# Is there any possibility to draw graphs for a maximally entangled state?

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?

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• Determining separability (i.e. if a state is entangled) is NP-hard. Even so, there is potentially some sort of bijection between multipartite qubit states and some sort of graphs. Just note that under any such mapping whatever the equivalent property of separability is, it's unlikely to be nice. Sep 13 at 18:44
• @Condo Thank you for your answer and time. The state is already entangled. The problem is that I am submitting an article, For a better understanding of the common audience, I want to add a graph of the state which visualize is as entangled. Such as on a Bloch Sphere if possible. yesterday

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.

• Thank you very much for the detailed answer. 2 days ago
• I don't really see how this "proves that the state is entangled graphically"
– glS
2 days ago
• @glS yes, and I am looking for some kind of graphical representation which can provide details of entanglement. 2 days ago
• @Attakhan just as a heads up: accepting an answer signals that you think it resolves the question. You can upvote an answer without accepting it if you think it has merit but doesn't fully address the question. That said, I don't think there is any particularly satisfying "graphical way" to do what you ask. The geometry of quantum states, with the exception of single qubits, is extremely complex, and involves many dimensions, making it hard to visualize
– glS
2 days ago
• @glS Thank you for your comment. This became now clear to me that there is no simple route to do this. yesterday