# Comparing discretization techniques

In Periodic problems and plane-wave discretisations we saw how simple 1D problems can be modelled using plane-wave basis sets. This example invites you to work out some details on these aspects yourself using a number of exercises. The solutions are given at the bottom of the page.

For this example we consider the discretization of

\[ H = - \frac12 Δ + V(x) \quad \text{with $V(x) = \cos(x)$}\]

on $[0, 2π]$ with periodic boundary conditions. The $\cos(x)$ takes the role of a lattice-periodic potential. We will be interested in the smallest eigenvalues of this discretized Hamiltonian. Of note the boundary condition matters: The spectrum we will get is different from e.g. the spectrum of $H$ on $\mathbb{R}$.

## Finite differences

We approximate functions $ψ$ on $[0, 2\pi]$ by their values at grid points $x_k = 2\pi \frac{k}{N}$, $k=1, \dots, N$. The boundary conditions are imposed by $ψ(x_0) = ψ(x_N), ψ(x_{N+1}) = ψ(x_0)$. We then have

\[ \big(Hψ\big)(x_k) \approx \frac 1 2 \frac{-ψ_{k-1} + 2 ψ_k - ψ_{k+1}}{2 δx^2} + V(x_k) ψ(x_k)\]

with $δx = \frac{2π}{N}$.

This can be put in matrix form in the following way:

```
# Finite differences Hamiltonian -1/2 Delta + V on [0, 2pi] with periodic bc.
# Pass it a function V.
using LinearAlgebra
function build_finite_differences_matrix(Vfunction, N::Integer)
δx = 2π/N
# Finite-difference approximation to -½Δ
T = 1/(2δx^2) * Tridiagonal(-ones(N-1), 2ones(N), -ones(N-1))
# The type Tridiagonal is efficient, but to establish the periodic boundary conditions
# we need to add elements not on the three diagonals, so convert to dense matrix
T = Matrix(T)
T[1, N] = T[N, 1] = -1 / (2δx^2)
# Finite-difference approximation to potential: We collect all coordinates ...
x_coords = [k * δx for k=1:N]
V = Diagonal(Vfunction.(x_coords)) # ... and evaluate V on each of the x_coords
T + V
end;
```

Show that the finite-difference approximation of -½Δ is indeed an approximation of the second derivative. Obtain an estimate of the first eigenvalue of $H$. *Hint:* Take a look at the `eigen`

function from `LinearAlgebra`

.

## Plane waves method

In this method, we expand states on the basis

\[ e_G(x) = \frac{1}{\sqrt{2\pi}} e^{iGx} \qquad \text{for $G=-N,\dots,N$}.\]

Show that

\[ \langle e_G, e_{G'}\rangle = ∫_0^{2π} e_G(x) e_{G'}(x) d x = δ_{G, G'}\]

and (assuming $V(x) = \cos(x)$)

\[ \langle e_G, H e_{G'}\rangle = \frac 1 2 \left(|G|^2 \delta_{G,G'} + \delta_{G, G'+1} + \delta_{G, G'-1}\right).\]

What happens for a more general $V(x)$?

Code this and check the first eigenvalue agrees with the finite-difference case. Compare accuracies at various basis set sizes $N$.

## Using DFTK

We now use DFTK to do the same plane-wave discretisation in this 1D system. To deal with a 1D case we use a 3D lattice with two lattice vectors set to zero.

```
using DFTK
a = 2π
lattice = a .* [[1 0 0.]; [0 0 0]; [0 0 0]];
```

Define Hamiltonian: Kinetic + Potential

```
terms = [Kinetic(),
ExternalFromReal(r -> cos(r[1]))] # r is a vector of size 3
model = Model(lattice; n_electrons=1, terms, spin_polarization=:spinless); # One spinless electron
```

Ecut defines the number of plane waves by selecting all those $G$, which satisfy the relationship $½ |G|^2 ≤ \text{Ecut}$.

```
Ecut = 500
basis = PlaneWaveBasis(model; Ecut, kgrid=(1, 1, 1))
```

```
PlaneWaveBasis discretization:
architecture : DFTK.CPU()
num. mpi processes : 1
num. julia threads : 1
num. blas threads : 2
num. fft threads : 1
Ecut : 500.0 Ha
fft_size : (135, 1, 1), 135 total points
kgrid : MonkhorstPack([1, 1, 1])
num. red. kpoints : 1
num. irred. kpoints : 1
Discretized Model(custom, 1D):
lattice (in Bohr) : [6.28319 , 0 , 0 ]
[0 , 0 , 0 ]
[0 , 0 , 0 ]
unit cell volume : 6.2832 Bohr
num. electrons : 1
spin polarization : spinless
temperature : 0 Ha
terms : Kinetic()
ExternalFromReal(Main.var"#3#4"())
```

