Time-reversal symmetry in tight-binding models

This document follows — mainly — Chapter 12 of Wooten's book. First, we present a general way to introduce a non-unitary transformation into the formalism that can be identified with TRS. Then, we specialize to TRS and, particularly, to bosonic TRS: $\Theta² = +\mathbb{I}$.

• Time-reversal symmetry in tight-binding models
• • Time-reversal representation: theory of corepresentations
  • • Construction of corepresentations (corep)
    • • Specialization to grey groups
      • Three scenarios for the co-representation
  • Quantization of TRS action on creation and annihilation operators
  • Finding an explicitly real form of irrep matrices
  • Theory of representations in crystalline systems
  • • Representation of symmetry operators using a basis
    • Action of symmetry operators on a Hamiltonian
    • • Quantization of the representations
    • Time reversal symmetry
  • Proof that physically real representations admit a real basis
  • Appendix A
  • References

Time-reversal representation: theory of corepresentations

We start by considering the combined action of anti-unitary operator $\Theta$ with another linear or nonlinear operator $\mathcal{O}$.

\[\Theta \mathcal{O} \psi_\mu = \Theta \sum \psi_\nu \Gamma_{\nu\mu}(\mathcal{O}) = \sum (\Theta \psi_\nu) \Gamma^*_{\nu\mu}(\mathcal{O}) = \sum_{\nu\mu} \psi_\lambda \Sigma_\lambda\nu(\Theta) \Gamma^*_{\nu\mu}(\mathcal{O})\]

This demonstrates that the product of the two operators does not lead to just a product of the corresponding matrix representatives, but leads, in addition, to a c-conjugation of the matrix representative of $\mathcal{O}$.

Construction of corepresentations (corep)

We consider a magnetic space group $\mathcal{M}$ which we write as

\[\mathcal{M} = \mathcal{N} + \mathcal{AN},\]

where $\mathcal{N}$ is a unitary subgroup of index 2 (normal subgroup), and $\mathcal{A} \notin \mathcal{N}$ an anti-unitary element of $\mathcal{M}$.

We start with applying $\Theta$ to a basis set $\{\psi_\mu\} \equiv \Psi$ which engenders an irrep $\Delta$ of $\mathcal{N}$, namely,

\[R \psi_\mu = \sum_\nu \psi_\nu \Delta_{\nu\mu}(R), \\ R \Psi = \Psi \Delta(R).\]

The effect of $\Theta$ on this basis was shown above, and is summarized by

\[\Theta R \Psi = \Theta (\Psi \Delta(R)) = (\Theta \Psi) \Delta^*(R)\]

Next, we define the generalized time reversed set $\Phi \equiv \{\phi_\mu\} = \{\mathcal{A} \psi_\mu\}$ such that

\[R \Phi = \Phi\quad {}^\mathcal{A}\Delta(R),\]

but, since $\Theta R = R \Theta$, we have

\[R (\mathcal{A} \Psi) = (\mathcal{A} \Psi) ^\mathcal{A}\Delta(R) = R S \Theta \Psi = S (S⁻¹ R S) \Theta \Psi \\ = S \Theta (S⁻¹ R S) \Psi = (S \Theta \Psi) \Delta^*(S⁻¹ R S) = (\mathcal{A} \Psi) \Delta^*(S⁻¹ R S). \\ \boxed{^\mathcal{A}\Delta(R) = \Delta^*(S⁻¹ R S),}\]

where $\Delta(S⁻¹ R S)$ is an irrep conjugate to $\Delta(R)$, since $\mathcal{N}$ is a normal subgroup.

We now construct the rep $\Gamma$ engendered by the combined basis $\mathbf{F} = [\Psi, \mathcal{A}\Psi]$, namely,

\[\boxed{R \mathbf{F} = \mathbf{F} \Gamma(R) = [\Psi \mathcal{A} \Psi] \begin{pmatrix} \Delta(R) & \mathbf{0} \\ \mathbf{0} & \Delta^*(S⁻¹ R S) \end{pmatrix}, \qquad \forall R \in \mathcal{N}.}\]

Next we apply an operation $\mathcal{B} = \mathcal{A} T \in \mathcal{AN}$, and obtain

\[\mathcal{B} \Psi = \mathcal{A} T \Psi = \mathcal{A} \Psi \Delta(T) = (\mathcal{A} \Psi) \Delta^*(T) = (\mathcal{A} \Psi) \Delta^*(\mathcal{A}⁻¹ \mathcal{B}), \\ \mathcal{B} (\mathcal{A} \Psi) = \mathcal{B} \mathcal{A} \Psi = \Psi \Delta(\mathcal{B} \mathcal{A}), \qquad \mathcal{BA} \in \mathcal{N}.\]

