Chapter 1 Mathemtical Introduction

1.1 Linear Vector Spaces: Basics

Exercise 1.1.1 Verify these claims. For the first consider $|0\rangle+|0^{\prime}\rangle$ and use the advertised properties of the two null vectors in turn. For the second start with $|0\rangle=(0+1)|V\rangle+|-V\rangle$. For the third, begin with $|V\rangle+(-|V\rangle)=0|V\rangle=|0\rangle$. For the last, let $|W\rangle$ also satisfy $|V\rangle+|W\rangle=|0\rangle$. Since $|0\rangle$ is unique, this means $|V\rangle+|W\rangle=|V\rangle+|-V\rangle$. Take it from here.

Solution.

(1) $|0\rangle$ is unique.

Proof. For an arbitrary state ket $|V\rangle$,

(a) $|V\rangle+|0\rangle=|V\rangle$

(b) $|V\rangle+|0^{\prime}\rangle=|V\rangle$

Set $|V\rangle=|0^{\prime}\rangle$ in (a) $\Rightarrow$ $|0^{\prime}\rangle+|0\rangle=|0^{\prime}\rangle$;

Set $|V\rangle=|0\rangle$ in (ii) $\Rightarrow$ $|0\rangle+|0^{\prime}\rangle=|0\rangle$;

Therefore, by commutativity of vector addition, we have

(2) $0|V\rangle=|0\rangle$

Proof. We have

where $1|V\rangle=|V\rangle$. Therefore

Since $|V\rangle$ is arbitrary here, compared with the definition of $|0\rangle$, which is $|V\rangle+|0\rangle=|V\rangle$ for any $|V\rangle$, plus $|0\rangle$ is unique, we can conclude that

(3) $|-V\rangle=-|V\rangle$

Proof. For arbitrary $|V\rangle$,

By definition of $|-V\rangle$, which is $|V\rangle+|-V\rangle=0$, we can conclude that

(4) $|-V\rangle$ is the unique addtive inverse of $|V\rangle$.

Proof. Suppose there exists another vector $|W\rangle$, satisfying $|W\rangle=-|V\rangle$, then

Add $|-V\rangle$ on the both sides, we have

Therefore,

Exercise 1.1.2 Consider the set of all entities of the form $(a, b, c)$ where the entries are real numbers. Addition and scalar multiplication are defined as follows:

Write down the null vector and inverse of $(a, b, c)$. Show that vectors of the form $(a, b, 1)$ do not form a vector space.

Solution.

• Null vector of $(a,b,c)$: By definition, for any $|V\rangle$,

Set $|0\rangle=(a_{0},b_{0},c_{0})$, $|V\rangle=(a,b,c)$, where $a,b,c$ are arbitrary numbers. Then

Therefore, the null vector of $(a,b,c)$ is $(0,0,0)$.

• Inverse vector of $(a,b,c)$: Suppose the inverse vector of $(a,b,c)$ is $(\bar{a},\bar{b},\bar{c})$. By definition,

Therefore, the inverse vector of $(a,b,c)$ is $(-a,-b,-c)$.

• $\{(a,b,1)\}$ does not form a vector space since

(a) It violates the closure under addition, i.e.

(b) It violates the closure under scalar multiplication, i.e.

as long as $\omega\neq 1$. (c) There is no null vector, i.e.

(d) The inverse does not exist, i.e.

Exercise 1.1.3 Do functions that vanish at the end points $x=0$ and $x=L$ form a vector space? How about periodic functions obeying $f(0)=f(L)$ ? How about functions that obey $f(0)=4$ ? If the functions do not qualify, list the things that go wrong.

Solution.

(1) $\{f(x)\}$, $f(0)=f(L)=0$, form a vector space.

(2) $\{f(x)\}$, periodic functions obeying $f(0)=f(L)$, form a vector space. If you want to prove the property of closure in this problem, please mention that $f(x)+g(x)$ and $\alpha f(x)$ are also periodic functions. That is, $f(0)+g(0)=f(L)+g(L)$, $\alpha f(0)=\alpha f(L)$.

