# Measurement of a qubit in the computational basis - Phase estimation

• I
• Peter_Newman
In summary, the conversation discusses the measurement of a qubit in the computational basis and how it relates to the probability of obtaining a measurement outcome. The question is then raised about a specific equation and its derivation, and the conversation ends with a potential explanation for the equation.
Peter_Newman
Hello,

I have a question about the measurement of a qubit in the computational basis. I would like to first state what I know so far and then ask my actual question at the end.What I know:
Let's say we have a qubit in the general state of ##|\psi\rangle = \alpha|0\rangle + \beta|1\rangle##. Now we can define the following measurement operators depending on whether we want to measure the qubit in state ##|0\rangle## or ##|1\rangle##. Let's say I am interested in the state ##|0\rangle##.

The corresponding operator would then be defined as follows ##M_0 = |0\rangle\langle 0|##. The probability of obtaining a measurement outcome ##0## is then defined by:

$$p(0)=\langle \psi|M_0^\dagger M_0|\psi\rangle = \langle\psi|M_0|\psi\rangle = |\alpha|^2$$.My Question:
I read the following in the Wikipedia article on Quantum Phase Estimation (Wiki, section measurement). We have now given there the following quantum state:

$$\frac{1}{2^n}\sum_{x=0}^{2^n-1}\sum_{k=0}^{2^n-1} e^{-\frac{2\pi i k}{2^n}}e^{2\pi i \delta k}|x\rangle|\psi\rangle$$

Now it is said that a measurement in the computational basis on the first register yields the result ##|a\rangle## with probability;

$$Pr(a) = \left|\left\langle a\left| \frac{1}{2^n}\sum_{x=0}^{2^n-1}\sum_{k=0}^{2^n-1} e^{-\frac{2\pi i k}{2^n}}e^{2\pi i \delta k}\right|x\right\rangle\right|^2 = \frac{1}{2^{2n}}\left| \sum_{k=0}^{2^n-1} e^{2\pi i \delta k} \right|^2$$

I am interested in the last equation here (##Pr(a) = ...##), how do you arrive at it? With what I know so far, I can't really derive the last equation, so I would be interested in knowing how the derivation is. Also the simplification does not open up to me. Maybe someone here can demystify it.

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It seems to me like there is a term (x-a) missing in the exponential function. Might that be the case?

Peter_Newman
Yes that is unfortunately correct!
I would like to improve my first post regarding this error. Unfortunately, I can no longer edit this one...

Correct it is:

$$\frac{1}{2^n}\sum_{x=0}^{2^n-1}\sum_{k=0}^{2^n-1} e^{-\frac{2\pi i k}{2^n}(x-a)}e^{2\pi i \delta k}|x\rangle|\psi\rangle$$

$$Pr(a) = \left|\left\langle a\left| \frac{1}{2^n}\sum_{x=0}^{2^n-1}\sum_{k=0}^{2^n-1} e^{-\frac{2\pi i k}{2^n}(x-a)}e^{2\pi i \delta k}\right|x\right\rangle\right|^2 = \frac{1}{2^{2n}}\left| \sum_{k=0}^{2^n-1} e^{2\pi i \delta k} \right|^2$$

Based on this, I would now assert the following as to why one come up with ##\frac{1}{2^{2n}}\left| \sum_{k=0}^{2^n-1} e^{2\pi i \delta k} \right|^2##.
So the scalar product of ##\langle a|x\rangle## is only 1 if ##a = x##, if this is the case, everything reduces to ##\frac{1}{2^{2n}}\left| \sum_{k=0}^{2^n-1} e^{2\pi i \delta k} \right|^2##, where we put out the constant ##\left|\frac{1}{2^n}\right|^2 = \frac{1}{2^{2n}}##and note that one of the exp terms is 1 since ##e^0## iff ##a = x##. Right? For all other ##a \neq x##, the scalar product is ##0##. Therefore, ## \frac{1}{2^{2n}}\left| \sum_{k=0}^{2^n-1} e^{2\pi i \delta k} \right|^2## then follows.

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That looks reasonable to me.

Peter_Newman

## 1. What is a qubit?

A qubit is the basic unit of quantum information, analogous to a classical bit in traditional computing. It can exist in multiple states at once, allowing for more complex calculations and processing in quantum computers.

## 2. What is the computational basis?

The computational basis is a set of basis states that can represent the different possible values of a qubit. In the case of a single qubit, the computational basis consists of two states: |0> and |1>, which represent the classical binary digits 0 and 1.

## 3. What is phase estimation in quantum computing?

Phase estimation is a quantum algorithm used to estimate the phase of a quantum state. It is an important component of many quantum algorithms, including Shor's algorithm for factoring large numbers and Grover's algorithm for searching unsorted databases.

## 4. How is a qubit measured in the computational basis?

To measure a qubit in the computational basis, the qubit is first prepared in a superposition state. Then, a measurement is performed using a quantum measurement device, which collapses the superposition state into one of the two computational basis states, |0> or |1>, with a certain probability determined by the amplitudes of the superposition state.

## 5. What is the purpose of phase estimation in quantum computing?

The purpose of phase estimation is to determine the phase of a quantum state, which can provide valuable information for various quantum algorithms. It allows for more precise calculations and can significantly improve the efficiency of quantum algorithms compared to classical algorithms.

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