1
$\begingroup$

I am using Qiskit.

I made two quantum circuits to generate $\cos\theta|00\rangle + \sin\theta|00\rangle$ for VQE calculations as follows. enter image description here

enter image description here

Apparently, the first one which has fewer CNOT gates is better ansatz for calculations. However, I would like to evaluate the efficiency of these circuits qualitatively.

Run time or something?

Thank you.

$\endgroup$
1
$\begingroup$

If you have one parameter (one $\theta$) for both circuits then I think the first one is better... they are doing the same job, but the second one is creating extra gates. So the first one will be faster and will have fewer errors because there are fewer gates in the first circuit.

But if you are obtaining the second circuit with two parameters (two different $\theta$s) it is harder to decide what ansatz is better. When you have two parameters you may have an infinite number of combinations of two parameters that match the optimal value that VQE should estimate. My intuition is that it will not be a good thing for VQE :). Why? Because imagine you have a parabola and the VQE should find the point in the middle that has minimal value:

enter image description here

Here $x$ corresponds to one $\theta$ and $y$ is the value that VQE obtains depending on $\theta$. For VQE this optimization is like "crossing the street perpendicularly" (the shortest way). You are restricting VQE in 1D space (You are restricting VQE "to cross the street only perpendicularly").

Now Imagine parabolic cylinder in 3D space:

enter image description here

Here $x$ and $y$ correspond to two $\theta$ parameters and $z$ is the value that VQE obtains depending on the two $\theta$s. Now you are giving freedom to VQE optimization to choose from a broad range of "paths" for finding to way to the minimal value. For example, now VQE can choose not the shortest path to the minimum (it may not "cross the street perpendicularly", it may "cross the street at an angle" XD).

Again, this is just my intuition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.