(3) $\{f(x)\}$, $f(0)=4$, do not form a vector space, since

(a) If $g(x), h(x)\in \{f(x)\}$, then $g(x)+h(x)\notin \{f(x)\}$, since $g(0)+h(0)=8\neq 4$.

(b) If $g(x)\in\{f(x)\}$, then $\lambda g(x)\notin \{f(x)\}$, since $\lambda g(0)=4\lambda\neq 4$, as long as $\lambda\neq 1$.

(c) No null vector. $g(x)\equiv 0\notin \{f(x)\}$, since $g(0)=0\neq 4$.

(d) If $g(x)\in \{f(x)\}$, then the inverse $-g(x)\notin \{f(x)\}$, since $-g(0)=-4\neq 4$.

Exercise 1.1.4 Consider three elements from the vector space of real $2 \times 2$ matrices:

Are they linearly independent? Support your answer with details. (Notice we are calling these matrices vectors and using kets to represent them to emphasize their role as elements of a vector space.)

Solution. Suppose $\alpha_{1}|1\rangle+\alpha_{2}|2\rangle+\alpha_{3}|3\rangle=0$. We have

It is not necessary for $\alpha_{1}, \alpha_{2}$ and $\alpha_{3}$ to be $0$ together. Therefore, $|1\rangle$, $|2\rangle$ and $|3\rangle$ are linearly dependent.

Exercise 1.1.5 Show that the following row vectors are linearly dependent: $(1,1,0)$, $(1,0,1)$, and $(3,2,1)$. Show the opposite for $(1,1,0),(1,0,1)$, and $(0,1,1)$.

Solution. Suppose $\alpha_{1}(1,1,0)+\alpha_{2}(1,0,1)+\alpha_{3}(3,2,1)=0$. Then

When $\alpha_{3}\neq 0$. $\alpha_{1}, \alpha_{2}, \alpha_{3}$ can have non-zero values. Therefore, $(1,1,0)$, $(1,0,1)$, $(3,2,1)$ are linearly dependent.

Suppose $\alpha_{1}(1,1,0)+\alpha_{2}(1,0,1)+\alpha_{3}(0,1,1)=0$. Then

Therefore, $(1,1,0)$, $(1,0,1)$, $(0,1,1)$ are linearly independent.

1.2 Inner Product Spaces

1.3 Dual Spaces and the Dirac Notation

Exercise 1.3.1 Form an orthonormal basis in two dimensions starting with $\vec{A}=3 \vec{i}+4 \vec{j}$ and $\vec{B}=2 \vec{i}-6 \vec{j}$. Can you generate another orthonormal basis starting with these two vectors? If so, produce another.

Solution. Using Gram-Schmidt process here, starting from $\vec{A}$.

Therefore, the new basis is $\vec{e}_{1}=\frac{3}{5}\vec{i}+\frac{4}{5}\vec{j}$, $\vec{e}_{2}=\frac{4}{5}\vec{i}-\frac{3}{5}\vec{j}$.

Exercise 1.3.2 Show how to go from the basis

to the orthonormal basis

Solution.

Exercise 1.3.3 When will this equality $\langle V \mid V\rangle=\frac{\langle W \mid V\rangle\langle V \mid W\rangle}{|W|^2}$ be satisfied? Does this agree with your experience with arrows?

Solution. When $|V\rangle=C|W\rangle$, we have

Also

Hence,

When two arrows are parallel or anti-parallel with each other, the square of their inner product equals to the product of their norms.

Exercise 1.3.4 Prove the triangle inequality starting with $|V+W|^{2}$. You must use $\mathrm{Re}\langle V|W\rangle\leqslant |\langle V|W\rangle|$ and the Schwarz inequality. Show that the final inequality becomes an equality only if $|V\rangle=a|W\rangle$ where $a$ is a real positive scalar.

Solution. $\mathrm{Re}\langle V\mid W\rangle\leqslant |\langle V\mid W\rangle|\leqslant |V||W|$

Add $|V|^{2}+|W|^{2}$ to both sides of the inequality above, we have

Therefore,

Attention: We are supposed to prove the equality holds only if $|V\rangle=\alpha|W\rangle$, where $\alpha$ is a real number.