We then find

\[\boxed{\mathcal{B} \mathbf{F} = \mathbf{F} \Gamma(\mathcal{B}) = [\Psi \mathcal{A} \Psi] \begin{pmatrix} \mathbf{0} & \Delta(\mathcal{BA}) \\ \Delta^*(\mathcal{A⁻¹B}) & \mathbf{0} \end{pmatrix}, \qquad \forall \mathcal{B} \in \mathcal{AN}.}\]

The set of unitary matrices obtained forms a corepresentation (corep) of $\mathcal{M}$, derived from the unitary irrep $\Delta$ of its normal subgroup $\mathcal{N}$.

Specialization to grey groups

We now specialize to grey groups (type II), the case relevant to us. Here $\mathcal{A} = \Theta$ and $\mathcal{N} = \mathcal{G}$, where $\mathcal{G}$ is the space group, so that

\[\mathcal{M} = \mathcal{G} \oplus \Theta \mathcal{G} = \mathcal{G} \otimes \{E, \Theta\}.\]

Constructing the rep $\Gamma$ for a transformation $R \in \mathcal{G}$, we obtain

\[\boxed{R \mathbf{F} = \mathbf{F} \Gamma(R) = [\Psi \quad \Theta \Psi] \begin{pmatrix} \Delta(R) & \mathbf{0} \\ \mathbf{0} & \Delta^*(R) \end{pmatrix}.}\]

Next we do it for any element $\mathcal{B} = \Theta T \in \Theta\mathcal{G}$, and obtain

\[\boxed{\mathcal{B} \mathbf{F} = \mathbf{F} \Gamma(\mathcal{B}) = [\Psi \Theta\Psi] \begin{pmatrix} \mathbf{0} & \Delta(T\Theta²) \\ \Delta^*(T) & \mathbf{0} \end{pmatrix} = [\Psi \Theta\Psi] \begin{pmatrix} \mathbf{0} & \Delta(T) \\ \Delta^*(T) & \mathbf{0} \end{pmatrix}.}\]

Three scenarios for the co-representation

Given the co-representation $\Gamma$ constructed above, three scenarios arise depending on the relationship between $\Psi$ and $\Theta\Psi$:

1. Real representation: $\Theta\Psi$ reproduces the same set as $\Psi$. The co-representation coincides with the original representation $\Delta$ and no new degeneracy is introduced.
2. Pseudo-real representation: $\Theta\Psi \neq \Psi$ but $\Theta\Psi$ also forms a basis for $\Delta$. The co-representation corresponds to $\Delta$ with doubled dimension. This case does not occur for site-symmetry groups (it requires even-dimensional representations).
3. Complex representation: $\Theta\Psi$ forms a basis for a representation $\Delta'$ inequivalent to $\Delta$. The co-representation pairs $\Delta$ and $\Delta'$, forcing them to become degenerate.

To work uniformly in scenario 1 (the most convenient case), we realify the representations beforehand using realify in Crystalline.jl, which constructs explicit co-representations in cases 2 and 3 so that the combined basis transforms under a single real representation.

Quantization of TRS action on creation and annihilation operators

Assuming a physically real basis (achieved via physically_realify in Crystalline.jl), $\Theta$ acts trivially on the real-space orbitals and therefore simply conjugates the Bloch phases. The action on creation operators follows directly:

\[\hat{\Theta} \hat{a}^\dagger_{I,\mathbf{k}} \hat{\Theta}^{-1} = \hat{a}^\dagger_{I,-\mathbf{k}}\]

For annihilation operators, we cannot use $\hat{\Theta}^\dagger = \hat{\Theta}^{-1}$ since $\Theta$ is anti-unitary. Instead, we use $\hat{\Theta}^2 = +1 \Rightarrow \hat{\Theta} = \hat{\Theta}^{-1}$ and act on a general single-particle state:

\[\hat{\Theta} \hat{a}_{I,\mathbf{k}} \hat{\Theta}^{-1} \ket{\varphi_{J,\mathbf{k}'}} = \hat{\Theta} \hat{a}_{I,\mathbf{k}} \ket{\varphi_{J,-\mathbf{k}'}} = \hat{\Theta} \left( \delta_{\mathbf{k},-\mathbf{k}'} \delta_{IJ} \ket{0} \right) = \delta_{\mathbf{k},-\mathbf{k}'} \delta_{IJ} \ket{0} = \hat{a}_{I,-\mathbf{k}} \ket{\varphi_{J,\mathbf{k}'}} \\ \Rightarrow \boxed{\hat{\Theta} \hat{a}_{I,\mathbf{k}} \hat{\Theta}^{-1} = \hat{a}_{I,-\mathbf{k}}.}\]

