Classification of Lie Algebra
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 mutually commuting generators, say ${\cal{T}}_1, \cdots,{\cal{T}}_m$, such that \begin{equation}\Big[{\cal{T}}_i\,,{\cal{T}}_j\Big]=0\end{equation} for $i,\,j=1,\cdots, m$.
- Compute a $m\times m$ matrix $\mathbf{G}$ whose matrix element is defined by \begin{equation}\mathbf{G}_{ij}\equiv\text{Tr}({\cal{T}}_i\,{\cal{T}}_j)\end{equation} for $i,\,j=1,\cdots, m$.
- Since $\mathbf{G}$ is a real symmetric matrix, it can be diagonalized \begin{equation}\mathbf{O}\,\mathbf{G}\,\mathbf{O}^T=\text{diag}[\lambda_1,\cdots, \lambda_m]\end{equation} by an orthonormal matrix $\mathbf{O}$.
- Finally, define a set of new generators \begin{equation}{\cal{H}}_i\equiv \sqrt{\frac{k_D}{\lambda_i}}\sum_{j=1}^m \mathbf{O}_{ij}\,{\cal{T}}_j\,,\end{equation} which satsifies all the properties in Eq. (1).
- Cartan generators can be simultaneously diagonalized as \begin{equation}{\cal{H}}_i |\mu\rangle =\mu_i|\mu\rangle\,,\tag{2}\end{equation} where $\mu$ is a $m$-component vector called weights.
- In Lie algebra $\mathfrak{su}(2)$, $m=1$, $\cal{H}_1=\cal{J}_3$ and $k_D=1/2$.
Raising and lowering operators
Raising operators
Out of the remaining generators, we can construct a set of raising operators $\cal{E}_{\alpha}$ such that \begin{equation}\Big[{\cal{H}}_i\,,{\cal{E}}_{\alpha}\Big]=\alpha_i\,\cal{E}_{\alpha}\,,\tag{3}\end{equation} where $\alpha$ is a $m$-component vector called roots.
Note that $\cal{E}_{\alpha}$ in Eq. (3) can be up to an arbitrary constant. So we impose the same normalization as Eq. (1) \begin{equation}\text{Tr}\left({\cal{E}}^{\dagger}_{\alpha}\,{\cal{E}}_{\alpha}\right)=k_D\,.\tag{4}\end{equation}
Remarks:
- The generator $\cal{E}_{\alpha}$ in Eq. (3) is called raising operator because the weight increases from $\mu$ to $\mu+\alpha$ when acting $\cal{E}_{\alpha}$ on the state $|\mu\rangle$ in Eq. (2): \begin{equation} {\cal{H}}_i\,\cal{E}_{\alpha}|\mu\rangle =\cal{E}_{\alpha}\,{\cal{H}}_i|\mu\rangle+\left[{\cal{H}}_i\,,\cal{E}_{\alpha}\right]|\mu\rangle = (\mu_i+\alpha_i)\,\cal{E}_{\alpha}|\mu\rangle\,. \end{equation}
- In Lie algebra $\mathfrak{su}(2)$, such $\cal{E}_{\alpha}$ is $\frac{1}{\sqrt{2}}{\cal{J}}_+\equiv \frac{1}{\sqrt{2}}\left(\cal{J}_1+i\cal{J}_2\right)$ and the commutator (3) is $[\cal{J}_3\,,\cal{J}_+]=\cal{J}_+$.
The construction of the raising operator in Eq. (3) is in the following steps:
- For each Cartan generator $\cal{H}_i$, find its adjoint representation $\cal{A}_i$ in the representation space spanned by ${\cal{T}}_1, \cdots,{\cal{T}}_N$.
- Find the eigenvalues and eigenvectors of $\cal{A}_i$ \begin{equation}{\cal{A}}_i\,\left| {\cal{E}}_{\alpha}\right\rangle=\alpha_i\,\left| {\cal{E}}_{\alpha}\right\rangle\,.\tag{5}\end{equation}
- Expand the eigenvectors in the basis \begin{equation}\left| {\cal{E}}_{\alpha}\right\rangle =c_{m+1}\left| {\cal{T}}_{m+1}\right\rangle+c_{m+2}\left| {\cal{T}}_{m+2}\right\rangle+\cdots+c_{N}\left| {\cal{T}}_{N}\right\rangle\,.\tag{6}\end{equation}
- Finally, the generator $\cal{E}_{\alpha}$ in Eq. (3) is \begin{equation} {\cal{E}}_{\alpha}\propto c_{m+1}\,{\cal{T}}_{m+1}+c_{m+2}\,{\cal{T}}_{m+2}+\cdots+c_{N}\,{\cal{T}}_{N}\,,\end{equation} up to a constant to in order to satisfy the normalization (3).
- The key insight for the classification of Lie algebra is to work with adjoint representation because of the property \begin{equation}{\cal{A}}_i\,\left|{\cal{T}}_j\right\rangle=\left|\left[{\cal{H}}_i, {\cal{T}}_j\right]\right\rangle\end{equation}for the adjoint representation $\cal{A}_i$. This is why the eigenvalue equation of the adjoint representation (4) leads to the commutator (3).
- The expansion (5) does not contain the terms of Cartan generators because ${\cal{A}}_i\,\left|{\cal{H}}_j\right\rangle=\left|\left[{\cal{H}}_i, {\cal{H}}_j\right]\right\rangle=\mathbf{0}$.
- From Eq. (2) and (4), roots are the weights in the adjoint representation.
Lowering operators
Each root forms a $\mathfrak{su}(2)$
In summary, starting with the generators ${\cal{T}}_1, \cdots,{\cal{T}}_N$, we find a different set of generators, $\cal{H}$s and $\cal{E}$s, satisfying the commutation relations \begin{equation} \color{red}{\Big[{\cal{H}}_i\,,{\cal{E}}_{\pm\alpha}\Big]=\pm\alpha_i\,{\cal{E}}_{\pm\alpha}\,,\quad \Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]=\sum_{j=1}^m \alpha_j {\cal{H}}_j}\,.\tag{8}\end{equation}
For each root $\alpha$, let \begin{equation}{\cal{E}}_{\pm}\equiv\frac{1}{|\alpha|}{\cal{E}}_{\pm\alpha}\,,\quad{\cal{E}}_{3}\equiv\sum_{j=1}^m \frac{\alpha_j}{|\alpha|^2} {\cal{H}}_j\,,\tag{9}\end{equation} where $|\alpha|\equiv\sqrt{\sum_{j=1}^m\alpha^2_j}$, the commutation relations (8) becomes $\mathfrak{su}(2)$: \begin{equation}\Big[{\cal{E}}_{3}\,,{\cal{E}}_{\pm}\Big] =\pm {\cal{E}}_{\pm}\,,\quad\Big[{\cal{E}}_{+}\,,{\cal{E}}_{-}\Big] = {\cal{E}}_{3}\,,\end{equation} which is identical to $\mathfrak{su}(2)$ with $\cal{E}_3=\cal{J}_3$ and $\cal{E}_{\pm}=\frac{1}{\sqrt{2}}\cal{J}_{\pm}$.
Comments
Post a Comment