The equality holds only if the following two equalities hold:

(a) $\langle V\mid W\rangle=|V||W|$

(b) $\mathrm{Re}\langle V\mid W\rangle=|\langle V\mid W\rangle|$

From the proof process of the Schwarz inequality in Shankar, we know that equality (a) holds only if

which means $|V\rangle$ must be able to expressed as $\alpha |W\rangle$, where $\alpha$ is a number. To prove that $\alpha$ must be real, we substitude $|V\rangle=\alpha |W\rangle$ into equality (b) above.

To satisfy equality (b), $\langle V\mid W\rangle$ must be real. Since $\langle V\mid V\rangle$ is real, $\alpha^{\star}$ must be real. Therefore, $\alpha$ must be a real number.

1.4 Subspaces

Exercise 1.4.1 In a space $\mathbb{V}^n$, prove that the set of all vectors $\left\{\left|V_{\perp}^1\right\rangle,\left|V_{\perp}^2\right\rangle, \ldots\right\}$, orthogonal to any $|V\rangle \neq |0\rangle$, form a subspace $\mathbb{V}^{n-1}$.

Solution. Given a vector space $\mathbb{V}^{n}$, one can start with an arbitrary vector $|V\rangle\neq 0$ and construct $n-1$ other vectors orthogonal to this $|V\rangle$ through Gram-Schmidt process. Since these $n-1$ vectors are linear independent, they span a $\mathbb{V}^{n-1}$ subspace. Now we prove that this subspace $\mathbb{V}^{n-1}$ is the set of all vectors orthogonal to $|V\rangle$.

(1) Every vector in $\mathbb{V}^{n-1}$ is orthogonal to $|V\rangle$: Since every vector in $\mathbb{V}^{n-1}$ can be expressed as a linear combination of the $n-1$ vectors we constructed above $|V_{\perp}\rangle=\sum\limits_{i=1}^{n-1}\alpha_{i}|V_{i}\rangle$, and $\langle V\mid V_{\perp}\rangle=\sum\limits_{i=1}^{n-1}\alpha_{i}\langle V\mid V_{i}\rangle=0$, because $\langle V\mid V_{i}\rangle=0$ for each $V_{i}$.

(2) Every vector in $\mathbb{V}^{n}$ but outside $\mathbb{V}^{n-1}$ is not orthogonal to $|V\rangle$: Since this kind of vectors can be expressed as $|W\rangle=\alpha|V\rangle+\sum\limits_{i=1}^{n-1}\alpha_{i}|V_{i}\rangle$, where $\alpha\neq 0$. Therefore $\langle V\mid W\rangle=\alpha\langle V\mid V\rangle=\alpha\neq 0$.

Exercise 1.4.2 Suppose $\mathbb{V}_1^{n_1}$ and $\mathbb{V}_2^{n_2}$ are two subspaces such that any element of $\mathbb{V}_1$ is orthogonal to any element of $\mathbb{V}_2$. Show that the dimensionality of $\mathbb{V}_1 \oplus \mathbb{V}_2$ is $n_1+n_2$. (Hint: Theorem 4.)

Solution. Since $\mathbb{V}_{1}^{n_{1}}$ and $\mathbb{V}_{2}^{n_{2}}$ are two subspace orthogonal to each other, we can take the $n_{1}$ basis vectors of $\mathbb{V}_{1}^{n_{1}}$ and the $n_{2}$ basis vectors of $\mathbb{V}_{2}^{n_{2}}$, and put them together. Because these $n_{1}+n_{2}$ vectors are orthogonal to each other, they can span a $\mathbb{V}^{n_{1}+n_{2}}$ subspace. We now prove that this $\mathbb{V}^{n_{1}+n_{2}}$ is nothing but the $\mathbb{V}_{1}\oplus \mathbb{V}_{2}$.

