Запутанность транзитивна?

Является ли запутанность транзитивной , в математическом смысле?

Более конкретно, мой вопрос заключается в следующем:

Рассмотрим 3 кубита и . Предположить, что $q_1, q_2$ $q_3$

$q_1$ и запутаны, и это $q_2$
$q_2$ and $q_3$ are entangled

Then, are $q_1$ and $q_3$ entangled? If so, why? If not, is there a concrete counterexample?

On my notion of entanglement:

qubits $q_1$ and $q_2$ are entangled, if after tracing out $q_3$ , the qbits $q_1$ and $q_2$ are entangled (tracing out $q_3$ corresponds to measuring $q_3$ and discarding the result).
qubits $q_2$ and $q_3$ are entangled, if after tracing out $q_1$ , the qbits $q_2$ and $q_3$ are entangled.
qubits $q_1$ and $q_3$ are entangled, if after tracing out $q_2$ , the qbits $q_1$ and $q_3$ are entangled.

Feel free to use any other reasonable notion of entanglement (not necessarily the one above), as long as you clearly state that notion.

entanglement

— Peter
источник

Can you confirm the last statement? After your question, I expected a similar statement but with the labels in a different order (a statement on the entanglement of q1 and q3 after measuring q2).

— agaitaarino

@agaitaarino i have updated the part on "entanglement", it should be clearer now...

— Peter

I've been regarding Latin squares as a probability matrix in which the elements for any one dimensional array are "entangled", in that the probabilities for any given expressed element are interdependent. When you add dimensions, those one dimensional arrays orthogonally intersect with other one dimensional arrays, extending the "entanglement". (My guess is this is about as far out in the weeds as one can get re: atypical notions entanglement, but I am not the first person to raise the idea some "similarities in spirit" between QT and Latin squares/Sudoku.) Thank you for this question!

— DukeZhou

Now that you have clarified that you are discarding the measurement result, this is not the localizable entanglement that I thought you were talking about, it is the more standard notion.. It's better to talk about "tracing out" the extra qubit instead of measuring and discarding the result.

— DaftWullie

@DaftWullie Thanks! I have updated the question accordingly

— Peter

Ответы:

TL;DR: It depends on how you choose to measure entanglement on a pair of qubits. If you trace out the extra qubits, then "No". If you measure the qubits (with the freedom to choose the optimal measurement basis), then "Yes".

Let $|\Psi\rangle$ be a pure quantum state of 3 qubits, labelled A, B and C. We will say that A and B are entangled if $\rho_{AB}=\text{Tr}_C(|\Psi\rangle\langle\Psi|)$ is not positive under the action of the partial transpose map. This is a necessary and sufficient condition for detecting entanglement in a two-qubit system. The partial trace formalism is equivalent to measuring qubit C in an arbitrary basis and discarding the result.

There's a class of counter-examples that show that entanglement is not transitive, of the form

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

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

| ϕ ⟩ \neq | 0 ⟩, | 1 ⟩

$|\phi\rangle\neq |0\rangle,|1\rangle$ . If you trace out qubit

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
источник

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
источник

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
источник