エンタングルメントは推移的ですか？

エンタングルメントは、数学的な意味で推移的ですか？

より具体的には、私の質問はこれです：

3つのキュービット $q_1, q_2$ 、考え $q_3$ ます。と仮定する

とが絡まっていて、 $q_1$ $q_2$
とが絡み合っている $q_2$ $q_3$

その後、あるともつれ $q_1$ $q_3$ ？もしそうなら、なぜですか？そうでない場合、具体的な反例はありますか？

私のもつれの概念について：

量子ビット及びアウトトレースした後ならば、絡み合っ、qbitsの及び（アウトトレース絡み合っている測定に相当すると結果を廃棄します）。 $q_1$ $q_2$ $q_3$ $q_1$ $q_2$ $q_3$ $q_3$
量子ビット及びアウトトレースした後ならば、交絡され、qbitsの及び絡み合っています。 $q_2$ $q_3$ $q_1$ $q_2$ $q_3$
量子ビットとアウトトレースした後ならば、交絡され、qbitsの及び絡み合っています。 $q_1$ $q_3$ $q_2$ $q_1$ $q_3$

あなたがその概念を明確に述べている限り、絡み合いの他の合理的な概念（必ずしも上記のものではない）を自由に使用してください。

entanglement

— ピーター
ソース

最後の声明を確認できますか？あなたの質問の後、私は同様のステートメントを期待していましたが、ラベルの順序は異なります（q2を測定した後のq1とq3の絡み合いに関するステートメント）。

— agaitaarino

@agaitaarino私は...それは今より明確にする必要があり、「エンタングルメント」の一部を更新しました

— ピーター

私はラテン方陣を、任意の1次元配列の要素が「絡み合っている」確率行列と見なしてきました。つまり、特定の表現された要素の確率は相互に依存しています。次元を追加すると、これらの1次元配列は他の1次元配列と直交し、「絡み合い」が広がります。（私の推測では1つが再取得することができ、これは限り雑草で出程度である：非定型概念の絡み合いが、私はQTとラテンの正方形/独間のアイデアいくつかの「精神の類似性」を高めるために最初の人ではないです。）ありがとうございましたこの質問をあなたに！

— 公爵

測定結果を破棄していることを明確にしたので、これはあなたが話していると思っていたローカライズ可能なエンタングルメントではなく、より標準的な概念です。結果を破棄します。

— -DaftWullie

@DaftWullieありがとう！それに応じて質問を更新しました

— ピーター

回答:

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".

しましょうば我々はAとBがもつれていると言うだろうA、BおよびCでラベルされた3つの量子ビットの純粋な量子状態、ことの作用で正ではありません部分転置マップ。これは、2キュービットシステムでエンタングルメントを検出するための必要十分条件です。部分トレース形式は、任意の基準でキュービットCを測定し、結果を破棄することと同等です。 $|\Psi\rangle$ $\rho_{AB}=\text{Tr}_C(|\Psi\rangle\langle\Psi|)$

絡み合いが推移的で はないことを示す反例のクラスがあります

| 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 ⟩

$|\phi\rangle\neq |0\rangle,|1\rangle$ . 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
ソース

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
ソース

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
ソース