(1) Every vector in $\mathbb{V}^{n_{1}+n_{2}}$ is in $\mathbb{V}_{1}\oplus \mathbb{V}_{2}$:

Since each vector in $\mathbb{V}^{n_{1}+n_{2}}$ can be expressed as $|U\rangle=\sum\limits_{i=1}^{n_{1}}\alpha_{i}|V_{i}\rangle+\sum\limits_{j=1}^{n_{2}}\beta_{j}|W_{j}\rangle$, where $\{|V_{i}\rangle\}$, $\{|W_{j}\rangle\}$ are basis vectors of $\mathbb{V}_{1}$ and $\mathbb{V}_{2}$ respectively. Notice that $\sum\limits_{i=1}^{n_{1}}\alpha_{i}|V_{i}\rangle$ is a vector in $\mathbb{V}_{1}$, and $\sum\limits_{j=1}^{n_{2}}b_{j}|W_{j}\rangle$ is a vector in $\mathbb{V}_{2}$. Therefore $|U\rangle$ can be expressed as a combination of vectors from $\mathbb{V}_{1}$ and $\mathbb{V}_{2}$. According to the definition of $\mathbb{V}_{1}\oplus \mathbb{V}_{2}$ (Definition 12 in Shankar), $|U\rangle\in \mathbb{V}_{1}\oplus \mathbb{V}_{2}$.

(2) Every vector in $\mathbb{V}_{1}\oplus \mathbb{V}_{2}$ is in $\mathbb{V}^{n_{1}+n_{2}}$:

Every vector in $\mathbb{V}_{1}\oplus \mathbb{V}_{2}$ can be expressed as $|Z\rangle=C_{1}|Z_{1}\rangle+C_{2}|Z_{2}\rangle$, where $|Z_{1}\rangle\in\mathbb{V}_{1}^{n_{1}}$, |$Z_{2}\rangle\in \mathbb{V}_{2}^{n_{2}}$. Therefore, $|Z_{1}\rangle=\sum\limits_{i=1}^{n_{1}}p_{i}|V_{i}\rangle$, and $|Z_{2}\rangle=\sum\limits_{j=1}^{n_{2}}q_{j}|W_{j}\rangle$. Thus, $|Z\rangle=\sum\limits_{i=1}^{n_{1}}C_{1}p_{i}|V_{i}\rangle+\sum\limits_{j=1}^{n_{2}}C_{2}q_{j}|W_{j}\rangle$, which lies in $\mathbb{V}^{n_{1}+n_{2}}$.

Therefore, by theorem 4 in Shankar, there are $n_{1}+n_{2}$ orthogonal vectors in $\mathbb{V}_{1}\oplus \mathbb{V}_{2}$, so the dimension of $\mathbb{V}_{1}\oplus \mathbb{V}_{2}$ is $n_{1}+n_{2}$.

1.5 Linear Operators

1.6 Matrix Elements of Linear Operators

Exercise 1.6.1 An operator $\Omega$ is given by the matrix

What is its action?

Solution. To see $\Omega$'s action, let's act on basis vectors:

This is a cyclic permutation of the three basis vectors.

It is equivalent to rotation of the coordinate axis along $(1,1,1)$ by $\frac{2\pi}{3}$.

Exercise 1.6.2 Given $\Omega$ and $\Lambda$ are Hermitian what can you say about (1) $\Omega \Lambda$; (2) $\Omega \Lambda+\Lambda \Omega$; (3) $[\Omega, \Lambda]$; and (4) $\mathrm{i}[\Omega, \Lambda]$ ?

Solution.

(1) Not Hermitian: $(\Omega\Lambda)^{\dagger}=\Lambda^{\dagger}\Omega^{\dagger}=\Lambda\Omega\neq \Omega\Lambda$

(2) Hermitian:

(3) Anti-Hermitian:

(4) Hermitian:

Exercise 1.6.3 Show that a product of unitary operator is unitary.

Solution. Suppose $U_{1}, U_{2}$ are unitary, which means that

Therefore,

Hence a product of unitary operator is unitary.

Exercise 1.6.4 It is assumed that you know (1) what a determinant is, (2) that $\det\Omega^\mathrm{T}=\det\Omega$ ($\mathrm{T}$ denotes transpose), (3) that the determinant of a product of matrices is the product of the determinants. [If you do not, verify these properties for a two-dimensional case

with $\det\Omega=(\alpha \delta-\beta \gamma)$.] Prove that the determinant of a unitary matrix is a complex number of unit modulus.

Solution. Suppose $U$ is the unitary matrix, which means that it satisfies

Take determinant of the both sides, we get

Therefore, $\det (U)$ is a complex number of unit modulus.

