2
$\begingroup$

So I was wondering about the following:

I can encode an algorithm in position $\hat x$ in quantum mechanics. However, in QFT position is not an operator it becomes a parameter $x$.

Has anyone expressed concern that quantum field theory (QFT) will become relevant in the scale-up when creating a quantum computer and thus one should be careful about the second quantization of operators?

Improve this question
$\endgroup$
1
  • $\begingroup$ "I can encode an algorothm in position" which algorithm did you have in mind? $\endgroup$
    – user1271772
    Jan 2, 2021 at 13:24

1 Answer 1

Reset to default
2
$\begingroup$

QFT (more specifically QED) is already at the heart of quantum computation. Second quantization formalism is used even in constructing toy models for quantum computers (see e.g., Section 7.3.2 of Nielsen and Chuang).

Many of the rapid advances in superconducting quantum processors over the past decade can be attributed to the introduction of circuit Quantum Electrodynamics (cQED, introduced here and here, full exposition, lecture notes), and the closely related introduction of the transmon (here). As suggested by the nomenclature these architectures are built on QED rather than non-relativistic QM.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I will go through the link but sooner or later aren't we going to hit the problem of how does "QM emerge from QFT"? Like they both can differ in an answer of a system trivially (by some decimal places) but non-trivally when scaled up? $\endgroup$
    – More Anonymous
    Oct 28, 2019 at 17:07
  • 3
    $\begingroup$ @MoreAnonymous I think it is somewhat analogous to the situation in classical computing, in that the answer will depend on where you are in the layers of abstraction. At the lowest level of abstraction (hardware) annihilation and creation operators are used pedagogically from the first toy examples, while at high levels of abstraction the same concepts may be entirely foreign in very sophisticated quantum algorithms. $\endgroup$
    – Jonathan Trousdale
    Oct 28, 2019 at 17:20
  • 1
    $\begingroup$ Distinction between cQED and QED. Questions like whether you are dealing with $a_k$/$a_k^\dagger$ for all wavevectors or not, is electron part of a spinor field? $\endgroup$
    – AHusain
    Oct 28, 2019 at 17:33

Your Answer

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

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