Let's say I have a system which one can use to compute an integral from $0$ to $X$ in time $2X$. Assuming it does so by taking time-steps $\epsilon$. What is the smallest time-step $\epsilon$ possible? What would such an $\epsilon$ correspond to physically in the case shown in the motivation?


Physics SE: Using a time-like boundary as a computer?


closed as off-topic by DaftWullie, Norbert Schuch, MEE was the missing bracket, Sanchayan Dutta, Niel de Beaudrap Nov 9 '18 at 9:41

  • This question does not appear to be about quantum computing or quantum information, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The question statement makes it sound like you are given input data as a computational state in $H=(\mathbb{C}^F)^{\otimes X/\epsilon}$ where $F$ is the number of states with each being a real number (like $2^{32}$ in single precision) then you have $\frac{X}{\epsilon}$ of them for grid points. The integration would be a unitary on $H \otimes \mathbb{C}^F$. But the description is not. Could you clarify? $\endgroup$ – AHusain Nov 7 '18 at 20:44
  • $\begingroup$ Sorry, I don't think I fully follow you :( ... By the "description is not" do you mean: the link I provided? It's description? $\endgroup$ – More Anonymous Nov 8 '18 at 0:30
  • 3
    $\begingroup$ I'm voting to close this question as off-topic because I don’t see what it has to do with quantum computation. $\endgroup$ – DaftWullie Nov 8 '18 at 6:06
  • $\begingroup$ Let's say the quickest algorithm for numerical integration takes time $T(\text{no. of digits}, \epsilon)$ where $\epsilon$ is the time-step. (I don't know what this function is except I can guess that it should be increasing with $\epsilon$). Then by fixing the number of digits in the calculation $T(\epsilon,\text{no. of digits}) = 2X \implies \epsilon = T^{-1} ( 2X )_{number of digits}$. I hope this makes it clearer what I was after and how it's related to quantum computing? $\endgroup$ – More Anonymous Nov 8 '18 at 8:38
  • $\begingroup$ I thought you were asking for something more akin to an adder circuit like to do midpoint method. By the description, I just meant the text/link of your question as opposed to the title. $\endgroup$ – AHusain Nov 8 '18 at 10:05

Browse other questions tagged or ask your own question.