7
$\begingroup$

Consider some state $|\psi\rangle$ of $n$ qubits. One can take any subsystem $A$ and compute its density matrix $\rho_A =Tr_{B} |\psi\rangle \langle\psi|$. The entanglement between subsystem $A$ and the rest of the system $B$ is quantified for example by von Nemumann entropy associated with the density matrix. For example, the subsystem $A$ can be first $k$ qubits and subsytem $B$ the remaining $n-k$.

Now, is it possible to associate in a similar way some measure of entaglment between two particular subsystems that do not sum up to the total system? Say, is it in any way reasonable to ask what is entanglement between qubit 1 and qubit 2? Intuitively, if the state of the system is something like $\sqrt{2}|\psi\rangle=|000000\dots\rangle+|110000\dots\rangle$ then I should be able to say that the first two qubits are maxiamally entangled with each other but not at all with the remaining qubits. But can one do this in general?

$\endgroup$
5
$\begingroup$

Of course you can. Take any entanglement measure that can be applied to a system whose overall description is a density matrix, and you can apply that to the density matrix describing your subsystem. (So, you wouldn't use the von Neumann entropy of one qubit because that assumes the overall state is pure.) A particularly good option for a pair of qubits is the partial transpose. You might also look at the concurrence.

$\endgroup$
3
  • $\begingroup$ Great, thank you! A follow-up question. Can I somehow relate the entanglement between qubit 1 and 2, 1 and 3,... 1 and n to the entanglement of qubit 1 with all of them (von Neumann entropy of qubit 1)? $\endgroup$ Jun 25 at 7:32
  • 1
    $\begingroup$ There are bounds (the "monogamy of entanglement") but not a strict relation. Yes, if 1 and 2 are maximally entangled, 1 is maximally entangled with (2 and 3). But if 1 is partially entangled with 2 and partially entangled with 3, there's a range of entanglement that 1 could have with (2 and 3). $\endgroup$
    – DaftWullie
    Jun 25 at 8:53
  • 1
    $\begingroup$ Take, for example $a|000\rangle+b|111\rangle$. You can freely choose how much entanglement there is between the $1|23$ partition by selecting $a$. But the reduced density matrices are $\rho_{12}=\rho_{13}=|a|^2|00\rangle\langle 00|+|b|^2|11\rangle\langle 11|$ and are therefore not at all entangled. Thus, the same entanglement properties of the pairs of qubits can lead to arbitrary entanglement of the composite system. $\endgroup$
    – DaftWullie
    Jun 25 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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