, starting from $|00\rangle$. The grader checks the full state vector.

Once you pass, try changing the starting X gates to reach the other three Bell states and observe how the amplitudes shift in the simulator.

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