Czy splątanie jest przechodnie?

Czy splątanie jest przechodnie w sensie matematycznym?

Mówiąc konkretniej, moje pytanie brzmi:

Rozważ 3 kubity i . Zakładać, że $q_1, q_2$ $q_3$

i są splątane i to $q_1$ $q_2$
i są splątane $q_2$ $q_3$

Zatem, czy i zaplątane $q_1$ $q_3$ ? Jeśli tak, dlaczego? Jeśli nie, to czy istnieje konkretny kontrprzykład?

Jeśli chodzi o moje pojęcie uwikłania:

kubity i są splątane, jeżeli po wykreśleniu qbit i są splątane ( wykasowanie odpowiada pomiarowi $q_1$ $q_2$ $q_3$ $q_1$ $q_2$ $q_3$ i odrzuceniu wyniku). $q_3$
kubity i są splątane, jeśli po wykreśleniu qbit i są splątane. $q_2$ $q_3$ $q_1$ $q_2$ $q_3$
kubity i są splątane, jeśli po wykreśleniu qbit i są splątane. $q_1$ $q_3$ $q_2$ $q_1$ $q_3$

Zachęcamy do skorzystania z jakiegokolwiek innego rozsądnego pojęcia splątania (niekoniecznie powyższego), o ile wyraźnie to określisz.

entanglement

— Piotr
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Czy możesz potwierdzić ostatnie oświadczenie? Po twoim pytaniu oczekiwałem podobnego stwierdzenia, ale z etykietami w innej kolejności (stwierdzenie o splątaniu q1 i q3 po pomiarze q2).

— agaitaarino

@agaitaarino zaktualizowałem część dotyczącą „splątania”, powinna być teraz jaśniejsza ...

— Peter

Traktuję kwadraty łacińskie jako macierz prawdopodobieństwa, w której elementy dowolnej tablicy jednowymiarowej są „splątane”, ponieważ prawdopodobieństwa dla dowolnego wyrażonego elementu są wzajemnie zależne. Po dodaniu wymiarów tablice jednowymiarowe przecinają się prostopadle z innymi tablicami jednowymiarowymi, rozszerzając „splątanie”. (Domyślam się, że jest to tak daleko w chwastach, jak to tylko możliwe: splątanie nietypowych pojęć, ale nie jestem pierwszą osobą, która podniosła pomysł „podobieństwa w duchu” między QT a łacińskimi kwadratami / Sudoku.) Dziękuję za to pytanie!

— DukeZhou

Teraz, gdy wyjaśniłeś, że odrzucasz wynik pomiaru, nie jest to możliwe do zlokalizowania splątanie, o którym myślałem, że mówisz, to bardziej standardowe pojęcie. Lepiej jest mówić o „wykreśleniu” dodatkowego kubitu zamiast pomiaru i odrzucając wynik.

— DaftWullie

@DaftWullie Thanks! Odpowiednio zaktualizowałem pytanie

— Peter

Odpowiedzi:

TL; DR: To zależy od tego, jak zdecydujesz się zmierzyć splątanie na parze kubitów. Jeśli wyśledzisz dodatkowe kubity, wtedy „Nie”. Jeśli mierzysz kubity (ze swobodą wyboru optymalnej podstawy pomiaru), wybierz „Tak”.

$|\Psi\rangle$ $\rho_{AB}=\text{Tr}_C(|\Psi\rangle\langle\Psi|)$

| Ψ ⟩ = \frac{1}{\sqrt{2}} (| 000 ⟩ + | 1 ϕ ϕ ⟩),

$|\Psi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|1\phi\phi\rangle),$

| ϕ ⟩ \neq | 0 ⟩, | 1 ⟩

$|\phi\rangle\neq |0\rangle,|1\rangle$

B

$B$ or qubit

C

$C$ , you'll get the same density matrix both times:

ρ_{A C} = ρ_{A B} = \frac{1}{2} (| 00 ⟩ ⟨ 00 | + | 1 ϕ ⟩ ⟨ 1 ϕ | + | 00 ⟩ ⟨ 1 ϕ | ⟨ ϕ | 0 ⟩ + | 1 ϕ ⟩ ⟨ 00 | ⟨ 0 | ϕ ⟩)

$\rho_{AC}=\rho_{AB}=\frac12\left(|00\rangle\langle 00|+|1\phi\rangle\langle 1\phi|+|00\rangle\langle 1\phi|\langle\phi|0\rangle+|1\phi\rangle\langle 00|\langle0|\phi\rangle\right)$ You can take the partial transpose of this (taking it on the first system is the cleanest):

ρ^{P T} = \frac{1}{2} (| 00 ⟩ ⟨ 00 | + | 1 ϕ ⟩ ⟨ 1 ϕ | + | 10 ⟩ ⟨ 0 ϕ | ⟨ ϕ | 0 ⟩ + | 0 ϕ ⟩ ⟨ 10 | ⟨ 0 | ϕ ⟩)

$\rho^{PT}=\frac12\left(|00\rangle\langle 00|+|1\phi\rangle\langle 1\phi|+|10\rangle\langle 0\phi|\langle\phi|0\rangle+|0\phi\rangle\langle 10|\langle0|\phi\rangle\right)$ Now take the determinant (which is equal to the product of the eigenvalues). You get

det (ρ^{P T}) = - \frac{1}{16} | ⟨ 0 | ϕ ⟩ |^{2} (1 - | ⟨ 0 | ϕ ⟩ |^{2})^{2},

$\text{det}(\rho^{PT})=-\frac{1}{16}|\langle 0|\phi\rangle|^2(1-|\langle 0|\phi\rangle|^2)^2,$ which is negative, so there must be a negative eigenvalue. Thus,

(A B)

$(AB)$ and

(A C)

$(AC)$ are entangled pairs. Meanwhile

ρ_{B C} = \frac{1}{2} (| 00 ⟩ ⟨ 00 | + | ϕ ϕ ⟩ ⟨ ϕ ϕ |) .

