API reference

Basis is an abstract type.

BasisSet(basis1, basis2, ...)

\[\{ \phi_1, \phi_2, \phi_3, \cdots \}\]

The basis set is the input for Rayleigh-Ritz method. You can define the basis set like this:

\[\begin{aligned} \phi_1(r) &= \exp(-13.00773 ~r^2), \\ \phi_2(r) &= \exp(-1.962079 ~r^2), \\ \phi_3(r) &= \exp(-0.444529 ~r^2), \\ \phi_4(r) &= \exp(-0.1219492 ~r^2). \end{aligned}\]

BS = BasisSet(
  SimpleGaussianBasis(13.00773),
  SimpleGaussianBasis(1.962079),
  SimpleGaussianBasis(0.444529),
  SimpleGaussianBasis(0.1219492),
)

ConstantPotential(constant=1)

\[+ c\]

ContractedBasis([c1, c2, ...], [basis1, basis2, ...])

\[\phi' = \sum_i c_i \phi_i\]

CoulombPotential(coefficient=1)

\[+ \frac{a}{r}\]

DeltaPotential(coefficient=1)

\[+ a δ(r)\]

ExponentialPotential(coefficient=1, exponent=1)

\[+ a \exp(- b r)\]

FiniteDifferenceMethod(Δr=0.1, rₘₐₓ=50.0, R=Δr:Δr:rₘₐₓ, l=0, direction=:c, solver=:LinearAlgebra)

FunctionPotential(f)

\[+ f(r)\]

GaussianBasis(a=1, l=0, m=0)

\[\phi_i(r, θ, φ) = N _{il} r^l \exp(-a_i r^2) Y_l^m(θ, φ)\]

GaussianPotential(coefficient=1, exponent=1)

\[+ a \exp(- b r^2)\]

GeometricBasisSet(basistype, r₁, rₙ, n; nₘᵢₙ=1, nₘₐₓ=n)

This is a basis set with exponentials generated by geometric().

Hamiltonian(operator1, operator2, ...)

\[\hat{H} = \sum_i \hat{o}_i\]

The Hamiltonian is the input for each solver. This is an example for the non-relativistic Hamiltonian of hydrogen atom in atomic units:

\[\hat{H} = - \frac{1}{2} \nabla^2 - \frac{1}{r}\]

H = Hamiltonian(
  NonRelativisticKinetic(ℏ =1 , m = 1),
  CoulombPotential(coefficient = -1),
)

KineticTerm <: Operator is an abstract type.

Laplacian(coefficient=1)

\[+ a\nabla^2\]

LinearPotential(coefficient=1)

\[+ ar\]

NonRelativisticKinetic(ℏ=1, m=1)

\[-\frac{\hbar^2}{2m} \nabla^2\]

Operator is an abstract type.

PotentialTerm <: Operator is an abstract type.

PowerLawPotential(coefficient=1, exponent=1)

\[+ ar^n\]

PrimitiveBasis <: Basis is an abstract type.

RelativisticCorrection(c=1, m=1, n=2) The p^{2n} term of the Taylor expansion:

\[\begin{aligned} \sqrt{p^2 c^2 + m^2 c^4} =& m \times c^2 \\ &+ 1 / 2 / m \times p^2 (n=1) \\ &- 1 / 8 / m^3 / c^2 \times p^4 (n=2) \\ &+ 1 / 16 / m^5 / c^4 \times p^6 (n=3) \\ &- 5 / 128 / m^7 / c^6 \times p^8 (n=4) \\ &+ \cdots \end{aligned}\]

Use c = 137.035999177 (from 2022 CODATA) in the atomic units.

RelativisticKinetic(c=1, m=1)

\[\sqrt{p^2 c^2 + m^2 c^4} - m c^2\]

Use c = 137.035999177 (from 2022 CODATA) in the atomic units.

RestEnergy(c=1, m=1)

\[m c^2\]

Use c = 137.035999177 (from 2022 CODATA) in the atomic units.

