# What is the meaning of approximating a phase to an accuracy of $2^{-n}$?

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.

• 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 $0.30102$ Feb 18, 2021 at 18:15
• @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? Feb 18, 2021 at 19:07
• You are completely right, does this answer your questions ? Feb 18, 2021 at 19:23
• @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. Feb 18, 2021 at 19:25

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 !