Group Theory Calculations

In this article I will summarize some simple examples as a quick introduction to Group Theory.

§1 Finite Group

If you wanna know about the full finite group, see Classification of finite simple groups. The FULL list is one of the greatest human achievements.

I think the best way to learn finite group is to start from various examples.

Here’s some general rules:

• Some rules to calculate permutations:
  • $(12)=(21)$, $(123)=(231)=(312)$
  • $(12)(23)=(123)$
  • $(12)^{-1}=(21)=(12)$, $(123)^{-1}=(321)$, $((12)(34))^{-1}=(43)(21)$, $((123)(45))^{-1}=(54)(321)$. The structure of the permutation is the same, the numbers in the permutation goes backwards.
  • $(12)(12)=e$
  • To calculate $((12)(34))(123)((12)(34))^{-1}$, first exchange $1$ and $2$ for $(123)$, than exchange $3$ and $4$. $(123)\xrightarrow{\text{exchange 1 and 2}}(213)\xrightarrow{\text{exchange 3 and 4}}(214)$
• For a $S_n$ group, permutations with the same structure form a conjugate class.
• The number of elements in a conjugate class is a factor of the order of the group.
• The order of the subgroup is a factor of the order of the group.
• An invariant subgroup must consist of several complete conjugate classes.

§1.1 C2

In the notation of permutation, $$C_2 = \lbrace e, (12) \rbrace \tag{1.1.1}$$It has $2$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (12) \rbrace \end{aligned} \tag{1.1.2}$$thus only has $2$ trivial invariant subgroups: $$\begin{aligned} H_1 &= \lbrace e\rbrace \\ H_2&=\lbrace e, (12) \rbrace (C_2 \space \text{itself}) \end{aligned} \tag{1.1.3}$$thus has $2$ quotient groups, which are trivial: $$\begin{aligned} C_2 / H_1 &= \lbrace H_1, (12)H_1 \rbrace \cong C_2 \\ C_2 / H_2 &= \lbrace H_2\rbrace \end{aligned} \tag{1.1.4}$$

In the character table, the first row is always all $1$, and the first column is always the dimension $n_{\mu}$ of the irreducible representation (irr. rep.) of the group, which satisfies $$n_G = \sum n_{\mu}^2 \tag{1.1.5}$$where $n_G$ is the order of original group, in this case $C_2$. So the character table is:

According to orthonormality, the $1^{\text{st}}$ row and the $2^{\text{nd}}$ row satisfy (the coefficient ahead is $\sqrt{\frac{\text{the number of elements in the conjugate class}}{n_G}}$ that each column has the same value):$$\begin{pmatrix} \sqrt{\frac{1}{2}}\cdot1 & \sqrt{\frac{1}{2}}\cdot1 \end{pmatrix} \begin{pmatrix} \sqrt{\frac{1}{2}}\cdot1 \\ \sqrt{\frac{1}{2}}\cdot x \end{pmatrix} = 0 \tag{1.1.6}$$so $x=-1$, thus complete the character table:

§1.2 C3

$$C_3 = \lbrace e, (123), (321) \rbrace \tag{1.2.1}$$It has $3$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (123) \rbrace \\ \zeta_3&=\lbrace (321) \rbrace \end{aligned} \tag{1.2.2}$$thus only has $2$ trivial invariant subgroups: $$\begin{aligned} H_1 &= \lbrace e\rbrace \\ H_2&=\lbrace e, (123), (321) \rbrace (C_3 \space \text{itself}) \end{aligned} \tag{1.2.3}$$thus has $2$ quotient groups, which are trivial: $$\begin{aligned} C_3 / H_1 &= \lbrace H_1, (123)H_1, (321)H_1 \rbrace \cong C_3 \\ C_3 / H_2 &= \lbrace H_2\rbrace \end{aligned} \tag{1.2.4}$$

According to Eq. (1.1.5), the character table is:

According to orthonormality, the character table is:

