2
$\begingroup$

I have a Hamiltonian $$H = aH_1 + bH_2\,,\tag{1}$$ with $a$ and $b$ being coefficients, and I want to apply each term/evolution separately, then recombine the measurement results to get the result for the original.

Is there any way to do that, like classical post-processing? Or am I breaking the laws of physics?

Please suggest some useful resources so I can read and understand this better.

$\endgroup$
4
  • $\begingroup$ What do you mean “separately?” Do $H1$ and $H2$ commute? $\endgroup$ Nov 12, 2023 at 18:56
  • $\begingroup$ @MarkSpinelli On two separate computers, they may not commute so i could trotter them in that case, right? , but the main question if i could recombine the results to get the result for the original $H$ (like circuit cutting perhaps? ) $\endgroup$
    – Moustafa
    Nov 12, 2023 at 19:30
  • $\begingroup$ I don't know what you mean by "on two separate computers" - but yes, if they don't commute then you will need to do something akin to trotterization to deal with the non-commutative cross-factors. I'm also unclear as to what you mean by circuit cutting in this context. But if $H1$ and $H2$ act on distinct qubits then surely they commute and you don't need to trotterize. $\endgroup$ Nov 12, 2023 at 23:02
  • $\begingroup$ @MarkSpinelli Oh, i am sorry for my unclear question,what i meant is that when you cut a circuit into two other smaller circuits,you can recombine the answers for the two small circuits to get the big one, then, if i had a big circuit with evolution H but that it's too big/deep for our QCs, and i find a way to take the unitary evolution of that circuit and separate it into $H1$ and $H2$, apply each subH separately(so two smaller circuits) then recombine the answers, or is this not logical? if you could give me something to read and clarify this too, that would be so nice of you, thanks a lot! $\endgroup$
    – Moustafa
    Nov 13, 2023 at 14:44

1 Answer 1

1
$\begingroup$

Community Wiki


It seems like you wish to perform a "Local Hamiltonian Simulation" where your Hamiltonian $H$ is the sum of two separate terms:

$$H=aH_1+bH_2.$$

That is, you wish to simulate $\exp(-iHt)$ where $H=aH_1+bH_2$ on a Hilbert space comprising, say, $n$ qubits, where $H_1$ and $H_2$ separately act on locally on, say, $k_1$ and $k_2$ of these qubits.

Certainly if the qubits acted upon by $H_1$ are distinct from those acted upon by $H_2$, then $H_1$ and $H_2$ necessarily commute and you can have a unitary that simulates $H_1$ as $\exp(-aiH_1t)$ for all of $t$, and another unitary that simulates all of $H_2$ as $\exp(-biH_2t)$ for all of $t$, and combine them together.

Also, if $H_1$ and $H_2$ commute with each other but act on the same or an overlapping number of qubits, then you can again evaluate all of $H_1$, then all of $H_2$, and combine together.

But if $[H_1,H_2]\ne 0$ then you'd probably be best to do some Trotterization to address the noncommutative overlap. The idea is to evolve unitarily for a little bit of time subject to $H_1$, then for a little bit of time for $H_2$, then back to $H_1$ and back to $H_2$, etc.

I really like Lloyd's analogy about parallel parking:

By going forward and backing up a sufficiently small distance a large enough number of times, it is possible to parallel park in a space only $\varepsilon$ longer than the length of the car.

That is, if your Trotterization factor is $r$, you evolve as $\exp(-aiH_1t/r)$ and $\exp(-biH_2/r)$ for $r$ iterations. You could similarly go forward and to the right for a short distance, then backwards to the left, then forward to the right... to parallel park.

Lloyd introduced Trotterization in 1995 as a response to Feynman's address from 1982, relying on the Baker-Campbell-Hausdorff to give a good bound on his method. All this was asked and answered even before we had a sufficiently formal framework for the problem, and Lloyd just affirmatively states that "Feynman's 1982 conjecture, that quantum computers can be programmed to simulate any local quantum system, is shown to be correct".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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