We now seek the ground state. To better separate the two steps (SCF outer loop and diagonalization inner loop), we set the diagtol (the tolerance of the eigensolver) to a small value.

```
diagtolalg = AdaptiveDiagtol(; diagtol_max=1e-8, diagtol_first=1e-8)
scfres = self_consistent_field(basis; tol=1e-4, diagtolalg)
scfres.energies
```

```
Energy breakdown (in Ha):
Kinetic 0.2090845
ExternalFromReal -0.7441493
total -0.535064852288
```

On this simple linear (non-interacting) model, the SCF converges in one step. The ground state energy of is simply the lowest eigenvalue; it should match the smallest eigenvalue of $H$ computed above.

## Plotting

We can also get the first eigenvector (in the plane wave basis) and plot it

```
using Plots
ψ_fourier = scfres.ψ[1][:, 1]; # first k-point, all G components, first eigenvector
# Transform the wave function to real space
ψ = ifft(basis, basis.kpoints[1], ψ_fourier)[:, 1, 1]
# Eigenvectors are only defined up to a phase. We fix it by imposing that psi(0) is real
ψ /= (ψ[1] / abs(ψ[1]))
plot(real(ψ); label="")
```

Again this should match with the result above.

**Exercise 4:** Look at the Fourier coefficients of $\psi$ ($\psi$_fourier) and compare with the result above.

## The DFTK Hamiltonian

We can ask DFTK for the Hamiltonian

```
E, ham = energy_hamiltonian(basis, scfres.ψ, scfres.occupation; ρ=scfres.ρ)
H = ham.blocks[1]
typeof(H)
```

`DFTK.DftHamiltonianBlock`

This is an opaque data structure, which encodes the Hamiltonian. What can we do with it?

```
using InteractiveUtils
methodswith(typeof(H), supertypes=true)
```

9-element Vector{Method}:- Array(block::
**HamiltonianBlock**) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/terms/Hamiltonian.jl:74 - *(H::
**HamiltonianBlock**, ψ) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/terms/Hamiltonian.jl:62 - (Matrix)(block::
**HamiltonianBlock**) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/terms/Hamiltonian.jl:75 - eltype(block::
**HamiltonianBlock**) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/terms/Hamiltonian.jl:63 - size(block::
**HamiltonianBlock**, i::**Integer**) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/terms/Hamiltonian.jl:64 - size(block::
**HamiltonianBlock**) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/terms/Hamiltonian.jl:65 - mul!(Hψ::
**AbstractArray**, H::**DFTK.DftHamiltonianBlock**, ψ::**AbstractArray**) in DFTK at /home/runner/.julia/packages/TimerOutputs/RsWnF/src/TimerOutput.jl:230 - PreconditionerNone(::
**HamiltonianBlock**) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/eigen/preconditioners.jl:21 - PreconditionerTPA(ham::
**HamiltonianBlock**;*kwargs...*) in DFTK at /home/runner/work/DFTK.jl/DFTK.jl/src/eigen/preconditioners.jl:45

This defines a number of methods. For instance, it can be used as a linear operator:

`H * DFTK.random_orbitals(basis, basis.kpoints[1], 1)`

```
63×1 Matrix{ComplexF64}:
0.08850946567064152 + 0.021385680388499813im
0.08893316969333047 + 0.024851810500206573im
0.15029554364765027 + 0.07611846242076661im
-0.7118210487877756 - 0.08205535590550457im
-1.100934486157015 - 0.5180797524069599im
-3.3707384685115755 - 2.746970537655698im
0.5086979696403797 + 1.425409741409736im
-3.2037491302909897 + 3.272229409650207im
-0.20355573371600033 + 6.144654327162291im
7.7205559235051355 - 1.9063415511283628im
⋮
5.5547074041958036 - 1.34821749223096im
-0.4333451841936238 + 0.4960358130179822im
3.1590340448074117 - 4.538882557963048im
0.7957291544788444 - 1.2070968151519466im
1.5810714619527642 - 0.932534890934508im
0.09425943036526939 - 0.4963306616081152im
-0.4120972336235175 + 0.0424977881182214im
-0.18127423391091008 + 0.17744734974077933im
-0.04455291106953063 + 0.03635409526145292im
```

