# Worked Problems in Physics

## Nielsen and Chuang Exercise 2.47

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.47 Suppose that ${A}$ and ${B}$ are Hermitian. Show that ${i[A,B]}$ is Hermitian.

$\displaystyle (i[A,B])^\dagger = [iA,iB]^\dagger = [(iB)^\dagger,(iA)^\dagger] = [-iB^\dagger,-iA^\dagger]=-i[B,A]=i[A,B].$

## Nielsen and Chuang Exercise 2.46

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.46 Show that ${[A,B]=-[B,A]}$.

$\displaystyle -[B,A]=-(BA-AB)=(AB-BA)=[A,B].$

## Nielsen and Chuang Exercise 2.45

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.45 Show that ${[A,B]^\dagger=[B^\dagger,A^\dagger]}$.

$\displaystyle [A,B]^\dagger=(AB-BA)^\dagger=(AB)^\dagger-(BA)^\dagger=B^\dagger A^\dagger-A^\dagger B^\dagger = [B^\dagger,A^\dagger]$

## Nielsen and Chuang Exercise 2.44

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.44 Suppose that ${[A,B]=0}$ and ${\{A,B\}=0}$ and ${A}$ is invertible. Show that ${B}$ must be 0.

If ${[A,B]=0}$, then ${AB=BA}$ and if ${\{A,B\}=0}$, then ${AB=-BA}$. The only way for ${AB=\pm BA}$ is for either ${A}$ or ${B}$ to be zero, and if ${A}$ is invertible it isn’t zero, and thus ${B}$ is zero.

## Nielsen and Chuang Exercise 2.43

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.43 Show that for ${j,k=1,2,3,}$

$\displaystyle \sigma_j\sigma_k = \delta_{jk}I + i\sum_{l=1,3}\epsilon_{jkl}\sigma_l.$

If ${j=k}$, ${\sigma_j^2=I}$, and the ${\sum_l}$ terms are zero. Otherwise, the terms are give by the matrix products from exercise 2.40.

## Nielsen and Chuang Exercise 2.42

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.42 Verify that ${AB=\frac{1}{2}([A,B]+\{A,B\})}$.

$\displaystyle \frac{1}{2}([A,B]+\{A,B\}) = \frac{1}{2}(AB-BA+AB+BA) = AB$

## Nielsen and Chuang Exercise 2.41

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.41 Anti-commutation relations for the Pauli matrices. Verify the anti-commutation relations ${\{\sigma_i,\sigma_j\}=0}$, where ${i\neq j}$ are both chosen from the set 1,2,3. Also verify that (${i=0,1,2,3}$) ${\sigma_i^2=I}$.

The anti-commutation relations follow from the Pauli products worked out in exercise 2.40. The Pauli squares were worked out in exercise 2.19.

## Nielsen and Chuang Exercise 2.40

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.40 Commutation relations for the Pauli matrices. Verify the commutation relations ${[X,Y]=2iZ}$, ${[Y,Z]=2iX}$, ${[Z,X]=2iY}$. There is an elegant way of writing this using ${\epsilon_{jkl}}$, the antisymmetric tensor on three indices, for which ${\epsilon_{jkl}=0}$, except for ${\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1}$ and ${\epsilon_{321}=\epsilon_{213}=\epsilon_{132}=-1}$:

$\displaystyle [\sigma_j,\sigma_k]=2i\sum_{l=1}^3\epsilon_{jkl}\sigma_l$

$\displaystyle XY = \left[\begin{array}{rr}0 & 1 \\ 1& 0\end{array}\right]\left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right] = \left[\begin{array}{rr}i & 0 \\ 0&-i\end{array}\right] =iZ$

$\displaystyle YX = \left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right]\left[\begin{array}{rr}0 & 1 \\1&0\end{array}\right] = \left[\begin{array}{rr}-i & 0 \\ 0&i\end{array}\right] =-iZ$

$\displaystyle YZ = \left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right]\left[\begin{array}{rr}1 & 0 \\ 0&-1\end{array}\right] = \left[\begin{array}{rr}0 & i \\ i&0\end{array}\right]=iX$

$\displaystyle ZY = \left[\begin{array}{rr}1 & 0 \\ 0&-1\end{array}\right] \left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right] = \left[\begin{array}{rr}0 & -i \\ -i&0\end{array}\right]=-iX$

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

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

From these it follows that:

$\displaystyle [\sigma_j,\sigma_k]=2i\sum_{l=1}^3\epsilon_{jkl}\sigma_l$

## Nielsen and Chuang Exercise 2.39

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.39 The Hilbert-Schmidt inner product on operators: the set ${L_V}$ of linear operators on a Hilbert space ${V}$ is obviously a vector space—the sum of two linear operators is a linear operator, ${zA}$ is a linear operator is ${A}$ is a linear operator and ${z}$ is a complex number, and there is a zero element 0. An important additional result is that the vector space ${L_V}$ can be given a natural inner product structure, turning it into a Hilbert space.

