# In VQE, what ansatz would be best used for a maximally entangled $n$-qubit $H$?

What ansatz would be best used for a maximally entangled n qubit H? I came across this question in the Qiskit textbook and I really don't know how I would answer it.

## 1 Answer

If we are given a Hamiltonian $$H$$ that can be represents with $$n$$-qubit, and promised that the ground state of $$H$$ is close to a maximally entangled $$n$$-qubit state, then we can use the fact that the maximal entangled $$n$$-qubit state $$|\psi\rangle$$ can be represent as: $$|\psi \rangle = \dfrac{|0\rangle^{\otimes n} + |1\rangle^{\otimes n} }{\sqrt{2}}$$ which can be created by using the circuit of the form: here I created the state $$|\psi^5 \rangle = \dfrac{|0\rangle^{\otimes 5} + |1\rangle^{\otimes 5} }{\sqrt{2}}$$ but you get the idea of how to create such a state for arbitrary $$n$$ qubit system.

Since we are promised that our state is near this state, you can create your Ansatze (parametrized circuit) based of the circuit above, and you can add a few additional rotation gates if you wish. For instance, something like this: Or if you think the coefficients might be imaginary, then you can add another rotation to your Ansatze, says the $$R_Z$$ rotation and have something like this: You can create these circuit quite easily through the Qiskit's TwoLocal circuit library. Within it, there is an option on selecting the type of "entanglement" you want to have for your Ansatze. There a few options, "Linear" or "Full" or "circular"... so you can use these pre-defined option as well. This way you don't have to worry about setting the CNOT gates by yourself.

Note that if your state is indeed the maximal entangled state $$|\psi^5 \rangle = \dfrac{|0\rangle^{\otimes 5} + |1\rangle^{\otimes 5} }{\sqrt{2}}$$ then all the angles will be $$0$$ or very close to it, except the first angle $$a$$, which will be $$\pi/2$$.