# Worked Problems in Physics

## Nielsen and Chuang Exercise 2.20

Posted in Nielsen/Chuang by rpmuller on March 2, 2010

Exercise 2.20 Basis change: Suppose ${A'}$ and ${A''}$ are matrix representations of an operator ${A}$ on a vector space ${V}$ with respect to two different orthonormal bases, ${\left|v_i\right>}$ and ${\left|w_i\right>}$. Then the elements of ${A'}$ and ${A''}$ are ${\left}$ and ${\left}$. Characterize the relationship between ${A'}$ and ${A''}$. $\displaystyle \left = \sum_{kl} \left\left\left \ \ \ \ \ (39)$

## Nielsen and Chuang Exercise 2.19

Posted in Nielsen/Chuang by rpmuller on March 2, 2010

Exercise 2.19 Show that the Pauli matrices are Hermetian and unitary.

Hermetian: $\displaystyle \left[ \begin{array}{rr} 0 & 1\\ 1 & 0 \end{array} \right]^\dagger = \left[ \begin{array}{rr} 0 & 1\\ 1 & 0 \end{array} \right] \ \ \ \ \ (33)$ $\displaystyle \left[ \begin{array}{rr} 0 & -i\\ i & 0 \end{array} \right]^\dagger = \left[ \begin{array}{rr} 0 & -i\\ i & 0 \end{array} \right] \ \ \ \ \ (34)$ $\displaystyle \left[ \begin{array}{rr} 1 & 0\\ 0 & -1 \end{array} \right]^\dagger = \left[ \begin{array}{rr} 1 & 0\\ 0 & -1 \end{array} \right] \ \ \ \ \ (35)$

Since we’ve now shown that the matrices are Hermetian, it’s enough to show that the matrices square to ${I}$: $\displaystyle X^2= \left[ \begin{array}{rr} 0 & 1\\ 1 & 0 \end{array} \right] \left[ \begin{array}{rr} 0 & 1\\ 1 & 0 \end{array} \right]=I \ \ \ \ \ (36)$ $\displaystyle Y^2 = \left[ \begin{array}{rr} 0 & -i\\ i & 0 \end{array} \right] \left[ \begin{array}{rr} 0 & -i\\ i & 0 \end{array} \right]=I \ \ \ \ \ (37)$ $\displaystyle Z^2 = \left[ \begin{array}{rr} 1 & 0\\ 0 & -1 \end{array} \right] \left[ \begin{array}{rr} 1 & 0\\ 0 & -1 \end{array} \right]=I \ \ \ \ \ (38)$

## Nielsen and Chuang Exercise 2.18

Posted in Nielsen/Chuang by rpmuller on March 2, 2010

Exercise 2.18 Show that all eigenvalues of a unitary matrix have modulus 1, that is, can be written ${e^{i\theta}}$ for some real ${\theta}$. $\displaystyle Ux = \lambda x \ \ \ \ \ (33)$ $\displaystyle x^\dagger U^\dagger = \lambda^\dagger x^\dagger \ \ \ \ \ (34)$

Multiplying the two $\displaystyle x^\dagger U^\dagger Ux = \lambda^\dagger x^\dagger \lambda x \ \ \ \ \ (35)$ $\displaystyle ||x||^2 = |\lambda|^2||x||^2 \ \ \ \ \ (36)$

or ${|\lambda|^2=1}$.

## Nielsen and Chuang Exercise 2.17

Posted in Nielsen/Chuang by rpmuller on March 2, 2010

Exercise 2.17 Show that a normal (i.e. diagonalizable) matrix is Hermitian if and only if it has real eigenvalues.

Part one: show that Hermiticity implies real eigenvalues. Since the matrix is diagonalizable, we can write $\displaystyle \lambda = U^\dagger AU \ \ \ \ \ (27)$

Thus $\displaystyle \lambda^\dagger = \left(U^\dagger AU\right)^\dagger = U^\dagger A^\dagger U \ \ \ \ \ (28)$

Since the matrix ${A}$ is Hermetian, ${A^\dagger=A}$, and thus ${\lambda^\dagger=\lambda}$, and thus the matrix has real eigenvalues.

Part two: Show that real eigenvalues imply Hermiticity. We may now write: $\displaystyle A = U\Lambda U^\dagger \ \ \ \ \ (29)$

and following a similar procedure show that $\displaystyle A^\dagger = \left(U\Lambda U^\dagger\right)^\dagger = U\Lambda^\dagger U^\dagger. \ \ \ \ \ (30)$

Now, since the eigenvalues are real, ${\Lambda^\dagger = \Lambda}$, and thus ${A^\dagger=A}$, and thus the matrix is Hermitian.