# Worked Problems in Physics

## Nielsen and Chuang Exercise 2.66

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.66 Show that the average value of the observable ${X_1Z_2}$ for a two-qubit system measured in the state ${(\left|00\right>+\left|11\right>)/\sqrt{2}}$ is zero.

Twice the average value is given by

$\displaystyle 2A=\left<00|XZ|00\right>+\left<00|XZ|11\right>+\left<11|XZ|00\right> +\left<11|XZ|11\right>$

$\displaystyle =\left<00|10-01+10-01\right>=0.$

## Nielsen and Chuang Exercise 2.65

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.65 Express the states ${(\left|0\right>+\left|1\right>)/\sqrt{2}}$ and ${(\left|0\right>-\left|1\right>)/\sqrt{2}}$ in a basis in which they are not the same up to a relative phase shift.

This is perhaps a trivial transformation, but, if we express the two states in the eigenbasis of the ${X}$ operator, the states ${(\left|0\right>+\left|1\right>)/\sqrt{2}}$ and ${(\left|0\right>-\left|1\right>)/\sqrt{2}}$ correspond to the ${\left|+\right>}$ and ${\left|-\right>}$ states, which no longer differ by merely a relative phase shift.

## Nielsen and Chuang Exercise 2.64

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.64 Suppose Bob is given a quantum state chosen from a set ${\left|\psi_1\right>,\dots,\left|\psi_m\right>}$ of linearly independent states. Construct a POVM ${\{E_1,E_2,\dots,E_{m+1}\}}$ such that if outcome ${E_i}$ occurs, ${1\leq i \leq m}$, then Bob knows with certainty that he was given the state ${\psi_i}$. The POVM must be such that ${\left<\psi_i|E_i|\psi_i\right>>0}$ for each ${i}$.

Since the states are linearly independent, we can define ${E_i=\left|i\right>\left, for ${i=1,\dots,m}$, and ${E_{m+1}=1-\sum_{i=1}^mE_i}$, which satisfies ${\sum_mE_m=1}$, and insures if ${E_i}$ is measured, that the system is in state ${\left|i\right>}$.

## Nielsen and Chuang Exercise 2.63

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.63 Suppose a measurement is described by measurement operators ${M_m}$. Show that there exist unitary operators ${U_m}$ such that ${M_m=U_m\sqrt{E_m}}$, where ${E_m}$ is the POVM associated to the measurement.

We can confirm that the proposed operator works by considering that ${M_m^\dagger M_m=\sqrt{E_m}U_m^\dagger U_m\sqrt{E_m}=E_m}$. Thus, if ${\sum_m E_m=1}$, which is the case if the ${\{E_m\}}$ represent a POVM, then ${\sum_m M_m^\dagger M_m=1}$, which demonstrates that the unitary transformation converts the measurement operators to the POVMs.