# Comment puis-je montrer qu'un état à deux qubits est un état intriqué?

19

L'État de Bell est un état intriqué. Mais pourquoi est-ce le cas? Comment puis-je prouver mathématiquement cela?$|{\mathrm{\Phi }}^{+}⟩=\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)$$|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$

Réponses:

20

Définition

Un état à deux qubits est un état intriqué si et seulement s'il n'existe deux états d' un qubit et de telle sorte que $|\psi ⟩\in {\mathbb{C}}^{4}$$|\psi\rangle \in \mathbb{C}^4$$|a⟩=\alpha |0⟩+\beta |1⟩\in {\mathbb{C}}^{2}$$|a\rangle = \alpha |0\rangle + \beta |1\rangle \in \mathbb{C}^2$$|b⟩=\gamma |0⟩+\lambda |1⟩\in {\mathbb{C}}^{2}$$|b\rangle = \gamma |0\rangle + \lambda |1\rangle \in \mathbb{C}^2$ , Où$|a⟩\otimes |b⟩=|\psi ⟩$$|a\rangle \otimes |b\rangle = |\psi\rangle$$\otimes$$\otimes$ désigne le produit tensoriel , et .$\alpha ,\beta ,\gamma ,\lambda \in \mathbb{C}$$\alpha, \beta, \gamma, \lambda \in \mathbb{C}$

Donc, pour montrer que l'état de Bell est un état intriqué, nous devons simplement montrer qu'il n'y existe pas deux étatsun qubitettelle que.$|{\mathrm{\Phi }}^{+}⟩=\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)$$|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$$|a⟩$$|a\rangle$$|b⟩$$|b\rangle$$|{\mathrm{\Phi }}^{+}⟩=|a⟩\otimes |b⟩$$|\Phi^{+}\rangle = |a\rangle \otimes |b\rangle$

Preuve

Supposer que

$\begin{array}{rl}|{\mathrm{\Phi }}^{+}⟩& =|a⟩\otimes |b⟩\\ & =\left(\alpha |0⟩+\beta |1⟩\right)\otimes \left(\gamma |0⟩+\lambda |1⟩\right)\end{array}$

Nous pouvons maintenant simplement appliquer la propriété distributive pour obtenir

$\begin{array}{rl}|{\mathrm{\Phi }}^{+}⟩& =\cdots \\ & =\left(\alpha \gamma |00⟩+\alpha \lambda |01⟩+\beta \gamma |10⟩+\beta \lambda |11⟩\right)\end{array}$

Cela doit être égal à $\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)$$\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$, that is, we must find coefficients $\alpha$$\alpha$, $\beta$$\beta$, $\gamma$$\gamma$ and $\lambda$$\lambda$, such that

$\begin{array}{rl}\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)& =\left(\alpha \gamma |00⟩+\alpha \lambda |01⟩+\beta \gamma |10⟩+\beta \lambda |11⟩\right)\end{array}$

Observe that, in the expression $\alpha \gamma |00⟩+\alpha \lambda |01⟩+\beta \gamma |10⟩+\beta \lambda |11⟩$$\alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle$, we want to keep both $|00⟩$$|00\rangle$ and $|11⟩$$|11\rangle$. Hence, $\alpha$$\alpha$ and $\gamma$$\gamma$, which are the coefficients of $|00⟩$$|00\rangle$, cannot be zero; in other words, we must have $\alpha \ne 0$$\alpha \neq 0$ and $\gamma \ne 0$$\gamma \neq 0$. Similarly, $\beta$$\beta$ and $\lambda$$\lambda$, which are the complex numbers multiplying $|11⟩$$|11\rangle$ cannot be zero, i.e. $\beta \ne 0$$\beta \neq 0$ and $\lambda \ne 0$$\lambda \neq 0$. So, all complex numbers $\alpha$$\alpha$, $\beta$$\beta$, $\gamma$$\gamma$ and $\lambda$$\lambda$ must be different from zero.

