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I am writing to double check that if have a hamiltonian of the form $H = I_1 \otimes I_2$, when I seek to find the unitary, $e^{-i\gamma I_1 \otimes I_2}$, there really is no need to convert this into a circuit given that it's just measuring two non-interacting qubits (despite their tensor product)?

I ask this because what throws me off is the $\alpha$ factor which, in the case of $H = \frac{1}{2} (Z_1 \otimes Z_2)$ it becomes a $U \approx e^{-i\frac{\alpha}{2} Z_1 \otimes Z_2}$ which becomes the following circuit:

Provided this circuit, one question I have is, do I need an Ansatz on top of this unitary, or is this unitary the ansatz I need to represent the Ising coupling?

Thank you.

Hi. Thank you for your feedback. I am wondering, does that also apply if I am trying implement VQE using ansatz (this circuit) ? – Enrique Segura – 2020-02-23T05:52:39.220

This circuit is not enough for VQE ansatz because it can rotate only Z axis. It means, it cannot flip the bits. QAOA ansatz is prefer for VQE for Ising model. – gyu-don – 2020-02-23T08:03:23.547

So if I want to use VQE I would have to have the ansatz first and then the circuit I showed above ? – Enrique Segura – 2020-02-24T05:32:17.853

Yes. RX(θ) or RY(θ) gate for each qubit is enough. And you may repeat RX, CNOT RZ CNOT, RX, CNOT RZ CNOT, ... – gyu-don – 2020-02-25T10:16:59.377

do you know of a good paper that illustrates this? – Enrique Segura – 2020-02-25T16:32:43.803

https://arxiv.org/pdf/1411.4028.pdf – gyu-don – 2020-02-26T07:59:37.030

funny! I meant a beginner paper! – Enrique Segura – 2020-02-26T08:04:13.527

Rigetti's document is good for beginners. https://grove-docs.readthedocs.io/en/latest/qaoa.html

– gyu-don – 2020-02-26T08:28:53.067