Finding an explicitly real form of irrep matrices

An explicit real, or physically real, form of a set of irrep matrices is one where the associated matrices $D(g)$ have the following property:

\[D(g) = D^*(g),\]

for all operations $g$ in the considered group $G$.

The standard listings of irreps are not explicitly real. However, if an irrep is either intrinsically real - or has been made into a corep in the complex or pseudoreal case - it is always equivalent to an intrinsically real form. That is, there exists a unitary transform $S$ such that:

\[S D(g) S^{-1} = S D(g) S^\dagger = D^*(g).\]

Suppose we can find this unitary transformation $S$ by some means. What we are interested in, is finding a related transform $W$, defining an explicitly real form of the irrep:

\[\tilde{D}(g) = W D(g) W^{-1} = W D(g) W^\dagger,\]

where $W$ is some other unitary transformation and where $\tilde{D}(g)$ is an intrinsically real form of $D(g)$, i.e., where

\[\tilde{D}(g) = \tilde{D}^*(g) \quad \forall g\in G.\]

Our aim is to find $W$, assuming we know $S$. For brevity, we will often write $D_g$ in place of $D(g)$. First, note that $S$ is not merely a unitary matrix: rather, since, by assumption $D(g)$ is a "real" matrix, what we really mean is that $S$ is also a symmetric unitary matrix, i.e., $S = S^{\mathrm{T}}$ and $S^{-1} = S^\dagger$ (implying, jointly, $S^* = S^\dagger = S^{-1}$); this is e.g., derived in Inui p. 74 (bottom) to 75 (top). Accordingly $S$ is also normal, i.e., $S S^* = S^* S$.

This property in turn implies that we can express $S$ as the square of another symmetric unitary matrix, say $W$, in the sense that $S = W^2$. This follows from the following manipulations (Inui, p. 75 bottom), involving the eigendecomposition $S = V Λ V^{-1}$ where $\Lambda$ is a diagonal matrix with unit-modulus values and $V$ are a set of real eigenvectors (real because $S$ is symmetric unitary) and $V^{-1} = V^\dagger = V^{\mathrm{T}}$ (since $S$ is normal).

\[S = VΛV^{-1} = VΛ^{1/2}Λ^{1/2}V^{\mathrm{T}} = (VΛ^{1/2}V^{\mathrm{T}}) (VΛ^{1/2}V^{\mathrm{T}}),\]

so we can pick $W = VΛ^{1/2}V^{\mathrm{T}}$ (note also that the square root of $\Lambda$ must exist and is well-defined since $S$ is invertible, i.e., has full rank). Hence $W^* = V(Λ^{1/2})^*V^{\mathrm{T}} = VΛ^{-1/2}V^{\mathrm{T}} = W^{-1}$ and $W^{\mathrm{T}} = (VΛ^{1/2}V^{\mathrm{T}})^{\mathrm{T}} = (V^{\mathrm{T}})^{\mathrm{T}}(Λ^{1/2})^{\mathrm{T}} V^{\mathrm{T}} = VΛ^{1/2}V^{\mathrm{T}} = W$. I.e., $W$ is also unitary symmetric and normal.

Now, let us rewrite $S D(g) S^{-1} = D^*(g)$ in terms of $W$:

\[WW D_g W^{-1}W^{-1} = D_g^* \\\]

Multiply from LHS by $W^*$ and from RHS by $W$:

\[W^*WW D_g W^{-1}W^{-1} W = W^*D_g^* W \\ \Leftrightarrow W D_g W^{-1} = W^*D_g^* W \\ \Leftrightarrow W D_g W^{-1} = W^* D_g^* (W^{-1})^*\]

where we have used the properties of $W$ to reduce the expressions.

Identifying $\tilde{D}(g) = W D(g) W^{-1}$ we clearly obtain the desired invariance under complex conjugation since $\tilde{D}^*(g) = (W D_g W^{-1})^* = W^* D_g^* (W^{-1})^* = W D_g W^{-1} = \tilde{D}(g)$.