But, to obtain the Bell state $|{\mathrm{\Phi }}^{+}⟩$$|\Phi^{+}\rangle$, we want to get rid of $|01⟩$$|01\rangle$ and $|10⟩$$|10\rangle$. So, one of the numbers (or both) multiplying $|01⟩$$|01\rangle$ (and $|10⟩$$|10\rangle$) in the expression $\alpha \gamma |00⟩+\alpha \lambda |01⟩+\beta \gamma |10⟩+\beta \lambda |11⟩$$\alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle$, i.e. $\alpha$$\alpha$ and $\lambda$$\lambda$ (and, respectively, $\beta$$\beta$ and $\gamma$$\gamma$), must be equal to zero. But we have just seen that $\alpha$$\alpha$, $\beta$$\beta$, $\gamma$$\gamma$ and $\lambda$$\lambda$ must all be different from zero. So, we cannot find a combination of complex numbers $\alpha$$\alpha$, $\beta$$\beta$, $\gamma$$\gamma$ and $\lambda$$\lambda$ such that

$\begin{array}{rl}\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)& =\left(\alpha \gamma |00⟩+\alpha \lambda |01⟩+\beta \gamma |10⟩+\beta \lambda |11⟩\right)\end{array}$

In other words, we are not able to express $|{\mathrm{\Phi }}^{+}⟩$$|\Phi^{+}\rangle$ as a tensor product of two one-qubit states. Therefore, $|{\mathrm{\Phi }}^{+}⟩$$|\Phi^{+}\rangle$ is a entangled state.

We can perform a similar proof for other Bell states or, in general, if we want to prove that a state is entangled.

2
Wow you answered your own question with a beautiful, understandable proof. Not something you see every day. This helped me thank you.
YungGun

12

A two qudit pure state is separable if and only if it can be written in the form 

$|\mathrm{\Psi }⟩=|\psi ⟩|\varphi ⟩$
for arbitrary single qudit states $$|ψ⟩|ψ⟩|\psi\rangle$$ and $$|ϕ⟩|ϕ⟩|\phi\rangle$$. Otherwise, it is entangled.

To determine if the pure state is entangled, one could try a brute force method of attempting to find satisfying states $$|ψ⟩|ψ⟩|\psi\rangle$$ and $$|ϕ⟩|ϕ⟩|\phi\rangle$$, as in this answer. This is inelegant, and hard work in the general case. A more straightforward way to prove whether this pure state is entangled is the calculate the reduced density matrix $$ρρ\rho$$ for one of the qudits, i.e. by tracing out the other. The state is separable if and only if $$ρρ\rho$$ has rank 1. Otherwise it is entangled. Mathematically, you can test the rank condition simply by evaluating $$Tr(ρ2)Tr(ρ2)\text{Tr}(\rho^2)$$. The original state is separable if and only if this value is 1. Otherwise the state is entangled.

For example, imagine one has a pure separable state $$|Ψ⟩=|ψ⟩|ϕ⟩|Ψ⟩=|ψ⟩|ϕ⟩|\Psi\rangle=|\psi\rangle|\phi\rangle$$. The reduced density matrix on $$AAA$$ is 

${\rho }_{A}={\text{Tr}}_{B}\left(|\mathrm{\Psi }⟩⟨\mathrm{\Psi }|\right)=|\psi ⟩⟨\psi |,$
and $$Tr(ρ2A)=Tr(|ψ⟩⟨ψ|⋅|ψ⟩⟨ψ|)=Tr(|ψ⟩⟨ψ|)=1.Tr(ρA2)=Tr(|ψ⟩⟨ψ|⋅|ψ⟩⟨ψ|)=Tr(|ψ⟩⟨ψ|)=1. \text{Tr}(\rho_A^2)=\text{Tr}(|\psi\rangle\langle\psi|\cdot |\psi\rangle\langle\psi|)=\text{Tr}(|\psi\rangle\langle\psi|)=1.$$ Thus, we have a separable state.

Meanwhile, if we take $$|Ψ⟩=12√(|00⟩+|11⟩)|Ψ⟩=12(|00⟩+|11⟩)|\Psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$, then 

${\rho }_{A}={\text{Tr}}_{B}\left(|\mathrm{\Psi }⟩⟨\mathrm{\Psi }|\right)=\frac{1}{2}\left(|0⟩⟨0|+|1⟩⟨1|\right)=\frac{1}{2}\mathbb{I}$
and $$Tr(ρ2A)=14Tr(I⋅I)=12Tr(ρA2)=14Tr(I⋅I)=12 \text{Tr}(\rho_A^2)=\frac14\text{Tr}(\mathbb{I}\cdot\mathbb{I})=\frac12$$ Since this value is not 1, we have an entangled state.

If you wish to know about detecting entanglement in mixed states (not pure states), this is less straightforward, but for two qubits there is a necessary and sufficient condition for separability: positivity under the partial transpose operation.

+1 This is a much more elegant method compared to the brute force algorithm.
Sanchayan Dutta

What are $A$$A$ and $B$$B$? Are these just the qudits themselves?
Dohleman

@Dohleman Yes, they're just labels for the two parts of the system, one part held by A (Alice), and the other by B (Bob). In this case it's the two qudits.
DaftWullie
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