Separable vs. Entangled States

Hello! Welcome to the next lesson in our journey through linear algebra for quantum mechanics and deep learning.

In the last two lessons, we established the framework for describing composite systems using tensor products. We learned how to combine vectors from two spaces to form a state in a larger, combined space, and how to combine operators to act on these new states. We've been working primarily with states that can be written as a simple tensor product, like $∣ ψ ⟩ \otimes ∣ ϕ ⟩$ .

Today, we address a profound question that lies at the heart of quantum mechanics: Is every state in a composite system just a simple product of its parts? The answer is a resounding "no," and this leads to one of the most fascinating and powerful concepts in all of physics.

Our goal is to define and distinguish between separable and entangled states in a tensor product space. Understanding this distinction is absolutely critical for your future study of quantum mechanics and quantum computing.

1. Separable vs. Entangled States: The Core Idea

Let's start with the definitions. A state in a tensor product space $V \otimes W$ can fall into one of two categories.

Separable States

A state $∣Ψ ⟩$ is called separable (or a product state) if it can be written as a single tensor product of a state from $V$ and a state from $W$ .

∣Ψ ⟩ = ∣ ψ ⟩ \otimes ∣ ϕ ⟩

Conceptually, this means the two subsystems are independent. The state of the first part, $∣ ψ ⟩$ , doesn't depend on the state of the second part, $∣ ϕ ⟩$ . All the information about the composite system is contained within the descriptions of its individual parts.

Entangled States

A state $∣Ψ ⟩$ is entangled if it is not separable. This means it can only be written as a sum of two or more product states, and this sum cannot be factored into a single tensor product.

∣Ψ ⟩ = c_{1} (∣ ψ_{1} ⟩ \otimes ∣ ϕ_{1} ⟩) + c_{2} (∣ ψ_{2} ⟩ \otimes ∣ ϕ_{2} ⟩) + \dots

Conceptually, entangled states describe systems where the parts are intrinsically linked. The state of one subsystem cannot be described independently of the other. Measuring a property of one part instantly influences the corresponding property of the other, no matter how far apart they are. This is the "spooky action at a distance" that famously puzzled Einstein.

This short article provides a clear summary of these definitions in the context of quantum computing.

What Is Entanglement in Quantum Computing & How It Works

The article 'What Is Entanglement in Quantum Computing & How It Works' from SpinQuanta gives a concise overview of the concept and its mathematical representation.

Please read the sections 'What is an Entangled State in Quantum Computing?' and 'Mathematical Representation'. Focus on the distinction it draws between the factorable form of a separable state and the non-factorable form of an entangled state like the Bell state.

2. The Mathematical Test for Entanglement

How can we prove a state is entangled? The method is surprisingly direct: we assume it is separable and show that this assumption leads to a mathematical contradiction.

Let's take the most famous example: the Bell state $∣ Φ^{+} ⟩$ , which is a state in the tensor product space of two qubits, $C^{2} \otimes C^{2}$ .

∣ Φ^{+} ⟩ = \frac{1}{2} (∣00 ⟩ + ∣11 ⟩) = \frac{1}{2} (∣0 ⟩ \otimes ∣0 ⟩ + ∣1 ⟩ \otimes ∣1 ⟩)

In column vector form, this is:

∣ Φ^{+} ⟩ = \frac{1}{2} 1000 + 0001 = 1/ 2 00 1/ 2

The Test: Can we find two single-qubit states, $∣ ψ ⟩ = (α β)$ and $∣ ϕ ⟩ = (γ δ)$ , such that $∣ ψ ⟩ \otimes ∣ ϕ ⟩ = ∣ Φ^{+} ⟩$ ?

Let's compute the product state $∣ ψ ⟩ \otimes ∣ ϕ ⟩$ :

(α β) \otimes (γ δ) = α (γ δ) β (γ δ) = α γ α δ β γ β δ

Now, we equate the components with our Bell state vector:

α γ α δ β γ β δ = 1/ 2 00 1/ 2

This gives us a system of four equations:

$α γ = 1/ 2$
$α δ = 0$
$β γ = 0$
$β δ = 1/ 2$

From equation (2), either $α = 0$ or $δ = 0$ .

If $α = 0$ , then equation (1) becomes $0 \cdot γ = 1/ 2$ , which is impossible.
If $δ = 0$ , then equation (4) becomes $β \cdot 0 = 1/ 2$ , which is also impossible.

Since our assumption (that the state is separable) leads to a contradiction, the assumption must be false. Therefore, the Bell state $∣ Φ^{+} ⟩$ is entangled.

The following video provides an excellent and clear walk-through of this exact process.

Bell States are Entangled States, Tensor Products of Single Qubit States

The video 'Bell States are Entangled States' by Elucyda demonstrates precisely how to prove non-separability by attempting to solve for the component vectors and arriving at a contradiction.

Watch from the beginning to 10:52. The first three minutes set up the problem by defining a generic separable two-qubit state. The core of the argument happens from 5:41 onwards, where the presenter equates the coefficients and uncovers the contradiction, proving the Bell states are entangled.