Exercise 1.6.5 Verify that $R\left(\frac{1}{2} \pi \mathbf{i}\right)$ is unitary (orthogonal) by examining its matrix.

Solution. We know from Example 1.6.1,

Therefore,

Hence, $R\left(\frac{1}{2}\pi \mathbf{i}\right)$ is unitary.

Exercise 1.6.6 Verify that the following matrices are unitary:

Verify that the determinant is of the form $e^{\mathrm{i} \theta}$ in each case. Are any of the above matrices Hermitian?

Solution.

• $\frac{1}{2^{1 / 2}}\begin{pmatrix} 1 & \mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix}$ is unitary, since

The determinant of $\frac{1}{2^{1 / 2}}\begin{pmatrix} 1 & \mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix}$ is of $\mathrm{e}^{\mathrm{i}\theta}$ form, since

$\frac{1}{2^{1 / 2}}\begin{pmatrix} 1 & \mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix}$ is not Hermitian, since

• $\frac{1}{2}\begin{pmatrix} 1+\mathrm{i} & 1-\mathrm{i} \\ 1-\mathrm{i} & 1+\mathrm{i} \end{pmatrix}$ is unitary, since

The determinant of $\frac{1}{2}\begin{pmatrix} 1+\mathrm{i} & 1-\mathrm{i} \\ 1-\mathrm{i} & 1+\mathrm{i} \end{pmatrix}$ is of $\mathrm{e}^{\mathrm{i}\theta}$ form, since

where $\theta=2k\pi+\frac{\pi}{2}$ for $k\in\mathbb{Z}$. $\frac{1}{2}\begin{pmatrix} 1+\mathrm{i} & 1-\mathrm{i} \\ 1-\mathrm{i} & 1+\mathrm{i} \end{pmatrix}$ is not Hermitian, since

1.7 Active and Passive Transformations

Exercise 1.7.1 The trace of a matrix is defined to be the sum of its diagonal matrix elements

Show that

(1) $\operatorname{Tr}(\Omega \Lambda)=\operatorname{Tr}(\Lambda \Omega)$.

(2) $\operatorname{Tr}(\Omega \Lambda \theta)=\operatorname{Tr}(\Lambda \theta \Omega)=\operatorname{Tr}(\theta \Omega \Lambda)$ (The permutations are \textit{cyclic}).

(3) The trace of an operator is unaffected by a unitary change of basis $|i\rangle \rightarrow U|i\rangle$. [Equivalently, show $\operatorname{Tr} \Omega=\operatorname{Tr}\left(U^{\dagger} \Omega U\right)$.]

Solution. (1) $\operatorname{Tr}(\Omega \Lambda)=\sum\limits_{i}(\Omega\Lambda)_{ii}=\sum\limits_{i}\sum\limits_{j}\Omega_{ij}\Lambda_{ji}=\sum\limits_{j}\sum\limits_{i}\Lambda_{ji}\Omega_{ij}=\sum\limits_{j}(\Lambda\Omega)_{jj}=\operatorname{Tr}(\Lambda\Omega)$.

(2)

(3) $\operatorname{Tr}(U^{\dagger}\Omega U)=\operatorname{Tr}(\Omega UU^{\dagger})=\operatorname{Tr}(\Omega\mathbb{I})=\operatorname{Tr}(\Omega)$.

Exercise 1.7.2 Show that the determinant of a matrix is unaffected by a unitary change of basis. [Equivalently show $\det \Omega=\det\left(U^{\dagger} \Omega U\right)$.]

Solution.

1.8 The Eigenvalue Problem

Exercise 1.8.1 (1) Find the eigenvalues and normalized eigenvectors of the matrix

(2) Is the matrix Hermitian? Are the eigenvectors orthogonal?

Solution.

(1) To find the eigenvalues and normalized eigenvectors of the matrix $\Omega$, we can compute the characteristic equation

So the eigenvalues are

The eigenvectors corresponding eigenvalues are

(2) Matrix $\Omega$ is not Hermitian, since

The eigenvectors are not orthogonal, since

Exercise 1.8.2 Consider the matrix

(1) Is it Hermitian?

(2) Find its eigenvalues and eigenvectors.

