Product states vs entangled states

In the last lesson, you saw how two independent qubits combine into a 2-qubit state via the tensor product. But there was a quiet assumption in that story — we assumed the qubits were independent.

What if they’re not?

Here’s a strange fact: not every 2-qubit state can be written as a tensor product of two single-qubit states. Some 2-qubit states refuse to factor. They describe a situation where the two qubits are so intimately correlated that neither one, on its own, has a definite state — but the pair, taken as a whole, does.

These non-factoring states are called entangled. Einstein hated them. They’re the reason quantum computing works.

The algebraic test

Suppose we have a general 2-qubit state:

$|\psi\rangle = c_{00}|00\rangle + c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle$

When is this a product state? When you can find single-qubit states $|\psi_A\rangle = \alpha_0|0\rangle + \beta_0|1\rangle$ and $|\psi_B\rangle = \alpha_1|0\rangle + \beta_1|1\rangle$ such that:

$|\psi\rangle = |\psi_B\rangle \otimes |\psi_A\rangle$

Expanding the right side (using the formula from Lesson 4.1):

$c_{00} = \alpha_1\alpha_0, \quad c_{01} = \alpha_1\beta_0, \quad c_{10} = \beta_1\alpha_0, \quad c_{11} = \beta_1\beta_0$

If you compute $c_{00}c_{11} - c_{01}c_{10}$ for a product state, you get:

$c_{00}c_{11} - c_{01}c_{10} = \alpha_1\alpha_0\beta_1\beta_0 - \alpha_1\beta_0\beta_1\alpha_0 = 0$

So here’s the test:

A 2-qubit state is a product state if and only if $c_{00}c_{11} = c_{01}c_{10}$.

If they’re not equal, the state is entangled. How entangled? The quantity

$C = 2\,|c_{00}c_{11} - c_{01}c_{10}|$

is called the concurrence, and it ranges from $0$ (product) to $1$ (maximally entangled).

The exemplar: a Bell state

Consider the state:

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

The amplitudes are $c_{00} = \frac{1}{\sqrt{2}}$, $c_{11} = \frac{1}{\sqrt{2}}$, $c_{01} = c_{10} = 0$.

Check the test: $c_{00}c_{11} - c_{01}c_{10} = \frac{1}{2} - 0 = \frac{1}{2} \neq 0$.

The concurrence is $C = 2 \cdot \frac{1}{2} = 1$. Maximally entangled.

If you try to write this state as $|\psi_B\rangle \otimes |\psi_A\rangle$, you’ll find it’s impossible. The |00⟩ term requires both qubits’ “zero” amplitudes to be nonzero, and the |11⟩ term requires both qubits’ “one” amplitudes to be nonzero — but then you’d also have nonzero |01⟩ and |10⟩ contributions, which you don’t. There’s no way to factor.

Play with the test

The widget below shows a full 2-qubit state. As you apply gates, watch the entanglement indicator below the state vector. It will toggle between “Product state” and “Entangled” as you cross the boundary.

Try:

1. Start from $|00\rangle$. Apply $H$ to qubit 0. The state is now $(|00\rangle + |01\rangle)/\sqrt{2}$ — still a product state ($|0\rangle \otimes |+\rangle$), the indicator says “Product state” with concurrence $0$.
2. Now apply $CNOT_{0 \to 1}$. The state becomes $(|00\rangle + |11\rangle)/\sqrt{2}$ — the Bell state. The indicator jumps to “Entangled” with concurrence $1$.
3. Reset. Apply $H$ to both qubits. The state is $(|00\rangle + |01\rangle + |10\rangle + |11\rangle)/2 = |+\rangle \otimes |+\rangle$ — still a product state, because all the qubits are still independent.
4. Starting from $|00\rangle$, apply only single-qubit gates (any of H, X, Y, Z, S, T) to either qubit. Try to create an entangled state. You will fail — single-qubit gates alone cannot create entanglement. You need a true 2-qubit gate like CNOT.

What entanglement isn’t

A lot of popular-science writing describes entanglement in ways that are close enough to be memorable but not quite right. Let’s clear up two of them:

Entanglement is not a signal. When two entangled qubits are separated by a large distance and one is measured, the other one’s state is “instantly” determined. It’s tempting to say this transmits information faster than light. But nobody can use this to send a message — the measurement outcomes are random, and the distant observer has no way to tell that anything has happened until classical information arrives (at light speed or slower). The “collapse” doesn’t carry information.

Entanglement is not shared randomness. It’s more than that. If two people each had a coin and agreed in advance to flip together, their results would be correlated — both heads or both tails every time. That’s shared randomness, and it can be explained classically. Entanglement is different: it produces correlations that cannot be explained by any pre-existing classical plan, no matter how clever. The proof is called Bell’s inequality, and we’ll see it in Module 5.

Why entanglement matters

The whole power of quantum computing comes from being able to create, manipulate, and measure entangled states. A quantum algorithm is essentially a recipe for:

1. Prepare a big entangled superposition over many qubits.
2. Apply gates that cause interference between the entangled parts, so the correct answer’s amplitude grows and the wrong answers’ amplitudes cancel.
3. Measure. With high probability, read out the answer.

Without entanglement, you just have a classical probabilistic computer — and those are no faster than regular ones at anything interesting. The moment you introduce entanglement, you unlock speedups for certain problems that no classical computer can match.

What’s next

You know what entangled states are. You’ve even seen one — the Bell state that appears when you apply H then CNOT starting from $|00\rangle$. The next lesson formally introduces CNOT along with the other two-qubit gates CZ and SWAP, and you’ll finally see how quantum operations create and transform entanglement.