$\rho_{BC}=\frac12(|00\rangle\langle 00|+|\phi\phi\rangle\langle\phi\phi |).$ Since this is a valid density matrix, it is non-negative. However, the partial transpose is just equal to itself. So, there are no negative eigenvalues and

(B C)

$(BC)$ is not entangled.

Localizable Entanglement

One might, instead, talk about the localizable entanglement. Before further clarification, this is what I thought the OP was referring to. In this case, instead of tracing out a qubit, one can measure it in a basis of your choice, and calculate the results separately for each measurement outcome. (There is later some averaging process, but that will be irrelevant to us here.) In this case, my response is specifically about pure states, not mixed states.

The key here is that there are different classes of entangled state. For 3 qubits, there are 6 different types of pure state:

a fully separable state
3 types where there is an entangled state between two parties, and a separable state on the third
a W-state
a GHZ state

Any type of quantum state can be converted into one of the standard representatives of each class just by local measurements and classical communication between the parties. Note that the conditions of $(q_1,q_2)$ and $(q_2,q_3)$ being entangled remove the first 4 cases, so we only have to consider the last 2 cases, W-state and GHZ-state. Both representatives are symmetric under exchange of the particles:

| W ⟩ = \frac{1}{\sqrt{3}} (| 001 ⟩ + | 010 ⟩ + | 100 ⟩) | G H Z ⟩ = \frac{1}{\sqrt{2}} (| 000 ⟩ + | 111 ⟩)

$|W\rangle=\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle)\qquad |GHZ\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)$ (i.e. if I swap qubits A and B, I still have the same state). So, these representatives must have the required transitivity properties: If A and B are entangled, then B and C are entangled, as are A and C. In particular, Both of these representatives can be measured in the X basis in order to localize the entanglement. Thus, any pure state that you're given must be such that you can include the measurement to convert it into the standard representative into the measurement for localizing the entanglement, and you're done!

— DaftWullie
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Thanks, this clears up quite a lot already. Could you point me to the "standard" measure of entanglement? I might want to use that explicitly in my question.

— Peter

@Peter: see if the edited version helps even more.

— DaftWullie

Thank you for this answer! Can I ask a naive question on symmetry means in this context "Both representatives are symmetric under exchange of the particles." (I'm very interested in different concepts of symmetry in general.)

— DukeZhou

@DaftWullie: given that your answer appears to be "no, entanglement is not transitive, even on three qubit systems", perhaps you should condense your answer to make this a bit more obvious?

— Niel de Beaudrap

@DukeZhou In this case, all it means is that if I swap particles, I still have the same state:

{SWAP}_{A, B} | Ψ ⟩ = | Ψ ⟩

$\text{SWAP}_{A,B}|\Psi\rangle=|\Psi\rangle$ . So, if I pick out one qubit to be a special qubit (e.g. C), and make some conclusion based on that, it doesn't matter which qubit I picked, because they are all equivalent.

— DaftWullie

This isn't an answer, but instead just some background facts that are important to know about in order to avoid "not even wrong" territory on these types of questions.

"Entanglement" is not all-or-nothing. Just saying "q1 is entangled with q2 and q2 is entangled with q3" is not enough information to determine the answer to questions like "if I measure q3, will q1 still be entangled with q2?". Entanglement gets complicated when dealing with larger systems. You really need to know the specific state, and the measurement, and whether you are permitted to condition on the result of the measurement.

It may be the case that q1,q2,q3 are entangled as a group but if you trace out any one of the qubits then the density matrix of the remaining two describes a mere classically correlated state. (E.g. this happens with GHZ states.)

You should be aware of the monogamy of entanglement. Past a certain threshold, increasing the strength of the entanglement between q1 and q2 must decrease the strength of entanglement between q1 and q3 (and equivalently q2 and q3).

— Craig Gidney
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yay for pointing out the monogamy of entanglement!

— agaitaarino

@agaitaarino which leads to "squashed entanglement" and Von Neumann entropy!

— DukeZhou

I read the following in Freudenthal triple classication of three-qubit entanglement:

"Dür et al. (Three qubits can be entangled in two inequivalent ways) used simple arguments concerning the conservation of ranks of reduced density matriceshere are only six three-qubit equivalence classes:

Null (The trivial zero entanglement orbit corresponding to vanishing states)
Separable (Another zero entanglement orbit for completely factorisable product states)
Biseparable (Three classes of bipartite entanglement: A-BC, B-AC, C-AB)
W (Three-way entangled states that do not maximally violate Bell-type inequalities) and
GHZ (maximally violate Bell-type inequalities)"

which as I understand it the answer to your question is yes: if A and B are entangled and B and C are entangled you necessarily are in one of the three-way entangled states so A and C are also entangled.

— agaitaarino
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