# What is a simple example of "weak entanglement" in a two-qubit system?

Bell states produce maximal entanglement between two qubits. On the other hand, two unentangled qubits provide no (i.e. minimal) entanglement at all.

However, I haven't seen any example of a weak entanglement and how it can be prepared.

The state $$|\psi\rangle = \dfrac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}}$$ is not separable/untangled since a pure state $$|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$$ is unentangled if and only if $$ad - bc = 0$$. But it is not a maximal entangled state either as @AdamZalcman pointed out in the comment as the Schmidt decomposition of this state do not have two equal Schmidt coefficient of $$1/\sqrt{2}$$. In fact, here the Schmidt decomposition of this state $$|\psi\rangle = \dfrac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}}$$ is:

$$|\psi\rangle = \sqrt{\dfrac{3 + \sqrt{5}}{6} }|u_1\rangle|v_1\rangle + \sqrt{\dfrac{3 - \sqrt{5}}{6} }|u_2\rangle|v_2\rangle$$

where $$|u_1 \rangle , |u_2\rangle$$ are the eigenvectors of the reduced density operator $$\rho^A = \dfrac{1}{3}\begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}$$ correspond to the eigenvalues $$\dfrac{3 + \sqrt{5}}{6}$$ and $$\dfrac{3 - \sqrt{5}}{6}$$ respectively. Similarly, $$|v_1 \rangle , |v_2\rangle$$ are the eigenvectors of the reduced density operator $$\rho^B = \dfrac{1}{3}\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}$$ correspond to the eigenvalues $$\dfrac{3 + \sqrt{5}}{6}$$ and $$\dfrac{3 - \sqrt{5}}{6}$$ respectively. Therefore, it is not a maximal entangled state.

You can prepare such state using this circuit in Quirk, which was also shown in Martin Vesely's answer here.

• +1 This is a nice example of a non-maximally-entangled state and a useful general rule for quickly checking for entanglement. Minor comment: A maximally entangled state of two qubits is one with two equal Schmidt coefficients (both $\frac{1}{\sqrt2}$). Therefore, Bell states aren't the only maximally entangled states of two qubits, e.g. $(|00\rangle+i|11\rangle)/\sqrt2$ is also maximally entangled. Dec 16, 2021 at 17:57
• @AdamZalcman Thanks for pointing that out. I edited my answer to incorporate in your very helpful (as always) comment. Dec 16, 2021 at 18:51

Another standard example are Werner states. These can be written as $$W_p\equiv p \frac{I}{4} + (1-p) \frac{|\Psi^-\rangle\!\langle\Psi^-|}{2},\qquad p\in[0,1],$$ and you can tune the entanglement by changing $$p$$. You can prove that the $$W_p$$ is entangled for $$p<2/3$$, and you go from a maximally entangled state at $$p=0$$ to a totally mixed state at $$p=1$$.