§1.3 S3

$$S_3 = \lbrace e, (12), (13), (23), (123), (321) \rbrace \tag{1.3.1}$$It has $3$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (12), (13), (23) \rbrace \\ \zeta_3&=\lbrace (123), (321) \rbrace \end{aligned} \tag{1.3.2}$$thus only has $1$ nontrivial invariant subgroup, composed of $\zeta_1$ and $\zeta_3$: $$H_1 = \lbrace e, (123), (321) \rbrace \tag{1.3.3}$$thus has $1$ quotient group: $$S_3 / H_1 = \lbrace H_1, (12)H_1 \rbrace \cong C_2 \tag{1.3.4}$$

According to Eq. (1.1.5), the character table is:

Because $H_1$ is composed of $\zeta_1$ and $\zeta_3$, and $S_3 / H_1 \cong C_2$, the column of $\zeta_1$ and $\zeta_3$ is the same as the column of $\zeta_1$ in the character table of $C_2$, when having the same dimension of irr. rep.; the column of $\zeta_2$ is the same as the column of $\zeta_2$ in the character table of $C_2$, when having the same dimension of irr. rep. Thus the character table is :

According to orthonormality, the character table is:

§1.4 S4

$$\begin{aligned}S_4 = \lbrace &e, \\ &(12), (13), (14), (23), (24), (34), \\ &(12)(34), (13)(24), (14)(23), \\ &(123), (124), (132), (134), (142), (143), (234), (243), \\ &(1234), (1243), (1324), (1342), (1423), (1432) \rbrace \end{aligned}\tag{1.4.1}$$It has $5$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (12), (13), (14), (23), (24), (34) \rbrace \\ \zeta_3&=\lbrace (12)(34), (13)(24), (14)(23) \rbrace \\ \zeta_4&=\lbrace (123), (124), (132), (134), (142), (143), (234), (243) \rbrace \\ \zeta_5&=\lbrace (1234), (1243), (1324), (1342), (1423), (1432) \rbrace \end{aligned} \tag{1.4.2}$$thus has $2$ nontrivial invariant subgroups: $$\begin{aligned} H_1 = \lbrace &e, (12)(34), (13)(24), (14)(23) \rbrace \cong V_4\\ H_2=\lbrace &e, \\ &(123), (321), (124), (421), (234), (432), \\ &(12)(34), (13)(24), (14)(23) \rbrace \cong A_4 \end{aligned} \tag{1.4.3}$$thus has $2$ quotient groups: $$\begin{aligned} S_4 / V_4 &= \lbrace V_4, (12)V_4, (13)V_4, (23)V_4, (123)V_4, (321)V_4 \rbrace \cong S_3 \\ S_4 / A_4 &\cong C_2 \end{aligned} \tag{1.4.4}$$

According to Eq. (1.1.5) and quotient group, the character table is:

According to orthonormality, the character table is:

According to orthonormality, the character table is:

§1.5 D2

$$D_2 = \lbrace e, (13), (24), (13)(24) \rbrace \tag{1.5.1}$$It has $4$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (13) \rbrace \\ \zeta_3&=\lbrace (24) \rbrace \\ \zeta_4&=\lbrace (13)(24) \rbrace\end{aligned} \tag{1.5.2}$$thus has $3$ nontrivial invariant subgroups: $$\begin{aligned} H_1 &= \lbrace e, (13)\rbrace \\ H_2&=\lbrace e, (24) \rbrace \\ H_3&=\lbrace e, (13)(24) \rbrace \end{aligned} \tag{1.5.3}$$thus has $3$ quotient groups: $$\begin{aligned} D_2 / H_1 &= \lbrace H_1, (13)H_1 \rbrace \cong C_2 \\ D_2 / H_2 &= \lbrace H_2, (24)H_2 \rbrace \cong C_2 \\ D_2 / H_3 &= \lbrace H_3, (13)(24)H_3 \rbrace \cong C_2\end{aligned} \tag{1.5.4}$$

According to Eq. (1.1.5) and quotient groups, the character table is:

§1.6 D3

$D_3 \cong S_3$, see §1.3.

