Measurement of a qubit in the computational basis - Phase estimation

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.
  • #1
Peter_Newman
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11
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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  • #2
It seems to me like there is a term (x-a) missing in the exponential function. Might that be the case?
 
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  • #3
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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  • #4
That looks reasonable to me.
 
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Related to Measurement of a qubit in the computational basis - Phase estimation

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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