- #1

Peter_Newman

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I'am interested in the notation of the approximation in quantum phase estimation algorithm.

In the literature there are different definitions, which I divide into two cases here. Both different in their definition of the ##\delta##. In both cases I start with a quote of the source and show an example of how I understand this in that context.

Let ##\phi_\text{exact} = \varphi_\text{exact} = 0.1011_2## in this scenario we limit our approximation of the phase (##\varphi,\phi##) to 2 Bits.

With ##m = 2## Bits e.g. best we can get with ##0 < |\delta| \leq \frac{1}{2^{m+1}}## is:

##\phi_\text{approx} = 0.10_2 + (-0.001_2 \leq \delta \leq 0.001_2) = 0.10_2 + 0.001_2##, since maximum value of ##\delta## is ##\frac{1}{2^3} = 0.001_2##, we leave out ##0.0001## in case of ##\delta## as defined above. I assumed ##0.10_2## is the best estimate we can get with two bits.

With ##t = 2## Bits e.g. best we can get with ##0 < \delta \leq \frac{1}{2^{t}}## is:

##\varphi_\text{approx} = 0.10_2 + (0 < \delta \leq 0.01_2)= 0.10_2 + 0.0011_2##, we see with ##\delta## defined in this way, we get a better approximation. We can at least describe the missing part of delta here exactly. I assumed ##0.10_2## is the best estimate we can get with two bits.My

In the literature there are different definitions, which I divide into two cases here. Both different in their definition of the ##\delta##. In both cases I start with a quote of the source and show an example of how I understand this in that context.

Let ##\phi_\text{exact} = \varphi_\text{exact} = 0.1011_2## in this scenario we limit our approximation of the phase (##\varphi,\phi##) to 2 Bits.

**Case 1:**... let ##\frac{a}{2^m} = 0.a_1...a_m## be the best ##m##-bit estimate of ##\phi##. Then ##\phi = \frac{a}{2^m} + \delta##, where ##0<|\delta|\leq \frac{1}{2^{m+1}}## [Cleve et al. from quant-ph/9708016, p11]

With ##m = 2## Bits e.g. best we can get with ##0 < |\delta| \leq \frac{1}{2^{m+1}}## is:

##\phi_\text{approx} = 0.10_2 + (-0.001_2 \leq \delta \leq 0.001_2) = 0.10_2 + 0.001_2##, since maximum value of ##\delta## is ##\frac{1}{2^3} = 0.001_2##, we leave out ##0.0001## in case of ##\delta## as defined above. I assumed ##0.10_2## is the best estimate we can get with two bits.

**Case 2:**Let ##b## be the integer in the range ##0## to ##2^t−1## such that ##b/2^t = 0.b_1 ... b_t## is the best ##t## bit approximation to ##\varphi## which is less than ##\varphi##. That is, the difference ##\delta ≡ \varphi − b/2^t## between

##\varphi## and ##b/2^t## satisfies ##0 ≤ \delta ≤ 2^{−t}##. [Nielsen and Chuang from QC, p223]

With ##t = 2## Bits e.g. best we can get with ##0 < \delta \leq \frac{1}{2^{t}}## is:

##\varphi_\text{approx} = 0.10_2 + (0 < \delta \leq 0.01_2)= 0.10_2 + 0.0011_2##, we see with ##\delta## defined in this way, we get a better approximation. We can at least describe the missing part of delta here exactly. I assumed ##0.10_2## is the best estimate we can get with two bits.My

**final question**is, why do people in the literature also use the first definition of delta (##0<|\delta|\leq \frac{1}{2^{m+1}}##), which would be more inaccurate according to my calculation?I hope that I have written my question understandably and I am very much looking forward to your opinions on this.
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