SimpleGaussianBasis(a=1)

Position-Space

\[\phi_i(\pmb{r}) = \exp(-a_i r^2)\]

Momentum-Space

\[\phi_{i}(\pmb{k}) = \frac{1}{(2a_i)^{\frac{3}{2}}} \exp(-k^2/4a_i)\]

Proof (Fourier Transform)

\[\begin{aligned} \phi_{n}(\pmb{k}) &= \frac{1}{\sqrt{2 \pi}^3} \int \phi_{n}(\pmb{r}) \mathrm{e}^{\mathrm{i} \pmb{k} \cdot \pmb{r}} \mathrm{d}\pmb{r} \\ &= \frac{1}{\sqrt{2 \pi}^3} \int \phi_{n}(\pmb{r}) \mathrm{e}^{\mathrm{i} \pmb{k} \cdot \pmb{r}} r^2 \sin (\theta) ~\mathrm{d}r \mathrm{d}\theta \mathrm{d} \varphi \\ &= \frac{1}{\sqrt{2 \pi}^3} \iiint \mathrm{e}^{-\alpha_i r^2} \sqrt{4\pi} Y_{00}(\hat{\pmb{r}}) \left[ 4 \pi \sum_{l'=0}^{\infty} \sum_{m=-l'}^{l'} \mathrm{i}^{l'} j_{l'}(pr) Y_{l'm'}(\hat{\pmb{k}}) Y_{l'm'}^*(\hat{\pmb{r}}) \right] r^2 \sin\theta~ \mathrm{d} r \mathrm{d} \theta \mathrm{d} \varphi \\ &= \frac{1}{\sqrt{2 \pi}^3} 4 \pi \sqrt{4\pi} \sum_{l'=0}^{\infty} \sum_{m=-l'}^{l'} \left[ \mathrm{i}^{l'} Y_{l'm'}(\hat{\pmb{k}}) \int_0^{2 \pi} \int_0^\pi Y_{00}(\hat{\pmb{r}}) Y_{l'm'}^*(\hat{\pmb{r}}) \sin (\theta)~ \mathrm{d} \theta \mathrm{d} \varphi \int_0^{\infty} j_{l'}(pr) \mathrm{e}^{-\alpha_i r^2} r^{2} \mathrm{d}r \right]\\ &= \frac{1}{\sqrt{2 \pi}^3} 4 \pi \sqrt{4\pi} \sum_{l'=0}^{\infty} \sum_{m=-l'}^{l'} \left[ \mathrm{i}^{l'} Y_{l'm'}(\hat{\pmb{k}}) \delta_{0l'} \delta_{0m'} \int_0^{\infty} j_{l'}(kr) \mathrm{e}^{-\alpha_i r^2} r^{2} \mathrm{d}r \right] \\ &= \frac{1}{\sqrt{2 \pi}^3} 4 \pi \sqrt{4\pi} \mathrm{i}^{0} Y_{00}(\hat{\pmb{k}}) \int_0^{\infty} j_{0}(kr) \mathrm{e}^{-\alpha_i r^2} r^{2} \mathrm{d}r \\ &= \frac{1}{2\pi\sqrt{2\pi}} 4 \pi \frac{\sqrt{4\pi}}{\sqrt{4\pi}} \sqrt{\frac{\pi}{2}} \sqrt{\frac{2}{\pi}} \int_0^{\infty} j_{0}(kr) \mathrm{e}^{-\alpha_i r^2} r^{2} ~\mathrm{d}r \\ &= \frac{1}{(2\alpha_i)^{\frac{3}{2}}} \mathrm{e}^{-\frac{k^2}{4 \alpha_i}} \end{aligned}\]

