Circuit of a very trivial thing

1

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:

Circuit resulting from Trotter

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.

Enrique Segura

Posted 2020-02-22T20:24:22.820

Reputation: 883

Answers

2

When $H = I_1 \otimes I_2$, $e^{-i\gamma I_1 \otimes I_2} = I_1 \otimes I_2$. Therefore, you don't have to apply any gates.

Your CNOT RZ CNOT circuit represents the time evolution of Ising coupling. However, if you want to solve the Ising model problem by QAOA, you need "mixing term" like RX rotation. Rigetti grove's document is very good to understanding QAOA.

gyu-don

Posted 2020-02-22T20:24:22.820

Reputation: 274

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

2

Remember that we can expand $$ e^{i\gamma(Z\otimes Z)}=I\cos\gamma+i(Z\otimes Z)\sin(\gamma). $$ Let's call this $U$. If I calculate $$ U|+\rangle|+\rangle=\cos\gamma|+\rangle|+\rangle+i\sin\gamma|-\rangle|-\rangle. $$ For most vales of $\gamma$, this state is entangled (indeed, for $\gamma=\pi/4$, you essentially have a Bell state in the Hadamard basis). So your statement "given that it's just measuring two non-interacting qubits" is false. The qubits are interacting.

For a point of comparison, perhaps you are thinking of a Hamiltonian $H=\gamma(I\otimes Z+Z\otimes I)$, because in this case, you have $$ e^{iH}=e^{i\gamma Z}\otimes e^{i\gamma Z}, $$ which is maybe what you thought was happening above?

The circuit that you show is what you need for an Ising interaction between two qubits. I'm not sure what "Ansatz" you could be referring to?

DaftWullie

Posted 2020-02-22T20:24:22.820

Reputation: 35 722

Thanks for replying. To clarify, I stated in regards with the simple case, $I \otimes I$ , not for $Z \otimes Z$.

As regard for the "ansatz", I think I might be using the wrong term. I am trying to find the lowest eigenvalue of a hamiltonian that I decomposed into a sum of Pauli products. One of them is this ZZ coupling. At first I applied SGD to the circuit above, but I do not think this is the right approach. – Enrique Segura – 2020-02-24T17:20:37.107

Oh, I see. Then your $I\otimes I$ is trivial. It only adds a phase $e^{-i\gamma}$ to the overall state, and that is irrelevant because it's a global phase. – DaftWullie – 2020-02-25T15:34:30.150

Thank you! I am wondering: why cannot I use this circuit to find the lowest eigenvalue? – Enrique Segura – 2020-02-25T16:26:41.840

Let $|\lambda_i\rangle$ be the eigenvectors of $H$. Any state $|\psi\rangle=\sum\alpha_i|\lambda_i\rangle$. Then, under the evolution, $e^{-iHt}|\psi\rangle=\sum_i\alpha_ie^{-i\lambda_it}|\lambda_i\rangle$: the weight in a given eigenvector never changes, so you cannot enhance the weight of the ground state. You probably need a combination of phase estimation and amplitude amplification to help produce the ground state. – DaftWullie – 2020-02-26T07:58:25.133

I see! Thank you! – Enrique Segura – 2020-02-26T08:04:48.517