How to tell if a state is entangled?

Question

I've read that to tell if a 2 qubit state is not entangled e.g. $$αγ|00⟩+αλ|01⟩+βγ|10⟩+βλ|11⟩$$ then multiplying the coefficients of $|00\rangle$ and $|11\rangle$ should equal the same as multiplying the coefficients of $|01\rangle$ and $|10\rangle$ i.e. $αβγλ$ is the same in both cases. I'm finding it hard to conceptualize this though. As far as I can understand this means all the basis states must be present in a state for it to not be entangled. But I'm sure that is wrong as then it would be trivial to know if a state is entangled. For example, is the state $$1/\sqrt{2}(|01\rangle + |11\rangle)$$ entangled because it does not contain $|00\rangle$ and $|10\rangle$? Or is it not entangled because $γ$ is not present so $αβγλ = 0$.

Another method I have seen to tell if a state is entangled is to check if it can be decomposed into a product state, so, for example, would it be sufficient to say that $$|01\rangle + |11\rangle$$ can be decomposed to $$(|0\rangle + |1\rangle)(|1\rangle)$$ i.e.$|+\rangle|1\rangle$ hence the state is not entangled? Thanks for any help :)

Answer 1

$\alpha\gamma|00\rangle + \alpha\lambda|01\rangle +\beta\gamma|10\rangle+\beta\lambda|11\rangle = (\alpha|0\rangle+\beta|1\rangle)\otimes(\gamma|0\rangle+\lambda|1\rangle)$ is a separable state, so it's not entangled.

$\frac{1}{\sqrt{2}}(|01\rangle + |11\rangle) = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |1\rangle$, so is also separable and not entangled.

If a state is separable (factorable), then by definition it is not entangled and the two states you factor into end up being completely independent.

Entanglement is a quantum phenomenon where the systems are highly (and non-classically) correlated. For example, the Bell state $|\Phi_+\rangle = \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ is unfactorable and so the two systems cannot be considered separately. This makes sense because measuring $|0\rangle$ (or $|1\rangle$) on one system means you'll always measure $|0\rangle$ (or $|1\rangle$) on the other.