How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

Question

Suppose that two spin-1/2 are in the state:

$$ |\psi \rangle = \frac{1}{\sqrt{2}} |+\rangle|+\rangle + a|+\rangle|x+\rangle + b|-\rangle|-\rangle $$

and we want to find values for a & b such that the state is entangled and separable.

So I started with rewriting state $|x+\rangle$ in z-representation:

$$ |\psi \rangle = \frac{1}{\sqrt{2}} |+\rangle|+\rangle + a|+\rangle \Bigl( \frac{1}{\sqrt{2}} \bigl( |+\rangle + |-\rangle \bigr) \Bigr) + b|-\rangle|-\rangle =$$ $$ = \frac{a+1}{\sqrt{2}} |+\rangle|+\rangle + \frac{a}{\sqrt{2}}|+\rangle|-\rangle + b|-\rangle|-\rangle $$

and I thought that we can get an entangled(bell) state by letting a=0:

$$ |\psi \rangle = \frac{1}{\sqrt{2}} |+\rangle|+\rangle + \frac{1}{\sqrt{2}}|-\rangle|-\rangle $$

But I can't directly think of an easy way to get a separable state! What's the easiest way to think about this? All advice appreciated

Answer 1

A pure bipartite state $$\left|\psi\right> = \sum_{i = 1}^{d_A} \sum_{j = 1}^{d_B} c_{ij} \left|i\right>\!\left|j\right> \in H_A \otimes H_B$$ is separable if and only if the matrix $(c_{ij})$ has rank $1$ (because $\left|\psi\right> = \left|\alpha\right>\!\left|\beta\right>$ means that $c_{ij} = a_i b_j$ where $a_i$ and $b_j$ are coefficients of the superpositions $\left|\alpha\right> = \sum_{i = 1}^{d_A} a_i \left|i\right>$ and $\left|\beta\right> = \sum_{j = 1}^{d_B} b_j \left|j\right>$).

For two qubits, we need to write down the $2\times 2$ matrix $(c_{ij})$ from the expression $$\left|\psi\right> = c_{00} \left|+\right>\!\left|+\right> + c_{01} \left|+\right>\!\left|-\right> + c_{10} \left|-\right>\!\left|+\right> + c_{11} \left|-\right>\!\left|-\right>$$ and compute the determinant to see when the rank is equal to 1.

Answer 2

The best way to quantify the entanglement for a 2-qubit state is something called the "concurrence"; it goes to zero for a separable state and 1 for a maximally entangled state (like a Bell state). Anything in between is a partially entangled state.

For the general two-qubit pure state

$$|\psi\rangle = A|++\rangle+B|+-\rangle+C|-+\rangle+D|--\rangle$$

the concurrence is 2|AD-BC|. There are generalizations of this to mixed states and to larger Hilbert spaces.

Plugging your problem into this format will reveal that if you want the state to be separable, then |AD-BC|=0, and this will fix your value of "b" to be zero.

Here is a reference which discusses both this and other measures of entanglement.