I'm trying to understand entanglement in terms of scarcity and abundance.

Given an arbitrary vector $v$ representing a pure quantum state of, say, dimension 4, i.e. $v \in \mathcal{H}^{\otimes 4}$,

Is $v$ more likely to be entangled than non-entangled (separable)?

By trying to answer it myself , I can see that the separability test is based on an existential quantifier, namely trying to prove that $\exists v_1, v_2 \in \mathcal{H}^{\otimes 2} $ such that $v_1 \otimes v_2 = v $.

The entanglement test on the other hand is based on a universal quantifier, $$\forall v_1, v_2 \in \mathcal{H}^{\otimes 2}, v_1 \otimes v_2 \neq v.$$ So, this reasoning could suggest that entangled vectors are much more scarce than separable ones because it is easier to find one simple example (existential) that satisfies the condition than to check for every single one (universal).

This result would make sense physically since entanglement is a valuable resource so, intuitively, it should be scarce.

Does this reasoning make any sense at all, or am I saying nonsense? Any help would be greatly appreciated.

PS: I would assume extending this reasoning to (density) matrices would be obvious.

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    $\begingroup$ "PS: I would assume extending this reasoning to (density) matrices would be obvious." --- Why would you assume so? In fact, while almost all pure states are entangled, only a finite fraction of all mixed states of a given dimension is entangled. $\endgroup$ Aug 22, 2016 at 13:03
  • $\begingroup$ @NorbertSchuch that's interesting, could you link to a paper showing this? $\endgroup$
    – glS
    Jan 15, 2020 at 10:10
  • $\begingroup$ @glS What is interesting? $\endgroup$ Jan 15, 2020 at 13:21
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    $\begingroup$ @glS Ok, for one thing is is known that in any dimension, there is a ball around the maximally mixed state which is separable, so the volume of separable states must be non-zero. On the other hand, you can see (e.g. using PPT) that around a pure entangled state there will be ball of entangled states, so they also have non-zero volume. But I'm sure this has been studied more quantitatively, but I don't know a reference off the top of my head. Reviews on entanglement might be a good starting point. $\endgroup$ Jan 15, 2020 at 13:28
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    $\begingroup$ @glS A search for "volume" "separable states" certainly gives some useful resources to start with. $\endgroup$ Jan 15, 2020 at 13:29

4 Answers 4


I'm assuming that you have a finite-dimensional base Hilbert space $\mathcal H_0$ and that you're building your full Hilbert space as $\mathcal H=\mathcal H_0\otimes \mathcal H_0$. In these conditions, the set of separable states has measure zero.

(It gets a bit more complicated if you have $\mathcal H_0^{\otimes 4}$ and you're allowed to split it any way you want among those two factors, and the answer is negative if you're allowed to look for any tensor-product structure in your space, as you can always take one factor along your given $|\psi⟩$.)

Consider, then, a given basis $\{|n⟩:n=1,\ldots,N\}$ for $\mathcal H_0$, which means that any arbitrary state $|\psi⟩\in\mathcal H$ can be written as $$ |\psi⟩=\sum_{n,m} \psi_{nm}|n⟩\otimes|m⟩. $$ If, in particular, $|\psi⟩$ can be written as a tensor product $|\psi⟩=|u⟩\otimes|v⟩$, then you have $$ |\psi⟩ =\left(\sum_n u_n |n⟩\right)\left(\sum_m v_m |m⟩\right) =\sum_{n,m} u_nv_m |n⟩\otimes|m⟩; $$ that is, the coefficient matrix $\psi_{nm}$ has the form $\psi_{nm}=u_n v_m$. This means that this matrix has rank one, which then means that it must have determinant equal to zero. Since the determinant is a continuous polynomial function $\det\colon \mathbb{C}^{N\times N}\to\mathbb C$, its zero set has Borel measure zero inside $\mathbb{C}^{N\times N}$, and therefore correspondingly inside $\mathcal H$.

This means, finally, that if you choose a random vector $|\psi⟩\in\mathcal H$ using a probability measure that is absolutely continuous with respect to the canonical Borel measure on $\mathcal H\cong\mathbb C^{N\times N}$, then it is almost certainly entangled. As an added bonus from exactly the same argument, such a vector will actually (almost certainly) have a full Schmidt rank.

