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View Code? Open in Web Editor NEWSymmetry-allowed k ⋅ p expansions
License: MIT License
Symmetry-allowed k ⋅ p expansions
License: MIT License
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Our check for isweyl
does agree with the results of Table S1 in https://doi.org/10.1016/j.scib.2021.10.023 for space groups 168-173 (see the supplement).
It seems that there is something off about the time-reversal constraint in our implementation, since all these cases are cases where we have irreps that are glued together.
E.g., for space group 168, we currently get:
julia> lgirsd = lgirreps(168)
julia> lgirsd_tr = Dict(klab=>realify(lgirs) for (klab, lgirs) in lgirsd)
julia> kdotp(lgirsd_tr["K"][2]; timereversal=true) # the K2K3 irrep
HamiltonianExpansion{3} up to degree 1 for 2D irrep (K₂K₃):
┌ MonomialHamiltonian{3} of degree 1 with 1 basis elements:
│ ₁₎ ┌ ┐
│ │ 1 · │z
│ │ · -1 │
└ └ ┘
Interestingly, if we pretend the irrep is glued together but not a corep - i.e., ignore the time-reversal constraint - we get a more Weyl-like result:
julia> kdotp(lgirsd["K"][2]+lgirsd["K"][3]; timereversal=false)
HamiltonianExpansion{3} up to degree 1 for 2D irrep (K₂⊕K₃):
┌ MonomialHamiltonian{3} of degree 1 with 4 basis elements:
│ ₁₎ ┌ ┐ ┌ ┐
│ │ · 1 │(-2x+y) + │ · -i │(-1.732y)
│ │ 1 · │ │ i · │
│ └ ┘ └ ┘
│ ₂₎ ┌ ┐ ┌ ┐
│ │ · 1 │1.732y + │ · -i │(-2x+y)
│ │ 1 · │ │ i · │
│ └ ┘ └ ┘
│ ₃₎ ┌ ┐
│ │ 1 · │z
│ │ · -1 │
│ └ ┘
│ ₄₎ ┌ ┐
│ │ 1 · │z
│ │ · 1 │
└ └ ┘
The last term disagrees with the model listed in the supplement of https://doi.org/10.1016/j.scib.2021.10.023 (p. 696) - but otherwise it seems about right.
The following is surprising - not sure if that makes sense ("G" is a nonmaximal k-point = [α, 1/2, γ])
julia> kdotp(realify(lgirreps(4)["G"])[1]; timereversal=true)
HamiltonianExpansion{3} up to degree 1 for 2D irrep (G₁G₁):
┌ MonomialHamiltonian{3} of degree 1 with 7 basis elements:
│ ₁₎ ┌ ┐
│ │ · 1 │x
│ │ 1 · │
│ └ ┘
│ ₂₎ ┌ ┐
│ │ 1 · │x
│ │ · -1 │
│ └ ┘
│ ₃₎ ┌ ┐
│ │ 1 · │x
│ │ · 1 │
│ └ ┘
│ ₄₎ ┌ ┐
│ │ · -i │y
│ │ i · │
│ └ ┘
│ ₅₎ ┌ ┐
│ │ · 1 │z
│ │ 1 · │
│ └ ┘
│ ₆₎ ┌ ┐
│ │ 1 · │z
│ │ · -1 │
│ └ ┘
│ ₇₎ ┌ ┐
│ │ 1 · │z
│ │ · 1 │
└ └ ┘
How can it have free coefficients along x
and z
when we know that it should remain 2-fold degenerate along [α, 1/2, γ]?
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