API reference

TwoBody.BasisSetType

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),
)
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TwoBody.DatabaseEntryType
DatabaseEntry(hamiltonian, energy)

A benchmark problem stored in the database. hamiltonian is ready to be passed to a solver and energy is its reference energy.

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TwoBody.ExponentialType

Exponential(coefficient=1, exponent=1)

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

ArgumentsSymbol
coefficient$a$
exponent$b$
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TwoBody.FiniteDifferenceMethodType

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

ArgumentsDefaultDescription
Δr::Real0.1Radial grid spacing. A uniform grid spacing is used, $r_{i+1} = r_{i} + \Delta r$.
rₘₐₓ::Real50.0The maximum value of the radial grid. This value is not directly used in the calculation, but it is used to determine the R.
R::StepRangeLenΔr:Δr:rₘₐₓRadial grid. The origin must be excluded from the grid to avoid divergence of the Coulomb potential and the centrifugal potential at the origin.
l::Int0Angular momentum quantum number. This is a positive integer, $0 \leq l$.
direction::Symbol:cThe direction of the finite difference, :c for central, :f for forward, :b for backward.
solver::Symbol:LinearAlgebraThe solver for eigenvalue problem, :LinearAlgebra or :ArnoldiMethod.
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TwoBody.GaussianType

Gaussian(coefficient=1, exponent=1)

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

ArgumentsSymbol
coefficient$a$
exponent$b$
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TwoBody.GeometricBasisSetType

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

This is a basis set with exponentials generated by geometric(r₁, rₙ, n; nₘₐₓ=n, nₘᵢₙ=1). You can define the same basis set as Table A2 in E. Hiyama, M. Kamimura, Front. Phys. 13, 132106 (2018) like this:

\[ r_1 = 0.1, r_{n_\mathrm{max}} = 80.0, n_\mathrm{max} = 20.\]

BS = GeometricBasisSet(SimpleGaussianBasis, 0.1, 80.0, 20)
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TwoBody.HamiltonianType

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(
  Kinetic(hbar = 1, m = 1),
  Coulomb(coefficient = -1),
)
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TwoBody.PowerLawType

PowerLaw(coefficient=1, exponent=1)

\[+ ar^n\]

ArgumentsSymbol
coefficient$a$
exponent$n$
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TwoBody.RelativisticCorrectionType

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.

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TwoBody.SimpleGaussianBasisType

SimpleGaussianBasis(a=1)

Note

This basis is not normalized and only for s-wave.

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:

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

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TwoBody.VariationalMonteCarloType

VariationalMonteCarlo(n_steps=10^5, burn_in=10^3, thinning=1, n_walkers=1, δ=0.5, r₀=[1.0, 0.0, 0.0])

Options for variational Monte Carlo with a symmetric, uniform Metropolis proposal. The sampler targets $|\psi(\mathbf{r})|^2$. n_steps is the number of retained samples per walker, burn_in is the number of initial transitions discarded from each walker, thinning is the number of transitions between retained samples, n_walkers is the number of Markov chains, δ is the proposal-box width, and r₀ is the initial position of every walker. Each walker performs burn_in + n_steps * thinning transitions.

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TwoBody.YukawaType

Yukawa(coefficient=1, exponent=1)

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

ArgumentsSymbol
coefficient$a$
exponent$b$
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Base.put!Method
put!(key, hamiltonian, energy)

Add a benchmark problem to the database. key may be a Symbol or string, hamiltonian must be a Hamiltonian, and energy must be real. Registering the same key twice throws an ArgumentError.

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TwoBody.dbMethod
db(key::Union{Symbol,AbstractString}) -> DatabaseEntry

Return the benchmark Hamiltonian and reference energy associated with key. The returned Hamiltonian is independent of the stored value and can safely be modified by callers.

Examples

entry = db(:hydrogen)
result = solve(entry.hamiltonian, method)
isapprox(result.values[1], entry.energy)
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TwoBody.dbkeysMethod
dbkeys() -> Vector{Symbol}

Return the available database keys in deterministic order.

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TwoBody.elementMethod

element(o::Constant, 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:

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

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TwoBody.elementMethod

element(o::Coulomb, 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:

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

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TwoBody.elementMethod

element(o::Gaussian, 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:

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

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TwoBody.elementMethod

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

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

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TwoBody.elementMethod

element(o::Kinetic, 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:

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

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TwoBody.elementMethod

element(o::Linear, 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:

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

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TwoBody.elementMethod

element(o::PowerLaw, 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:

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

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TwoBody.elementMethod

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:

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

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TwoBody.elementMethod

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:

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

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TwoBody.elementMethod

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

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

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TwoBody.geometricMethod

geometric(r₁, rₙ, n::Int; nₘₐₓ::Int=n, nₘᵢₙ::Int=1)

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
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TwoBody.local_energyFunction

local_energy(hamiltonian, wavefunction, position)

Evaluate

\[E_\mathrm{loc}(\mathbf{r}) = \frac{\hat{H}\psi(\mathbf{r})}{\psi(\mathbf{r})}.\]

The Laplacian of a real-valued trial wavefunction is evaluated with ForwardDiff.hessian. Non-relativistic kinetic, Laplacian, rest-energy, and potential terms with a defined V method are supported.

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TwoBody.matrixMethod

matrix(basisset::BasisSet)

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

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TwoBody.matrixMethod

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

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

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TwoBody.matrixMethod

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

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

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

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TwoBody.matrixMethod

matrix(o::Kinetic, 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

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

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TwoBody.matrixMethod

matrix(operator::Operator, basisset::BasisSet)

Note

This function is used for the expectation values and is not used in computing the Hamiltonian matrix.

This function returns the matrix corresponding to the operator in the given basis set. The element is written as $O_{ij} = \langle \phi_{i} | \hat{o} | \phi_{j} \rangle$.

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TwoBody.matrixMethod

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

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

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

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TwoBody.matrixMethod

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

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

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

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TwoBody.optimizeMethod

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.

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

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TwoBody.optimizeMethod

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...).

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TwoBody.optimizeMethod

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.

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

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TwoBody.solveFunction

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

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TwoBody.solveMethod

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 $S_{ij} = \langle \phi_{i} | \phi_{j} \rangle$.

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TwoBody.solveMethod

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).

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TwoBody.solveMethod

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,

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

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TwoBody.solveMethod

solve(hamiltonian, wavefunction, method::VariationalMonteCarlo; rng=Random.MersenneTwister(123), info=0)

Estimate the variational energy by averaging the local energy over Metropolis samples from $|\psi|^2$. A seeded random-number generator can be supplied with rng; by default, a new MersenneTwister(123) is used for a reproducible calculation. Set info to a positive value to print a result summary.

The returned named tuple echoes the input hamiltonian and method and contains E, variance, the naive standard_error, acceptance_rate, the sampling counts n_accepted, n_attempted, n_burn_in_discarded, n_samples, and n_discarded, as well as local_energies and samples. Data from multiple walkers are stored consecutively, with positions in the columns of samples. Non-finite local energies, which can occur at a measure-zero singularity such as the origin of a Coulomb potential, are excluded from the energy statistics and counted in n_discarded.

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TwoBody.solveMethod

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

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

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