Formula

plane-wave expansion in spherical harmonics:

\[\mathrm{e}^{\mathrm{i} \pmb{k} \cdot \pmb{r}} = 4 \pi \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \mathrm{i}^{l} j_{l}(pr) Y_{lm}(\hat{\pmb{k}}) Y_{lm}^*(\hat{\pmb{r}})\]

special case of spherical harmonics:

\[Y_{00}(\hat{\pmb{r}}) = \frac{1}{\sqrt{4\pi}}\]

orthonormality of spherical harmonics:

\[\int_0^{2\pi} \int_0^\pi Y_{lm}(\hat{\pmb{r}})^* Y_{l'm'}(\hat{\pmb{r}}) \sin(\theta) ~ \mathrm{d} \theta \mathrm{d} \varphi = \delta_{ll'} \delta_{mm'}\]

citation needed:

\[\sqrt{\frac{2}{\pi}} \int r^{l} j_l(kr) \mathrm{e}^{-\alpha r^2} r^{2} \mathrm{d} r = \frac{1}{(2\alpha)^{l+\frac{3}{2}}} k^l e^{-\frac{k^2}{4\alpha}}\]

UniformGridPotential(R, V)

YukawaPotential(coefficient=1, exponent=1)

\[+ \frac{a}{r} \exp(- b r)\]

element(o::ConstantPotential, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} \langle \phi_{i} | c | \phi_{j} \rangle &= c \langle \phi_{i} | \phi_{j} \rangle \\ &= c \iiint \phi_{i}^*(r) \phi_{j}(r) ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= c \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \sin\theta ~\mathrm{d}\theta \int_0^\infty r^{2} \mathrm{e}^{-(\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= c \times 2\pi \times 2 \times \frac{1!!}{2^{2}} \sqrt{\frac{\pi}{a^{3}}} \\ &= \underline{c \left( \frac{\pi}{\alpha_i + \alpha_j} \right)^{3/2}} \end{aligned}\]

Integral Formula:

\[\int_0^{\infty} r^{2n} \exp \left(-a r^2\right) ~\mathrm{d}r = \frac{(2n-1)!!}{2^{n+1}} \sqrt{\frac{\pi}{a^{2n+1}}}\]

element(o::CoulombPotential, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} \langle \phi_{i} | \frac{1}{r} | \phi_{j} \rangle &= \iiint \phi_{i}^*(r) \times \frac{1}{r} \times \phi_{j}(r) ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \sin\theta ~\mathrm{d}\theta \int_0^\infty r \mathrm{e}^{-(\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= 2\pi \times 2 \times \frac{0!}{2 (\alpha_i + \alpha_j)} \\ &= \underline{\frac{2\pi}{\alpha_i + \alpha_j}} \end{aligned}\]

Integral Formula:

\[\int_0^{\infty} r^{2n+1} \exp \left(-a r^2\right) ~\mathrm{d}r = \frac{n!}{2 a^{n+1}}\]

element(o::GaussianPotential, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} \langle \phi_{i} | \exp(-br^2) | \phi_{j} \rangle &= \iiint \phi_{i}^*(r) \times \exp(-br^2) \times \phi_{j}(r) ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \sin\theta ~\mathrm{d}\theta \int_0^\infty r^2 \mathrm{e}^{-(b+\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= 2\pi \times 2 \times \frac{1!!}{2^{2}} \sqrt{\frac{\pi}{(b + \alpha_i + \alpha_j)^{2\cdot1+1}}} \\ &= \underline{\left( \frac{\pi}{b + \alpha_i + \alpha_j} \right)^{3/2}} \end{aligned}\]

Integral Formula:

\[\int_0^{\infty} r^{2n} \exp \left(-a r^2\right) ~\mathrm{d}r = \frac{(2n-1)!!}{2^{n+1}} \sqrt{\frac{\pi}{a^{2n+1}}}\]

element(o::Hamiltonian, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} H_{ij} &= \langle \phi_{i} | \hat{H} | \phi_{j} \rangle \\ &= \langle \phi_{i} | \sum_k \hat{o}_k | \phi_{j} \rangle \\ &= \sum_k \langle \phi_{i} | \hat{o}_k | \phi_{j} \rangle \\ \end{aligned}\]

element(o::LinearPotential, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} \langle \phi_{i} | r | \phi_{j} \rangle &= \iiint \phi_{i}^*(r) \times r \times \phi_{j}(r) ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \sin\theta ~\mathrm{d}\theta \int_0^\infty r^3 \mathrm{e}^{-(\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= 2\pi \times 2 \times \frac{1!}{2 (\alpha_i + \alpha_j)^{2}} \\ &= \underline{\frac{2\pi}{(\alpha_i + \alpha_j)^2}} \end{aligned}\]

