# Worked Problems in Physics

## Nielsen and Chuang Exercise 2.26

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.26 Let ${\left|\psi\right> = \frac{1}{\sqrt{2}}(\left|0\right>+\left|1\right>)}$. Write out ${\left|\psi\right>^{\otimes 2}}$ and ${\left|\psi\right>^{\otimes 3}}$ explicitly, both in terms of tensor products like ${\left|01\right>}$, and using the Kronecker product.

In terms of tensor products, ${\left|\psi\right>^{\otimes 2}=\frac{1}{2}(\left|00\right>+\left|10\right>+\left|01\right>+\left|11\right>)}$. Writing this as a Kronecker product yields: $\displaystyle \left|\psi\right>^{\otimes 2}= \frac{1}{2}\left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array} \right].$

Similarly, ${\left|\psi\right>^{\otimes 3}=\frac{1}{2\sqrt{2}}(\left|000\right>+\left|100\right>+\left|010\right>+\left|110\right>+\left|001\right>+\left|101\right>+\left|011\right>+\left|111\right>)}$. Writing this as a Kronecker product yields: $\displaystyle \left|\psi\right>^{\otimes 2}= \frac{1}{2\sqrt{2}}\left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1\\1 \\ 1 \\ 1 \\ 1 \end{array} \right].$

## Nielsen and Chuang Exercise 2.25

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.25 Show that for any operator ${A}$, ${A^\dagger A}$ is positive.

Because ${B=A^\dagger A=B^\dagger}$, ${BB^\dagger=B^\dagger B}$ trivially, and thus ${A^\dagger A}$ is normal, and thus has an eigendecomposition. We can therefore write: $\displaystyle A^\dagger Au=\lambda u.$

Multiplying both sides on the left by ${u^\dagger}$: $\displaystyle u^\dagger A^\dagger Au=\lambda u^\dagger u=\lambda.$

Defining ${v=Au}$ yields $\displaystyle v^\dagger v = |v|^2 = \lambda$

thus any eigenvalue may be taken to be nonnegative, and thus the matrix ${A^\dagger A}$ is positive.

## Nielsen and Chuang Exercise 2.24

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 24 Show that a positive operator is necessarily Hermitian. Hint: show that an arbitrary operator ${A}$ can be written ${A=B+iC}$, where ${B}$ and ${C}$ are Hermitian.

I don’t really get the point of this—obviously if a matrix has all nonnegative eigenvalues they are trivially real, and the matrix is Hermitian. However, I think this is something along the lines of what they want:

By defining the matrices ${B=\frac{1}{2}(A+A^\dagger)}$ and ${C=\frac{-i}{2}(A-A^\dagger)}$ we can write ${A=B+iC}$ where ${B}$ and ${C}$ is Hermitian. We can then show that the eigenvalues of ${B,C}$ are given by: $\displaystyle Bu_i = \lambda_iu_i$ $\displaystyle Cv_i = \mu_iv_i$

where ${\lambda_i, \mu_i}$ are real. Thus the eigenvalues of ${A}$ can be written as the sum of ${\lambda_i + i\mu_i}$. If the matrix is positive, the imaginary part must be zero, so that the eigenvalues are all real and the matrix is Hermitian

## Nielsen and Chuang Exercise 2.23

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.23 Show that the eigenvalues of a projector are all either 0 or 1.

We can write a projector as $\displaystyle P=\sum_i \left|i\right>\left

We may also define the projector of the orthogonal space ${Q=(1-P)}$ in a diagonal basis (exercise 2.22) $\displaystyle Q = \sum_j \left|j\right>\left

and can thus expand ${P}$ in an eigenbasis with either eigenvalue 1 (for the ${P}$ space), or eigenvalue 0 (for the ${Q}$ space) $\displaystyle P = \sum_i 1\left|i\right>\left\left

and thus the eigenvalues are either 0 or 1.

## Nielsen and Chuang Exercise 2.22

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.22 Prove that two eigenvectors of a Hermitian operator with different eigenvalues are necessarily orthogonal.

A Hermitian operator can act to the right or the left. Thus, if we have $\displaystyle Av_i = \lambda_i v_i \ \ \ \ \ (44)$ $\displaystyle Av_j = \lambda_j v_j, \ \ \ \ \ (45)$

then $\displaystyle \left = \lambda_i \left, \ \ \ \ \ (46)$

(acting to the right), or $\displaystyle \left = \lambda_j \left, \ \ \ \ \ (47)$

(acting to the left). Subtracting the two: $\displaystyle 0=\left-\left=(\lambda_i-\lambda_j)\left \ \ \ \ \ (48)$

so that either ${\lambda_i=\lambda_j}$ or ${\left=0}$