How would a quantum computer calculate a definite integral (without resorting to approximations) of a function?


According to a post of mine (in the context of classical field theory) seems to make the following premises:

  1. One can always infer the value of a field at a point.
  2. There can exist a Robinson Boundary Condition.
  3. This (Klien-Gordon) field obeys Lorentz transformations.

The conclusion seems to be:

Using classical calculations and the Robin boundary condition I show that one calculate the anti-derivative of a function within time $2X$ (I can compute an integral below)

$$\frac{2 \alpha}{\beta} e^{ \frac{\alpha}{\beta}X} \int_0^{X} e^{- \frac{\alpha}{\beta}\tau} \tilde f_0(\tau) d \tau$$

where $f_0$ is an arbitrary function whose integral above converges.

Note: It is unclear to me how a quantum computer would stimulate a classical field in a way which would tell you the definite integral?

  • $\begingroup$ I suggest to elaborate more. There is a VtC as unclear for your post. $\endgroup$ – peterh Nov 27 '19 at 14:58
  • $\begingroup$ @peterhsaysreinstateMonica is this better? $\endgroup$ – More Anonymous Nov 29 '19 at 9:00
  • 1
    $\begingroup$ Yes, I think it will survive the vote (no new close vote is arrived). $\endgroup$ – peterh Nov 29 '19 at 11:15
  • $\begingroup$ So the field theory is not important for this question. You are just asking for an ordinary definite integral now with us being able to forget that this came from solving a pde $\endgroup$ – AHusain Dec 11 '19 at 6:05
  • $\begingroup$ @AHusain Yes. I only mention the field theory for comparison purposes: time taken is $2X$ and the space of functions (to integrate) is not limited to the original space of solutions of the PDE (Klein Gordon Equation). $\endgroup$ – More Anonymous Dec 11 '19 at 6:38

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