Integral Formula:

\[\int_0^{\infty} r^{2n+1} \exp \left(-a r^2\right) ~\mathrm{d}r = \frac{n!}{2 a^{n+1}}\]

element(o::NonRelativisticKinetic, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

Derivation (without Green's identity)

\[\begin{aligned} T_{ij} = \langle \phi_{i} | \hat{T} | \phi_{j} \rangle &= \iiint \mathrm{e}^{-\alpha_i r^2} \left[ -\frac{\hbar^2}{2\mu} \nabla^2 \right] \mathrm{e}^{-\alpha_j r^2} ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= -\frac{\hbar^2}{2\mu} \iiint \mathrm{e}^{-\alpha_i r^2} \left[ \nabla^2 \right] \mathrm{e}^{-\alpha_j r^2} ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= -\frac{\hbar^2}{2\mu} \iiint \mathrm{e}^{-\alpha_i r^2} \left[ -6\alpha_j + 4\alpha_j^2 r^2 \right] \mathrm{e}^{-\alpha_j r^2} ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= -\frac{\hbar^2}{2\mu} \iint \sin\theta ~\mathrm{d}\theta \mathrm{d}\varphi \int \left[ -6\alpha_j + 4\alpha_j^2 r^2 \right] r^2 \mathrm{e}^{-(\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= -\frac{\hbar^2}{2\mu} \cdot 4\pi \left[ -6\alpha_j \mathrm{GGI}(2, \alpha_i + \alpha_j) +4\alpha_j^2 \mathrm{GGI}(4, \alpha_i + \alpha_j) \right] \\ &= -\frac{\hbar^2}{2\mu} \cdot 4\pi \left[ -6\alpha_j \frac{\Gamma\left( \frac{3}{2} \right)}{2 (\alpha_i + \alpha_j)^{\frac{3}{2}}} +4\alpha_j^2 \frac{\Gamma\left( \frac{5}{2} \right)}{2 (\alpha_i + \alpha_j)^{\frac{5}{2}}} \right] \\ &= -\frac{\hbar^2}{2\mu} \cdot 4\pi \left[ -6\alpha_j \frac{ \sqrt{\pi}/2}{2 (\alpha_i + \alpha_j)^{\frac{3}{2}}} +4\alpha_j^2 \frac{3\sqrt{\pi}/4}{2 (\alpha_i + \alpha_j)^{\frac{5}{2}}} \right] \\ &= -\frac{\hbar^2}{2\mu} \cdot 4\pi \left[ \frac{\alpha_j}{\alpha_i + \alpha_j} - 1 \right] \cdot 6 \alpha_j \cdot \frac{\sqrt{\pi}/2}{2 (\alpha_i + \alpha_j)^{\frac{3}{2}}} \\ &= -\frac{\hbar^2}{2\mu} \cdot 4\pi \left[ - \frac{\alpha_i}{\alpha_i + \alpha_j} \right] \cdot 6 \alpha_j \cdot \frac{\sqrt{\pi}/2}{2 (\alpha_i + \alpha_j)^{\frac{3}{2}}} \\ &= \underline{ \frac{\hbar^2}{2\mu} \cdot 6 \cdot \frac{\alpha_i \alpha_j \pi^{\frac{3}{2}}}{(\alpha_i + \alpha_j)^{\frac{5}{2}}} } \end{aligned}\]

