# How many $n$-qubit states are entangled?

My question is a bit broad, but my concern is mainly on understand the ratio between the number of possible linear combinations that can be decomposed in a direct product of states and the number of possible entangled states in that system. The system is a just an $$n$$-qubits set, as an usual vector space, with the assumption that all the states in a certain superposition can be expected as equally probable. As an example to clarify what I mean take:

$$\left| \psi \right\rangle = \frac{\sqrt{2}}{2} (\left| 00 \right\rangle + \left| 11 \right\rangle)$$

I know that the ratio in a 2-qubits system should be 50%, but what about $$n=100$$?

To improve the clarity of the question I would ask, how do you interpret this kind of graph?

Where Product = a state that can be decomposed uniquely by a tensor product. Entangled = it's own defition. Is this a true picture of what's going on in an $$n$$-qubit system or not?

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• As far as I am aware you can not import the physics package (which includes the bra-ket commands) (or any other package, for that matter). However, the commands \rangle and \langle will work as a substitute. – JSdJ Jan 13 at 15:48
• – Nat Jan 14 at 5:35
• I don't understand the question. Are you asking what is the probability of finding an entangled state when sampling random pure $n$-qubit pure states? Or the probability of finding an entangled state when sampling among the $n$-qubit pure balanced states? Or something else? – glS Jan 14 at 19:57
• Thanks JSdJ for your help. – Kamui9610 Jan 14 at 20:15
• then the answer should be that the probability is $1$: almost all states are entangled. See this answer. See also this paper – glS Jan 14 at 20:41

The probability of finding an entangled state while randomly sampling (from the uniform distribution) over pure states is one: almost all pure states are entangled.

This is shown in this answer on physics.SE. This paper might also be of interest.

The gist is that separable states correspond to rank-1 matrices, which have zero measure in the set of all matrices. This is also discussed in this post on math.SE.

Another less rigorous argument could be to observe that separable states are always infinitesimally close to entangled out: given an arbitrary separable pure state $$|\psi\rangle\otimes|\phi\rangle$$, take $$|u\rangle,|v\rangle$$ such that $$|\Psi\rangle\equiv\langle u|\psi\rangle=\langle v|\phi\rangle=0$$. Then, the state $$\frac{1}{\sqrt{1+\epsilon^2}}\left[|\Psi\rangle+\epsilon(|u\rangle\otimes|v\rangle)\right]$$ is entangled for all $$\epsilon>0$$ and can be made to be infinitesimally close to $$|\Psi\rangle$$. This hints towards the set of separable states having empty interior (although technically, having empty interior is not in itself enough to prove having zero measure, see e.g. this blog post).

Even for two qubits, there are infinite number of entangled states:

$$\left| \psi \right\rangle_2 = \alpha \left| 00 \right\rangle + \beta \left| 11 \right\rangle$$

where $$\left| \alpha \right|^2$$ and $$\left| \beta \right|^2$$ can be any probabilities that satisfy $$\left| \alpha \right|^2 + \left| \beta \right|^2 = 1$$. And even this $$\left| \psi \right\rangle_2$$ is not describing all possible entangled states for two qubits.

For $$n=100$$ you can have a possible entangled quantum state:

$$\left| \psi \right\rangle_{100} = \alpha \left| 000...0 \right\rangle + \beta \left| 111...1 \right\rangle$$

and you can choose $$\alpha = \beta = 2/\sqrt{2}$$ or any other combination. A way of constructing this state:

1. Prepare the first qubit in a superposition state : $$\left| \psi \right\rangle_1 = \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle$$
2. Apply 99 CNOTs on the rest of the qubits controlled from the first qubit.
• thank you, but I think that there is a method to simplify this infinity of possible linear combinations because usually we deal with precisely defined entangled states, like triplet states or Bell states that have precise relative phase between the two (of course this is up to some overall phase factor $\gamma$, but we are interested only to relative phase factors in this case). – Kamui9610 Jan 14 at 20:44
• Even if you ignore the global phase you still will have an infinite number of possible entangled states. – Davit Khachatryan Jan 15 at 7:22
• Bell state is maybe the nicest entangled state, but in QC we use also other entangled states. For example in HHL algorithm after QPE subroutine you are obtaining some entangled state depending on the given system of linear equations. And that states are not Bell-like states. – Davit Khachatryan Jan 15 at 7:26
• @Kamui9610 After reading the conversation above and the added parts in your question I understand that my answer wasn't what you were looking for :) – Davit Khachatryan Jan 15 at 7:59