This note is part of [[quantum/Practical Quantum Information System]]. > [!info] Course code > Use the companion repository for this lecture's runnable lab, helper functions, and regression checks: > - [notebooks/by_concept/separable_vs_entangled_states.ipynb](https://github.com/montekkundan/quantum-code/blob/main/notebooks/by_concept/separable_vs_entangled_states.ipynb) > - [qcourse/entanglement.py](https://github.com/montekkundan/quantum-code/blob/main/qcourse/entanglement.py) > - [tests/test_states_density_entanglement.py](https://github.com/montekkundan/quantum-code/blob/main/tests/test_states_density_entanglement.py) [TODO: add video - Separable vs. Entangled States] ## What This Concept Is Take two qubits. Sometimes the joint state is just "one state for the first qubit and one state for the second qubit, multiplied together." That is a [[quantum/Glossary#Separable state|separable state]]. Sometimes the joint state cannot be written that way at all. That is an [[quantum/Glossary#Entangled state|entangled state]]. This distinction is one of the conceptual centerpieces of the course. If you only remember that "entanglement means correlation," you will miss what makes it special. Classical systems can also be correlated. Entanglement is the stronger statement that the quantum state itself does not factor across the chosen partition. ## Foundation Terms You Need First A [[quantum/Glossary#Tensor product|tensor product]] combines subsystem state spaces into one larger joint space. A [[quantum/Glossary#Separable state|separable state]] factors across that partition. An [[quantum/Glossary#Entangled state|entangled state]] does not. A [[quantum/Glossary#Bell state|Bell state]] is the standard first example of a maximally entangled two-qubit state. The important word here is partition. A state can be entangled across one split of a system and not across another. You always need to say which subsystems you are talking about. ## How The Idea Actually Works For pure bipartite states, the first test is structural. Can you write the state as $ |\psi\rangle = |\alpha\rangle_A \otimes |\beta\rangle_B? $ If yes, the state is separable. If not, it is entangled. The Bell state $ \frac{|00\rangle + |11\rangle}{\sqrt{2}} $ is the classic example because it cannot be factored into one state for qubit $A$ and one for qubit $B$. Why does that matter physically? Because entanglement changes what subsystem descriptions look like and what kinds of correlations are possible. A globally pure entangled state can produce locally mixed reduced states. It can also support protocols such as teleportation and superdense coding that have no classical analog with the same resource story. This is also where many students make a bad shortcut. Strong correlation is not the same as entanglement. A classical coin-flip procedure can create perfectly correlated outcomes without creating an entangled quantum state. The real question is not "are the outputs related?" but "does the joint quantum state factor?" ## Worked Example: The Two-Qubit Determinant Test For a pure two-qubit state $ |\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle, $ reshape the amplitudes into $ M= \begin{bmatrix} a&b\\ c&d \end{bmatrix}. $ The state is separable exactly when $M$ has rank 1. For a $2\times2$ matrix, that is the same as $ ad-bc=0. $ For the Bell state, $a=d=1/\sqrt2$ and $b=c=0$, so $ad-bc=1/2\ne0$. It is entangled. For $(|0\rangle+|1\rangle)\otimes|0\rangle/\sqrt2=(|00\rangle+|10\rangle)/\sqrt2$, $a=c=1/\sqrt2$ and $b=d=0$, so $ad-bc=0$. It is separable. ## Source Reading Lens Use `qclec.pdf` Lectures 5 and 11 for tensor products, Bell states, and entanglement. Use *Quantum Computing since Democritus* Chapter 12 as the conceptual warning: the important claim is not merely that outcomes correlate, but that the quantum state cannot be reduced to local classical hidden data. ## Why It Matters - It is the resource language behind many core QIS protocols. - It prepares you for reduced density matrices, Bell tests, and entropy-based entanglement measures. - It prevents you from confusing ordinary correlation with genuinely nonfactorizable quantum structure. ## Related Questions > [!question] Why is "correlated" not the same as "entangled"? >> [!answer] Classical systems can