Derivation (with Green's identity)

\[\begin{aligned} T_{ij} = \langle \phi_{i} | \hat{T} | \phi_{j} \rangle &= \iiint \mathrm{e}^{-\alpha_i r^2} \left[ -\frac{\hbar^2}{2\mu} \nabla^2 \right] \mathrm{e}^{-\alpha_j r^2} ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= -\frac{\hbar^2}{2\mu} \iiint \mathrm{e}^{-\alpha_i r^2} \nabla^2 \mathrm{e}^{-\alpha_j r^2} ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \frac{\hbar^2}{2\mu} \iiint \left[ \nabla \mathrm{e}^{-\alpha_i r^2} \right] \left[ \nabla \mathrm{e}^{-\alpha_j r^2} \right] ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \frac{\hbar^2}{2\mu} \iiint \left[ -2 \alpha_i r \mathrm{e}^{-\alpha_i r^2} \right] \left[ -2 \alpha_j r \mathrm{e}^{-\alpha_j r^2} \right] ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \frac{\hbar^2}{2\mu} \cdot 4 \alpha_i \alpha_j \iiint \left[ r \mathrm{e}^{-\alpha_i r^2} \right] \left[ r \mathrm{e}^{-\alpha_j r^2} \right] ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \frac{\hbar^2}{2\mu} \cdot 4 \alpha_i \alpha_j \iint \sin\theta ~\mathrm{d}\theta \mathrm{d}\varphi \int r^4 \mathrm{e}^{- (\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= \frac{\hbar^2}{2\mu} \cdot 4 \alpha_i \alpha_j \cdot 4 \pi \cdot \mathrm{GGI}(4, \alpha_i + \alpha_j) \\ &= \frac{\hbar^2}{2\mu} \cdot 4 \alpha_i \alpha_j \cdot 4 \pi \cdot \frac{\Gamma\left( \frac{5}{2} \right)}{2 (\alpha_i + \alpha_j)^{\frac{5}{2}}} \\ &= \frac{\hbar^2}{2\mu} \cdot 4 \alpha_i \alpha_j \cdot 4 \pi \cdot \frac{3\sqrt{\pi}/4}{2 (\alpha_i + \alpha_j)^{\frac{5}{2}}} \\ &= \underline{ \frac{\hbar^2}{2\mu} \cdot 6 \cdot \frac{\alpha_i \alpha_j \pi^{\frac{3}{2}}}{(\alpha_i + \alpha_j)^{\frac{5}{2}}} } \end{aligned}\]

Formula

Green's first identity:

\[\begin{aligned} \iiint_V f \pmb{\nabla}^2 g ~ \mathrm{d}V + \iiint_V \pmb{\nabla} f \cdot \pmb{\nabla} g ~ \mathrm{d}V = \iint_{\partial V} f \pmb{\nabla} g \cdot \pmb{n} ~ \mathrm{d}S \end{aligned}\]

generalized Gaussian integral:

\[\begin{aligned} \mathrm{GGI}(n,b) = \int_0^{\infty} x^{n} \exp \left(-b x^2\right) \mathrm{d}x = \frac{\Gamma\left( \frac{n+1}{2} \right)}{2 b^{\frac{n+1}{2}}} \end{aligned}\]

element(o::PowerLawPotential, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} \langle \phi_{i} | r^n | \phi_{j} \rangle &= \iiint \phi_{i}^*(r) \times r^n \times \phi_{j}(r) ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \sin\theta ~\mathrm{d}\theta \int_0^\infty r^{n+2} \mathrm{e}^{-(\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= 2\pi \times 2 \times \frac{\Gamma\left( \frac{n+3}{2} \right)}{2 (\alpha_i + \alpha_j)^{\frac{n+3}{2}}} \\ &= \underline{2\pi\frac{\Gamma\left( \frac{n+3}{2} \right)}{(\alpha_i + \alpha_j)^{\frac{n+3}{2}}}} \end{aligned}\]

