Understanding Entanglement

Classical computing typically features bits that operate independently unless explicitly linked by a logic gate during a specific operation. When two bits are on a hard drive, for example, the state of one provides no information about the state of the other; they are distinct entities. Quantum mechanics introduces a phenomenon where this independence vanishes entirely. This is entanglement.

Entanglement occurs when a group of qubits share a single quantum state. In this scenario, the quantum state of each particle cannot be described independently of the state of the others, even when the particles are separated by a large distance. Instead of having "Qubit A" and "Qubit B," you have a single system "Qubit A+B."

Separability vs. Entanglement

To understand entanglement technically, we must look at the mathematical definition of separability.

A multi-qubit state is considered separable if it can be written as the tensor product of individual single-qubit states. For a two-qubit system, if we can find two vectors $|\psi\rangle$ and $|\phi\rangle$ such that the total state $|\Psi\rangle$ equals $|\psi\rangle \otimes |\phi\rangle$, the qubits are not entangled.

Consider the state where both qubits are in the $|0\rangle$ state:

$|\Psi\rangle = |00\rangle = |0\rangle \otimes |0\rangle$

This is clearly separable. We can describe the first qubit as being in state $|0\rangle$ and the second qubit as being in state $|0\rangle$.

Now consider the state created by combining a superposition and a CNOT gate, which we will analyze in depth later as a "Bell State." The mathematical representation is:

$|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$

This state describes a scenario where the system is in an equal superposition of $|00\rangle$ and $|11\rangle$. There is no probability amplitude for $|01\rangle$ or $|10\rangle$.

If we try to describe this as two independent qubits, we run into a mathematical contradiction. Let us attempt to factor $|\Phi^+\rangle$ into a product of two generic single-qubit states:

$(a_0|0\rangle + a_1|1\rangle) \otimes (b_0|0\rangle + b_1|1\rangle)$

When we expand this tensor product, we get:

$a_0b_0|00\rangle + a_0b_1|01\rangle + a_1b_0|10\rangle + a_1b_1|11\rangle$

For this expanded equation to match our entangled state $|\Phi^+\rangle$, the middle terms $|01\rangle$ and $|10\rangle$ must have a coefficient of zero. This implies:

1. $a_0b_1 = 0$
2. $a_1b_0 = 0$

However, for the outer terms $|00\rangle$ and $|11\rangle$ to exist (which they do in our entangled state), $a_0b_0$ and $a_1b_1$ must be non-zero. This creates an impossible system of equations. If $a_0$ is non-zero, then $b_1$ must be zero (from equation 1). But if $b_1$ is zero, then the term $a_1b_1|11\rangle$ would vanish, which contradicts the target state.

Because no such coefficients exist, the state $|\Phi^+\rangle$ is non-separable. The qubits are entangled.

Correlation and Measurement

The physical implication of non-separability is strictly correlated measurement outcomes.

In a separable system, measuring one qubit does not affect the probability distribution of the other. If you have a system in the state $\frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$, measuring the first qubit and finding it in state $|0\rangle$ leaves the second qubit in a superposition of $|0\rangle$ and $|1\rangle$. The second qubit retains its own randomness.

In an entangled system like $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, the outcomes are linked.

1. You measure the first qubit. You have a 50% chance of observing $|0\rangle$ and a 50% chance of observing $|1\rangle$.
2. Suppose you measure $|0\rangle$. The system collapses to the state consistent with this measurement. The only part of the superposition compatible with the first qubit being $|0\rangle$ is the $|00\rangle$ term.
3. The system state instantly becomes $|00\rangle$.
4. If you subsequently measure the second qubit, you are guaranteed to observe $|0\rangle$ with 100% certainty.

The randomness exists only in the first measurement. Once that outcome is determined, the second outcome is fixed.

Flow of measurement collapse in a maximally entangled system.

Visualizing the Probability Distribution

To visualize the difference between a uniform superposition (separable) and an entangled state, we can look at the probability histograms. In quantum computing SDKs like Qiskit or Cirq, this is often the primary way we verify entanglement.

In a uniform superposition of two qubits, all four outcomes ($00, 01, 10, 11$) are equally likely. In a Bell state, the mixed states ($01, 10$) are forbidden.

Probability distribution comparison showing how entanglement eliminates specific basis state combinations.

The Role of Interference

It is important to note that correlation alone is not unique to quantum mechanics. If you take a red card and a blue card, put them in envelopes, and mail them to two different locations, opening one envelope reveals the color of the other. That is classical correlation.

Quantum entanglement differs because of the superposition and interference inherent in the state before measurement. The state $|\Phi^+\rangle$ is not just a statistical mixture where "it is either 00 or 11 but we do not know which yet." It is a coherent linear combination of both.

This distinction becomes practical when we apply quantum gates to entangled qubits. Operations performed on one half of an entangled pair can affect the global state in ways that purely classical correlation cannot replicate. This property is utilized in protocols like superdense coding and quantum teleportation, where manipulating the entanglement resource allows for information transfer capabilities exceeding classical limits.

In the next section, we will construct these states programmatically using the CNOT gate to see how this mathematical non-separability is implemented in code.