§1.7 D4

$$\begin{aligned}D_4 = \lbrace &e, \\ &(13), (24), \\ &(12)(34), (14)(23), \\ &(1234), (13)(24), (4321) \rbrace \end{aligned}\tag{1.7.1}$$where $(13)(24)=(1234)^2$, $(4321)=(1234)^3$. It has $5$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (13)(24) \rbrace \\ \zeta_3&=\lbrace (13), (24) \rbrace \\ \zeta_4&=\lbrace (12)(34), (14)(23) \rbrace \\ \zeta_5&=\lbrace (1234), (4321) \rbrace \end{aligned} \tag{1.7.2}$$thus has $4$ nontrivial invariant subgroups: $$\begin{aligned} H_1 &= \lbrace e, (13)(24)\rbrace \\ H_2&=\lbrace e, (13), (24), (13)(24) \rbrace \\ H_3&=\lbrace e, (12)(34), (14)(23), (13)(24) \rbrace \\ H_4&=\lbrace e, (1234), (4321), (13)(24) \rbrace \end{aligned} \tag{1.7.3}$$thus has $4$ quotient groups: $$\begin{aligned} D_4 / H_1 &= \lbrace H_1, (13)H_1, (12)(34)H_1, (1234)H_1 \rbrace \cong D_2 \\ D_4 / H_2 &= \lbrace H_2, (12)(34)H_2\rbrace \cong C_2 \\ D_4 / H_3 &= \lbrace H_3, (13)H_3\rbrace \cong C_2 \\ D_4 / H_4 &= \lbrace H_4, (13)H_4\rbrace \cong C_2 \end{aligned} \tag{1.7.4}$$

According to Eq. (1.1.5) and quotient groups, the character table is:

According to orthonormality, the character table is:

§1.8 D5

$$\begin{aligned} D_5 = \lbrace &e, \\ &(12345), (13524), (42531), (54321), \\ &(25)(34), (13)(45), (15)(24), (12)(35), (14)(23) \rbrace \end{aligned} \tag{1.8.1}$$where $(13524)=(12345)^2$, $(42531)=(12345)^3$, $(54321)=(12345)^4$. It has $4$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (25)(34), (13)(45), (15)(24), (12)(35), (14)(23) \rbrace \\ \zeta_3&=\lbrace (12345), (54321) \rbrace \\ \zeta_4&=\lbrace (13524), (42531) \rbrace \end{aligned} \tag{1.8.2}$$thus has $1$ nontrivial invariant subgroup: $$ H_1 = \lbrace e, (12345), (13524), (42531), (54321)\rbrace \cong C_5\tag{1.8.3}$$thus has $1$ quotient group: $$D_5 / C_5 = \lbrace C_5, (25)(34)C_5 \rbrace \cong C_2\tag{1.8.4}$$

According to Eq. (1.1.5) and quotient groups, the character table is:

According to orthonormality, the character table is:

By the way, $2\cos(\frac{2 \pi}{5}) \approx 0.618$, is the Golden Ratio.

§1.9 A4

$$\begin{aligned}A_4 = \lbrace &e, \\ &(123), (321), (124), (421), (134), (431), (234), (432), \\ &(12)(34), (13)(24), (14)(23) \rbrace \end{aligned} \tag{1.9.1}$$It has $4$ conjugate classes: $$\begin{aligned} \zeta_1&=\lbrace e \rbrace \\ \zeta_2&=\lbrace (12)(34), (13)(24), (14)(23) \rbrace \\ \zeta_3&=\lbrace (123), (421), (134), (432) \rbrace \\ \zeta_4&=\lbrace (321), (124), (431), (234) \rbrace \end{aligned} \tag{1.9.2}$$thus has $1$ nontrivial invariant subgroup: $$H_1 = \lbrace e, (12)(34), (13)(24), (14)(23)\rbrace \cong V_4\tag{1.9.3}$$thus has $1$ quotient group: $$A_4 / V_4 = \lbrace V_4, (123)V_4, (321)V_4 \rbrace \cong C_3 \tag{1.9.4}$$

According to Eq. (1.1.5) and quotient groups, the character table is:

According to orthonormality, the character table is:

According to orthonormality, the character table is:

§2 Yang Tableaux of S3

Drawing the Yang Tableaux with html is so simple and elegant:

<table>
    <tr>
        <td>1</td>
        <td>2</td>
    </tr>
    <tr>
        <td>3</td>
    </tr>
</table>

In this section we use normal Yang Tableaux to construct the Irr. Rep. matrices of S3.