Theory of representations in crystalline systems

This section describes how to express a general Hamiltonian in a symmetry-adapted basis and derives the resulting symmetry constraints. We first study the action of a symmetry transformation on a basis set in $k$-space, and then derive the constraints that the Hamiltonian matrix must satisfy.

Representation of symmetry operators using a basis

Following the deductions made by Bradlyn et al. in Ref. [1].

Let us start with a basis set in real space $\{ψ_{iα} (\mathbf{r})\}$, where $i$ indicates the internal degrees of freedom of the orbital, $α$ indicates the site $\mathbf{q}_α$ inside the Wyckoff position. Notice that by construction we assume each function $ψ_{iα} (\mathbf{r})$ is localized on $\mathbf{q}_α$. Intuitively, these can be thought of as Wannier functions.

We focus on a particular orbital $ψ_{i1}(\mathbf{r})$ localized in the site $\mathbf{q}_1 \equiv \mathbf{q}$. This orbital will transform under the representation $ρ$ of the site-symmetry group $G_\mathbf{q}$, associated with $\mathbf{q}$. Then, for each $h \in G_\mathbf{q}$:

\[h ψ_{i1}(\mathbf{r}) = [ρ(h)]_{ji} ψ_{j1}(\mathbf{r})\]

Within the primitive unit cell, an orbital localized on each $\mathbf{q}_α$ can be defined as:

\[ψ_{iα}(\mathbf{r}) = g_α ψ_{i1}(\mathbf{r}) = ψ_{i1}(g_α^{-1} \mathbf{r}),\]

where $g_α$, with translations, generates the coset decomposition of $G_\mathbf{q}$ in $G$. In other words, we can assign for each $\mathbf{q}_α$ a space group element $g \in G$, such that $\mathbf{q}_α = g_α\mathbf{q}$ and:

\[G = \bigcup_{α=1}^n g_α (G_\mathbf{q} \ltimes T).\]

By extension, translated counterparts in other unit cells can be defined by:

\[\{E|\mathbf{t}\} ψ_{iα}(\mathbf{r}) = ψ_{iα}(\mathbf{r}-\mathbf{t}),\]

where $\mathbf{t}$ is a lattice translation. The set of $n \times \text{dim}(ρ) \times \mathcal{N}$ functions $ψ_{iα}(\mathbf{r}-\mathbf{t})$, where $\mathcal{N}$ is the number of unit cells of the system, will be the basis set on which the induced representation $D$ will act.

Specifically, given $g = \{R|\mathbf{v}\} \in G$, the coset decomposition implies that for each $g_α$, there is a unique operation $g_β$ such that:

\[g g_α = \{E|\mathbf{t}_{αβ}\} g_β h,\]

where $h \in G_\mathbf{q}$ and $t_{αβ} \equiv g\mathbf{q}_α - \mathbf{q}_β$.

Taking all of this into consideration, we can deduce how our basis set will transform under the action of every $g \in G$:

\[g ψ_{iα}(\mathbf{r}-\mathbf{t}) = g \{E|\mathbf{t}\} ψ_{iα}(\mathbf{r}) = \\ \{E|R\mathbf{t}\} g ψ_{iα}(\mathbf{r}) = \\ \{E|R\mathbf{t}\} \{E|\mathbf{t}_{αβ}\} g_β h ψ_{i1}(\mathbf{r}) = \\ \{E|R\mathbf{t}\} \{E|\mathbf{t}_{αβ}\} g_β [ρ(h)]_{ji} ψ_{j1}(\mathbf{r}) = \\ \{E|R\mathbf{t}\} \{E|\mathbf{t}_{αβ}\} [ρ(h)]_{ji} ψ_{jβ}(\mathbf{r}) = \\ \{E|R\mathbf{t}\} [ρ(h)]_{ji} ψ_{jβ}(\mathbf{r}-\mathbf{t}_{αβ}) = \\ [ρ(h)]_{ji} ψ_{jβ}(\mathbf{r}-R\mathbf{\mathbf{t}-\mathbf{t}_{αβ}})\]

While it is natural to define the representation in real space, it will be more useful to view it in reciprocal space. This is more evident when $\mathcal{N} \to \infty$. To this end, we define the Fourier transform of our basis:

\[φ_{iα,\mathbf{k}}(\mathbf{r}) = \sum_\mathbf{t} e^{i\mathbf{k}\cdot(\mathbf{t}+\mathbf{q}_α)} ψ_{iα}(\mathbf{r}-\mathbf{t}),\]

where the sum is over all lattice vectors $\mathbf{t} \in T$.

