Introductory textbooks I've read define entanglement as when your product state cannot be factored into the tensor product of individual quantum states. But consider a three-qbit system:

$$C_{2,0}H_2|000\rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \\ 0 \end{bmatrix}.$$

In this system, the most- and least- significant qubits are entangled. However, a naive "cannot factor" analysis would also lead you to believe the middle qubit is entangled with the rest. The definition of entanglement must thus include permutation of qubit order in some way. So, what is this definition?


2 Answers 2


There is far more structure in entanglement than your simplistic definition seems to be implying. Yes, you could ask a question

Is there any entanglement in this pure state $|\psi\rangle$ of $n$ qubits?

Which can be answered by whether or not the entire state can be expressed as $$ |\psi\rangle=|\phi_1\rangle|\phi_2\rangle\ldots|\phi_n\rangle $$ but as you are rightly realising, you can ask far more structured questions of

is qubit $i$ entangled?

or even

is qubit $i$ entangled with qubit $j$?


What sort of entanglement does qubit $i$ share with different qubits, and how much?

The first is answered in much the same way as your original strategy: if it can be written as $|\phi\rangle_i\otimes|\psi\rangle_{1,2,\ldots,n\setminus i}$, then qubit $i$ is not entangled with anything. In your example, qubit 2 is not entangled, but qubits 1 and 3 are.

The second is far more subtle, and perhaps requires a more precise definition of entanglement. Should we say that qubits 1 and 2 in the state $(|000\rangle+|111\rangle)/\sqrt{2}$ are entangled? If you look at just those two qubits, they are in a maximally mixed state, i.e. not entangled, but they are definitely part or an entangled state that cannot be localised to any pair of qubits. Hence the question about the type of entanglement.

Ultimately, this really comes down to a question of why are you interested in entanglement. Usually, it's to do something with the entangled state, so you have a measure of a given state's usefulness (and that is often some form of entanglement measure), and you can ask questions with respect to that particular measure.


One definition for whether a state $\psi \in V_1 \otimes \cdots V_n$ for $n$ qudits with various values of $d_i = \text{dim} V_i$:

$\psi$ is not entangled if it can be written as a single summand $\psi = v_1 \otimes \cdots v_n$. Conversely it is entangled if it requires more than 1 summand.

Let $\sigma \in S_n$ be a permutation and $R_\sigma$ be the corresponding operator $V_1 \otimes \cdots V_n \to V_{\sigma (1)} \otimes \cdots V_{\sigma (n)}$. If we apply $R_\sigma$ to $\psi$ and it was not entangled then we could write it as $v_{\sigma (1)} \otimes \cdots v_{\sigma (n)}$. Conversely if $R_{\sigma} \psi$ is not entangled then we get $\psi$ is not either. That means this notion of entanglement does not depend on the ordering.

One problem with this is that it is very sensitive. Consider the state $|00\rangle + \epsilon | 11 \rangle$ for small $\epsilon$.

For further information see the notion of tensor rank and border rank. Substitute $\mathbb{C}$ for the field $F$ everywhere. The statistics applications more use the $\mathbb{R}$ case. The border rank is the phenomenon underlying the W-state.

Another thing one can do is do redefinition using any set partition of $n$ into $k$ parts. For example, if $n=4$ a set partition might be $1,2,4$ and $3$ into $k=2$ parts. Then from this we define $W_1 = V_1 \otimes V_2 \otimes V_4$ and $W_2 = V_3$. Then we get the image of $\psi$ as an element of $W_1 \otimes \cdots W_k$ by an appropriate permutation and taking tensor products. One can then compute the rank in this context instead of the $V$ context. With some permutation, we can get the state to become unentangled. This is when all the entanglement has been absorbed into the set partition.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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