İki kubit durumun birbirine dolanmış bir durum olduğunu nasıl gösterebilirim?

Bell eyaleti sarmalanmış bir durumdur. Ama neden böyle? Bunu matematiksel olarak nasıl kanıtlayabilirim?$|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$

Tanım

---

İki-kubit bir devlet olan sarmalanmış bir devlet ancak ve orada yalnızca değil iki tek qubit durumlarını var | Bir ⟩ = alfa | 0 ⟩ + p | 1 ⟩ ∈ C 2 ve | b ⟩ = y | 0 ⟩ + Â | 1 ⟩ ∈ Cı 2 bu şekilde | Bir ⟩ ⊗ | 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$ , Burada ⊗ belirtmektedirtensör ürünve a , p , y , A, ∈ Cı .$|a\rangle \otimes |b\rangle = |\psi\rangle$$\otimes$$\alpha, \beta, \gamma, \lambda \in \mathbb{C}$

Yani, Bell devlet göstermek için , basitçe hiçbir iki tek qubit durumlarını orada var olduğunu göstermek için bir dolaşmış devlet var olduğu| Bir⟩ve| b⟩öyle ki| Φ+⟩=| Bir⟩⊗| b⟩.$|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$$|a\rangle$$|b\rangle$$|\Phi^{+}\rangle = |a\rangle \otimes |b\rangle$

Kanıt

---

Farz et ki

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

Artık dağıtım özelliğini yalnızca

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

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.

$$|\Psi\rangle=|\psi\rangle|\phi\rangle$$ for arbitrary single qudit states $|\psi\rangle$ and $|\phi\rangle$. Otherwise, it is entangled.

$|\psi\rangle$ 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 $\text{Tr}(\rho^2)$. Orijinal durum ancak bu değer 1 ise ayrılabilir. Aksi takdirde durum birbirine dolanır.

$|\Psi\rangle=|\psi\rangle|\phi\rangle$$A$

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