The Fourier transform amounts to a unitary transformation that exchanges $\mathcal{N}$ unit cells for $\mathcal{N}$ distinct $\mathbf{k}$ points. The action of $g \in G$ in reciprocal space becomes:

\[g φ_{iα,\mathbf{k}}(\mathbf{r}) = \sum_\mathbf{t} e^{i\mathbf{k}\cdot(\mathbf{t}+\mathbf{q}_α)} g ψ_{iα}(\mathbf{r}-\mathbf{t}) = \\ \sum_\mathbf{t} e^{i\mathbf{k}\cdot(\mathbf{t}+\mathbf{q}_α)} [ρ(h)]_{ji} ψ_{jβ}(\mathbf{r}-R\mathbf{\mathbf{t}-\mathbf{t}_{αβ}}) = \\ \sum_\mathbf{t}' e^{i\mathbf{k}\cdot R^{-1}(\mathbf{t}'+\mathbf{q}_β-\mathbf{v})} [ρ(h)]_{ji} ψ_{jβ}(\mathbf{r}-\mathbf{t}') = \\ e^{-i([R⁻¹]ᵀ \mathbf{k}) \cdot \mathbf{v}} [ρ(h)]_{ji} \sum_\mathbf{t}' e^{i([R⁻¹]ᵀ \mathbf{k}) \cdot (\mathbf{t}'+\mathbf{q}_β)} ψ_{jβ}(\mathbf{r}-\mathbf{t}') = \\ e^{-i([R⁻¹]ᵀ \mathbf{k}) \cdot \mathbf{v}} [ρ(h)]_{ji} φ_{jβ,[R⁻¹]ᵀ\mathbf{k}}(\mathbf{r}),\]

where we have made the substitution: $\mathbf{t}' = R\mathbf{t} + \mathbf{t}_{αβ} = R\mathbf{t} + g\mathbf{q}_α - \mathbf{q}_β = R\mathbf{t} + R\mathbf{q}_α + \mathbf{v} - \mathbf{q}_β = R(\mathbf{t}+\mathbf{q}_α) + \mathbf{v} - \mathbf{q}_β \Rightarrow (\mathbf{t}+\mathbf{q}_α) = R^{-1} (\mathbf{t}'+\mathbf{q}_β-\mathbf{v})$.

In reciprocal space, the matrix representation can be interpreted as a $\mathcal{N} \times \mathcal{N}$ matrix of $n\dim(ρ) \times n\dim(ρ)$ blocks, each block can be labeled by $\mathbf{k},\mathbf{k}'$. Most of the blocks are zero: given $g = \{R| \mathbf{v}\} \in G$, there is only one non-zero block in each row and column, corresponding to $\mathbf{k}' = R\mathbf{k}$. Mathematically, we can express this as:

\[g φ_{iα,\mathbf{k}}(\mathbf{r}) = \sum_{jβ\mathbf{k}'} D_{jβ\mathbf{k}',iα\mathbf{k}}(g) φ_{jβ,\mathbf{k}'}(\mathbf{r}),\]

where we have that:

\[D_{jβ\mathbf{k}',iα\mathbf{k}}(g) = e^{-i(g\mathbf{k}) \cdot \mathbf{v}} ρ_{ji}(h) \delta_{g\mathbf{k},\mathbf{k}'} \delta_{g\mathbf{q}_α - \mathbf{q}_β \mod T},\]

We will use the following notation:

\[Ρ_{jβ,iα}(g) = e^{-i(g\mathbf{k}) \cdot \mathbf{v}} ρ_{ji}(h) \delta_{g\mathbf{q}_α - \mathbf{q}_β \mod T},\]

where the dependence on $\mathbf{k}$ is left implicit.

We can vectorize the previous equation as:

\[\boxed{g Φ_\mathbf{k}(\mathbf{r}) = Ρ^T(g) Φ_{g\mathbf{k}}(\mathbf{r}),}\]

where $Φ_\mathbf{k}(\mathbf{r})$ is a column vector formed by $\{φ_{iα,\mathbf{k}}(\mathbf{r})\}$, and, $Ρ(g)$ is an $n \times n$ matrix of $\dim(ρ) \times \dim(ρ)$ blocks, each of them can be labelled by $α,β$. Most of the blocks are zero: given $g \in G$, there is only one non-zero block in each row and column, corresponding to $g\mathbf{q}_α - \mathbf{q}_β = 0 \mod T$, and is equal to:

\[Ρ_{jβ,iα}(g) = e^{-i(g\mathbf{k}) \cdot \mathbf{v}} [ρ(h)]_{ji} \delta_{g\mathbf{q}_α - \mathbf{q}_β \mod T}.\]