(3) Verify that $U^{\dagger} \Omega U$ is diagonal, $U$ being the matrix of eigenvectors of $\Omega$.

Solution. (1) Matrix $\Omega$ is Hermitian since

(2) To find the eigenvalues and eigenvectors of the matrix $\Omega$, we can compute the characteristic equation

Therefore, eigenvalues are

The eigenvectors corresponding to eigenvalues are

(3) If $U$ is the matrix of eigenvectors of $\Omega$, then

We can compute

This is a diagonal matrix.

Exercise 1.8.3 Consider the Hermitian matrix

(1) Show that $\omega_{1}=\omega_{2}=1$; $\omega_{3}=2$.

(2) Show that $|\omega=2\rangle$ is any vector of the form

(3) Show that the $\omega=1$ eigenspace contains all vectors of the form

either by feeding $\omega=1$ into the equations or by requiring that the $\omega=1$ eigenspace be orthogonal to $|\omega=2\rangle$.

Solution. (1) The characteristic equation is

Then the eigenvalues are

(2) To get the eigenvector corresponding to eigenvalue $\omega=2$, we need to solve the equation

Set $x_{2}=a$, we have $x_{3}=-a$. Therefore,

(3) For $\omega=1$:

$x_{1}$ is arbitrary, set $x_{1}=b$. $x_{2}-x_{3}=0$, set $x_{2}=c$, then $x_{3}=c$. Therefore, the eigenvector corresponding $\omega=1$ is of the form

Exercise 1.8.4 An arbitrary $n \times n$ matrix need not have $n$ eigenvectors. Consider as an example

(1) Show that $\omega_1=\omega_2=3$.

(2) By feeding in this value show we get only one eigenvector of the form

We cannot find another one that is linear independent.

Solution. (1) The characteristic equation is

Thus the eigenvalues are

(2) By feeding this eigenvalue $\omega=3$, we get the equation

Set $x_{1}=a$, we have $x_{2}=-a$. Therefore, the eigenvector is of the form

This is the only eigenvector we can find.

Exercise 1.8.5 Consider the matrix

(1) Show that it is unitary.

(2) Show that its eigenvalues are $\mathrm{e}^{\mathrm{i} \theta}$ and $\mathrm{e}^{-\mathrm{i} \theta}$.

(3) Find the corresponding eigenvectors; show that they are orthogonal.

(4) Verify that $U^{\dagger} \Omega U=$ (diagonal matrix), where $U$ is the matrix of eigenvectors of $\Omega$.

Solution. (1) Matrix $\Omega$ is unitary, since

(2) Solve the characteristic equation

By Euler's formula, we get the eigenvalues

(3) By feeding this eigenvalue, we get the equations

Thus the corresponding eigenvectors are

They are orthogonal since

(4) The matrix of eigenvectors of $\Omega$ is

Then

is diagonal.

Exercise 1.8.6 (1) We have seen that the determinant of a matrix is unchanged under a unitary change of basis. Argue now that

for a Hermitian or unitary $\Omega$.

(2) Using the invariance of the trace under the same transformation, show that

Solution. (1) Suppose $U$ is the unitary matrix that transforms $\Omega$ into a diagonal matrix $D$ with $\Omega$'s eigenvalues $\omega_{i}$ on its diagonal. Then

(2) By using the same transformation, we have

Exercise 1.8.7 By using the results on the trace and determinant from the last problem, show that the eigenvalues of the matrix

are $3$ and $-1$. Verify this by explicit computation. Note that the Hermitian nature of the matrix is an essential ingredient.

Solution. According to Exercise 1.8.6, we have

Solving the equation, we get

For verification, we can calculate the characteristic equation

We can get the eigenvalues

Exercise 1.8.8 Consider Hermitian matrices $M^1, M^2, M^3, M^4$ that obey

(1) Show that the eigenvalues of $M^i$ are $\pm 1$. (Hint: go to the eigenbasis of $M^i$, and use the equation for $i=j$.)

(2) By considering the relation

show that $M^i$ are traceless. [Hint: $\operatorname{Tr}(ACB)=\operatorname{Tr}(CBA)$.]

(3) Show that they cannot be odd-dimensional matrices.

