Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the defining property for two quantum states to become entangled? Is it just that the combined system cannot be in a product state? Why is this?

share|cite|improve this question
up vote 1 down vote accepted

No, the definition of entanglement is not only just that the combined system is not in a product state. This is the case only if the combined system is in a pure state.

For the general definition of entanglement consider two systems with respective Hilbert spaces $H_1$ and $H_2$. The composed system has Hibert space $H_1 \otimes H_2$. The state of the full system is described by a density matrix, i.e. a Hermitian positive operator of trace 1 on $H_1 \otimes H_2$. Let $\rho_{12}$ be such a density matrix, and suppose it can be written as a convex combination

$$ \rho_{12}=\sum_k p_k \rho_k^{(1)} \otimes \rho_k^{(2)},$$

where $p_k \ge 0$ and $\sum_k p_k =1$, and where the $\rho_k^{(1)}$'s and $\rho_k^{(2)}$'s are density matrices on $H_1$ and $H_2$, respectively. Then we call the state separable . Otherwise it is called entangled.

share|cite|improve this answer
To have true entanglement then do both subsystems have to be mixed? – user32462 Jan 30 '14 at 10:54
If the total system of the two parties is in a pure state, then the two subsystems are entangled if and only if the restricted states of the subsystems are mixed. However, if the total system is already in a mixed state, then also the subsystems can be in a mixed state without being entangled, e.g., they can be in a mixed product state. So to answer your question: to have entanglement, the restricted states of the subsystems must be in a mixed state (but this is an "if and only if" condition when the total system is in a pure state). – Zoltan Zimboras Jan 30 '14 at 15:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.