My question is how is Entanglement Entropy (EE) and Entanglement Negativity (N) related to the combinations of pure/mixed and separable/entangled states? That is for pure separable (PS), pure entangled (PE), mixed separable (MS) and mixed entangled (ME) states?

Beginning with Entanglement Entropy, we know that it is zero if and only if it is a pure state (or am I talking about Von Neumann Entropy at this point, I am a bit lost), and non-zero if it is mixed. So that means that

$$EE = 0 \iff pure \text{ (PS or PE)} $$
so, $$\text{(PS)} \implies EE = 0$$ which is fine, but also, $$\text{(PE)} \implies EE = 0$$ It is the second one that I am confused about, because EE is supposed to measure the entanglement(?)

A clarification on this would be extremely helpful, since I am trying to understand some things in the particular context of product states.

Coming to the case of Entanglement Negativity, a similar question arises. The following is known:

$$ separable \text{ (PS or MS ?)} \implies N = 0 $$ which thus means that: $$ N \neq 0 \implies entangled \text{ (PE or ME ?)} $$

Is the above correct to begin with? Also, are there no if and only if statements in the above?

What would be a known complete classification of these 4 types of states PE,PS,ME,MS versus the two parameters mentioned EE and N ?

The issue is also with the following from wikipedia: "he entropy of entanglement is the Von Neumann entropy of the reduced density matrix for any of the subsystems. If it is non-zero, i.e. the subsystem is in a mixed state, it indicates the two subsystems are entangled. More mathematically; if a state describing two subsystems A and B is a separable state, then the reduced density matrix is a pure state. Thus, the entropy of the state is zero."

This seems to say that pure=separable and mixed=entangled, which seems to deny the possibility of MS and PE (?) Where am i going wrong?

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    $\begingroup$ Do you mean entanglement entropy instead of von Neumann entropy? $\endgroup$ Jan 22, 2023 at 19:21
  • $\begingroup$ @TobiasFünke in the wiki page above it states that "If it is non-zero, i.e. the subsystem is in a mixed state, it indicates the two subsystems are entangled" which seems to state that mixed implies entangled automatically. But that is not true as we can have MS as well, yes? $\endgroup$
    – Brian M.
    Jan 22, 2023 at 19:57
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    $\begingroup$ There's too many questions here, without sufficient structure. This is very hard to answer succinctly. Try to make the question more focused. -- Also, your abbreviations make this very hard too read, there's just too many of them. $\endgroup$ Jan 22, 2023 at 20:19
  • $\begingroup$ That being said, you have a misconception what the "entropy of entanglement" is, when used as an entanglement measure. (The term is indeed misused sometimes -- but not in the quote you give. Do you understand what a reduced density matrix is?) $\endgroup$ Jan 22, 2023 at 20:19
  • $\begingroup$ Regarding your last paragraphs: Have you read the sentence just before the one you quote? And even the sentence you quote after "More mathematically [...]" contains more information... -See also the answer of @NorbertSchuch $\endgroup$ Jan 22, 2023 at 23:06

1 Answer 1


Entanglement entropy only makes sense for pure states. In that case, a state is entangled if and only if its entanglement entropy is non-zero.

Entanglement negativity makes sense both for pure and for mixed states.

  1. For pure states, as well as for mixed states in $2\times 2$ and $2\times 3$ dimensions, a state is entangled if and only if its entanglement negativity is non-zero.
  2. For mixed states in higher dimensions (i.e. at least $2\times 4$ or $3\times 3$), non-zero entanglement negativity implies that the state is entangled, but not conversely: There are entangled states with zero entanglement negativity as well (so-called bound entangled states).
  • $\begingroup$ That clarifies a lot actually! This raises two questions: 1) For EE, if we start with an overall pure state, and then if the subsystem is also pure then the Von Neumann entropy is 0, thus EE =0, meaning the initial pure state is separable? Or if the subsystem is mixed then the Von Neumann entropy is not 0, thus EE is not 0, meaning the initial pure state is entangled? So the fact that the subsystem is pure/mixed is equivalent to the overall initial pure system being separable/entangled respectively, correct? 2) For N, why have you omitted higher dimension pure states? Is for trivial reasons? $\endgroup$
    – Brian M.
    Jan 23, 2023 at 1:05
  • $\begingroup$ 1. Yes. 2. I haven't. $\endgroup$ Jan 23, 2023 at 6:29

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