# Worked Problems in Physics

## Nielsen and Chuang Exercise 2.56

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.56 Use the spectral decomposition to show that ${K=-i\log(U)}$ is Hermitian for any unitary ${U}$, and thus ${U=\exp(iK)}$ for some Hermitian ${K}$.

In Exercise 2.18 we showed that the eigenvalues of a unitary matrix can be written ${e^{i\theta}}$ for some real ${\theta}$. Thus,

$\displaystyle K=-i\log(U)=V-i\log(\exp(i\theta))V^\dagger=V\theta V^\dagger.$

It then follows that

$\displaystyle K^\dagger=V\theta^\dagger V^\dagger=V\theta V^\dagger=K.$

## Nielsen and Chuang Exercise 2.55

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.55 Prove that ${U(t_1,t_2)}$ defined in Eq (2.91) is unitary.

$\displaystyle U(t_1,t_2)=\exp\left(\frac{-i}{\hbar}H(t_1-t_2)\right).$

Then,

$\displaystyle U(t_1,t_2)^\dagger=\exp\left(\frac{i}{\hbar}H^\dagger(t_1-t_2)\right).$

Because ${H}$ is Hermitian, ${H^\dagger=H}$, and thus ${[U,U^\dagger]=0}$, since ${H}$ commutes with itself. By the previous exercise,

$\displaystyle U(t_1,t_2)^\dagger U(t_1,t_2)=\exp\left(\frac{i}{\hbar}H^\dagger(t_1-t_2)\right)\exp\left(\frac{-i}{\hbar}H(t_1-t_2)\right)$

$\displaystyle =\exp\left(\frac{i}{\hbar}H^\dagger(t_1-t_2)+\frac{-i}{\hbar}H(t_1-t_2)\right)$

$\displaystyle =\exp\left(\frac{i}{\hbar}H(t_1-t_2)+\frac{-i}{\hbar}H(t_1-t_2)\right)=\exp(0)=I.$

## Nielsen and Chuang Exercise 2.54

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.54 Suppose that ${A}$ and ${B}$ are commuting Hermitian operators. Verify that ${\exp(A)\exp(B)=\exp(A+B)}$.

Because ${[A,B]=0}$, we can find a simultaneous eigenbasis, such that ${A=ULU^\dagger}$ and ${B=UMU^\dagger}$. Thus ${\exp(A)=U\exp(L)U^\dagger}$, ${\exp(B)=U\exp(M)U^\dagger}$ and ${\exp(A)\exp(B)=U\exp(L)\exp(M)U^\dagger=U\exp(L+M)U^\dagger=\exp(A+B)}$

## Nielsen and Chuang Exercise 2.53

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.53 What are the eigenvalues and eigenvectors of ${H}$?

There is no doubt a more elegant representation, but I worked these out numerically. The eigenvalues are ${(-1,1)}$, and the eigenvectors are given by the columns of

$\displaystyle \left[\begin{array}{rr}0.3827 & -0.9239 \\ -0.9239 & -0.3827\end{array}\right].$

## Nielsen and Chuang Exercise 2.52

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.52 Verify that ${H^2=I}$.

This was already proven in the previous example, since the Hadamard gate is Hermitian and unitary, it follows that ${H^2=I}$.

## Nielsen and Chuang Exercise 2.51

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.51 Verify that the Hadamard gate is unitary.

In this case, since the Hadamard gate is Hermitian, ${H^\dagger=H}$, and thus we only have to prove that ${H^\dagger H=I}$, since the other form ${HH^\dagger = H^\dagger H}$. Thus,

$\displaystyle \frac{1}{2} \left[\begin{array}{rr}1 & 1 \\ 1 & -1\end{array}\right] \left[\begin{array}{rr}1 & 1 \\ 1 & -1\end{array}\right] = \left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right].$

## Nielsen and Chuang Exercise 2.50

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.50 Find the left and right polar decomposition of the matrix

$\displaystyle \left[\begin{array}{rr} 1 & 0 \\ 1 & 1\end{array}\right]$

For the left polar decomposition,

$\displaystyle A=UJ= \left[\begin{array}{rr} 0.89 & -0.45\\ 0.45 & 0.89 \end{array}\right] \left[\begin{array}{rr} 1.34 & 0.45\\ 0.45 & 0.89 \end{array}\right].$

For the right polar decomposition

$\displaystyle A=KU= \left[\begin{array}{rr} 0.89 & 0.45\\ 0.45 & 1.34 \end{array}\right] \left[\begin{array}{rr} 0.89 & -0.45\\ 0.45 & 0.89 \end{array}\right].$

I worked out ${K}$ and ${J}$ from the spectral decomposition of ${A^\dagger A}$ and ${AA^\dagger}$, respectively, and formed ${U}$ from ${AJ^{-1}}$.

## Nielsen and Chuang Exercise 2.49

Posted in Nielsen/Chuang by rpmuller on March 15, 2010

Exercise 2.49 Express the polar decomposition of a normal matrix in outer product representation.

A normal matrix is a matrix for which ${A^\dagger A=AA^\dagger}$, and is guaranteed to have a spectral decomposition. A normal matrix also will have ${J=K}$, so that the right and left polar decompositions are equal.

In outer product form,

$\displaystyle J=\sum_i|\lambda_i|\left|i\right>\left

where ${\left|i\right>}$ are the eigenvectors, and ${\lambda_i}$ are the corresponding eigenvalues. Similarly,

$\displaystyle U=J^{-1}H=\sum_i{\lambda_i\over |\lambda_i|}\left|i\right>\left