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 :)

  • $\begingroup$ I don't quite understand the question. Separable (and thus product) pure two-qubit states are all and only those whose coefficients have the structure in your first equation, yes. Which means that yes, if you know the coefficients of the state, in such cases, it is trivial to figure out whether the state is separable or not. Checking for that structure of the coefficients is the same as checking whether the state "can be decomposed into a product state". One is the condition for the other. See also physics.stackexchange.com/a/643655/58382 for relations between separable/product states $\endgroup$
    – glS
    Jun 2, 2022 at 17:16

1 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.


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