Life with Physics

Recall that the Lie group representation only depends a linear combination of generators as in its  exponential form   ${\cal{D}}(\alpha)= e^{i\,\alpha^a {\cal{T}}_a}$, we are able to choose a particular set of generators for the classification of Lie algebra. We consider the compact Lie groups whose representation can always be unitary and thus their generators are Hermitian ${\cal{T}}_a={\cal{T}}^{\dagger}_a$. Cartan generators Out of all the generators, we can construct a maximal subset of mutually commuting Hermitian generators (Cartan generators), ${\cal{H}}_i$ for $i=1,\cdots, m$, such that \begin{equation}{\cal{H}}_i={\cal{H}}^{\dagger}_i\,,\quad \Big[{\cal{H}}_i\,,{\cal{H}}_j\Big]=0\,,\quad\text{Tr}\left({\cal{H}}_i{\cal{H}}_j\right)=k_D\,\delta_{ij}\,\tag{1}\end{equation} for some positive constant $k_D$. The construction of such Cartan generators is in the following steps:  From the generators ${\cal{T}}_1, \cdots,{\cal{T}}_N$, we can pick a maximal subset of mu...

This is a simple exercise before jumping into the general theory of Lie algebra classification. The commutator relations in Lie algebra $\mathfrak{su}(2)$ are \begin{equation}\Big[{\cal{J}}_a\,,{\cal{J}}_b\Big]=i\,\epsilon_{abc}\,{\cal{J}}_c\end{equation} for $a,b,c=1,2,3$. We construct an irreducible representation in the following steps:  Pick any operator, say ${\cal{J}}_3$, and denote its eigenvector by $|m\rangle$: \begin{equation}{\cal{J}}_3\,|m\rangle = m\,|m\rangle\,.\tag{1}\end{equation} For the remaining ${\cal{J}}_1$ and ${\cal{J}}_2$, construct a set of new operators \begin{equation}{\cal{J}}_{\pm}\equiv {\cal{J}}_1 \pm i {\cal{J}}_2\,.\tag{2}\end{equation} The commutator relations become \begin{eqnarray}\Big[{\cal{J}}_3\,,{\cal{J}}_{\pm}\Big]=\pm {\cal{J}}_{\pm}\,,\quad \Big[{\cal{J}}_+\,,{\cal{J}}_-\Big]=2\,{\cal{J}}_3\,.\tag{3}\end{eqnarray} From the relation ${\cal{J}}_3\,{\cal{J}}_{\pm}\,|m\rangle = \Big[{\cal{J}}_3\,,{\cal{J}}_{\pm}\Big]\,|m\rangle + {\cal{J}}_{\p...

Exponential Map Intuitively, we can simply view Lie groups $G$ as the groups of elements $g(\alpha)\in G$ that are parameterized continuously by a set of real numbers $\alpha$. We denote their representation s by $\cal{D}(\alpha)$, which can be constructed in the following steps: We choose $\alpha$ such that $g(0)=e$. As a result, ${\cal{D}}(0)=\cal{I}$ is the identity operator. For small parameters $\epsilon$, we are able to Taylor expand to the first order of each component  $\epsilon^a$: \begin{equation}{\cal{D}}(\epsilon)={\cal{I}}+i\sum_{a=1}^N\epsilon^a \cal{T}_a\,\end{equation} where ${\cal{T}}_1,\cdots , {\cal{T}}_N$ are operators called generators . For finite parameters $\alpha$, the group property allows to make $k$ sequential transformations, each with small parameters $\alpha/k$. As a result, \begin{equation}{\cal{D}}(\alpha)=\lim_{k\rightarrow\infty}{\cal{D}}^k(\alpha/k)=\lim_{k\rightarrow\infty}\left({\cal{I}}+\frac{i}{k}\sum_{a=1}^N\alpha^a {\cal{T}}_a\rig...

The representation of a group $G$ is a mapping from any $g\in G$ to a linear operator $\cal{D}(g)$ that preserves the group multiplication. Note: Recall that operators are linear transformations that map one vector to another vector. So the multiplication between two operators $\cal{D}(g_1)\cal{D}(g_2)$ is defined as two sequential transformations \begin{equation}\cal{D}(g_1)\cal{D}(g_2)\,\,|\psi\rangle:= \cal{D}(g_1)\,\Big(\cal{D}(g_2)|\psi\rangle\Big)\end{equation} when acting on any vector $|\psi\rangle$. The requirement of preserving the group multiplication means \begin{equation}\cal{D}(g_1)\cal{D}(g_2)=\cal{D}(g_1g_2)\,\tag{1}\end{equation} for any $g_1,g_2\in G$. [Exercise 1] Prove that the representation forms a group. [Solution]  Closure: By Eq. (1). Associativity: \begin{eqnarray}\cal{D}(g_1)\Big(\cal{D}(g_2)\cal{D}(g_3)\Big)=\cal{D}(g_1)\cal{D}(g_2g_3)&=&\cal{D}(g_1(g_2g_3))\\&=&\cal{D}((g_1g_2)g_3)=\cal{D}(g_1g_2)\cal{D}(g_3)=\Big(\cal{D}(g_1)\cal{D}(g_...

