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<i\right|

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<i\right|.