1. normal Yang tableaux $\Theta_1$:

According to Hook length formula, the irr. rep. is one-dimensional.

We have$$\begin{aligned} s_1 &= e+(12)+(23)+(13)+(123)+(321) \\ a_1 &= e \\ e_1 &= e+(12)+(23)+(13)+(123)+(321) \end{aligned} \tag{2.1}$$Define $$\ket{\nu} = \ket{e}+\ket{(12)}+\ket{(23)}+\ket{(13)}+\ket{(123)}+\ket{(321)} \tag{2.2}$$Thus we have $$p\ket{\nu}=\ket{\nu}\tag{2.3}$$where $p=e, (12), (23), (13), (123), (321)$. So $$p=1 \tag{2.4}$$

2. normal Yang tableaux $\Theta_2$:

According to Hook length formula, the irr. rep. is one-dimensional.

We have$$\begin{aligned} s_2 &= e \\ a_2 &= e-(12)-(23)-(13)+(123)+(321) \\ e_2 &= e-(12)-(23)-(13)+(123)+(321) \end{aligned} \tag{2.5}$$Define $$\ket{\nu} = \ket{e}-\ket{(12)}-\ket{(23)}-\ket{(13)}+\ket{(123)}+\ket{(321)} \tag{2.6}$$Thus when $p$ is an odd permutation, $$p\ket{\nu}=(12)\ket{\nu}=(23)\ket{\nu}=(13)\ket{\nu}=-\ket{\nu} \tag{2.7}$$when $p$ is a even permutation, $$p\ket{\nu}=e\ket{\nu}=(123)\ket{\nu}=(321)\ket{\nu}=\ket{\nu} \tag{2.8}$$So $$p=(-1)^p \tag{2.9}$$

3. normal Yang tableaux $\Theta_3$:

According to Hook length formula, the irr. rep. is two-dimensional.

We have$$\begin{aligned} s_3 &= e+(12) \\ a_3 &= e-(13) \\ e_3 &= e+(12)-(13)-(12)(13) \\ &=e+(12)-(13)-(321) \end{aligned} \tag{2.10}$$ Define $$\ket{\nu_1} = \ket{e}+\ket{(12)}-\ket{(13)}-\ket{(321)} \tag{2.11}$$Thus$$\begin{aligned} e\ket{\nu_1}&=\ket{\nu_1} \\ (12)\ket{\nu_1}&=\ket{(12)}+\ket{e}-\ket{(321)}-\ket{(13)}&=\ket{\nu_1}\\ (23)\ket{\nu_1}&=\ket{(23)}+\ket{(321)}-\ket{(123)}-\ket{(12)}&=\ket{\nu_2}\\ (13)\ket{\nu_1}&=\ket{(13)}+\ket{(123)}-\ket{e}-\ket{(23)}&=-\ket{\nu_1}-\ket{\nu_2}\\ (123)\ket{\nu_1}&=\ket{(123)}+\ket{(13)}-\ket{(23)}-\ket{e}&=-\ket{\nu_1}-\ket{\nu_2}\\ (321)\ket{\nu_1}&=\ket{(321)}+\ket{(23)}-\ket{(12)}-\ket{(123)}&=\ket{\nu_2}\\ e\ket{\nu_2}&=\ket{\nu_2}\\ (12)\ket{\nu_2}&=\ket{(123)}+\ket{(13)}-\ket{(23)}-\ket{e}&=-\ket{\nu_1}-\ket{\nu_2}\\ (23)\ket{\nu_2}&=\ket{e}+\ket{(12)}-\ket{(13)}-\ket{(321)}&=\ket{\nu_1}\\ (13)\ket{\nu_2}&=\ket{(321)}+\ket{(23)}-\ket{(12)}-\ket{(123)}&=\ket{\nu_2}\\ (123)\ket{\nu_2}&=\ket{(12)}+\ket{e}-\ket{(321)}-\ket{(13)}&=\ket{\nu_1}\\ (321)\ket{\nu_2}&=\ket{(13)}+\ket{(123)}-\ket{e}-\ket{(23)}&=-\ket{\nu_1}-\ket{\nu_2} \end{aligned} \tag{2.12}$$So $$\begin{aligned} D(e) &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ D(12) &= \begin{pmatrix} 1 & -1 \\ 0 & -1 \end{pmatrix}, D(23) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, D(13) = \begin{pmatrix} -1 & 0 \\ -1 & 1 \end{pmatrix} \\ D(123) &= \begin{pmatrix} -1 & 1 \\ -1 & 0 \end{pmatrix}, D(321) = \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} \\ \end{aligned} \tag{2.13}$$

