# Crystal symmetries

## Theory

In this discussion we will only describe the situation for a monoatomic crystal $\mathcal C \subset \mathbb R^3$, the extension being easy. A symmetry of the crystal is an orthogonal matrix $\widetilde{S}$ and a real-space vector $\widetilde{\tau}$ such that

\[\widetilde{S} \mathcal{C} + \widetilde{\tau} = \mathcal{C}.\]

The symmetries where $\widetilde{S} = 1$ and $\widetilde{\tau}$ is a lattice vector are always assumed and ignored in the following.

We can define a corresponding unitary operator $U$ on $L^2(\mathbb R^3)$ with action

\[ (Uu)(x) = u\left( \widetilde{S} x + \widetilde{\tau} \right).\]

We assume that the atomic potentials are radial and that any self-consistent potential also respects this symmetry, so that $U$ commutes with the Hamiltonian.

This operator acts on a plane-wave as

\[\begin{aligned} (U e^{iq\cdot x}) (x) &= e^{iq \cdot \widetilde{\tau}} e^{i (\widetilde{S}^T q) x}\\ &= e^{- i(S q) \cdot \tau } e^{i (S q) \cdot x} \end{aligned}\]

where we set

\[\begin{aligned} S &= \widetilde{S}^{T}\\ \tau &= -\widetilde{S}^{-1}\widetilde{\tau}. \end{aligned}\]

(these equations being also valid in reduced coordinates).

It follows that the Fourier transform satisfies

\[\widehat{Uu}(q) = e^{- iq \cdot \tau} \widehat u(S^{-1} q)\]

In particular, if $e^{ik\cdot x} u_{k}(x)$ is an eigenfunction, then by decomposing $u_k$ over plane-waves $e^{i G \cdot x}$ one can see that $e^{i(S^T k) \cdot x} (U u_k)(x)$ is also an eigenfunction: we can choose

\[u_{Sk} = U u_k.\]

This is used to reduce the computations needed. For a uniform sampling of the Brillouin zone (the *reducible $k$-points*), one can find a reduced set of $k$-points (the *irreducible $k$-points*) such that the eigenvectors at the reducible $k$-points can be deduced from those at the irreducible $k$-points.

## Example

Let us demonstrate this in practice. We consider silicon, setup appropriately in the `lattice`

and `atoms`

objects as in Tutorial and to reach a fast execution, we take a small `Ecut`

of `5`

and a `[4, 4, 4]`

Monkhorst-Pack grid. First we perform the DFT calculation disabling symmetry handling

```
model = model_LDA(lattice, atoms)
basis_nosym = PlaneWaveBasis(model; Ecut, kgrid, use_symmetry=false)
scfres_nosym = @time self_consistent_field(basis_nosym, tol=1e-8)
```

n Energy Eₙ-Eₙ₋₁ ρout-ρin α Diag --- --------------- --------- -------- ---- ---- 1 -7.868428668666 NaN 1.88e-01 0.80 2.9 2 -7.872623855515 -4.20e-03 2.85e-02 0.80 1.0 3 -7.872877536219 -2.54e-04 2.59e-03 0.80 2.9 4 -7.872878198005 -6.62e-07 4.20e-04 0.80 2.3 5 -7.872878211476 -1.35e-08 3.34e-05 0.80 1.3 6 -7.872878213074 -1.60e-09 6.48e-06 0.80 2.8 5.755119 seconds (1.58 M allocations: 555.039 MiB, 2.88% gc time, 3.03% compilation time)

and then redo it using symmetry (the default):

```
basis_sym = PlaneWaveBasis(model; Ecut, kgrid)
scfres_sym = @time self_consistent_field(basis_sym, tol=1e-8)
```

n Energy Eₙ-Eₙ₋₁ ρout-ρin α Diag --- --------------- --------- -------- ---- ---- 1 -7.868244716888 NaN 1.88e-01 0.80 3.6 2 -7.872629383821 -4.38e-03 2.89e-02 0.80 1.0 3 -7.872877307474 -2.48e-04 2.55e-03 0.80 2.9 4 -7.872878194166 -8.87e-07 4.47e-04 0.80 2.5 5 -7.872878211379 -1.72e-08 4.18e-05 0.80 1.7 6 -7.872878213103 -1.72e-09 4.70e-06 0.80 2.8 1.037127 seconds (222.88 k allocations: 103.025 MiB, 3.36% gc time)

Clearly both yield the same energy but the version employing symmetry is faster, since less $k$-points are explicitly treated:

`(length(basis_sym.kpoints), length(basis_nosym.kpoints))`

(10, 64)

Both SCFs would even agree in the convergence history if exact diagonalization was used for the eigensolver in each step of both SCFs. But since DFTK adjusts this `diagtol`

value adaptively during the SCF to increase performance, a slightly different history is obtained. Try adding the keyword argument `determine_diagtol=(args...; kwargs...) -> 1e-8`

in each SCF call to fix the diagonalization tolerance to be `1e-8`

for all SCF steps, which will result in an almost identical convergence history.

We can also explicitly verify both methods to yield the same density:

```
(norm(scfres_sym.ρ - scfres_nosym.ρ),
norm(values(scfres_sym.energies) .- values(scfres_nosym.energies)))
```

(1.3875354518984363e-5, 4.453426851872815e-6)

To demonstrate the mapping between `k`

-points due to symmetry, we pick an arbitrary `k`

-point in the irreducible Brillouin zone:

```
ikpt_irred = 2
kpt_irred_coord = basis_sym.kpoints[ikpt_irred].coordinate
basis_sym.ksymops[ikpt_irred]
```

6-element Vector{Tuple{StaticArrays.SMatrix{3, 3, Int64, 9}, StaticArrays.SVector{3, Float64}}}: ([1 0 0; 0 1 0; 0 0 1], [0.0, 0.0, 0.0]) ([0 0 1; 1 0 0; 0 1 0], [0.0, 0.0, 0.0]) ([0 1 0; 0 0 1; 1 0 0], [0.0, 0.0, 0.0]) ([0 0 -1; 0 -1 0; -1 0 0], [0.0, 0.0, 0.0]) ([0 -1 0; -1 0 0; 0 0 -1], [0.0, 0.0, 0.0]) ([-1 0 0; 0 0 -1; 0 -1 0], [0.0, 0.0, 0.0])

This is a list of all symmetries operations $(S, \tau)$ that can be used to map this irreducible $k$-point to reducible $k$-points. Let's pick the third symmetry operation of this $k$-point and check.

```
S, τ = basis_sym.ksymops[ikpt_irred][3]
kpt_red_coord = S * basis_sym.kpoints[ikpt_irred].coordinate
ikpt_red = findfirst(kcoord -> kcoord ≈ kpt_red_coord,
[k.coordinate for k in basis_nosym.kpoints])
[scfres_sym.eigenvalues[ikpt_irred] scfres_nosym.eigenvalues[ikpt_red]]
```

7×2 Matrix{Float64}: -0.12437 -0.12437 0.0951515 0.0951517 0.176057 0.176058 0.231686 0.231686 0.366275 0.366275 0.397683 0.397684 0.405285 0.405285