Product States vs Entangled States

Overview

When you combine qubits, not every joint state behaves like “a state of qubit A” plus “a state of qubit B.” Some joint states can be separated into two single-qubit states; others cannot.

This page makes that distinction precise:

• Product (separable) states factor as $|\psi\rangle\otimes|\phi\rangle$.
• Entangled states do not factor, meaning the joint system contains information that cannot be assigned to either qubit alone.

Intuition

A product state is the quantum analogue of independence: you can describe each subsystem on its own, and the full description is just the combination.

Entanglement is different. It does not mean “mystery” or “spooky behavior.” It means:

• the joint system is in a perfectly well-defined pure state, but
• there is no way to assign a pure state to each qubit that reproduces the joint state.

Entanglement is therefore about structure of the joint state, not about faster-than-light signals or hidden messages.

It is also not “just correlation.” Classical systems can be correlated because of shared randomness. Entanglement is stronger: it is correlation that cannot be explained by saying “each subsystem had its own (possibly random) state all along.”

Formal Description

A general two-qubit pure state is

$$|\Psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle.$$

Product (separable) states

A two-qubit state is a product state if there exist single-qubit states

$$|a\rangle = a_0|0\rangle + a_1|1\rangle,\quad |b\rangle = b_0|0\rangle + b_1|1\rangle$$

such that

$$|\Psi\rangle = |a\rangle\otimes|b\rangle.$$

Expand the tensor product:

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

Using distributivity, this becomes

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

So a product state has coefficients of the special form

$$\alpha_{00}=a_0b_0,\ \alpha_{01}=a_0b_1,\ \alpha_{10}=a_1b_0,\ \alpha_{11}=a_1b_1.$$

Entangled states

A state is entangled if it is not a product state—meaning there do not exist single-qubit states $|a\rangle$ and $|b\rangle$ that produce it.

A simple factorization test (two-qubit pure states)

For a two-qubit pure state, a useful criterion is:

$$\alpha_{00}\alpha_{11} = \alpha_{01}\alpha_{10}\quad\text{(product state)}.$$

If this equality fails, the state cannot factor and is entangled.

In words: for product states, the “cross-multiplication” of coefficients matches because all four coefficients are built from just two pairs of numbers $(a_0,a_1)$ and $(b_0,b_1)$.

Worked Example

Consider the state

$$|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle+|11\rangle).$$

Here the coefficients are:

• $\alpha_{00}=\tfrac{1}{\sqrt{2}}$
• $\alpha_{11}=\tfrac{1}{\sqrt{2}}$
• $\alpha_{01}=0$
• $\alpha_{10}=0$

Compute the factorization test:

$$\alpha_{00}\alpha_{11} = \tfrac{1}{2},\quad \alpha_{01}\alpha_{10} = 0.$$

They are not equal, so $|\Phi^+\rangle$ is entangled.

For contrast, take

$$|\Psi\rangle = \left(\tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\right)\otimes|0\rangle.$$

Expanding gives

$$|\Psi\rangle = \tfrac{1}{\sqrt{2}}|00\rangle + \tfrac{1}{\sqrt{2}}|10\rangle,$$

which is clearly a product state (because we constructed it as a tensor product).

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Now that we have the definition of entanglement, we’ll study the most important examples: the four Bell states, which are maximally entangled two-qubit states with crisp, testable measurement correlations.