Representation of symmetry operators in second quantuzation
Hamiltonian invariant under a symmetry- The action of a group $G$ on the set of Bloch momentum is given by a linear representation $T_g: k \to T_g k \equiv k_g $. Now say that a fermionic Bloch Hamiltonian is invariant under the action of such Group. In second quantization this means
\begin{equation} \tag{1}\label{1} \mathcal U_g \mathcal H \mathcal U_g^{-1} = \mathcal H, \end{equation} Here, $\mathcal U_g$ represents the group action of $G$ on the hilbert space in second quantization, as \begin{align} \tag{2} \label{2} U_g c_{k\alpha} U_g^{-1} = c_{k_g\alpha}\ \tag{3} \label{3} U_g c^\dagger_{k\alpha} U_g^{-1} = c^\dagger_{k_g\alpha} \end{align} The Hamiltonian is a non-interacting fermionic Hamiltonian, such as \begin{equation} \tag{4}\label{4} \mathcal H = \sum_{k \alpha \beta} c^\dagger_{k \alpha} H(k){\alpha \beta} c{k \beta} \end{equation} Where, $\alpha, \beta$ are internal quantum numbers and $H(k)$ is the matrix valued Hamiltonian (also known as first quantized Hamiltonian) written in the basis $c^\dagger_{k\alpha}\left|0\right>$. 656
Due of the symmetry \eqref{1} of the second quantized Hamiltonian $\mathcal H$, a constraint is imposed on the first quantized matrix-valued Hamiltonian $H(k)$ at momentum $k$ as follows,
\begin{align} \tag{5}\label{5} \mathcal U_g \mathcal H \mathcal U_g^{-1} &= \sum_{k \alpha \beta}\mathcal U_g c^\dagger_{k \alpha} H(k){\alpha \beta} c{k \beta}\mathcal U_g^{-1},\ &= \sum_{k \alpha \beta}\mathcal U_g c^\dagger_{k \alpha}\mathcal U_g^{-1}\mathcal U_g H(k){\alpha \beta}\mathcal U_g^{-1}\mathcal U_g c{k \beta}\mathcal U_g^{-1},\tag{6}&&{\text{(Inserting $\mathcal U_g \mathcal U_g^{-1} = \mathbb I$)}}\ &= \sum_{k \alpha \beta} c^\dagger_{k_g \alpha}\mathcal U_g H(k){\alpha \beta}\mathcal U_g^{-1}\mathcal c{k_g \beta},\tag{7}&&{\text{(using \eqref{2} and \eqref{3}}})\ &= \sum_{k \alpha \beta} c^\dagger_{k \alpha}\mathcal U_g H(T_g^{-1} k){\alpha \beta}\mathcal U_g^{-1}\mathcal c{k \beta},\tag{8}\label{8}&&\text{(change of variable $k_g \to k$)} \end{align} Eq. \eqref{8} along with \eqref{1} and \eqref{4} implies that \begin{align} \tag{9} \label{9} H(k) = \mathcal U_g H(T_g^{-1} k)\mathcal U_g^{-1} \end{align}
This works out perfectly. I want to know that given a group $G$ and a representation of action of $G$ on momentum, can I construct the operators $\mathcal U_g$ (unitary representation) that defines the action of the group on the <PERSON> space.
I have a possible solution of this question, - If the Hamiltonian is invariant under some symmetry $G$ then there must exist matrix $\mathcal U_g$ such that \eqref{9} will hold. 616