Life with Physics

This post is for knowledge sharing only. It is not intended to be investment or tax advice. 0. Account Overview  Table 1 Common retirement  accounts in U.S. I summarizes the common retirement accounts in Table 1:  The second row in Table 1 lists the company-sponsored accounts that are offered through employers. The third row in Table 1 lists the individual retirement accounts (IRA) that is unrelated to employments. The columns in Table 1 are about different tax advantages as described below. WARNING: because of the tax advantages, there are also limitations and specific rules imposed by IRS for early withdrawal before the retirement age. For accounts listed in the second column,  Contributions are made with before-tax dollars . The money taken out will be taxed as regular income in the year of withdrawal. For accounts listed in the third column,  Contributions are made with after-tax dollars . The money taken out is  tax free . For after-tax 401K in the fou...

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...