§3 Irr. Rep. of SO(3) Lie Algebra

§3.1 j=1/2

When $j = \frac{1}{2}$, $\frac{3}{2}$, $\frac{5}{2}$, …, it’s a double-valued representation. Let’s take $j=\frac{1}{2}$ for example.

Define $$\ket{\frac{1}{2},\frac{1}{2}}=\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \ket{\frac{1}{2},-\frac{1}{2}}=\begin{pmatrix} 0 \\ 1 \end{pmatrix} \tag{3.1.1}$$According to $$\begin{aligned} J_3 \ket{j,m} &= m \ket{j,m} \\ J_{\pm} \ket{j,m} &= \sqrt{j(j+1)-m(m \pm 1)} \ket{j,m \pm 1}\end{aligned} \tag{3.1.2}$$we have $$J_3=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2} \end{pmatrix}, J_+=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, J_-=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \tag{3.1.3}$$According to $$J_{\pm}=J_1 \pm i J_2 \tag{3.1.4}$$we have $$J_1=\begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix}, J_2=\begin{pmatrix} 0 & -\frac{i}{2} \\ \frac{i}{2} & 0 \end{pmatrix}, J_3=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2} \end{pmatrix} \tag{3.1.5}$$According to $$J_k=\frac{\sigma_k}{2}, k=1,2,3 \tag{3.1.6}$$we have $$\sigma_1=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \sigma_2=\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \sigma_3=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \tag{3.1.7}$$So $$\begin{aligned} d^{\frac{1}{2}}(\beta) &=e^{-i \beta J_2} \\ &= e^{-\frac{i \beta \sigma_2}{2}} \\ &=E+(-\frac{i \beta}{2})\sigma_2+\frac{1}{2!}(-\frac{i \beta}{2})^{2}\sigma_2^2+\frac{1}{3!}(-\frac{i \beta}{2})^{3}\sigma_2^3+ \cdots \\ &=E\cos{\frac{\beta}{2}}-i \sigma_2 \sin{\frac{\beta}{2}} \\ &= \begin{pmatrix} \cos{\frac{\beta}{2}} & -\sin{\frac{\beta}{2}} \\ \sin{\frac{\beta}{2}} & \cos{\frac{\beta}{2}} \end{pmatrix}\end{aligned} \tag{3.1.8}$$So $$D^{\frac{1}{2}}(\alpha, \beta, \gamma)=\begin{pmatrix} e^{-\frac{i\alpha}{2}}\cos{\frac{\beta}{2}}e^{-\frac{i\gamma}{2}} & -e^{-\frac{i\alpha}{2}}\sin{\frac{\beta}{2}}e^{\frac{i\gamma}{2}} \\ e^{\frac{i\alpha}{2}}\sin{\frac{\beta}{2}}e^{-\frac{i\gamma}{2}} & e^{\frac{i\alpha}{2}}\cos{\frac{\beta}{2}}e^{\frac{i\gamma}{2}}\end{pmatrix} \tag{3.1.9}$$So $$\begin{aligned} D[R_n(2n \pi)] &= D[R] e^{in \pi \sigma_2} D[R]^{-1} \\ &= D[R] (-E)^n D[R]^{-1} \\ &=(-1)^n E \end{aligned} \tag{3.1.10}$$Thus it’s a double-valued representation.

