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_...

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...

What is a particle? S. Weinberg described a particle simply as "a physical system that has no continuous degrees of freedom except for its total momentum".  Recall that the spacetime admits the so-called Poincare symmetry \begin{equation} {x'}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,x^{\nu} + a^{\mu}\,.\end{equation} By Noether's theorem , each continuous symmetry indicates a conservation. Thus, the total momentum $p^{\mu}$ is a good quantum because of the spacetime translation invariance. As we shall see below, there are other discrete degrees of freedom, denoted by $\sigma$, that are associated with the Lorentz invariance. So the quantum state of a particle is described as $|p, \sigma\rangle$.  Under the Lorentz transformation $\Lambda$, the one-particle state $|p, \sigma\rangle$ changes to a new state $U(\Lambda)|p,\sigma\rangle$ by a unitary operator $U(\Lambda)$. Note that the Lorentz transformation only changes the reference frames, elementary particles should remain...

Problem:  Show that the Lorentz transformation $W$ that keeps $k^{\mu}\equiv[1, 0,  0,  1]^T$ unchanged forms a two-dimensional Euclidean group SE(2). Solution: As shown in this post , the general solution of the Lorentz transformation $W$ that satisfies $W^{\mu}_{\,\,\,\,\nu}\,k^{\nu}=k^{\mu}$ can be parametrized in three parameters $\alpha,\beta,\theta$ as \begin{equation}W(\alpha, \beta, \theta)=\left[\begin{matrix} 1+(\alpha^2+\beta^2)/2  & -\alpha & -\beta & -(\alpha^2+\beta^2)/2  \\  -\alpha & 1 & 0 & \alpha \\ -\beta & 0 &  1 & \beta \\ (\alpha^2+\beta^2)/2 & -\alpha & -\beta & 1-(\alpha^2+\beta^2)/2 \end{matrix}\right]\left[\begin{matrix} 1 &  &  & \\  & \cos\theta & -\sin\theta & \\ & \sin\theta & \cos\theta  &   \\ & &  & 1\end{matrix}\right]\,.\tag{1}\end{equation} To work out the Lie algebra, we expand $W(\...