Basics of Group Representation
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_2)\Big)\cal{D}(g_3)\end{eqnarray}
- Identity element is \cal{D}(e) because \cal{D}(g)\cal{D}(e)=\cal{D}(ge)=\cal{D}(g) and \cal{D}(e)\cal{D}(g)=\cal{D}(eg)=\cal{D}(g).
- Inverse element for any \cal{D}(g) is \cal{D}(g^{-1}) because \cal{D}(g^{-1})\cal{D}(g)=\cal{D}(g^{-1}g)=\cal{D}(e) and \cal{D}(g)\cal{D}(g^{-1})=\cal{D}(gg^{-1})=\cal{D}(e).
Note:
- From the above exercise, we can see that Eq. (1) already contains the requirements that \begin{equation}\cal{D}(e)=\cal{I}\,.\end{equation} and that all the operators in the representation must be invertible \begin{equation}\cal{D}^{-1}(g):=\cal{D}(g^{-1})\,.\end{equation}
As shown in this post, given a basis |\alpha_1\rangle, \cdots, |\alpha_n\rangle, there is one-to-one correspondence between an operator and its matrix form: \begin{equation}\cal{D}(g) \Big[|\alpha_1\rangle, \cdots, |\alpha_n\rangle\Big]=\Big[|\alpha_1\rangle, \cdots, |\alpha_n\rangle\Big]\,\mathbf{D}(g)\,.\end{equation} So in some textbook, the representation of a group G is directly defined as a mapping from any g\in G to a n\times n invertible matrix of complex numbers \mathbf{D}(g).
Remarks:
- All the invertible matrices form a group under matrix multiplication called general linear group \text{GL}\left(n, \mathbb{C}\right).
- There is no requirement for the dimension n. In other word, a group can has representations in any dimensions.
- The same operator \cal{D}(g) has different but similar matrix forms under different basis \begin{equation}\mathbf{D'}(g)=\mathbf{S}^{-1}\,\mathbf{D}(g)\,\mathbf{S}\,.\end{equation} The central task of representation theory is to find a suitable basis within which the matrix form \mathbf{D'}(g) becomes block diagonal if \mathbf{D}(g) is not block diagonal.
- In an orthonormal basis, \mathbf{D}(e)=\mathbf{I}. Then in any basis, \mathbf{D'}(e)=\mathbf{I} by similar transformation.
Adjoint Representation
For any finite group G, its group elements can form a basis |g_1\rangle, \cdots, |g_n\rangle of a representation space. There is a so-called adjoint representation \cal{A} defined by \begin{equation}\cal{A}(g)\,|h\rangle :=| g hg^{-1}\tag{2}\rangle\end{equation} for any g, h\in G.
[Exercise 2] Verify that Eq. (2) forms a representation.
[Solution] We only need to check Eq. (2) preserves the group multiplication. Indeed, for any g_1, g_2, h\in G,
\begin{equation}\cal{A}(g_1)\cal{A}(g_2)\, |h\rangle=\cal{A}(g_1) |g_2hg^{-1}_2\rangle= |g_1g_2hg^{-1}_2g^{-1}_1\rangle=|(g_1g_2)h(g_1g_2)^{-1}\rangle=\cal{A}(g_1g_2)|h\rangle\,,\end{equation} and thus \cal{A}(g_1)\cal{A}(g_2)=\cal{A}(g_1g_2).
Note:
- If the basis |g_1\rangle, \cdots, |g_n\rangle is orthonormal, i.e. \langle g_i| g_j\rangle=\delta_{ij} for any g_i,g_j\in G, then the matrix element of \cal{A}(g) is \begin{equation}\mathbf{A}_{ij}(g)=\langle g_i|\cal{A}(g)|g_j\rangle\,.\tag{3}\end{equation}
- If the basis |g_1\rangle, \cdots, |g_n\rangle is not orthonormal, the matrix form \mathbf{A'}(g) differs from the above \mathbf{A}(g) by a similar transformation.
- To find irreducible representation, one can start with any matrix form and diagonalize it. So for convenience, we usually use the matrix form (3) as the adjoint representation.
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