Action of symmetry operators on a Hamiltonian

Let us start with the most general non-interacting Hamiltonian:

\[\hat{H} = \sum_{IJ,\mathbf{R}\mathbf{R}'} h_{IJ,\mathbf{R}-\mathbf{R}'} \; \hat{c}_{I,\mathbf{R}}^\dagger \hat{c}_{J,\mathbf{R}'},\]

where $I,J$ collect the internal degrees of freedom of the orbitals and the sites of the WP, i.e., $I = (i, α)$; and $\mathbf{R},\mathbf{R}'$ run over the lattice translations.

To be consistent with the Fourier transform convention above, the creation operator transforms as:

\[ĉ_{I,𝐑}^† = \frac{1}{\sqrt{N}} \sum_{𝐤} e^{-i𝐤·(𝐑+𝐪_α)} â_{I,𝐤}^†,\]

obtaining:

\[\hat{H} = \frac{1}{N} \sum_{IJ,\mathbf{R}\mathbf{R}'} h_{IJ,\mathbf{t}} \sum_{\mathbf{k}\mathbf{k}'} e^{-i\mathbf{k}·(\mathbf{R}+\mathbf{q}_α)} e^{i\mathbf{k}'·(R'+\mathbf{q}_β)} \hat{a}^\dagger_{I,\mathbf{k}} \hat{a}_{J,\mathbf{k}'} \\ = \frac{1}{N} \sum_{IJ,\mathbf{t},\mathbf{k}\mathbf{k}'} h_{IJ,\mathbf{t}} \left[ \sum_{\mathbf{R}'} e^{i(\mathbf{k}'-\mathbf{k})·\mathbf{R}} \right] e^{i\mathbf{k}·(\mathbf{t}-\mathbf{q}_α)} e^{i\mathbf{k}'·\mathbf{q}_β} \hat{a}^\dagger_{I,\mathbf{k}} \hat{a}_{J,\mathbf{k}'} \\ = \sum_{IJ,\mathbf{t},\mathbf{k}\mathbf{k}'} h_{IJ,\mathbf{t}} \delta_{\mathbf{k},\mathbf{k}'} e^{i\mathbf{k}·(\mathbf{t}-\mathbf{q}_α)} e^{i\mathbf{k}'·\mathbf{q}_β} \hat{a}^\dagger_{I,\mathbf{k}} \hat{a}_{J,\mathbf{k}'} \\ = \sum_{IJ,\mathbf{t},\mathbf{k}} h_{IJ,\mathbf{t}} e^{i\mathbf{k}·(\mathbf{t}+\mathbf{q}_β-\mathbf{q}_α)} \hat{a}^\dagger_{I,\mathbf{k}} \hat{a}_{J,\mathbf{k}} \\ = \sum_{IJ,\mathbf{k}} h_{IJ,\mathbf{k}} \hat{a}^\dagger_{I,\mathbf{k}} \hat{a}_{J,\mathbf{k}},\]

where we have defined: $h_{IJ,\mathbf{k}} = \sum_\mathbf{t} h_{IJ,\mathbf{t}} e^{i\mathbf{k}·(\mathbf{t}+\mathbf{q}_β-\mathbf{q}_α)}$.

Quantization of the representations

The quantization of the previous (classical) theory of representations can be written using "braket" notation as:

\[\hat{g} \ket{φ_{I,\mathbf{k}}} = Ρ_{JI}(g) \ket{φ_{J,g\mathbf{k}}},\]

where $\ket{φ_{I,\mathbf{k}}} \equiv a^\dagger_{I,\mathbf{k}} \ket{0}$. Then:

\[\hat{g} \hat{a}^\dagger_{I,\mathbf{k}} \hat{g}^{-1} \ket{0} = \hat{g} \hat{a}^\dagger_{I,\mathbf{k}} \ket{0} = Ρ_{JI}(g) \hat{a}^\dagger_{J,g\mathbf{k}} \ket{0} \\ \Rightarrow \boxed{\hat{g} \hat{a}^\dagger_{I,\mathbf{k}} \hat{g}^{-1} = Ρ_{JI}(g) \hat{a}^\dagger_{J,g\mathbf{k}}.}\]

