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

[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.

 

[Cthugha]

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.

 

[Cthugha]

That looks reasonable to me.