मैं कैसे दिखाऊं कि दो-तिहाई राज्य एक उलझा हुआ राज्य है?

19

बेल राज्य एक उलझ राज्य है। लेकिन ऐसा क्यों है? मैं गणितीय रूप से यह कैसे साबित करूं? $|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$

entanglement tensor-product

— nbro
स्रोत

20

परिभाषा

एक दो-राज्य है एक उलझ राज्य यदि और केवल यदि वहाँ नहीं दो एक qubit राज्यों मौजूद और ऐसी है कि $|\psi\rangle \in \mathbb{C}^4$ $|a\rangle = \alpha |0\rangle + \beta |1\rangle \in \mathbb{C}^2$ $|b\rangle = \gamma |0\rangle + \lambda |1\rangle \in \mathbb{C}^2$ , जहां $|a\rangle \otimes |b\rangle = |\psi\rangle$ $\otimes$ को दर्शाता है टेन्सर उत्पाद और । $\alpha, \beta, \gamma, \lambda \in \mathbb{C}$

तो, यह दिखाने के लिए कि बेल राज्य एक उलझ राज्य, हम बस दिखाने के लिए वहाँ मौजूद है कि कोई दो एक qubit राज्यों है हैऔरऐसा है कि। $|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$ $|a\rangle$ $|b\rangle$ $|\Phi^{+}\rangle = |a\rangle \otimes |b\rangle$

प्रमाण

मान लो कि

$\begin{aligned} | Φ^{+} ⟩ & = | a ⟩ \otimes | b ⟩ \\ = (α | 0 ⟩ + β | 1 ⟩) \otimes (γ | 0 ⟩ + λ | 1 ⟩) \end{aligned}$ $\begin{align} |\Phi^{+}\rangle &= |a\rangle \otimes |b\rangle \\ &= \left( \alpha |0\rangle + \beta |1\rangle \right) \otimes \left( \gamma |0\rangle + \lambda |1\rangle \right) \end{align}$
अब हम केवल प्राप्त करने के लिए वितरण योग्य संपत्ति को लागू कर सकते हैं

$\begin{aligned} | Φ^{+} ⟩ & = \dots \\ = (α γ | 00 ⟩ + α λ | 01 ⟩ + β γ | 10 ⟩ + β λ | 11 ⟩) \end{aligned}$ $\begin{align} |\Phi^{+}\rangle &= \cdots \\ &= \left( \alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle \right) \end{align}$
यह बराबर होना चाहिए, है कि, हम पाते हैं चाहिए गुणांक,,और, ऐसा है कि $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$ $\alpha$ $\beta$ $\gamma$ $\lambda$

$\begin{aligned} \frac{1}{\sqrt{2}} (| 00 ⟩ + | 11 ⟩) & = (α γ | 00 ⟩ + α λ | 01 ⟩ + β γ | 10 ⟩ + β λ | 11 ⟩) \end{aligned}$ $\begin{align} \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle ) &= \left( \alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle \right) \end{align}$
Observe that, in the expression $\alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle$ , we want to keep both $|00\rangle$ and $|11\rangle$ . Hence, $\alpha$ and $\gamma$ , which are the coefficients of $|00\rangle$ , cannot be zero; in other words, we must have $\alpha \neq 0$ and $\gamma \neq 0$ . Similarly, $\beta$ and $\lambda$ , which are the complex numbers multiplying $|11\rangle$ cannot be zero, i.e. $\beta \neq 0$ and $\lambda \neq 0$ . So, all complex numbers $\alpha$ , $\beta$ , $\gamma$ and $\lambda$ must be different from zero.

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

$\begin{aligned} \frac{1}{\sqrt{2}} (| 00 ⟩ + | 11 ⟩) & = (α γ | 00 ⟩ + α λ | 01 ⟩ + β γ | 10 ⟩ + β λ | 11 ⟩) \end{aligned}$ $\begin{align} \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle ) &= \left( \alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle \right) \end{align}$
In other words, we are not able to express $|\Phi^{+}\rangle$ as a tensor product of two one-qubit states. Therefore, $|\Phi^{+}\rangle$ 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.

— nbro
स्रोत

2

वाह आपने अपने प्रश्न का उत्तर एक सुंदर, समझने योग्य प्रमाण के साथ दिया। ऐसा कुछ नहीं जो आप रोज देखते हैं। इससे मुझे धन्यवाद देने में मदद मिली।

— युंगगुन

12

| Ψ ⟩ = | ψ ⟩ | ϕ ⟩

$|\Psi\rangle=|\psi\rangle|\phi\rangle$ for arbitrary single qudit states

| ψ ⟩

$|\psi\rangle$ and

| 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 $|\psi\rangle$ and $|\phi\rangle$ , 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 $\rho$ for one of the qudits, i.e. by tracing out the other. The state is separable if and only if $\rho$ has rank 1. Otherwise it is entangled. Mathematically, you can test the rank condition simply by evaluating $\text{Tr}(\rho^2)$ . 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 $|\Psi\rangle=|\psi\rangle|\phi\rangle$ . The reduced density matrix on $A$ is

ρ_{A} = {Tr}_{B} (| Ψ ⟩ ⟨ Ψ |) = | ψ ⟩ ⟨ ψ |,

$\rho_A=\text{Tr}_B(|\Psi\rangle\langle\Psi|)=|\psi\rangle\langle\psi|,$ and

Tr (ρ_{A}^{2}) = Tr (| ψ ⟩ ⟨ ψ | \cdot | 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 $|\Psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ , then

ρ_{A} = {Tr}_{B} (| Ψ ⟩ ⟨ Ψ |) = \frac{1}{2} (| 0 ⟩ ⟨ 0 | + | 1 ⟩ ⟨ 1 |) = \frac{1}{2} I

$\rho_A=\text{Tr}_B(|\Psi\rangle\langle\Psi|)=\frac12\left(|0\rangle\langle 0|+|1\rangle\langle 1|\right)=\frac12\mathbb{I}$ and

Tr (ρ_{A}^{2}) = \frac{1}{4} Tr (I \cdot I) = \frac{1}{2}

$\text{Tr}(\rho_A^2)=\frac14\text{Tr}(\mathbb{I}\cdot\mathbb{I})=\frac12$ 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.

— DaftWullie
स्रोत

+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