Circuit of a very trivial thing

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.

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.

• 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 Feb 23 at 5:52
• 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 Feb 23 at 8:03
• So if I want to use VQE I would have to have the ansatz first and then the circuit I showed above ? – Enrique Segura Feb 24 at 5:32
• 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 Feb 25 at 10:16
• do you know of a good paper that illustrates this? – Enrique Segura Feb 25 at 16:32

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?

• 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 Feb 24 at 17:20
• 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 Feb 25 at 15:34
• Thank you! I am wondering: why cannot I use this circuit to find the lowest eigenvalue? – Enrique Segura Feb 25 at 16:26
• 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 Feb 26 at 7:58
• I see! Thank you! – Enrique Segura Feb 26 at 8:04