The Bell state $|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$ is an entangled state. But why is that the case? How do I mathematically prove that?



A two-qubit state $|\psi\rangle \in \mathbb{C}^4$ is an entangled state if and only if there not exist two one-qubit states $|a\rangle = \alpha |0\rangle + \beta |1\rangle \in \mathbb{C}^2$ and $|b\rangle = \gamma |0\rangle + \lambda |1\rangle \in \mathbb{C}^2$ such that $|a\rangle \otimes |b\rangle = |\psi\rangle$, where $\otimes$ 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 $|\Phi^{+}\rangle = |a\rangle \otimes |b\rangle$.


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.

  • 3
    $\begingroup$ Wow you answered your own question with a beautiful, understandable proof. Not something you see every day. This helped me thank you. $\endgroup$
    – ICW
    Oct 5 '18 at 16:49

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.

  • $\begingroup$ +1 This is a much more elegant method compared to the brute force algorithm. $\endgroup$ Jun 8 '18 at 10:03
  • $\begingroup$ What are $A$ and $B$? Are these just the qudits themselves? $\endgroup$
    – Dohleman
    Jul 10 '19 at 9:08
  • 1
    $\begingroup$ @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. $\endgroup$
    – DaftWullie
    Jul 15 '19 at 7:21

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