# 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

$$$$\tag{1}\label{1} \mathcal U_g \mathcal H \mathcal U_g^{-1} = \mathcal H,$$$$ Here, $$\mathcal U_g$$ represents the group action of $$G$$ on the hilbert space in second quantization, as \begin{align} \tag{2} \label{2} \mathcal U_g c_{k\alpha} \mathcal U_g^{-1} = c_{k_g\alpha}\\ \tag{3} \label{3} \mathcal U_g c^\dagger_{k\alpha} \mathcal U_g^{-1} = c^\dagger_{k_g\alpha} \end{align} The Hamiltonian is a non-interacting fermionic Hamiltonian, such as $$$$\tag{4}\label{4} \mathcal H = \sum_{k \alpha \beta} c^\dagger_{k \alpha} H(k)_{\alpha \beta} c_{k \beta}$$$$ Where, $$\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>$$. Due of the symmetry \eqref{1} of the second quantized Hamiltonian $$\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 hilbert 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. Given that $$\mathcal U_g$$ exist, then \eqref{9} provides us with a set of equation, that needs to be solved to find the exact form of $$\mathcal U_g$$. Although this approach might work, it seems naive.

Is there a better way to find $$\mathcal U_g$$?

• Is the different notation $U_g$ instead of $\mathcal{U}_g$ in (2) and (3) on purpose or is it a typo? It seems to me these should be the same, as they are both acting on Fock space. Commented Apr 11, 2023 at 16:19
• @LucasBaldo Yes, that was indeed a typo. Fixed it. Thanks for noticing. Commented Apr 12, 2023 at 11:37
• You have the action of $\mathcal{U}_g$ on $c_{k\alpha}^\dagger$ and the states $\left(\prod_i c_{k_i\alpha_i}^\dagger\right)|0\rangle$ form a basis for the Fock space, so doesn't that, at least in principle, tell you everything? Commented Apr 12, 2023 at 12:14
• Or to put it another way, why try to solve eq.(9) rather than eqs.(2) and (3)? Commented Apr 12, 2023 at 13:03
• If all you want are the single particle states then you can more or less read of eq.(3) $\mathcal{U}_g|\mathbf{k}\alpha\rangle = \mathcal{U}_g c_{k\alpha}^\dagger|0\rangle = \mathcal{U}_g c_{k\alpha}^\dagger \mathcal{U}_g^{-1} \mathcal{U}_g|0\rangle = c_{k_g\alpha}^\dagger|0\rangle = |\mathbf{k}_g\alpha\rangle$ Commented Apr 12, 2023 at 16:55