Integral Formula:

\[\int_0^{\infty} r^{n} \exp \left(-a r^2\right) ~\mathrm{d}r = \frac{\Gamma\left( \frac{n+1}{2} \right)}{2 a^{\frac{n+1}{2}}}\]

element(o::RestEnergy, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} \langle \phi_{i} | mc^2 | \phi_{j} \rangle &= mc^2 \langle \phi_{i} | \phi_{j} \rangle \\ &= mc^2 \iiint \phi_{i}^*(r) \phi_{j}(r) ~r^2 \sin\theta ~\mathrm{d}r \mathrm{d}\theta \mathrm{d}\varphi \\ &= mc^2 \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \sin\theta ~\mathrm{d}\theta \int_0^\infty r^{2} \mathrm{e}^{-(\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= mc^2 \times 2\pi \times 2 \times \frac{1!!}{2^{2}} \sqrt{\frac{\pi}{a^{3}}} \\ &= \underline{mc^2 \left( \frac{\pi}{\alpha_i + \alpha_j} \right)^{3/2}} \end{aligned}\]

Integral Formula:

\[\int_0^{\infty} r^{2n} \exp \left(-a r^2\right) ~\mathrm{d}r = \frac{(2n-1)!!}{2^{n+1}} \sqrt{\frac{\pi}{a^{2n+1}}}\]

element(SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} S_{ij} = \langle \phi_{i} | \phi_{j} \rangle &= \int \phi_{i}^*(r) \phi_{j}(r) \mathrm{d} \pmb{r} \\ &= \iiint \mathrm{e}^{-\alpha_i r^2} \mathrm{e}^{-\alpha_j r^2} ~r^2 \sin\theta ~ \mathrm{d} r \mathrm{d} \theta \mathrm{d} \varphi \\ &= \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \sin\theta ~\mathrm{d}\theta \int_0^\infty r^{2} \mathrm{e}^{-(\alpha_i + \alpha_j) r^2} ~\mathrm{d}r \\ &= 2\pi \times 2 \times \frac{1!!}{2^{2}} \sqrt{\frac{\pi}{a^{3}}} \\ &= \underline{\left( \frac{\pi}{\alpha_i + \alpha_j} \right)^{3/2}} \end{aligned}\]

Integral Formula:

\[\int_0^{\infty} r^{2n} \exp \left(-a r^2\right) ~\mathrm{d}r = \frac{(2n-1)!!}{2^{n+1}} \sqrt{\frac{\pi}{a^{2n+1}}}\]

element(o::Laplacian, SGB1::SimpleGaussianBasis, SGB2::SimpleGaussianBasis)

\[\begin{aligned} \langle \phi_{i} | \nabla^2 | \phi_{j} \rangle = \underline{ -6 \frac{\alpha_i \alpha_j \pi^{\frac{3}{2}}}{(\alpha_i + \alpha_j)^{\frac{5}{2}}} } \end{aligned}\]

Exponents of Gaussian basis functions are given by geometric progression:

\[\begin{aligned} & v_i = \frac{1}{r_i^2}, \\ & r_i = r_1 a^{i-1}. \end{aligned}\]

This function return array of $\nu_i$:

\[(r_1, r_{n}, n, n_\mathrm{max}) \mapsto (\nu_1, \nu_2, \cdots, \nu_{n-1}, \nu_n, \nu_{n+1}, \cdots, \nu_{n_\mathrm{max}})\]

Usually $n = n_\mathrm{max}$. Set $n<n_\mathrm{max}$ if you want to extend the geometric progression.

Examples:

julia> ν = TwoBody.geometric(0.1, 10.0, 5)
5-element Vector{Float64}:
 100.0
  10.0
   0.9999999999999997
   0.09999999999999996
   0.009999999999999995

julia> ν = TwoBody.geometric(0.1, 10.0, 5, nₘₐₓ = 10)
10-element Vector{Float64}:
 100.0
  10.0
   0.9999999999999997
   0.09999999999999996
   0.009999999999999995
   0.0009999999999999994
   9.999999999999994e-5
   9.999999999999992e-6
   9.999999999999991e-7
   9.999999999999988e-8

matrix(basisset::BasisSet)