A bit more intuitively, what this argument is saying is that separable states form a very thin manifold inside the full Hilbert space, and this is caught quite well by the spirit of zeldredge's answer. In particular, to describe an arbitrary separable state, you need $2N-1$ complex parameters ($N$ each for the components of $|u⟩$ and $|v⟩$, minus a shared normalization), so roughly speaking the separable states will form a submanifold of dimension $2N-1$. However, this is embedded inside a much bigger manifold $\mathcal H$ of dimension $N^2$, which requires many more components to describe, so for $N$ bigger than two the separable states are a very thin slice indeed.

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    $\begingroup$ Of course this answer shows even more --- the non-entangled states have positive codimension in the state space, which is considerably stronger than having measure zero. $\endgroup$
    – WillO
    Jul 19, 2016 at 14:26

No, in fact, "most states" are entangled. (This is meant to be a heuristic; I freely admit this is probably a sticky thing to get into formally as far as randomly picking a state.) My intuition/reason for saying this is that there exist approximation methods which work by restricting themselves to a low-entanglement subspace, such as Matrix Product States (MPS) or Projected Entangled Pair State (PEPS) which can be used to efficiently represent the ground states and dynamics of certain Hamiltonians.

Here's another, more concrete example. Suppose we have a string of qubits. The product states can all be specified with by storing two numbers per qubit (the Bloch angles or the $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ amplitudes). Therefore for $n$ qubits, we can store $2n$ numbers which capture any product state. However, the full Hilbert space has $2^{n}$ states, which means a general (non-product) state requires specification by $2^n$ different complex numbers. Obviously there are infinite numbers of each since the amplitudes are continuous, but I think it should be clear that the latter case admits many more states, especially if you discretize the space.

The reason why entanglement is a resource is not that entangled states are rare; rather it's that we want to restrict entanglement to be solely between a few systems of interest, whereas most states rapidly entangle with their environment.

  • $\begingroup$ Most states are indeed entangled (at least by the standard measure). On the other hand, most states are still much more entangled than MPS states (full Schmidt rank vs a rather limited rank). $\endgroup$ Jul 19, 2016 at 14:20
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    $\begingroup$ @EmilioPisanty Right; it's just what my first thought was upon seeing the question. I suppose this could be seen by a more general version of my argument about the cost of parametrizing a state (i.e., the cost to represent higher Schmidt ranks grows exponentially). But I think your answer was complete enough not to bother extending mine greatly. $\endgroup$
    – zeldredge
    Jul 19, 2016 at 14:22

The set of entangled states is open and dense in the space of all states (for a given system). In that sense, almost every state is entangled.

  • $\begingroup$ Hi. Could you include a suitable reference? $\endgroup$ Sep 19, 2023 at 6:17
  • $\begingroup$ @TobiasFünke : Just think about the map $V\times V\rightarrow V\otimes V$ and note that the image lies in a quadratic hypersurface. $\endgroup$
    – WillO
    Sep 19, 2023 at 16:07
  • $\begingroup$ Sorry, I don't understand. $\endgroup$ Sep 19, 2023 at 17:15
  • $\begingroup$ @TobiasFünke : Pick a basis $ (v_i) $ for $ V $. Then $ (v_i\otimes v_j) $ is a basis for $ V\otimes V $. For a product state $(\sum a_i v_i)\otimes (\sum b_i v_i)=\sum c_{ij}(v_i\otimes v_j)$, check that $c_{ii}c_{jj}=c_{ij}c_{ji}$. For any $i\neq j$ this is a nontrivial polynomial condition. $\endgroup$
    – WillO
    Sep 19, 2023 at 18:56
  • $\begingroup$ @TobiasFünke : For added clarity if needed,V is the Hilbert space for a single particle and "open" means Zariski open (hence also open in the usual topology). $\endgroup$
    – WillO
    Sep 19, 2023 at 21:20

I know from this talk by M Horodecki that entangled states are more abundant than separable states. Also see this paper on how separable states form a set of measure zero in pure state space.

As for your argument, I think it can also be seen this way.

$|\psi\rangle = |\psi_1\rangle \otimes|\psi_2\rangle$ can be viewed as a constraint on the structure of the state. For an entangled state there is no such constraint on the structure, therefore one might expect more such states than separable states.

One more argument can be that the reduced density matrix of a party from an entangled state is a mixed state. Now, there are more mixed states than pure states.

However, I would advise caution against trusting such arguments too much.


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