Bell States

Combining the superposition of a single qubit with the conditional logic of a CNOT gate creates something entirely new. A CNOT gate's conditional logic involves acting on computational basis states such as $|00\rangle$ or $|10\rangle$ to flip bits deterministically. However, when uncertainty is introduced into the control qubit before applying the CNOT, entanglement is generated.

The specific states that result from this process are known as Bell states. These are the simplest examples of entanglement and serve as the standard resource for many quantum protocols, including quantum teleportation and superdense coding. There are four specific Bell states, which represent the four maximally entangled states for a two-qubit system.

Constructing the Circuit

To create a Bell state, you need a circuit with two specific components: a Hadamard gate (H) and a Controlled-NOT gate (CNOT).

1. Preparation: Start with two qubits initialized to the zero state: $|00\rangle$.
2. Superposition: Apply a Hadamard gate to the first qubit (the control qubit). This places it into a superposition of $|0\rangle$ and $|1\rangle$.
3. Entanglement: Apply a CNOT gate using the first qubit as the control and the second qubit as the target.

The following diagram illustrates the flow of this circuit.

Circuit logic for generating a Bell state. The top wire represents the control qubit and the bottom wire represents the target qubit.

Mathematical Derivation

Let us walk through the mathematics of this circuit to prove that the qubits become entangled. We use the standard tensor product notation where $|ab\rangle = |a\rangle \otimes |b\rangle$.

Step 1: Initialization We begin with both qubits in the ground state. $|\psi_0\rangle = |0\rangle \otimes |0\rangle = |00\rangle$

Step 2: Superposition We apply the Hadamard gate to the first qubit. Recall that $H|0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}$. The second qubit remains unchanged. $|\psi_1\rangle = (H \otimes I) |00\rangle = \left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right) \otimes |0\rangle$

Expanding the tensor product, we get: $|\psi_1\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |10\rangle)$

At this stage, the system is in an equal superposition of $|00\rangle$ and $|10\rangle$. The first qubit is random (50/50 chance of 0 or 1), while the second qubit is deterministically 0. They are not yet entangled.

Step 3: The CNOT Operation Now we apply the CNOT gate. The first qubit is the control.

• For the term $|00\rangle$, the control is 0, so the target remains 0. Result: $|00\rangle$.
• For the term $|10\rangle$, the control is 1, so the target flips from 0 to 1. Result: $|11\rangle$.

Combining these results yields the final state vector: $|\text{Bell}\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$

Analyzing the Result

This specific state is denoted as $|\Phi^+\rangle$ (Phi-plus). Look closely at the final equation: $|\Phi^+\rangle = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$

This describes a system that, upon measurement, has a 50% probability of collapsing to $|00\rangle$ and a 50% probability of collapsing to $|11\rangle$.

The defining characteristic of this state is the perfect correlation between the two qubits. If you measure the first qubit and find it is 0, the state collapses to $|00\rangle$, meaning the second qubit is guaranteed to be 0 immediately. Conversely, if you measure 1, the second qubit is guaranteed to be 1. It is impossible to find the system in state $|01\rangle$ or $|10\rangle$.

We can visualize the probability distribution of this state compared to a random unentangled state.

Probability outcomes when measuring the $|\Phi^+\rangle$ Bell state. Note that mixed outcomes like |01> and |10> are impossible.

This correlation holds regardless of the physical distance between the qubits. Once entangled, they act as a single mathematical object. You cannot describe the state of one qubit without referencing the other. If you attempt to factor $|\Phi^+\rangle$ into a product $(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle)$, you will find there are no coefficients $a, b, c, d$ that satisfy the equation. This mathematical inseparability is the definition of entanglement.

The Four Bell States

Depending on the initial state of the two qubits before applying the Hadamard and CNOT gates, you can generate four different maximally entangled states. These constitute the Bell Basis.

If we change the input from $|00\rangle$ to the other basis states ($|01\rangle, |10\rangle, |11\rangle$), the resulting Bell states differ by phase signs or bit flips.

1. Input $|00\rangle \rightarrow |\Phi^+\rangle$: $|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$ Outcome: Correlated (00 or 11). Same phase.

2. Input $|01\rangle \rightarrow |\Psi^+\rangle$: We start with $|01\rangle$. The Hadamard on the first qubit gives $\frac{|01\rangle + |11\rangle}{\sqrt{2}}$. The CNOT (control 1 flips target 1 to 0) results in: $|\Psi^+\rangle = \frac{|01\rangle + |10\rangle}{\sqrt{2}}$ Outcome: Anti-correlated (01 or 10). If qubit A is 0, qubit B is 1.

3. Input $|10\rangle \rightarrow |\Phi^-\rangle$: We start with $|10\rangle$. The Hadamard creates a phase difference: $\frac{|00\rangle - |10\rangle}{\sqrt{2}}$. The CNOT results in: $|\Phi^-\rangle = \frac{|00\rangle - |11\rangle}{\sqrt{2}}$ Outcome: Correlated (00 or 11) but with a phase difference. Probabilities look identical to $\Phi^+$ upon standard measurement, but the phase affects interference patterns in subsequent operations.

4. Input $|11\rangle \rightarrow |\Psi^-\rangle$: We start with $|11\rangle$. Result: $|\Psi^-\rangle = \frac{|01\rangle - |10\rangle}{\sqrt{2}}$ Outcome: Anti-correlated with a phase difference.

In Python frameworks like Qiskit, generating these states is a standard exercise. You simply initialize your qubits to the desired starting state (using X gates to flip $|0\rangle$ to $|1\rangle$ if necessary), apply the H gate, and then apply the CX (CNOT) gate. We will implement this directly in the practice section following this explanation.