§3.2 j=1

When $j = 0$, $1$, $2$, …, it’s a irreducible representation. Let’s take $j=1$ for example.

$$J_3=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}, J_+=\begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix}, J_-=\begin{pmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{pmatrix} \tag{3.2.1}$$And$$e^{-i \beta J_2} = e^{\beta K} =E+\sin{\beta} K+(1-\cos{\beta})K^2 \tag{3.2.2}$$

The rest is left as an exercise for the reader.

Eventually, $$d^1(\beta)= \begin{pmatrix} \frac{1+\cos{\beta}}{2} & -\frac{\sin{\beta}}{\sqrt{2}} & \frac{1-\cos{\beta}}{2} \\ \frac{\sin{\beta}}{\sqrt{2}} & \cos{\beta} & -\frac{\sin{\beta}}{\sqrt{2}} \\ \frac{1-\cos{\beta}}{2} & \frac{\sin{\beta}}{\sqrt{2}} & \frac{1+\cos{\beta}}{2} \end{pmatrix} \tag{3.2.3}$$

§4 Clebsch–Gordan Coefficients

In this section we decompose the direct product of Irr. Rep. of SU(2) into the direct sum of Irr. Rep. Here’s an example of $j_1 = \frac{1}{2}$, $j_2=1$. You can check Table of Clebsch–Gordan coefficients to make sure it’s right.

1. $$\ket{\frac{3}{2}, \frac{3}{2}}=\ket{\frac{1}{2}, \frac{1}{2}} \otimes \ket{1, 1} \tag{4.1}$$

2. $$J_-\ket{\frac{3}{2}, \frac{3}{2}}=\sqrt{\frac{3}{2}(\frac{3}{2}+1)-\frac{3}{2}(\frac{3}{2}-1)}\ket{\frac{3}{2}, \frac{1}{2}}=\sqrt{3}\ket{\frac{3}{2}, \frac{1}{2}} \tag{4.2}$$ $$\begin{aligned}J_-(\ket{\frac{1}{2}, \frac{1}{2}} \otimes \ket{1, 1})&=(J_-\ket{\frac{1}{2}, \frac{1}{2}})\otimes\ket{1, 1}+\ket{\frac{1}{2}, \frac{1}{2}} \otimes J_-(\ket{1, 1}) \\ &=\ket{\frac{1}{2}, -\frac{1}{2}} \otimes \ket{1, 1}+\sqrt{2}\ket{\frac{1}{2}, \frac{1}{2}} \otimes \ket{1, 0} \end{aligned} \tag{4.3}$$ So $$\ket{\frac{3}{2}, \frac{1}{2}}=\sqrt{\frac{1}{3}}\ket{\frac{1}{2}, -\frac{1}{2}} \otimes \ket{1, 1}+\sqrt{\frac{2}{3}}\ket{\frac{1}{2}, \frac{1}{2}} \otimes \ket{1, 0} \tag{4.4}$$

3. Similarly, $$\ket{\frac{3}{2}, -\frac{1}{2}}=\sqrt{\frac{2}{3}}\ket{\frac{1}{2}, -\frac{1}{2}} \otimes \ket{1, 0}+\sqrt{\frac{1}{3}}\ket{\frac{1}{2}, \frac{1}{2}} \otimes \ket{1, -1} \tag{4.5}$$

