# [Physics] Are there more entangled states or non-entangled ones

density-operatorquantum mechanicsquantum-entanglement

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.

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.