2 큐 비트 상태가 얽힌 상태임을 어떻게 표시합니까?

벨 주 | Φ + ⟩ = 1√2 (|00⟩+|11⟩)얽힌 상태이다. 그러나 왜 그런가? 수학적으로 어떻게 증명합니까?$|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$

정의

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2 큐빗 상태 | ψ ⟩ ∈ C 4 인 얽힌 상태 경우 거기에만 있지 이 하나의 큐 비트 상태를 존재 | ⟩ = α | 0 ⟩ + β | 1 ⟩ ∈ C 2 및 | B ⟩ = γ | 0 ⟩ + λ | 1 ⟩ ∈ C 2 등이 | ⟩ ⊗ | B ⟩ = | $|\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$⟩ 여기서$|a\rangle \otimes |b\rangle = |\psi\rangle$$\otimes$ denotes the tensor product and $\alpha, \beta, \gamma, \lambda \in \mathbb{C}$.

So, to show that the Bell state $|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$ is an entangled state, we simply have to show that there exist no two one-qubit states $|a\rangle$ and $|b\rangle$ such that $|\Phi^{+}\rangle = |a\rangle \otimes |b\rangle$.

Proof

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Suppose that

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

We can now simply apply the distributive property to obtain

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

This must be equal to $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$, that is, we must find coefficients $\alpha$, $\beta$, $\gamma$ and $\lambda$, such that

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

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

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

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.

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

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

$$\rho_A=\text{Tr}_B(|\Psi\rangle\langle\Psi|)=|\psi\rangle\langle\psi|,$$ and $$\text{Tr}(\rho_A^2)=\text{Tr}(|\psi\rangle\langle\psi|\cdot |\psi\rangle\langle\psi|)=\text{Tr}(|\psi\rangle\langle\psi|)=1.$$ Thus, we have a separable state.

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

$$\rho_A=\text{Tr}_B(|\Psi\rangle\langle\Psi|)=\frac12\left(|0\rangle\langle 0|+|1\rangle\langle 1|\right)=\frac12\mathbb{I}$$ and $$\text{Tr}(\rho_A^2)=\frac14\text{Tr}(\mathbb{I}\cdot\mathbb{I})=\frac12$$ 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.