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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.

glS
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sitems
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2 Answers2

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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.

KAJ226
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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 $p=1$.

glS
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