2
$\begingroup$

Given a state of n qubits, how can I find if it is entangled? Specifically, I want to find all qubits (if any) that can be separated from a state. I have seen many answers for two-qubit states but how do I find this for more qubits?

$\endgroup$
0

3 Answers 3

2
$\begingroup$

If you have an $n$-qubit pure state $|\Psi\rangle$, then you can easily proceed to find out of an individual qubit is separable from the rest. Pick a particular $i$ and compute $$ \rho_i=\text{Tr}_{1,\ldots,N\setminus i}|\Psi\rangle\langle\Psi|. $$ If $\rho_i$ is rank 1, that qubit is separable. This is just the same as you would do in the two-qubit case.

If there is ever a qubit which is not separable, your overall state is entangled.

$\endgroup$
1
$\begingroup$

Given a state of n qubits, how can I find if it is entangled?

This is a hard problem in general, so what is commonly used is a technique known as 'witnessing' entanglement. Depending on how much fine-grained information you want you may use different techniques.

In the case you have the classical information about the state and just want to certify some entanglement you may use monotone information. Here is a simple one to calculate, valid only for entangled pure states:

$$ \mathcal{E}\left(\vert \psi \rangle \right) := 2\left(1 - \frac{1}{m}\sum_{k=0}^{m-1}\text{Tr} \left(\rho_k^2\right)\right),$$

Any value $$\mathcal{E}(\vert \psi \rangle)>0$$ witnesses the fact that the state is entangled. Above, $\rho_k$ is constructing by tracing all qubits expect the $k$-th one. You then make this process, find the purity of this partial state, and sum over all $m$ qubits defining your system.

Specifically, I want to find all qubits (if any) that can be separated from a state.

I am not sure I have followed this part, but it seems that you are now interested into the complete characterization of n-qubit entanglement classes? This is extremely hard, and only known for some small number of qubits. Also, you may need more advanced techniques for analysing this problem.

$\endgroup$
-1
$\begingroup$

Entanglement means only that the correlation between qubits is not completely independent (like the outcome of 3 dices being rolled). So for instance in the 3 qubits case (unnormalized),

$$ |000\rangle = 1$$ $$ |001\rangle = 1$$ $$ |010\rangle = 1$$ $$ |011\rangle = 1$$ $$ |100\rangle = 1$$ $$ |101\rangle = 1$$ $$ |110\rangle = 1$$ $$ |111\rangle = 1$$

is an non-entangled/uncorrelated state, whereas

$$ |000\rangle = 1$$ $$ |001\rangle = 0$$ $$ |010\rangle = 0$$ $$ |011\rangle = 1$$ $$ |100\rangle = 1$$ $$ |101\rangle = 1$$ $$ |110\rangle = 1$$ $$ |111\rangle = 1$$

is an entangled state, simply because when $q_1 = |0\rangle$, there is this weird correlation between $q_2,q_3$ being always synchronized in the same state if measured. We call this correlation between different qubits an "entanglement".

A hunch is that if we take a messy non-symmetric-looking states density like

$$ |000\rangle = 0.34$$ $$ |001\rangle = 0.66$$ $$ |010\rangle = 0.91$$ $$ |011\rangle = 0.23$$ $$ |100\rangle = 0.55$$ $$ |101\rangle = 0.92$$ $$ |110\rangle = 0.18$$ $$ |111\rangle = 0.77$$

there is probably some correlation/entanglement hidden somewhere in there because of the non-symmetry...

$\endgroup$
2
  • 1
    $\begingroup$ "entanglement" and "correlation" are related concepts but they are not the same thing. You can have highly correlated qubits without any entanglement. $\endgroup$
    – glS
    Commented Jul 10, 2023 at 8:14
  • $\begingroup$ @glS thank you for pointing this out. The distinction is still not yet fully clear to me, I guess, say for 2 qubits $|00\rangle = a, |01\rangle = b, |10\rangle = c, |11\rangle = d$, could you give example numerical values $a,b,c,d$ where there is correlation, but not entanglement, please? $\endgroup$
    – James
    Commented Jul 10, 2023 at 9:23

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