be strongly correlated because they share a common preparation history. Entanglement is a statement about the quantum state not factoring across a partition, not merely about related measurement outcomes. > [!question] How do you test separability for a pure two-qubit state by inspection? >> [!answer] Write the amplitudes as a $2\times2$ coefficient matrix. If that matrix has rank 1, the state factors into one-qubit states. If it has rank greater than 1, the state is entangled. > [!question] Why does the partition matter? >> [!answer] Entanglement is always relative to a split into subsystems. A state can be product across one grouping and entangled across another, especially in multi-qubit systems. ## Additional Study Notes **Check yourself: superpositions can change separability** A superposition of product states can be entangled, such as $(|00\rangle+|11\rangle)/\sqrt{2}$. A superposition of entangled states can also become product, so neither direction is automatic. **Check yourself: mixtures behave differently** A mixture of separable states is separable by definition. A mixture of entangled states is not guaranteed to remain entangled, because classical averaging can wash out the entanglement. **Key fact: classical correlation is not entanglement** The state $\frac12|00\rangle\langle00|+\frac12|11\rangle\langle11|$ is classically correlated but separable. It is a mixture of product projectors, not a nonfactorizable pure state. **Check yourself: determinant test for a pure two-qubit state** For $\frac12(|00\rangle+i|01\rangle+i|10\rangle+|11\rangle)$, the coefficient matrix has determinant $(1)(1)-(i)(i)=2$ up to the common factor, so the state is entangled. **Question: How can a state have perfect same-basis correlations but still fail to be entangled?** Perfect correlations can arise from a classical random choice. If a source flips a coin and prepares either $|00\rangle$ or $|11\rangle$, the outcomes match in the computational basis, but the density matrix is still a separable mixture. ## Study Checks Use these after the explanation, not before it. ### Quick Checks - MCQ: For a pure two-qubit state with coefficient matrix $M$, what rank indicates separability? A. rank 0 B. rank 1 C. rank 2 D. rank 4 **Answer:** B. rank 1. - What determinant condition tests separability for $a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$? **Answer:** The state is separable exactly when ad - bc = 0. - MCQ: Which state is classically correlated but separable? A. $(|00\rangle+|11\rangle)/\sqrt2$ B. $\frac12|00\rangle\langle00|+\frac12|11\rangle\langle11|$ C. $(|01\rangle-|10\rangle)/\sqrt2$ D. A GHZ state **Answer:** B. The 50-50 mixture of |00><00| and |11><11|. ### Oral Exam Anchors > [!question] Oral exam anchor > Give the pure two-qubit separability test and apply it to a Bell state. A good answer should say: For |psi> = a|00> + b|01> + c|10> + d|11>, reshape the amplitudes into the coefficient matrix M = [[a, b], [c, d]]. The state is separable if and only if M has rank 1. For a 2 by 2 matrix, this is equivalent to ad - bc = 0. For the Bell state (|00>+|11>)/sqrt(2), we have a = d = 1/sqrt(2) and b = c = 0. Then ad - bc = 1/2, which is nonzero. Therefore the Bell state is entangled. ## Practical Lab Use concrete two-qubit states so separability stops feeling like a vague label. - Prepare at least one product state and one Bell state in Qiskit. - Compare their measurement correlations in more than one basis. - Try to factor each state's amplitudes into one-qubit components and record where that factorization fails. ## Homework The homework should force a mathematical separability test rather than a visual guess. - For several two-qubit states, decide whether each one is separable or entangled and justify the answer. - Write one product-state example and one entangled-state example in ket notation. - Explain why correlation by itself is not enough to prove entanglement. ## References - Scott Aaronson, [Introduction to Quantum Information Science](https://www.scottaaronson.com/qclec.pdf), Lecture 5. - Scott Aaronson, [Introduction to Quantum Information Science](https://www.scottaaronson.com/qclec.pdf), Lecture 11. - [[Resources]] - IBM Quantum Learning, [entanglement learning resources](https://quantum.cloud.ibm.com/learning/en).