I Notation of the approximation in quantum phase estimation algorithm

[Peter_Newman]

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.

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.

 

[Peter_Newman]

Unfortunately, I haven't made any progress myself, otherwise I would have presented a solution here. Therefore, I am still interested in helpful tips and hints. Is the question so far clear, or is there a need to concretize it a bit?