More about Orthonormal Basis First of all, consider any vector $|\psi\rangle$ expanded within an orthonormal basis $|\psi\rangle=\sum_{i=1}^n\psi_i|\epsilon_i\rangle$. Because of the orthonormal property \begin{equation}\langle \epsilon_i |\epsilon_j\rangle=\delta_{ij}\,,\tag{1}\end{equation} by applying $\langle\epsilon_i|$ on both sides of the expansion , we obtain $\psi_i = \langle \epsilon_i |\psi\rangle$. Since the relation \begin{equation}|\psi\rangle =\sum_{i=1}^n\psi_i|\epsilon_i\rangle = \sum_{i=1}^n \langle \epsilon_i |\psi\rangle\,|\epsilon_i\rangle = \sum_{i=1}^n |\epsilon_i\rangle\langle\epsilon_i |\psi\rangle\end{equation} holds for any $|\psi\rangle$, we obtain the completeness property \begin{equation} \sum_{i=1}^n |\epsilon_i\rangle\langle\epsilon_i |= \cal{I}\,,\tag{2}\end{equation} where $\cal{I}$ is the identity  transformation  that leaves any vector unchanged.    Secondly, let's write an operator within the orthonormal basis \beg...

Inner Product  Given a  vector space  $V$ over the field $\mathbb{C}$, we define an inner product, i.e, a map $(\,,\,): \, V\times V \rightarrow \mathbb{C}$ that satisfies the following axioms: $\Big(|\psi\rangle\,,\,  |\phi\rangle\Big)=\Big(|\phi\rangle\,,\,  |\psi\rangle\Big)^*$ $\Big(|\psi\rangle\,,\,  |\phi\rangle+|\chi\rangle\Big) = \Big(|\psi\rangle\,,\,  |\phi\rangle\Big)+\Big(|\psi\rangle\,,\,  |\chi\rangle\Big)$ $\Big(|\psi\rangle\,,\,  c|\phi\rangle\Big)=c\Big(|\psi\rangle\,,\, |\phi\rangle\Big)$ for any $c\in\mathbb{C}$ $\Big(|\psi\rangle\,,\, |\psi\rangle\Big)\rangle \geq 0$, equality holds iff $|\psi\rangle=\mathbf{0}$ Note: In the  dual vector space , $\langle\, | \,\rangle$ is bilinear. In contrast, here the inner product $(,)$ is linear in its second argument and antilinear in its first argument.  The definition of inner product leads to the  projection theo...

Dual Vector Space Given a  vector space  $V$, its dual $V^*$ is a set of linear maps $f:\,V\rightarrow \mathbb{C}$. Following the  Dirac notation , we denote such map $f$ by $\langle f|$. When acting on a vector $|\psi\rangle$, it leads to a complex number denoted by $\langle f|\psi\rangle$. The map is linear such that \begin{equation}\langle f|\,\Big(c_1|\psi\rangle + c_2|\phi\rangle\Big)=c_1\,\langle f|\psi\rangle+c_2\langle f|\phi\rangle\,.\end{equation} $V^*$ becomes a vector space when the linear maps are equipped with the addition and scaling operations: \begin{eqnarray}\langle f+g| \psi\rangle&=& \langle f| \psi\rangle + \langle g| \psi\rangle\,,\\ \langle c\,f|\psi\rangle&=& c \langle f|\psi\rangle\,,\end{eqnarray} for all $f, g \in V^*$, $|\psi\rangle\in V$ and $c\in\mathbb{C}$. Note: $\langle f|\psi\rangle$ is bilinear in each of the two arguments. We will use Greek letters for vectors in the vector space and use English letters for maps in the d...