Consequently:

\[\hat{g} \hat{a}_{I,\mathbf{k}} \hat{g}^\dagger = Ρ^*_{JI}(g) \hat{a}_{J,g\mathbf{k}} \Rightarrow \boxed{\hat{g} \hat{a}_{I,\mathbf{k}} \hat{g}^{-1} = Ρ^*_{JI}(g) \hat{a}_{J,g\mathbf{k}}.}\]

If the Hamiltonian is invariant under a symmetries $\{g\}$, we must require that:

\[\hat{H} = \hat{g} \hat{H} \hat{g}^{-1},\]

for every $g$. Then, recalling that $\hat{H} = \sum_{IJ,\mathbf{k}} \hat{a}^\dagger_{I,\mathbf{k}} h_{IJ,\mathbf{k}} \hat{a}_{J,\mathbf{k}}$, this requirement can be translated to a requirement on the Bloch Hamiltonian $H_{\mathbf{k}}$ associated with $\hat{H}$:

\[\hat{g} \hat{H} \hat{g}^{-1} = \sum_{IJ,\mathbf{k}} \hat{g} \hat{a}^\dagger_{I,\mathbf{k}} h_{IJ,\mathbf{k}} \hat{a}_{J,\mathbf{k}} \hat{g}^{-1} \\ = \sum_{IJ,\mathbf{k}} \hat{g} \hat{a}^\dagger_{I,\mathbf{k}} \hat{g}^{-1} h_{IJ,\mathbf{k}} \hat{g} \hat{a}_{J,\mathbf{k}} \hat{g}^{-1} \\ = \sum_{IJ,\mathbf{k},I'J'} \hat{a}^\dagger_{I',g\mathbf{k}} Ρ_{I'I}(g) h_{IJ,\mathbf{k}} Ρ^*_{J'J}(g) \hat{a}_{J',g\mathbf{k}} \\ = \sum_{IJ,\mathbf{k}} \hat{a}^\dagger_{I,g\mathbf{k}} \left[ Ρ(g) H_{\mathbf{k}} Ρ^\dagger(g) \right]_{IJ} \hat{a}_{J,g\mathbf{k}},\]

where we have defined $H_\mathbf{k} \equiv h_{IJ,\mathbf{k}}$ and made the substitution $I',J' \to I,J$. Comparing the first and final rows we obtain the following relation for the Hamiltonian to be invariant under symmetries:

\[\boxed{H_\mathbf{k} = Ρ(g) H_{g^{-1}\mathbf{k}} Ρ^\dagger(g).}\]

Time reversal symmetry

For TRS, a similar computation can be performed. Let us assume that the action of TRS over our basis is the following:

\[Θ \ket{φ_{I,\mathbf{k}}} = \ket{φ_{I,\mathbf{-k}}},\]

then, we obtain the following relations:

\[\boxed{\hat{Θ} \hat{a}^\dagger_{I,\mathbf{k}} \hat{Θ}^{-1} = \hat{a}^\dagger_{I,-\mathbf{k}}, \qquad \hat{Θ} \hat{a}_{I,\mathbf{k}} \hat{Θ}^{-1} = \hat{a}_{I,-\mathbf{k}}.}\]

Then, the invariance under TRS of the Hamiltonian is simply reduced to:

\[\hat{Θ} \hat{H} \hat{Θ}^{-1} = \sum_{IJ,\mathbf{k}} \hat{Θ} \hat{a}^\dagger_{I,\mathbf{k}} h_{IJ,\mathbf{k}} \hat{a}_{J,\mathbf{k}} \hat{Θ}^{-1} \\ = \sum_{IJ,\mathbf{k}} \hat{Θ} \hat{a}^\dagger_{I,\mathbf{k}} \hat{Θ}^{-1} h^*_{IJ,\mathbf{k}} \hat{Θ} \hat{a}_{J,\mathbf{k}} \hat{Θ}^{-1} \\ = \sum_{IJ,\mathbf{k}} \hat{a}^\dagger_{I,-\mathbf{k}} h^*_{IJ,\mathbf{k}} \hat{a}_{J,-\mathbf{k}} = \\ \hat{H} = \sum_{IJ,\mathbf{k}} \hat{a}^\dagger_{I,\mathbf{k}} h_{IJ,\mathbf{k}} \hat{a}_{J,\mathbf{k}}.\]

Obtaining the following relation:

\[\boxed{H_\mathbf{k} = H^*_{-\mathbf{k}}.}\]