4. Similarly, $$\ket{\frac{3}{2}, -\frac{3}{2}}=\ket{\frac{1}{2}, -\frac{1}{2}} \otimes \ket{1, -1} \tag{4.6}$$

5. Because $\ket{\frac{3}{2}, \frac{1}{2}}$ and $\ket{\frac{1}{2}, \frac{1}{2}}$ are orthogonal, and by normalization, we have $$\ket{\frac{1}{2}, \frac{1}{2}} = \sqrt{\frac{2}{3}}\ket{\frac{1}{2}, -\frac{1}{2}} \otimes \ket{1, 1}-\sqrt{\frac{1}{3}}\ket{\frac{1}{2}, \frac{1}{2}} \otimes \ket{1, 0} \tag{4.7}$$

6. Similarly, $$\ket{\frac{1}{2}, -\frac{1}{2}} = \sqrt{\frac{1}{3}}\ket{\frac{1}{2}, -\frac{1}{2}} \otimes \ket{1, 0}-\sqrt{\frac{2}{3}}\ket{\frac{1}{2}, \frac{1}{2}} \otimes \ket{1, -1} \tag{4.8}$$

Thus complete the Clebsch–Gordan coefficients for $j_1 = \frac{1}{2}$, $j_2=1$.

§5 Box-Weight Diagram of SU(3) Irr. Rep.

§5.1 Box-Weight Diagram

We have $$\begin{cases} \vec{\alpha^1} = (\frac{1}{2},\frac{\sqrt{3}}{2}) \\ \vec{\alpha^2} = (\frac{1}{2},-\frac{\sqrt{3}}{2}) \end{cases} \tag{5.1.1}$$and according to the definition of Cartan matrix $$A_{ij} = \frac{2 \vec{\alpha^i} \cdot \vec{\alpha^j}}{|\vec{\alpha^j}|^2} \tag{5.1.2}$$we have:$$A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \tag{5.1.3}$$

When drawing the box-weight diagram, there are some rules to follow:

• When minus the number on the left of $\boxed{x \space y}$, the $x$,
  1. minus the $1^{\text{st}}$ row of eq. (5.1.3);
  2. draw a line down-right from $x$ to the $x’$ in the next box;
  3. until it’s $-x$.
• When minus the number on the right of $\boxed{x \space y}$, the $y$,
  1. minus the $2^{\text{nd}}$ row of eq. (5.1.3);
  2. draw a line down-left from $y$ to the $y’$ in the next box;
  3. until it’s $-y$.
• If you start from $\boxed{a \space b}$ on the top, the final result should be $\boxed{-b \space -a}$ in the bottom.

For example, when the highest weight is $3$ and $0$, or $\boxed{3 \space 0}$, the box-weight diagram should look like this:

§5.2 All Possible Roots

According to $$\frac{2 \vec{\alpha^j} \cdot \vec{\mu^k}}{|\vec{\alpha^j}|^2}=\delta_{jk} \tag{5.2.1}$$we have $$\begin{cases} \vec{\mu^1}=(\frac{1}{2}, \frac{\sqrt{3}}{6}) \\ \vec{\mu^2}=(\frac{1}{2}, -\frac{\sqrt{3}}{6}) \end{cases} \tag{5.2.2}$$

Starting from $3 \vec{\mu^1}$, from box-weight diagram, we have $$\begin{aligned} &3 \vec{\mu^1}-\vec{\alpha^1}, \\ &3 \vec{\mu^1}-\vec{\alpha^1}-\vec{\alpha^2}, 3 \vec{\mu^1}-2\vec{\alpha^1}, \\ &3 \vec{\mu^1}-2\vec{\alpha^1}-\vec{\alpha^2}, 3 \vec{\mu^1}-3\vec{\alpha^1}, \\ &3 \vec{\mu^1}-2\vec{\alpha^1}-2\vec{\alpha^2}, 3 \vec{\mu^1}-3\vec{\alpha^1}-\vec{\alpha^2}, \\ & 3 \vec{\mu^1}-3\vec{\alpha^1}-2\vec{\alpha^2}, \\ &3 \vec{\mu^1}-3\vec{\alpha^1}-3\vec{\alpha^2} \end{aligned} \tag{5.2.3}$$ (Or you can do it more easily by using the fact that the figure below is symmetric with respect to $H_2$ axis.)

Drawing on the $H_1 - H_2$ plane, we have: