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I am working on quantum entanglement swapping and I have derived that:
$|\psi\rangle=\frac{1}{2}(|0001\rangle+|0010\rangle+|0100\rangle+|1000\rangle)$
Is there any possibility that I can make a graphical review of my result for this state? In other words, is there any procedure to prove that the state is entangled graphically?
In the specific $|\psi \rangle$ there is symmetry on the order of bits, so we must only show that one bit is entangled and it then follows that they all are.
Notice we may write:
$$|\psi \rangle = \frac{1}{2}\Big(|0 \rangle \big(|001 \rangle +|010 \rangle +|100 \rangle \big) + |1 \rangle \big(|000 \rangle \big)\Big)=\frac{\sqrt3 |0\rangle |A \rangle + |1\rangle |B \rangle }{\sqrt2}$$
Where:
$$|A \rangle = \frac{1}{\sqrt3}\big( |001 \rangle +|010 \rangle +|100 \rangle \big) \text{ , } |B \rangle = |000\rangle$$
Are orthonormal states. From here it is easy to show the state is entangled, and for a graphical argument you way consider $| A \rangle$ and $| B \rangle$ to be projections onto a Bloch sphere; since technically they are higher bit systems a Bloch sphere doesn't apply (a single Bloch Sphere cannot accommodate the full Hilbert space of 3 bits), but because of the simplification, we are only interested in 2 eigenstates of the 3 bit Hilbert space namely $| A \rangle$ and $| B \rangle$.
So in fact state may be written as such:
$$|\psi \rangle = \frac{1}{\sqrt2}\big( \sqrt3 |0\rangle |\tilde{0} \rangle + |1\rangle |\tilde{1} \rangle \big)$$
Now either using partial traces or comparing coefficients one can see the first bit is clearly entangled to the the last three. And by the symmetry in the order of bits; all bits in the state are entangled.