Proof that physically real representations admit a real basis

In the presence of time-reversal symmetry, i.e., $Θ g = g Θ$, and for an explicitly real representation $Ρ(g) = Ρ^*(g)$, we now show that this implies the existence of a real basis for the representation. First, we recall the definition of the representation as it acts on a basis element $φ_{I,\mathbf{k}}$: $g φ_{I,\mathbf{k}} = Ρ_{JI}(g) φ_{J,g\mathbf{k}}$. On the other hand, in the presence of time-reversal symmetry, we also have $Θ g = g Θ$. Accordingly:

\[g Θ φ_{I,\mathbf{k}} = Θ (g φ_{I,\mathbf{k}}) = Θ (Ρ_{JI}(g) φ_{J,g\mathbf{k}}) = Ρ^*_{JI}(g) (Θ φ_{J,g \mathbf{k}}) = Ρ_{JI}(g) (Θ φ_{J,g\mathbf{k}}).\]

Then $Θ φ_{I,\mathbf{k}}$ is another basis element for the same (i.e., identical) representation $Ρ_{JI}(g)$ as $φ_{I,\mathbf{k}}$. As a result, they can be chosen such that $φ_{I,\mathbf{k}} = Θ φ_{I,\mathbf{k}}$.

Appendix A

For completeness, we derive the commutatation relations of a symmetry $g ∈ G$ with the creation and annihilation operators, as an alternative to the conjugation relations used above.

\[[\hat{g}, \hat{a}^\dagger_{I,\mathbf{k}}] \ket{0} = \hat{g} \hat{a}^\dagger_{I,\mathbf{k}} \ket{0} - \hat{a}^\dagger_{I,\mathbf{k}} \hat{g} \ket{0} = \hat{g} \ket{φ_{I,\mathbf{k}}} - \hat{a}^\dagger_{I,\mathbf{k}} \ket{0} \\ = Ρ_{JI}(g) \ket{φ_{J,g\mathbf{k}}} - \hat{a}^\dagger_{I,\mathbf{k}} \ket{0} = \left( Ρ_{JI}(g) \hat{a}^\dagger_{J,g\mathbf{k}} - \hat{a}^\dagger_{I,\mathbf{k}} \right) \ket{0} \\ \Rightarrow \boxed{[\hat{g}, \hat{a}^\dagger_{I,\mathbf{k}}] = Ρ_{JI}(g) \hat{a}^\dagger_{J,g\mathbf{k}} - \hat{a}^\dagger_{I,\mathbf{k}}.}\]

Now we want to do a similar computation for the annihilation operator. However, since $[a,a^\dagger] ≠ 0$, we cannot do the previous trick. We will use a more general single-particle state $\ket{φ_{I',\mathbf{k}'}}$:

\[\left[\hat{g}, \hat{a}_{I,\mathbf{k}} \right] \ket{φ_{I',\mathbf{k}'}} = \hat{g} \hat{a}_{I,\mathbf{k}} \ket{φ_{I',\mathbf{k}'}} - \hat{a}_{I,\mathbf{k}} \hat{g} \ket{φ_{I',\mathbf{k}'}} = δ_{II'} δ_{\mathbf{k}\mathbf{k}'} \hat{g} \ket{0} - P_{JI'}(g) \hat{a}_{I,\mathbf{k}} \ket{φ_{J,\mathbf{gk}}} \\ \Rightarrow \boxed{\left[\hat{g}, \hat{a}_{I,\mathbf{k}} \right] = 0.}\]

Then:

\[[\hat{g}, \hat{H}] = \sum_{IJ,\mathbf{k}} h_{IJ,\mathbf{k}} [\hat{g}, \hat{a}^\dagger_{I,\mathbf{k}} \hat{a}_{J,\mathbf{k}}] = \sum_{IJ,\mathbf{k}} h_{IJ,\mathbf{k}} \left( [\hat{g}, \hat{a}^\dagger_{I,\mathbf{k}}] \hat{a}_{J,\mathbf{k}} + \hat{a}^\dagger_{I,\mathbf{k}} [\hat{g}, \hat{a}_{J,\mathbf{k}}] \right) \\ = \sum_{IJ,\mathbf{k}, I'} h_{IJ,\mathbf{k}} \left[ Ρ_{I'I}(g) \hat{a}^\dagger_{I',g\mathbf{k}} - \hat{a}^\dagger_{I,\mathbf{k}} \right] \hat{a}_{J,\mathbf{k}}.\]

References

[1] Band Representations and Topological Quantum Chemistry by Bradlyn et al. (2021) https://doi.org/10.1146/annurev-conmatphys-041720-124134