How do I show that a two-qubit state is an entangled state?

Question

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?

Answer 1

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.

Answer 2

Actually an even simpler way is as follows (reusing @nbro's notations). We have:

\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}

Now if we multiply the coefficients of $|00\rangle $ and $|11\rangle $, we get $\alpha \beta \gamma \lambda$.

Also, if we multiply the coefficients of $|01\rangle $ and $ |10\rangle $, we get $\alpha \beta \gamma \lambda$

If these two are equal, then the qubits are in a product state. Else, it automatically means the qubits are in an entangled state.

For the Bell state above, the first product = $1/2$ and the second product = 0. Since they are unequal they are in an entangled state.

I find this is the quickest and easiest way to tell if a pair of qubits are entangled or not.

Answer 3

Take a look at "entanglement of formation". It's a nice metric for how much is a state entangled. the "units" are (a log of) how many bell states required to get into this state. See here for more info, and also here.

Example:

from qiskit.quantum_info import entanglement_of_formation
import numpy as np
li = np.sqrt(0.5) #shortcut for 1/sqrt(2)

Bell state:

bell = [li, 0, 0, li]
entanglement_of_formation(bell)
#returns 0.9999999999999999 (practically 1...)

|00> state:

just00 = [1, 0, 0, 0] 
entanglement_of_formation(just00)
#returns 0.0

Not really Bell (|+0>):

plus0 = [li, li, 0, 0]
entanglement_of_formation(plus0)
#returns 1.6017132519074586e-16 (practically zero...)

Answer 4

There are several ways. Easiest of the is:

1. Find $\rho = |\Psi\rangle\langle\Psi|$
2. Find $\rho^2$
3. Check if $\rho^2 = \rho$. If yes, it is a pure state and entangled. This state $|\Psi\rangle$ can not be written as $|\psi\rangle\otimes|\phi\rangle$ form.

Another way is to find Tr($\rho^2$). If it $1$, then the state is entangled.

Third way would be checking the Schmidt decomposition method.