This function returns the overlap matrix $\pmb{S}$. The element is written as $S_{ij} = \langle \phi_{i} | \phi_{j} \rangle$.

matrix(hamiltonian::Hamiltonian, basisset::BasisSet)

This function returns the Hamiltonian matrix $\pmb{H}$. The element is written as $H_{ij} = \langle \phi_{i} | \hat{H} | \phi_{j} \rangle$.

matrix(o::Hamiltonian, method::FiniteDifferenceMethod)

The matrix for the Hamiltonian is a sum of matrices for each term,

\[\pmb{H} = \sum_i \pmb{O}_i.\]

matrix(o::NonRelativisticKinetic, method::FiniteDifferenceMethod)

We use the shorthand notation $\psi'(r) = \frac{\mathrm{d}\psi}{\mathrm{d}r}(r)$ and $\psi''(r) = \frac{\mathrm{d}^{2}\psi}{\mathrm{d}r^{2}}(r)$. For the uniform grid spacing ($r_{i+1} = r_{i} + \Delta r$), the finite difference for the first derivative,

\[\frac{\mathrm{d}\psi}{\mathrm{d}r}(r) = \frac{\psi(r+\Delta r) - \psi(r-\Delta r)}{2\Delta r} + O(\Delta r^{2})\]

is written as

\[\left(\begin{array}{ccccc} \psi'(r_1) \\ \psi'(r_2) \\ \psi'(r_3) \\ \psi'(r_4) \\ \vdots \end{array}\right) \simeq \frac{1}{2\Delta r} \left(\begin{array}{ccccc} 0 & 1 & 0 & 0 &\ldots \\ -1 & 0 & 1 & 0 &\ldots \\ 0 & -1 & 0 & 1 &\ldots \\ 0 & 0 & -1 & 0 &\ldots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}\right) \left(\begin{array}{ccccccc} \psi(r_1) \\ \psi(r_2) \\ \psi(r_3) \\ \psi(r_4) \\ \vdots \end{array}\right),\]

and the finite difference for the second derivative,

\[\frac{\mathrm{d}^{2}}{\mathrm{d}r^{2}}(r) = \frac{\psi(r+\Delta r) - 2f(r) + \psi(r-\Delta r)}{\Delta r^{2}} + O(\Delta r^{2}).\]

is written as

\[\left(\begin{array}{ccccc} \psi''(r_1) \\ \psi''(r_2) \\ \psi''(r_3) \\ \psi''(r_4) \\ \vdots \end{array}\right) \simeq \frac{1}{\Delta r^2} \left(\begin{array}{ccccccc} -2 & 1 & 0 & 0 & \ldots \\ 1 & -2 & 1 & 0 & \ldots \\ 0 & 1 & -2 & 1 & \ldots \\ 0 & 0 & 1 & -2 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right) \left(\begin{array}{ccccccc} \psi(r_1) \\ \psi(r_2) \\ \psi(r_3) \\ \psi(r_4) \\ \vdots \end{array}\right).\]

Similarly, the matrix for the kinetic energy,

\[\hat{T} = -\frac{\hbar^2}{2\mu} \left[ \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} - \frac{l(l+1)}{r^2} \right]\]

is written as

\[\pmb{T} = - \frac{\hbar^2}{2\mu} \left[ \frac{1}{{\Delta r}^2} \left(\begin{array}{ccccccc} -2 & 1 & 0 & \ldots \\ 1 & -2 & 1 & \ldots \\ 0 & 1 & -2 & \ldots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right) + \left(\begin{array}{ccccccc} 2/r_1 & 0 & 0 & \ldots \\ 0 & 2/r_2 & 0 & \ldots \\ 0 & 0 & 2/r_3 & \ldots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right) \frac{1}{\Delta r} \left(\begin{array}{ccccccc} 0 & 1 & 0 & \ldots \\ -1 & 0 & 1 & \ldots \\ 0 & -1 & 0 & \ldots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right) - l(l+1) \left(\begin{array}{ccccccc} 1/{r_1}^2 & 0 & 0 & \ldots \\ 0 & 1/{r_2}^2 & 0 & \ldots \\ 0 & 0 & 1/{r_3}^2 & \ldots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right) \right].\]

matrix(o::RestEnergy, method::FiniteDifferenceMethod)