This procedure is the fundamental tool for distinguishing separable and entangled states. The same logic applies to any state in any tensor product space.

3. The Bigger Picture: Most States Are Entangled

It's crucial to realize that separable states are the exception, not the rule. A general state in a tensor product space is a linear combination of basis vectors. For two qubits, a general state $∣ χ ⟩$ is:
$∣ χ ⟩ = a_{00} ∣00 ⟩ + a_{01} ∣01 ⟩ + a_{10} ∣10 ⟩ + a_{11} ∣11 ⟩$
For this state to be separable, the four coefficients $a_{ij}$ must satisfy the condition $a_{00} a_{11} - a_{01} a_{10} = 0$ . (This condition, known as the Schmidt criterion for 2x2 systems, is a more advanced topic).

If you pick four complex numbers $a_{ij}$ at random, it is virtually certain that this condition will not be met. Entanglement is not a rare or exotic property; it's the generic state of a composite quantum system.

This video provides a slightly more abstract perspective on why entanglement is the general case.

Tensor product state spaces

Let's revisit 'Tensor product state spaces' by Professor M does Science. This clip generalizes the idea we just explored.

Please watch from 4:18 to 7:28. The presenter asks if a general state, described by coefficients a_ij, can always be factored such that a_ij = c_i * d_j. He demonstrates with a counter-example that this is not always possible, leading to the definition of entangled states.

Test your understanding!

Consider the state $∣Ψ ⟩ = \frac{1}{2} (∣00 ⟩ - ∣01 ⟩ + ∣10 ⟩ - ∣11 ⟩)$ . Can this state be written as a product state $∣ ψ ⟩ \otimes ∣ ϕ ⟩$ ? If so, find $∣ ψ ⟩$ and $∣ ϕ ⟩$ .

Show answer

Yes, this state is separable. Let's try to find its factors. We are looking for $∣ ψ ⟩ = α ∣0 ⟩ + β ∣1 ⟩$ and $∣ ϕ ⟩ = γ ∣0 ⟩ + δ ∣1 ⟩$ such that their tensor product equals $∣Ψ ⟩$ .

Expanding the product gives $α γ ∣00 ⟩ + α δ ∣01 ⟩ + β γ ∣10 ⟩ + β δ ∣11 ⟩$ .

Equating coefficients with $∣Ψ ⟩$ :

$α γ = 1/2$
$α δ = - 1/2$
$β γ = 1/2$
$β δ = - 1/2$

From (1) and (2), we can see that $γ = - δ$ .
From (1) and (3), we can see that $α = β$ .

Let's choose $α = 1/ 2$ . Since $α = β$ , we have $β = 1/ 2$ .
Substituting $α = 1/ 2$ into equation (1) gives $(1/ 2) γ = 1/2$ , so $γ = 1/ 2$ .
Since $γ = - δ$ , we have $δ = - 1/ 2$ .

Let's check if these values work for all four equations:

$(1/ 2) (1/ 2) = 1/2$ . Correct.
$(1/ 2) (- 1/ 2) = - 1/2$ . Correct.
$(1/ 2) (1/ 2) = 1/2$ . Correct.
$(1/ 2) (- 1/ 2) = - 1/2$ . Correct.

The solution is consistent! We have found the factors:

∣ ψ ⟩ = \frac{1}{2} (∣0 ⟩ + ∣1 ⟩) and ∣ ϕ ⟩ = \frac{1}{2} (∣0 ⟩ - ∣1 ⟩)

So, the original state is the separable state $∣Ψ ⟩ = \frac{1}{2} (∣0 ⟩ + ∣1 ⟩) \otimes \frac{1}{2} (∣0 ⟩ - ∣1 ⟩)$ .

Conclusion

Today we've uncovered a deep and essential feature of composite systems. While we build our tensor product spaces from simple product states, the space itself is vastly richer, containing states that defy separation into independent parts.

Key Takeaways:

Separable states (or product states) can be written as a single tensor product, $∣ ψ ⟩ \otimes ∣ ϕ ⟩$ , and represent independent subsystems.
Entangled states are states that are not separable. They represent systems where the parts are fundamentally correlated, and the state of the whole system contains more information than the sum of its parts.
To determine if a state is entangled, we can try to write it as a product state and solve for its components. If the resulting system of equations leads to a contradiction, the state is entangled.
Entanglement is the generic case, not a special one. Most states in a tensor product space are entangled.

Preview of the Next Lesson:

So far in this course, we've focused on the algebraic structure of vector spaces—adding vectors, scaling them, and transforming them. We are now about to enter a new module, "Inner Products and Orthogonality," where we will introduce a geometric structure. In the next lesson, we will define the inner product, a tool that allows us to generalize the dot product to any vector space. This will let us define concepts like length, distance, and angles, which are fundamental for calculating probabilities in quantum mechanics and for finding optimal solutions in machine learning.