Constructing symmetry-adapted basis

A typical procedure to use irreps is as follows:

1. Define an action of symmetry operations of a space group \(\mathcal{G}\) on your interested objects

2. Fourier-transform your selected basis \(\{ \phi^{(\mathbf{k})}_{i} \}_{\mathbf{k}, i}\) such that

   \[\begin{split} g \phi^{(\mathbf{k})}_{j} &= \sum_{i} \phi^{(\mathbf{k})}_{i} \Gamma^{(\mathbf{k})}_{ij} \quad (g \in \mathcal{G}^{\mathbf{k}}) \\ \mathbf{\Gamma}^{(\mathbf{k})}((\mathbf{E}, \mathbf{t})) &= e^{ -i\mathbf{k}\cdot\mathbf{t} } \mathbf{1} \quad ( (\mathbf{E}, \mathbf{t}) \in \mathcal{G}^{\mathbf{k}}) \\\end{split}\]

3. Compute unitary small representations \(\Gamma^{ (\mathbf{k}, \alpha) }\) of little group \(\mathcal{G}^{\mathbf{k}}\) by spgrep.get_spacegroup_irreps_from_primitive_symmetry() in primitive cell

4. Apply projection operator by spgrep.representation.project_to_irrep()

Projection operator

Let \(\Delta^{(\alpha)}\) be unitary projective irrep of group \(G\) with \(\mu(E, E)=1\). The projection operator can be defined as the same form with the ordinary representation [Alt89] [1],

\[ P^{(\alpha)}_{ij} := \frac{ d_{\alpha} }{|G|} \sum_{ R \in G } \Delta^{ (\alpha) }(R)_{ij}^{\ast} R. \]

Let \(\phi^{(\alpha,j)}_{i} := P^{(\alpha)}_{ij} \phi\). Basis vectors \(\{ \phi^{(\alpha,j)}_{i} \}_{i}\) forms \(\Delta^{ (\alpha) }\).

\[\begin{split} P^{(\alpha)}_{ij} P^{(\alpha)}_{i'j'} &= \frac{ d_{\alpha}^{2} }{|G|^{2}} \sum_{ R, R' \in G } \Delta^{ (\alpha) }(R)_{ij}^{\ast} \Delta^{ (\alpha) }(R)_{i'j'}^{\ast} RR' \\ &= \frac{ d_{\alpha}^{2} }{|G|^{2}} \sum_{ R, T \in G } \sum_{l} \frac{ \mu(R, R^{1})\mu(R, R^{-1}T) }{ \mu(R^{-1}, T) } \Delta^{ (\alpha) }(R)_{ij}^{\ast} \Delta^{ (\alpha) }(R)_{li'} \Delta^{ (\alpha) }(T)_{lj'}^{\ast} T \\ &= \delta_{ji'} \frac{ d_{\alpha} }{|G|} \sum_{ T \in G } \Delta^{ (\alpha) }(T)_{ij'}^{\ast} T \quad (\because \mu(E, T) = 1) \\ &= \delta_{ji'} P^{(\alpha)}_{ij'} \\ P^{(\alpha) \dagger}_{ij} &= P^{(\alpha)}_{ji} \\ \end{split}\]

Basis vectors \(\{ \phi^{(\alpha,j)}_{i} \}_{i}\) are mutually orthogonal:

\[\begin{split} ( \phi^{(\alpha,j)}_{i}, \phi^{(\alpha,j)}_{i'} ) &= (P^{(\alpha)}_{ij}\phi, P^{(\alpha)}_{i'j}\phi) \\ &= (\phi, P^{(\alpha)}_{ji}P^{(\alpha)}_{i'j}\phi) \\ &= \delta_{ii'}(\phi, P^{(\alpha)}_{jj}\phi) \\ &= \delta_{ii'}(\phi, \phi^{(\alpha, j)}_{j}). \end{split}\]

References

[Alt89]

S. L. Altmann. Projection operators and clebsch–gordan coefficients for projective representations. Int. J. Quantum Chem., 35(3):441–456, 1989. URL: https://onlinelibrary.wiley.com/doi/abs/10.1002/qua.560350309, doi:https://doi.org/10.1002/qua.560350309.

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[1]

\(\{ \phi^{ (\alpha,j,n) }_{i} \}_{i=1}^{d_{\alpha}}\) forms basis for \(\Delta^{ (\alpha) }\) if \(\phi^{ (\alpha,j,n) }_{i} \neq 0\). The derivation is as follows:

\[\begin{split} S \phi^{ (\alpha,j,n) }_{i} &= \frac{ d_{\alpha} }{|G|} \sum_{ R \in G } \Delta^{ (\alpha) }(R)_{ij}^{\ast} R \phi \\ &= \frac{ d_{\alpha} }{|G|} \sum_{ T \in G } \Delta^{ (\alpha) }(S^{-1}T)_{ij}^{\ast} T \phi \\ &= \frac{ d_{\alpha} }{|G|} \sum_{ T \in G } \sum_{l} \frac{\mu(T^{-1}, S) \mu(T, T^{-1})}{\mu(S^{-1}T, T^{-1}S)} \Delta^{ (\alpha) }(S)_{li} \Delta^{ (\alpha) }(T)_{lj}^{\ast} T \phi \\ &= \sum_{l} \phi^{ (\alpha,j,n) }_{l} \Delta^{ (\alpha) }(S)_{li}.\end{split}\]

Here we assume the factor system is chosen as \(\mu(E, E) = 1\). Then, we use \(\mu(S, S^{-1}) = \mu(S^{-1}, S)\).