In order to find the matrix representation of $$\mathcal{U}_g$$ we need first to define a basis for this operator. Since the $$\mathcal{U}_g$$ acts on Fock space, the basis elements will be Fock states \begin{align} |\mathbf{n}\rangle \equiv \prod_{\mathbf{k}, \alpha} \left(c^{\dagger}_{\mathbf{k},\alpha} \right)^{n_{\mathbf{k}, \alpha}} | 0 \rangle. \end{align}
The matrix of $$\mathcal{U}_g$$ elements are then \begin{align} \left(\mathcal{U}_g\right)_{\mathbf{n},\mathbf{n}'} &= \langle \mathbf{n}|\mathcal{U}_g|\mathbf{n}'\rangle \\ &= \langle \mathbf{n} | \mathcal{U}_g \left[ \prod_{\mathbf{q}, \beta} \left(c^{\dagger}_{\mathbf{q},\beta} \right)^{n^{\prime}_{\mathbf{q}, \beta}} \right] | 0 \rangle \\ &= \langle \mathbf{n} | \mathcal{U}_g \mathcal{U}_g^{-1} \left[ \prod_{\mathbf{q}, \beta} \mathcal{U}_g \left(c^{\dagger}_{\mathbf{q},\beta} \right)^{n^{\prime}_{\mathbf{q}, \beta}} \mathcal{U}_g^{-1} \right] \mathcal{U}_g | 0 \rangle \\ &= \langle \mathbf{n} | \left[ \prod_{\mathbf{q}, \beta} \left(c^{\dagger}_{\mathbf{q}_g,\beta} \right)^{n^{\prime}_{\mathbf{q}, \beta}} \right] \mathcal{U}_g | 0 \rangle \\ &= \langle 0 | \left[ \prod_{\mathbf{k}, \alpha} \left(c_{\mathbf{k},\alpha} \right)^{n_{\mathbf{k}, \alpha}} \right] \left[ \prod_{\mathbf{q}, \beta} \left(c^{\dagger}_{\mathbf{q}_g,\beta} \right)^{n^{\prime}_{\mathbf{q}, \beta}} \right] | 0 \rangle \\ &= \langle 0 | \left[ \prod_{\mathbf{q}_g,\beta} \left(c_{\mathbf{q}_g,\beta} \right)^{n_{\mathbf{q}_g,\beta}} \left(c^{\dagger}_{\mathbf{q}_g,\beta} \right)^{n^{\prime}_{\mathbf{q}, \beta}} \right] | 0 \rangle (-1)^{N} \\ &= \langle 0 | { \prod_{\mathbf{q}_g,\beta} \left[ (1-\delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}}) \left(c^{\dagger}_{\mathbf{q}_g,\beta} \right)^{n^{\prime}_{\mathbf{q}, \beta}} \left(c_{\mathbf{q}_g,\beta} \right)^{n_{\mathbf{q}_g,\beta}} + \delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}} (\delta_{\mathbf{q}_g, \mathbf{q}_g} \delta_{\beta, \beta} -c^{\dagger}_{\mathbf{q}_g,\beta} c_{\mathbf{q}_g,\beta}\right]}) | 0 \rangle (-1)^{N} \\ &= \langle 0 | { \prod_{\mathbf{q}_g,\beta} \left[ (1-\delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}}) \left(c^{\dagger}_{\mathbf{q}_g,\beta} \right)^{n^{\prime}_{\mathbf{q}, \beta}} \left(c_{\mathbf{q}_g,\beta} \right)^{n_{\mathbf{q}_g,\beta}} + \delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}} \right]} | 0 \rangle (-1)^{N} \\ &= \langle 0 | { \prod_{\mathbf{q}_g,\beta} \left[ \delta_{0,n^{\prime}_{\mathbf{q}, \beta}}\delta_{0,n_{\mathbf{q}_g,\beta}} + \delta_{1,n^{\prime}_{\mathbf{q}, \beta}}\delta_{0,n_{\mathbf{q}_g,\beta}} c^{\dagger}_{\mathbf{q}_g,\beta} + \delta_{0,n^{\prime}_{\mathbf{q}, \beta}}\delta_{1,n_{\mathbf{q}_g,\beta}} c_{\mathbf{q}_g,\beta} + \delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}} \right]} | 0 \rangle (-1)^{N} \\ &= \langle 0 | { \prod_{\mathbf{q}_g,\beta} \left[ \delta_{0,n^{\prime}_{\mathbf{q}, \beta}}\delta_{0,n_{\mathbf{q}_g,\beta}} + \delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}} \right]} | 0 \rangle (-1)^{N} \\ &= { \prod_{\mathbf{q}_g,\beta} \left[ \delta_{0,n^{\prime}_{\mathbf{q}, \beta}}\delta_{0,n_{\mathbf{q}_g,\beta}} + \delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}} \right]} \langle 0 | 0 \rangle (-1)^{N} \\ &= { \prod_{\mathbf{q}_g,\beta} \left[ \delta_{0,n^{\prime}_{\mathbf{q}, \beta}}\delta_{0,n_{\mathbf{q}_g,\beta}} + \delta_{1,n^{\prime}_{\mathbf{q},\beta}} \delta_{1,n_{\mathbf{q}_g,\beta}} \right]} (-1)^{N} \\ \end{align}