Quantum phase estimation - Question regarding rewriting the state

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• Peter_Newman
In summary, the conversation discusses the use of a new notation, ##|(b+l)(\text{mod } 2^t)\rangle##, introduced in the context of applying the inverse quantum Fourier transform. The purpose of this notation is to shift the labels of the states by a fixed offset ##b##, which is useful when dealing with periodic functions. This offset can be seen as noise in the measurement process, but it becomes purely phase information when taking the Fourier transform. The conversation also mentions that the offset is not explicitly explained in the source material and the explanation provided in the article is helpful in understanding its significance.
Peter_Newman
In Nielsen and Chuang p.223 we have the following situation:

$$\frac{1}{2^t} \sum\limits_{k,l=0}^{2^t-1} e^{\frac{-2\pi i k l}{2^t}} e^{2 \pi i \varphi k} |l\rangle$$

Which results from applying the inverse quantum Fourier transform to state ##\frac{1}{2^{t/2}} \sum\limits_{k=0}^{2^t-1} e^{-2\pi i \varphi k} |l\rangle##. We have more or less a sum over the basis states ##|l\rangle##. This is clear so far.

Next, the following new notation is then introduced ##|(b+l)(\text{mod } 2^t)\rangle##. Were we know from ##b## that it is an integer in the range from ##0## to ##2^t -1##.

What does this ##b+l## mean exactly, how is this to interpret? Why does one not leave it with the basic state ##|l\rangle##?

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You have computational basis states ##|0\rangle##, ##|1\rangle##, ##|2\rangle##, etc.

Instead of writing ##|2\rangle## you could write ##|1+1\rangle## because ##1+1=2##.

Instead of writing ##|2\rangle## you could write ##|3+7 \pmod{8}\rangle## because ##(3+7) \bmod 8 = 2##.

Get the idea? From that you can see that looking at ##\sum_\ell |b+\ell \pmod{N}\rangle## instead of ##\sum_\ell |\ell\rangle## is shifting the labels of the states over by some fixed offset ##b##, wrapping around just before you get to ##|N\rangle##.

Peter_Newman
Hello, thank you for the answer!

Yes because of the modular arithmetic one always remains in the "limits" of the index. One could also have added 42 to each state and calculate ##\text{mod } 2^t##, I suppose ;)

But what is the advantage of this? I mean, one does not introduce a new notation, in order to have nothing of it at the end?

Ah, it's not a description of a solution it's a description of a problem. When you have a register x that contains a uniform superposition and you measure f(x), where f is a periodic function period p, you are left with a superposition where every p'th state is present... with some uncontrollable offset b. The book is then going to show that when you take the Fourier transform of this register, the offset b becomes purely phase information; the magnitudes of the amplitudes of the superposition don't depend on b. Measuring in the frequency basis avoids the unknown offset problem.

Image is from here: https://algassert.com/post/1718

Peter_Newman

So let me understand this correctly. So with the offset ##b## in ##l+b## you are anticipating something at the point (anticipating shor algorithm)? I find problematic that at the point it is not pointed out where the offset comes from and why it occurs. Mathematically this is equivalent whether I write ##\sum_\ell |\ell\rangle## or ##\sum_\ell |b+\ell \pmod{N}\rangle##, the index remains the same. I can only explain this in such a way that one would like to measure ideally ##l##, but this does not always go so that one then "measures around" ##l## with a distance of ##b##.

I take from your article that this offset is caused by the sampling and can be dismissed/interpreted quasi as noise. And exactly this noise is picked up by considering this as offset?! This is a good explanation, which I would have liked to see in the Nielsen as well.

Is my reasoning right so far?

1. What is quantum phase estimation?

Quantum phase estimation is a quantum algorithm used to estimate the phase of a quantum state. It is a key component in many quantum algorithms, including Shor's algorithm for factoring large numbers.

2. How does quantum phase estimation work?

In quantum phase estimation, the input state is first prepared and then transformed using a series of controlled operations and measurements. The resulting measurements are then used to estimate the phase of the input state.

3. What is the significance of quantum phase estimation?

Quantum phase estimation is significant because it allows for the efficient estimation of the phase of a quantum state, which is a crucial step in many quantum algorithms. It also has applications in quantum metrology and quantum simulation.

4. Can quantum phase estimation be used for any quantum state?

Quantum phase estimation can be used for any quantum state, as long as the state can be prepared and manipulated using quantum operations. However, the accuracy of the estimation may vary depending on the complexity of the state.

5. Are there any challenges or limitations to quantum phase estimation?

One of the main challenges of quantum phase estimation is the need for precise and accurate control of quantum operations and measurements. It also requires a large number of qubits to achieve high accuracy, making it difficult to implement on current quantum computers. Additionally, errors and noise in the quantum system can affect the accuracy of the estimation.

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