Prove whether the state $|0\rangle\otimes|0\rangle+|1\rangle\otimes|+\rangle$ is entangled

Question

I'm trying to figure out whether or not the following state is entangling:

$$|Ψ⟩ = 1/\sqrt2 (|0⟩ ⊗ |0⟩ + |1⟩ ⊗ |+⟩)$$

Expanding it out I get:

$$1/\sqrt2 (|0⟩ ⊗ |0⟩ + 1/\sqrt 2(|1⟩ ⊗ |1⟩ + |1⟩ ⊗ |0⟩))$$

Now I know that a state is a product state if for any pair of states $|ψ_1⟩, |ψ_2⟩$ we have

$$|ψ′_1⟩ := \langle ψ_1| ⊗ I)|Ψ \rangle, \qquad\text{and}\qquad |ψ′_2 \rangle := (\langle ψ_2| ⊗ I)|Ψ \rangle.$$

I tried applying this to the expanded form to get a contradiction but it didn't lead anywhere

Answer 1

For a two-qubit state, there's a very straightforward method. If the state is separable, that means you can write it as $$ |\psi\rangle\otimes|\phi\rangle=\alpha|0\rangle\otimes|\phi\rangle+\beta|1\rangle\otimes|\phi\rangle. $$ So, if you look at your state, you look at the $|0\rangle\otimes |0\rangle$ term and say "$|\phi\rangle$ must be $|0\rangle$". But that is inconsistent with the $|1\rangle\otimes |+\rangle$ term. So, the state is not separable, so it is entangled.

If you have a more complex state, I suggest another method. Write out the coefficients of the state in a matrix. In this case, $$ C=\begin{bmatrix} \frac{1}{\sqrt{2}} & 0 \\ \frac12 & \frac12 \end{bmatrix} $$ Now if you calculate $\text{Tr}(CC^\dagger CC^\dagger)$ and you get the answer 1, the state is separable. Otherwise, it's entangled. (Why? $CC^\dagger$ is the same as the reduced density matrix on the first qubit, so you're measuring the purity of the first qubit. If it's pure, the state is separable. If it's mixed, the overall state is entangled.)

Answer 2

One way to check if your state is entangled or not is PPT criterion or with method related to Schmidt decomposition(Nielsen and Chuang's book, chap2.5).

Another way for pure state is that $\rho_{AB}=\rho_A\otimes\rho_B$ iff $\rho _{AB}$ is product state(also separable state for pure state), where $\rho_i$ is the reduced state.

In your case, we have $$\rho _{AB}=\left( \begin{matrix} \frac{1}{2}& 0& \frac{1}{2\sqrt{2}}& \frac{1}{2\sqrt{2}}\\ 0& 0& 0& 0\\ \frac{1}{2\sqrt{2}}& 0& \frac{1}{4}& \frac{1}{4}\\ \frac{1}{2\sqrt{2}}& 0& \frac{1}{4}& \frac{1}{4}\\ \end{matrix} \right) $$ and $$\rho _A=\left( \begin{matrix} \frac{1}{2}& \frac{1}{2\sqrt{2}}\\ \frac{1}{2\sqrt{2}}& \frac{1}{2}\\ \end{matrix} \right) ,\rho _B=\left( \begin{matrix} \frac{3}{4}& \frac{1}{4}\\ \frac{1}{4}& \frac{1}{4}\\ \end{matrix} \right) .$$ Then it's very easy to see that $\rho _{AB}\ne \rho _A\otimes \rho _B$,hence entangled.