The matrix for the rest energy $mc^2$ is a diagonal matrix,

\[mc^2 \left(\begin{array}{ccccccc} 1 & 0 & 0 & \ldots \\ 0 & 1 & 0 & \ldots \\ 0 & 0 & 1 & \ldots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right).\]

matrix(o::PotentialTerm, method::FiniteDifferenceMethod)

The matrix for the potential energy $V(r)$ is a diagonal matrix,

\[\pmb{V} = \left(\begin{array}{ccccccc} V(r_1) & 0 & 0 & \ldots \\ 0 & V(r_2) & 0 & \ldots \\ 0 & 0 & V(r_3) & \ldots \\ \vdots & \vdots & \vdots & \ddots \\ \end{array}\right).\]

function optimize(hamiltonian::Hamiltonian, basisset::BasisSet; perturbation=Hamiltonian(), info=4, progress=true, optimizer=Optim.NelderMead(), options...)

This function minimizes the energy by changing the exponents of the basis functions using Optim.jl.

\[\frac{\partial E}{\partial a_i} = 0\]

optimize(hamiltonian::Hamiltonian, basis::Basis; perturbation=Hamiltonian(), info=4, optimizer=Optim.NelderMead())

This a optimizer for 1-basis calculations. This function returns optimize(hamiltonian, BasisSet(basis); perturbation=perturbation, info=info, progress=progress, optimizer=optimizer, options...).

optimize(hamiltonian::Hamiltonian, basisset::GeometricBasisSet; perturbation=Hamiltonian(), info=4, optimizer=Optim.NelderMead())

This function minimizes the energy by optimizing $r_1$ and $r_n$ using Optim.jl.

\[\frac{\partial E}{\partial r_1} = \frac{\partial E}{\partial r_n} = 0\]

solve(hamiltonian::Hamiltonian, wavefunction::Function, method::FiniteDifferenceMethod, info=4, nₘₐₓ=4)

solve(hamiltonian::Hamiltonian, basisset::BasisSet)

This function returns the eigenvalues $E$ and eigenvectors $\pmb{c}$ for

\[\pmb{H} \pmb{c} = E \pmb{S} \pmb{c}.\]

The Hamiltonian matrix is defined as $H_{ij} = \langle \phi_{i} | \hat{H} | \phi_{j} \rangle$. The overlap matrix is defined as $S_{ij} = \langle \phi_{i} | \phi_{j} \rangle$.

solve(hamiltonian::Hamiltonian, basis::Basis; perturbation=Hamiltonian(), info=4)

This a solver for 1-basis calculations. This function returns solve(hamiltonian, BasisSet(basis); perturbation=perturbation, info=info).

solve(hamiltonian::Hamiltonian, method::FiniteDifferenceMethod; perturbation=Hamiltonian(), info=4, nₘₐₓ=4)

This method solve the eigenvalue problem for the Hamiltonian discretized as a sparse matrix with finite difference approximation,

\[\pmb{H} \pmb{\psi} = E \pmb{\psi}.\]

The eigenvalue $E$ is an approximation of the exact energy and the eigenvector $\pmb{\psi}$ is a vector of the approximated values of the exact wavefunction $\psi(r)$ on points of the grid,

\[\pmb{\psi} = \left(\begin{array}{c} \psi(r_1) \\ \psi(r_2) \\ \psi(r_3) \\ \vdots \\ \end{array}\right).\]

solve(hamiltonian::Hamiltonian, basisset::GeometricBasisSet; perturbation=Hamiltonian(), info=4)

This function is a wrapper for solve(hamiltonian::Hamiltonian, basisset::BasisSet, ...).