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zh.m.wikipedia.org/zh-hans/%E9%85%89%E7%BE%A4
youtube.com/watch?v=yeK2N7FbkVI&t=12s
J2. Unitary Groups. First, we define the group of unitary $n \times n$ matrices where the binary operation is matrix multiplication. A matrix is unitary if
$$U^{\dagger}=U^{-1}$$
1. $\rm U(1)$, i.e., $1 \times 1$ Matrices
$$
\begin{gathered}
U=[u ] \quad\text { and }\quad U^{\dagger}=\left[u^{*}\right] \\
U U^{\dagger}=[u ]\left[u^{*}\right]=\left[u u^{*}\right]=[1] \\
u(\theta)=e^{i \theta}
\end{gathered}
$$
Closure: $u(\alpha) u(\beta)=[e^{i \alpha} e^{i \beta}]=[e^{i(\alpha+\beta)}]$
Association: $\big(u(\alpha) u(\beta)\big) u(\gamma)=u(\alpha)\big(u(\beta) u(\gamma)\big)$
Identity: $I=u(0)=[1]$
Inverse: $u^{-1}(\theta)=[e^{-i \theta}]$, consistent with $U^{-1}=U^{\dagger}$
2. $\mathrm{SU}(1)$
The "S" stands for "special". We must have the determinant equal to $1 .$
$$
\operatorname{det} U=\operatorname{det}[u ]=u=1
$$
So the group $\rm SU(1)$ contains the single matrix: $A=[1]$.
3. $\mathrm{SU}(2)$
This group consists of the special unitary $2 \times 2$ matrices.
$$
A=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right], A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]
$$
As the determinant must be 1, we have
$$
A=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \quad \text { and } \quad A^{-1}=\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]
$$
For the unitary matrices we must have
$$
A^{\dagger}=A^{-1} \quad \text { which means } \quad A^{-1}=\left[\begin{array}{ll}
a^{*} & c^{*} \\
b^{*} & d^{*}
\end{array}\right] \text {. }
$$
The conditions are $a^{*}=d$ and $c=-b^{*}$ along with $|A|=a d-b c=1$.
Let's write the real and imaginary components out, applying these conditions. Then,
$$
A=\left[\begin{array}{cc}
a_{r}+i a_{i} & b_{r}+i b_{i} \\
-b_{r}+i b_{i} & a_{r}-i a_{i}
\end{array}\right]
$$
for the general special unitary $2 \times 2$ matrix, where $a_{r}^{2}+a_{i}^{2}+b_{r}^{2}+b_{i}^{2}=1$
Our general form the a matrix in the group $\rm SU(2)$.
$$
A=\left[\begin{array}{cc}
a_{r}+i a_{i} & b_{r}+i b_{i} \\
-b_{r}+i b_{i} & a_{r}-i a_{i}
\end{array}\right]
$$
Recall how you can write a vector in terms of basis unit vectors:
$$
\vec{A}=A_{x} \hat{i}+A_{y} \widehat{j}+A_{z} \hat{k}
$$
Check out the same trick with matrices:
$$
A=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=a\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]+b\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]+c\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right]+d\left[\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right]
$$
But there is a natural expansion for our $\rm SU(2)$ matrices:
$$
\begin{gathered}
A=\left[\begin{array}{cc}
a_{r}+i a_{i} & b_{r}+i b_{i} \\
-b_{r}+i b_{i} & a_{r}-i a_{i}
\end{array}\right] \\
A=a_{r}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+i a_{i}\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]+b_{r}\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right]+i b_{i}\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right] \\
A=a_{r}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+i a_{i}\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]+i b_{r}\left[\begin{array}{cc}
0 & -i \\
i & 0
\end{array}\right]+i b_{i}\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]
\end{gathered}
$$
The Pauli matrices are $\sigma_{x}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \quad \sigma_{y}=\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right] \quad \sigma_{z}=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$
The anticommutator of $A$ and $B$ is defined as $\{A, B\} \equiv A B+B A$
The Kronecker Delta symbol is defined as
$$
\delta_{i j}= \begin{cases}1, & \text { if } i=j \\ 0, & \text { if } i \neq j\end{cases}
$$
and named after the German mathematician Leopold Kronecker. It is a symmetric symbol.
PJ4 (Practice Problem). Show
$$\left\{\sigma_{j}, \sigma_{k}\right\}=2 \delta_{j k} I$$
The commutator of $A$ and $B$ is defined as $[A,B]=AB-BA$.
The Levi-Civita or permutation symbol is defined below. It is an antisymmetric symbol.
$$
\varepsilon_{i j k}= \begin{cases}+1 & \text { if }(i, j, k) \text { is }(1,2,3),(3,1,2) \text { or }(2,3,1) \\ -1 & \text { if }(i, j, k) \text { is }(1,3,2),(3,2,1) \text { or }(2,1,3) \\ 0 & \text { if } i=j \text { or } j=k \text { or } k=i\end{cases}
$$
PJ5 (Practice Problem). Show
$$
\left[\sigma_{j}, \sigma_{k}\right]=2 i \varepsilon_{j k l} \sigma_{l} \text { and thus } \mathrm{SU}(2) \text { is non-abelian. }
$$
4. $\mathrm{SO}(2)$
This group consists of the special unitary orthogonal matrices. These are matrices where the columns and rows, thought of as vectors, are orthogonal. Remember rotation matrix in 2D?
$$
\begin{gathered}
R(\theta)=\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] \\
A=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]=\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]
\end{gathered}
$$
These matrices satisfy our orthogonal conditions since
$$
\begin{gathered}
a_{11} a_{12}+a_{21} a_{22}=\cos \theta \sin \theta-\cos \theta \sin \theta=0 \\
a_{11} a_{21}+a_{12} a_{22}=-\cos \theta \sin \theta+\cos \theta \sin \theta=0
\end{gathered}
$$
Note that now the transpose is the inverse.
$$
\begin{gathered}
A^{-1}=A^{T} \\
A A^{-1}=\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{gathered}
$$
PJ7 (Practice Problem). Show $\rm SO(2)$ is abelian. |
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