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In quantum phase estimation we often see that approximating $\phi$ ito an accuracy of $2^{-n}$. Can anybody explain what is the meaning of that?

Does that mean after decimal we can only believe the n values? So if n=4, we can think of it as $0.1234567$ only the 4 values are correct. So we should consider $0.1234$.

I am asking this question because we face this situation when applying the phase estimation algorithm.

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  • $\begingroup$ Pay attention, 4 decimals correct would mean an accuracy of $10^{-4}$. Since $2^{-4} \approx 0.0625 \approx 10^{-2}$, this is a pretty bad accuracy ! The way to know how many decimals are correct would be to put $n$ in this formula : $\lfloor \log_{10} (2^n) \rfloor$. This is equal to taking the integer part of $n$ multiplied by $0.30102$ $\endgroup$ Feb 18, 2021 at 18:15
  • $\begingroup$ @BrockenDuck So as per your comment, $0.1234$ up to $2^{-4}$ accuracy would be decimal $4*0.30102 = 1.20408$ or taking floor its 1 decimal bit. Right? So up to $0.1$ we are super confident and rest are shaky? $\endgroup$
    – user27286
    Feb 18, 2021 at 19:07
  • $\begingroup$ You are completely right, does this answer your questions ? $\endgroup$ Feb 18, 2021 at 19:23
  • $\begingroup$ @BrockenDuck Yes it does. Should I close the question ? If you have time to write an answer maybe I can upvote select and close it. $\endgroup$
    – user27286
    Feb 18, 2021 at 19:25

1 Answer 1

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As I said in the comments, the exponential showing the accuracy is base 2. To get the number of decimals correct you need to convert it to base 10 : $$ 2^n = 10^{n'} \Leftrightarrow n' = log_{10} (2^n) \Leftrightarrow n' = n * log_{10} (2) \approx n * 0.30102 $$

As you can see we do not get a integer value, so we need to take the floor function ($\lfloor n' \rfloor$) of the result. This conversion is very current, please tell in the comments if you need clarifications on why we need base 10 or specific functions, tell me in the comments !

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  • $\begingroup$ We need base 10 because decimal system is state 10. (which we are familiar with). We used the binary arithmetic because that would make the analysis easier with the qubit state representations. I guess these are the reasons? $\endgroup$
    – user27286
    Feb 19, 2021 at 0:40
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    $\begingroup$ The precision given with a base 2 exponential function is because of the algorithm's properties. I just converted it to base 10, since we use base 10, just as you said $\endgroup$ Feb 19, 2021 at 20:49
  • $\begingroup$ Thanks for clarifying $\endgroup$
    – user27286
    Feb 19, 2021 at 20:57

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