- #1

Peter_Newman

- 155

- 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.

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:

$$p(0)=\langle \psi|M_0^\dagger M_0|\psi\rangle = \langle\psi|M_0|\psi\rangle = |\alpha|^2$$.

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.

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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