We can also get its full matrix representation:

`Array(H)`

```
63×63 Matrix{ComplexF64}:
-3.5247e-17+0.0im … 0.5-4.87133e-17im
0.5-4.87133e-17im 1.93395e-17+1.19856e-17im
1.93395e-17+1.19856e-17im -1.76211e-18+3.43427e-18im
-1.76211e-18+3.43427e-18im -5.27235e-18-1.56473e-17im
-5.27235e-18-1.56473e-17im 1.48093e-17-1.93685e-17im
1.48093e-17-1.93685e-17im … 8.85398e-18+4.44201e-18im
8.85398e-18+4.44201e-18im -6.82427e-18+6.26594e-18im
-6.82427e-18+6.26594e-18im 6.5791e-18+9.27972e-18im
6.5791e-18+9.27972e-18im 1.9934e-18+3.73282e-18im
1.9934e-18+3.73282e-18im 2.96007e-18-6.58852e-18im
⋮ ⋱
1.9934e-18-3.73282e-18im 1.84038e-17+2.28001e-19im
1.84038e-17+2.28001e-19im … -6.82427e-18-6.26594e-18im
-6.82427e-18-6.26594e-18im 8.85398e-18-4.44201e-18im
8.85398e-18-4.44201e-18im 1.48093e-17+1.93685e-17im
1.48093e-17+1.93685e-17im -5.27235e-18+1.56473e-17im
-5.27235e-18+1.56473e-17im -1.76211e-18-3.43427e-18im
-1.76211e-18-3.43427e-18im … 1.93395e-17-1.19856e-17im
1.93395e-17-1.19856e-17im 0.5+6.88035e-17im
0.5+6.88035e-17im 0.5+0.0im
```

Compare this matrix with the one you obtained previously, get its eigenvectors and eigenvalues. Try to guess the ordering of $G$-vectors in DFTK.

Increase the size of the problem, and compare the time spent by DFTK's internal diagonalization algorithms to a full diagonalization of `Array(H)`

. *Hint:* The `@time`

macro is handy for this task.

## Solutions

### Exercise 1

If we consider a function $f : [0, 2π] → \mathbb{R}$, to first order

\[f(x + δx) = f(x) + δx f'(x) + O(δx^2)\]

therefore after rearrangement

\[f'(x) = \frac{f(x + δx) - f(x)}{δx} + O(δx).\]

Similarly

\[f''(x) = \frac{f'(x + δx) - f'(x)}{δx} + O(δx),\]

such that overall

\[f''(x) \simeq \frac{f(x + 2δx) - f(x + δx) - f(x + δx) + f(x)}{δx^2} = \frac{f(x + 2δx) - 2f(x + δx) f(x)}{δx^2}\]

In finite differences we consider a stick basis of vectors

\[\left\{ e_i = (0, …, 0, \underbrace{δx}_\text{$i$-th position}, 0, …, 0) \middle| i = 1, … N \right\}.\]

Keeping in mind the periodic boundary conditions (i.e. $e_0 = e_N$) projecting the Hamiltonian $H$ onto this basis thus yields the proposed structure.

More details needed

We start off with $N = 500$ to obtain

```
Hfd = build_finite_differences_matrix(cos, 100)
L, V = eigen(Hfd)
L[1:5]
```

```
5-element Vector{Float64}:
-0.5351640695081414
0.3429576842587494
0.8532206682511969
2.053679933442362
2.078512110605267
```

This is already pretty accurate (to about 4 digits) as can be estimated looking at the following convergence plot:

```
using Plots
function fconv(N)
L, V = eigen(build_finite_differences_matrix(cos, N))
first(L)
end
Nrange = 10:10:100
plot(Nrange, abs.(fconv.(Nrange) .- fconv(200)); yaxis=:log, legend=false)
```

### Exercise 2

This solution has not yet been written. Any help with a PR is appreciated.

### Exercise 3

This solution has not yet been written. Any help with a PR is appreciated.

### Exercise 4

This solution has not yet been written. Any help with a PR is appreciated.

### Exercise 5

This solution has not yet been written. Any help with a PR is appreciated.

### Exercise 6

This solution has not yet been written. Any help with a PR is appreciated.