Solution. (1) Start with equation

Take $i=j$, we get

Apply $M^{i}M^{i}$ to eigenvector $|\omega\rangle$ of $M^{i}$, we have

Therefore,

(2) From the relation

We can take the trace of $M^{i}$ to get

$M^{i}$ is traceless.

(3) According to 1.8.6,

where $n$ is the dimension of the matrix. Since $\omega_{k}=\pm 1$, $\operatorname{Tr} M^{i}$ can be zero only if $n$ is even. (The sum of an odd number of odd numbers is still odd, and cannot be zero.)

Exercise 1.8.9 A collection of masses $m_\alpha$, located at $\mathbf{r}_\alpha$ and rotating with angular velocity $\boldsymbol{\omega}$ around a common axis has an angular momentum

where $\mathbf{v}_\alpha=\boldsymbol{\omega} \times \mathbf{r}_\alpha$ is the velocity of $m_\alpha$. By using the identity

show that each Cartesian component $l_i$ of $\mathbf{l}$ is given by

where

or in Dirac notation

(1) Will the angular momentum and angular velocity always be parallel?

(2) Show that the moment of inertia matrix $M_{i j}$ is Hermitian.

(3) Argue now that there exist three directions for $\boldsymbol{\omega}$ such that $\mathbf{l}$ and $\boldsymbol{\omega}$ will be parallel. How are these directions to be found?

(4) Consider the moment of inertia matrix of a sphere. Due to the complete symmetry of the sphere, it is clear that every direction is its eigendirection for rotation. What does this say about the three eigenvalues of the matrix $M$?

Solution. Start from the angular momentum

Writing in components, we get

where $M_{ij}\equiv\sum_{\alpha} m_{\alpha}[r_{\alpha}^{2}\,\delta_{ij}-(\mathbf{r}_{\alpha})_{i}\,(\mathbf{r}_{\alpha})_{j}]$. Or in Dirac notation,

(1) No. The angular momentum and angular velocity are not parallel unless $|\omega\rangle$ is an eigenvector of $M

(2) $M_{ji}^{\star}=(\sum_{\alpha} m_{\alpha}[r_{\alpha}^{2}\,\delta_{ji}-(\mathbf{r}_{\alpha})_{j}\,(\mathbf{r}_{\alpha})_{i}])^{\star}=\sum_{\alpha} m_{\alpha}[r_{\alpha}^{2}\,\delta_{ij}-(\mathbf{r}_{\alpha})_{i}\,(\mathbf{r}_{\alpha})_{j}]=M_{ij}$.

(3) Since $M$ is Hermitian, we can always find three eigenvectors which are orthogonal to each other by solving the eigen-problem $M|\omega\rangle=\omega|\omega\rangle$. And these three eigenvectors denote the three directions for $\boldsymbol{\omega}$ we want to find in the 3-dimensional Euclidean space.

(4) The complete symmetry of sphere means all directions are equivalent eigendirections. Therefore, the eigenvalues are degenerate.

Exercise 1.8.10 By considering the commutator, show that the following Hermitian matrices may be simultaneously diagonalized. Find the eigenvectors common to both and verify that under a unitary transformation to this basis, both matrices are diagonalized.

Since $\Omega$ is degenerate and $\Lambda$ is not, you must be prudent in deciding which matrix dictates the choice of basis.

Solution. Since the two Hermitian matrices commute

They can be diagonalized simultaneously. We choose $\Lambda$'s characteristic equation

The eigenvalues are

Then the eigenvectors corresponding the eigenvalues are

Then the matrix of eigenvectors of $\Lambda$ is

To verify $\Omega$ and $\Lambda$ are simultanelously diagonalized:

Exercise 1.8.11 Consider the coupled mass problem discussed above.

(1) Given that the initial state is $|1\rangle$, in which the first mass is displaced by unity and the second is left alone, calculate $|1(t)\rangle$ by following the algorithm.

(2) Compare your result with that following from Eq. (1.8.39).

Solution.

(1) Equation of motion

Set

we have

The corresponding eigenvectors are

Then the matrix of eigenvectors is

It can diagonalize the original matrix

In eigenbasis,

In this problem,

We first transform it into eigenbasis

In the eigenbasis, the state evolves according to ($\star$).