Basis A set of vectors $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ is a basis of a  vector space  if $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ are linearly independent. That is, the only solution of the equation \begin{equation}x_1 |\alpha_1\rangle+x_2 |\alpha_2\rangle+\cdots + x_n |\alpha_n\rangle =\mathbf{0}\end{equation} is $x_1=\cdots=x_n=0$. Any vector $|\psi\rangle$ in the vector space is a linear combination of $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$, i.e. \begin{equation}|\psi\rangle =c_1 |\alpha_1\rangle+c_2 |\alpha_2\rangle+\cdots + c_n |\alpha_n\rangle\end{equation} for $c_1,\cdots, c_n\in \mathbb{C}$. [Exercise 1] If there is vector $|\psi\rangle$ that can not be a linear combination of linearly independent vectors $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$, then $|\alpha_1\rangle, \cdots, |\alpha_n\rangle, |\psi\rangle$ are linearly independent. [Solution] Consider the equation $x_1 |\alpha_1\rangle+\cdots + x_n |\alpha_n\rangle +d|\psi\rangle =\ma...

This series of posts is just a refresh of linear algebra basics in the language of quantum mechanics. One of the motivations is that in many books like Prof. Georgi's "Lie Algebra in Particle Physics" , linear algebra is directly presented in quantum mechanics notations. Vector Space The fundamental concept in linear algebra is the so-called vector space (or linear space). It is a set of elements called vectors that can be added together or multiplied by scalars  in the field $\mathbb{F}$. When the scalars are complex numbers, we can denote the vectors by  kets  $|\,\rangle$. So in the language of quantum mechanics, the basic definition of vector space over $\mathbb{C}$ can be formulated as follows: $|\psi\rangle + |\phi\rangle = |\phi\rangle + |\psi\rangle$. $\Big(|\psi\rangle + |\phi\rangle\Big)+|\chi\rangle=|\psi\rangle+\Big(|\phi\rangle + |\chi\rangle\Big)$. There exists an $\mathbf{0}$ such that $|\psi\rangle + \mathbf{0}=|\psi\rangle$ for every $|\psi\rangle$....

Inspired by Einstein's quote, "If Euclid failed to kindle your youthful enthusiasm, then you were not born to be a scientific thinker," I recently purchased the book "Euclid's Elements" for my son. With this esteemed work on our bookshelf, I hope it brings an opportunity for my children to one day open its pages and delve into a new world of wonder. This post is to share two interesting discoveries after I read the first three pages of the book. Euclid's fourth postulate At the beginning of the book, Euclid proposes five postulates for plane geometry, which I quote directly from this web : A straight line segment may be drawn from any given point to any other.  A straight line may be extended to any finite length.  A circle may be described with any given point as its center and any distance as its radius.  All right angles are congruent.  If a straight line intersects two other straight lines, and so makes the two interior angles on one side of it toget...

I occasionally browsed a youtube video on "2023 MIT Integration Bee". I thought it is a competition between students taking the course of calculus. I took a try and it turns out that I was not able to even solve the first problem in the allowed 4 mins.  The first problem is to solve \begin{equation} I \equiv \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\sin x + \cos x)^2}dx\,.\tag{1}\end{equation} It is obvious that \begin{eqnarray}I &=&  \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\tan x + 1)^2} \frac{dx}{\cos^2 x} =  \int_0^{\infty} \frac{\sqrt[3]{x}}{(x+1)^2}dx\\&=&-\left. \frac{\sqrt[3]{x}}{x+1}\right|_0^{\infty} +  \int_0^{\infty} \frac{d\,\sqrt[3]{x}}{x+1}\\&=&  \int_0^{\infty}\frac{dx}{x^3+1}\,.\tag{2}\end{eqnarray} I know how to solve the indefinite integral $\int\frac{dx}{x^3+1}$, but I was not able to accomplish the entire calculation within 4 mins. Here is another youtube video  that provides step-by-step calculat...

In mathematics, the first homotopy group of a pathwise-connected topological space $\cal{M}$, denoted by $\pi_1(\cal{M})$, is about the classification of all the loops in $\cal{M}$: The relations between loops that can be continuously deformed into one another is an equivalence relation. Then all loops in $\cal{M}$ can be partitioned into disjoint equivalent classes. Finally, all equivalent classes of loops form a group.  Note: Let $\alpha$ and $\beta$ be two loops in $\cal{M}$, the group multiplication between the equivalent classes $[\alpha]$ and $[\beta]$ is defined as $[\alpha]\cdot[\beta]=[\alpha * \beta]$, where $\alpha * \beta$ is a concatenated loop that we first traverse through the loop $\alpha$ and then through the loop $\beta$. In general, the n-th homotopy group, $\pi_n(\cal{M})$ is about the classification of the n-dimensional sphere $S^n$ in $\cal{M}$. Examples: $\pi_1\left(S^1\right)=\mathbb{Z}$ suggests that we can label each equivalent class of loops in $S^1...