1. Show that the function (,) on ${L_V\times L_V}$ defined by ${(A,B)\equiv \mathrm{tr}(A^\dagger B)}$ is an inner product function. This inner product is known as the Hilbert-Schmidt or trace inner product.
2. If ${V}$ has ${d}$ dimensions, show that ${L_V}$ has dimension ${d^2}$.
3. Find an orthonormal basis of Hermitian matrices for the Hilbert space ${L_V}$.

1. Show that the function (,) on ${L_V\times L_V}$ defined by ${(A,B)\equiv \mathrm{tr}(A^\dagger B)}$ is an inner product function. This inner product is known as the Hilbert-Schmidt or trace inner product.

An inner product must satisfy:

1. ${(.,.)}$ is linear in the second argument:

$\displaystyle \left(\left|v\right>,\sum_i\lambda_i\left|w_i\right>\right) = \sum_i\lambda_i\left(\left|v\right>,\left|w_i\right>\right)$

We can show this via:

$\displaystyle (A,zB)=\mathrm{tr}(A^\dagger zB)=\sum_{ik}A_{ik}^\dagger zB_{ki}=z\sum_{ik}A_{ik}^\dagger B_{ki}=z(A,B)$

2. ${(\left|v\right>,\left|w\right>) = (\left|w\right>,\left|v\right>)^*}$

$\displaystyle (A,B)=\sum_{ik}A_{ik}^\dagger B_{ki}=\sum_{ik}A_{ki}^*B_{ki}=\sum_{ik}B_{ik}^{*\dagger}A_{ki}^*=(B,A)^*$

3. ${(\left|v\right>,\left|v\right>) \geq 0}$, with equality iff ${\left|v\right>=0}$.

$\displaystyle (A,A)=\sum_{ik}A^\dagger_{ik}A_{ki}=\sum_{ik}A^*_{ki}A_{ki}=\sum_{ik}|A_{ik}|^2.$

This is a sum of strictly positive values, so that the sum is ${\geq 0}$, and is equal to zero iff all entries are zero.

2. If ${V}$ has ${d}$ dimensions, show that ${L_V}$ has dimension ${d^2}$: For ${L_V}$ to fully span ${V}$, it must be able to map any vector ${v_i}$ into itself and into any other vector ${v_j}$. Thus there must be ${d^2}$ different transformation.
3. Find an orthonormal basis of Hermitian matrices for the Hilbert space ${L_V}$. Here we can do something along the lines of the Gramm-Schmidt orthogonalization. Taking the ${L_V}$ linear operators, we can Hermitize each one via ${\tilde{A_i}=(A_i+A_i^\dagger)/2}$, and then orthogonalize each matrix ${\tilde{A_i}}$ to all of the ${\tilde{A_j}}$ for ${j=1,\dots,i-1}$ by subtracting the inner product representation given by ${(A,B)}$, above.

I think my answers to 2 and 3 are still incomplete. Please comment if you have suggestions.

## Nielsen and Chuang Exercise 2.38

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.38 Linearity of the trace: if ${A}$ and ${B}$ are two linear operators, show that ${\mathrm{tr}(A+B)=\mathrm{tr}(A)+\mathrm{tr}(B)}$, and, if ${z}$ is an arbitrary complex number, show that ${\mathrm{tr}(zA)=z\mathrm{tr}(A)}$.

$\displaystyle \mathrm{tr}(A+B)=\sum_i(A_{ii}+B_{ii}) = \sum_iA_{ii}+\sum_iB_{ii}=\mathrm{tr}(A)+\mathrm{tr}(B)$

$\displaystyle \mathrm{tr}(zA)=\sum_izA_{ii}=z\sum_{i}A_{ii}=z\mathrm{tr}(A).$