Then we transform it back to the original basis:

(2) By subsitituting $\begin{pmatrix} x_{1}(0)\\ x_{2}(0) \end{pmatrix}=\begin{pmatrix} 1\\0 \end{pmatrix}$ in equation (1.8.39), we can get the same solution:

Exercise 1.8.12 Consider once again the problem discussed in the previous example. (1) Assuming that

has a solution

find the differential equation satisfied by $U(t)$. Use the fact that $|x(0)\rangle$ is arbitrary.

(2) Assuming (as is the case) that $\Omega$ and $U$ can be simultaneously diagonalized, solve for the elements of the matrix $U$ in this common basis and regain Eq. (1.8.43). Assume $|\dot{x}(0)\rangle=0$.

Solution. (1) Assuming that

has a solution

Then we can get

Since $|x(0)\rangle$ is arbitrary, we get the differential equation

(2) From Exercise 1.8.11, we know the $\Lambda=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)$ can diagonalize $\Omega$, and therefore, it can also diagonalize $U$.

In this common basis, we have

Then

Since $|x(0)\rangle$ is arbitrary, we have

which means

To satisfy that $U$ is unitary, we have

Therefore,

which is the same as equation (1.8.43).

1.9 Functions of Operators and Related Concepts

Exercise 1.9.1 We know that the series

may be equated to the function $f(x)=(1-x)^{-1}$ if $|x|<1$. By going to the eigenbasis, examine when the $q$ number power series

of a Hermitian operator $\Omega$ may be identified with $(1-\Omega)^{-1}$.

Solution. In the eigenbasis,

where $\omega_{i}$ are eigenvalues.

The third equality holds if and only if $|\omega_{i}|<1$ for $i=1,\ldots,m$. Therefore, $f(\Omega)$ can be defined as $\frac{1}{1-\Omega}$ if and only if the absolute value of each of $\Omega$'s eigenvalues is less than $1$.

Exercise 1.9.2 If $H$ is a Hermitian operator, show that $U=\mathrm{e}^{\mathrm{i} H}$ is unitary. (Notice the analogy with $c$ numbers: if $\theta$ is real, $u=\mathrm{e}^{\mathrm{i} \theta}$ is a number of unit modulus.)

Solution. Since $H$ is Hermitian, it satisfies

We can compute

Then(The second equality holds only for commuting operators.)

Therefore $U$ is unitary.

Exercise 1.9.3 For the case above, show that $\det U=\mathrm{e}^{\mathrm{i}\operatorname{Tr} H}$.

Solution. In the eigenbasis of $H$,

where $\epsilon_{1},\ldots,\epsilon_{m}$ are eigenvalues of $H$, i.e.

Therefore,

1.10 Generalization to Infinite Dimensions

Exercise 1.10.1 Show that $\delta(a x)=\delta(x) /|a|$. [Consider $\int \delta(a x)\,\mathrm{d}(a x)$. Remember that $\delta(x)=\delta(-x)$.] Solution. Since $\delta(x)=\delta(-x)$, we have

Therefore,

Thus,

Exercise 1.10.2 Show that

where $x_i$ are the zeros of $f(x)$. Hint: Where does $\delta(f(x))$ blow up? Expand $f(x)$ near such points in a Taylor series, keeping the first nonzero term.

Solution. Expand $f(x)$ around $x_{i}$, where $f(x_{i})=0$:

Introduce a test function $g(x)$,\footnote{We use the Exercise 1.10.1 at the last equality.}

Therefore,

Exercise 1.10.3 Consider the theta function $\theta(x-x^{\prime})$ which vanishes if $x-x^{\prime}$ is negative and equals 1 if $x-x^{\prime}$ is positive. Show that $\delta(x-x^{\prime})=\frac{\mathrm{d}}{\mathrm{d} x} \theta(x-x^{\prime})$.

Solution. Introduce a test function\footnote{$g(-\infty)$ and $g(\infty)$ are finite} $g(x)$, we have

Therefore,

Exercise 1.10.4 A string is displaced as follows at $t=0$ :

Show that

Solution. We start from equation (1.10.55)

Then

From equation (1.10.55), we have

Therefore,

where

Then

Therefore,