Quantum circuit simplification using classical computers

Quantum circuit simplification using classical computers Ask Question Asked 3 years, 3 months ago Modified today Viewed 198 times 2 Suppose that we have this kind of circuit where the first unitary operator U is used for the state preparation while the Hadamard operator is used of state detection. Let's say we try to run this circuit experimentally, but we don't have enough resources to operate two operators together. Let's imagine that we can use perform one arbitrary single unitary operation M as below: Compared to the first circuit, the operator M is equivalent to the multiplication of U and H such as M=H×U 𝑀 = 𝐻 × 𝑈 . To generate the matrix M 𝑀 as H×U 𝐻 × 𝑈 , the classical computer is required to calculate the structure of the matrix, and afterwards we can run the same circuit in the second figure as H×U 𝐻 × 𝑈 . But I wonder if the usage of classical computers for this kind of simplification is generally allowable. quantum-gatecircuit-constructionrandom-quantum-circuit Share Improve this question Follow edited Jan 20, 2023 at 13:25 asked Jan 20, 2023 at 2:02 William 1829 9 bronze badges Add a comment 1 Answer Sorted by: Highest score (default) Date modified (newest first) Date created (oldest first) 0 Of course the "simplification" you are doing here is always possible and, more in general, any quantum circuit can be represented as a single 2 n × 2 n 2 𝑛 × 2 𝑛 unitary matrix M 𝑀 (acting on a n 𝑛 -qubits quantum state). However, what you are probably missing here is that the linear algebra of unitary operators is just the mathematical tool that we use to describe quantum computation or, at most, to simulate quantum computers. Once you come to the actual realization of your operator M 𝑀 on a real quantum device, you have to consider that the hardware is constrained to use a quite limited set of quantum gates that are actually implemented on physical qubits (e.g. as microwave pulses on superconducting qubits). This means that, from the quantum hardware's point of view, applying first the U 𝑈 gate and then the H 𝐻 gate is perfectly equivalent to applying the resulting gate M=H×U 𝑀 = 𝐻 × 𝑈 and there is no such a thing as a "simplification". Share Improve this answer Follow edited Mar 25, 2025 at 8:29 answered Jan 20, 2023 at 14:33 SimoneGasperini 1,6441 1 gold badge 4 4 silver badges 18 18 bronze badges I know that the circuit simplification itself would be fine. But my questions is the process of the simplification here. If I want to build the M operator to be H x U, then I need to know how H x U is built by classical computers, and I wonder if this simplication by classical computers before running quantum circuits would be generic method. Isnt's there any speedup problem for this kind of classical-computer-based simplification? –  William Commented Jan 20, 2023 at 17:12 Taking a stab although I'm still not 100% sure what you're asking: decomposing an arbitrary, desired operation M 𝑀 into its component operations that I can physically run using the relevant gate-set on my computer is not only a necessary, but also challenging and doing so efficiently/optimally is an open problem –  Chris E Commented Jan 20, 2023 at 20:54 @Chris E Thank you for your comment. What I'm saying is similar to what you are saying, which is decomposing the M operator to the H x U operation using classical computer. You said that it is an open problem and I wonder if this kind of decomposition using classical computer is reasonable generally. –  William Commented Jan 20, 2023 at 21:06 Oh, got it. I believe there are some schemes that work reasonably well for N 𝑁 qubits, but the required number 1- and 2-qubit gates scales exponentially with N 𝑁 . Classical algorithms that try to minimize the number of quantum gates are fairly complicated/expensive though (but spending the classical time to run more efficiently on quantum hardware is still a huge net benefit). For more details, you could look into quantum circuit compiling like arxiv.org/pdf/2101.02993.pdf –  Chris E Commented Jan 20, 2023 at 21:33 Add a comment Your Answer Sign up or log in Sign up using Google Sign up using Email and Password Post as a guest Name Email Required, but never shown Post Your Answer By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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