Is there any simple algorithm to determine the entanglement of a tri-partite state?

In particular, what is the proof for entanglement of $ |GHZ\rangle $ and $ |W\rangle $ states?

$ |GHZ\rangle =\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle) $
$ |W\rangle =\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle) $

  • $\begingroup$ Notice that in the first state if you measure one of the particles to be in $0$ or $1$ state then you know that it they are also $0$ or $1$ respectively. For the second case if you measure something to be in $1$ state, then you know for sure that the others are $0$. $\endgroup$
    – Gonenc
    Commented May 16, 2015 at 18:26
  • $\begingroup$ This is my intuition as well, but I seek for a more rigorous proof $\endgroup$
    – Alexander
    Commented May 16, 2015 at 18:29
  • $\begingroup$ What do you mean by "determine the entanglement"? Show that the state is non-separable? The general procedure is: Form the most general separable state and see what restrictions are imposed on its coefficients, then check whether your state obeys these restrictions or not. $\endgroup$
    – ACuriousMind
    Commented May 16, 2015 at 18:44
  • $\begingroup$ I mean to determine wheter the state is entangled or not. The process you suggest is not so simple and not very practical. The most general state is a sum of 64 components (since a most general single state is a superposition of 4 pauli matrices) $\endgroup$
    – Alexander
    Commented May 16, 2015 at 18:59

2 Answers 2


$\newcommand{\ket}[1]{\left| #1 \right>}$A state $ \ket \psi \in H_1 \otimes H_2 \otimes H_3$ is said to be entangled if there exist no coefficients $a_i,b_i,c_i$ such that:

$$\ket \psi = \sum_{ijk} d_{ijk} \ket{e^1_i} \otimes \ket{e^2_j} \otimes \ket{e^3_j} = \sum_i a_i \ket{e^1_i} \otimes \sum_j b_j \ket{e^2_j} \sum_k c_k \ket{e^3_k} \tag{1}$$

where $\ket{e^i_j}$ is the $j$th basis vector of $H_i$.

If there exists such coefficients than the state said to be separable and the state is not entangled.

People tend to condense the notation by defining $\ket{e_1e_2e_3}\equiv \ket{e_1} \otimes \ket{e_2} \otimes \ket{e_3}$ so I shall do so in what is next to come.

Assuming that your states are doublet states, let's start calculating. I'll only do the first one, which is easier to do and leave the second one to you, which is analogous to what I'll do now.

Assume that you can write your state in the separable form (1)

\begin{align} \ket{GHZ}&= \frac{1}{\sqrt{2}} (\ket{000}+\ket{111})\\&\overset{?}{=} \big(a_1 \ket 0 + a_2 \ket 1 \big) \otimes \big(b_1 \ket 0 + b_2 \ket 1 \big) \otimes \big(c_1 \ket 0 + c_2 \ket 1 \big) \\ &= a_1b_1c_1\ket{000}+ a_2b_2c_2 \ket{111} + \text{other linear combinations} \end{align}

Notice that in order to get your state you cannot have any of the $a_i,b_i$ or $c_i$ to be zero, if this would be the case you cannot get either $\ket{000}$ or $\ket{111}$. However these are all the coefficients you have, therefore if none of them zero, then you cannot get any of the other linear combinations to vanish, which is a contradiction to the assumption that you can write your state in the separable form (1). Thus this state is an entangled state.

Note that the intuition that I talked about in my comment above saves you a lot of time and effort but as Ellis Cooper said, "Rigor cleans the window through which intuition shines.", so it is up to you how much rigour you want.

  • $\begingroup$ Note that this answer shows the state is not fully separable. In general there is the possibility that the state is partially separable, which can be disproved in a similar manner. $\endgroup$
    – Mebert
    Commented May 17, 2015 at 14:31

You can calculate the density matrix of the state $$\rho = |\varphi\rangle\langle\varphi|$$ And then, the reduced density matrix for one of the particles taking the partial trace $$ \rho_A = \textrm{tr}_{B,C}(\rho)$$ The state is entangled if and only if the reduced matrix is a mixed state. This can be checked, for example, calculating the von Neumann entropy $$ S = -\textrm{tr} (\rho_A \log \rho_A)$$

  • $\begingroup$ This seems excellent, though too easy. Can you refer to any proof of the "if and only if" statement (for multipartite states)? $\endgroup$
    – Alexander
    Commented May 16, 2015 at 19:16
  • $\begingroup$ @Alexander Sure. See, for example, physics.stackexchange.com/a/70455 $\endgroup$
    – Bosoneando
    Commented May 16, 2015 at 20:48

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