Hartree-Fock methods

Morten Hjorth-Jensen, National Superconducting CyclotronLaboratory and Department of Physics and Astronomy, MichiganState University, East Lansing, MI 48824, USA & Department of

Physics, University of Oslo, Oslo, Norway

July 6-24 2015

Contentsparagraph>18 paragraph>18 paragraph>18 paragraph>18 paragraph>20 para-graph>20 paragraph>21 paragraph>21 paragraph>22 paragraph>22

Why Hartree-Fock? Derivation of Hartree-Fock equationsin coordinate spaceHartree-Fock (HF) theory is an algorithm for finding an approximative expressionfor the ground state of a given Hamiltonian. The basic ingredients are

• Define a single-particle basis {ψα} so that

hHFψα = εαψα

with the Hartree-Fock Hamiltonian defined as

hHF = t+ uext + uHF

• The term uHF is a single-particle potential to be determined by the HFalgorithm.

• The HF algorithm means to choose uHF in order to have

〈H〉 = EHF = 〈Φ0|H|Φ0〉

that is to find a local minimum with a Slater determinant Φ0 being the ansatzfor the ground state.

• The variational principle ensures that EHF ≥ E0, with E0 the exact groundstate energy.

We will show that the Hartree-Fock Hamiltonian hHF equals our definition of theoperator f discussed in connection with the new definition of the normal-orderedHamiltonian (see later lectures), that is we have, for a specific matrix element

〈p|hHF|q〉 = 〈p|f |q〉 = 〈p|t+ uext|q〉+∑i≤F

〈pi|V |qi〉AS ,

meaning that〈p|uHF|q〉 =

∑i≤F

〈pi|V |qi〉AS .

The so-called Hartree-Fock potential uHF brings an explicit medium dependencedue to the summation over all single-particle states below the Fermi level F . Itbrings also in an explicit dependence on the two-body interaction (in nuclearphysics we can also have complicated three- or higher-body forces). The two-body interaction, with its contribution from the other bystanding fermions,creates an effective mean field in which a given fermion moves, in addition to theexternal potential uext which confines the motion of the fermion. For systems likenuclei, there is no external confining potential. Nuclei are examples of self-boundsystems, where the binding arises due to the intrinsic nature of the strong force.For nuclear systems thus, there would be no external one-body potential in theHartree-Fock Hamiltonian.

Variational Calculus and Lagrangian MultipliersThe calculus of variations involves problems where the quantity to be minimizedor maximized is an integral.

In the general case we have an integral of the type

E[Φ] =∫ b

a

f(Φ(x), ∂Φ∂x

, x)dx,

where E is the quantity which is sought minimized or maximized. The problemis that although f is a function of the variables Φ, ∂Φ/∂x and x, the exactdependence of Φ on x is not known. This means again that even though theintegral has fixed limits a and b, the path of integration is not known. In ourcase the unknown quantities are the single-particle wave functions and we wishto choose an integration path which makes the functional E[Φ] stationary. Thismeans that we want to find minima, or maxima or saddle points. In physicswe search normally for minima. Our task is therefore to find the minimum ofE[Φ] so that its variation δE is zero subject to specific constraints. In our casethe constraints appear as the integral which expresses the orthogonality of thesingle-particle wave functions. The constraints can be treated via the techniqueof Lagrangian multipliers

Let us specialize to the expectation value of the energy for one particle inthree-dimensions. This expectation value reads

E =∫dxdydzψ∗(x, y, z)Hψ(x, y, z),

2

with the constraint ∫dxdydzψ∗(x, y, z)ψ(x, y, z) = 1,

and a HamiltonianH = −1

2∇2 + V (x, y, z).

We will, for the sake of notational convenience, skip the variables x, y, z below,and write for example V (x, y, z) = V .

The integral involving the kinetic energy can be written as, with the functionψ vanishing strongly for large values of x, y, z (given here by the limits a and b),∫ b

a

dxdydzψ∗(−1

2∇2)ψdxdydz = ψ∗∇ψ|ba +

∫ b

a

dxdydz12∇ψ

∗∇ψ.

We will drop the limits a and b in the remaining discussion. Inserting thisexpression into the expectation value for the energy and taking the variationalminimum we obtain

δE = δ

{∫dxdydz

(12∇ψ

∗∇ψ + V ψ∗ψ

)}= 0.

The constraint appears in integral form as∫dxdydzψ∗ψ = constant,

and multiplying with a Lagrangian multiplier λ and taking the variationalminimum we obtain the final variational equation

δ

{∫dxdydz

(12∇ψ

∗∇ψ + V ψ∗ψ − λψ∗ψ)}

= 0.

We introduce the function f

f = 12∇ψ

∗∇ψ + V ψ∗ψ − λψ∗ψ = 12(ψ∗xψx + ψ∗yψy + ψ∗zψz) + V ψ∗ψ − λψ∗ψ,

where we have skipped the dependence on x, y, z and introduced the shorthandψx, ψy and ψz for the various derivatives.

For ψ∗ the Euler-Lagrange equations yield

∂f

∂ψ∗− ∂

∂x

∂f

∂ψ∗x− ∂

∂y

∂f

∂ψ∗y− ∂

∂z

∂f

∂ψ∗z= 0,

which results in−1

2(ψxx + ψyy + ψzz) + V ψ = λψ.

We can then identify the Lagrangian multiplier as the energy of the system.The last equation is nothing but the standard Schroedinger equation and thevariational approach discussed here provides a powerful method for obtainingapproximate solutions of the wave function.

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Definitions and notationsBefore we proceed we need some definitions. We will assume that the interactingpart of the Hamiltonian can be approximated by a two-body interaction. Thismeans that our Hamiltonian is written as the sum of some onebody part and atwobody part

H = H0 + HI =A∑i=1

h0(xi) +A∑i<j

v(rij), (1)

with

H0 =A∑i=1

h0(xi). (2)

The onebody part uext(xi) is normally approximated by a harmonic oscillatorpotential or the Coulomb interaction an electron feels from the nucleus. However,other potentials are fully possible, such as one derived from the self-consistentsolution of the Hartree-Fock equations to be discussed here.

Our Hamiltonian is invariant under the permutation (interchange) of twoparticles. Since we deal with fermions however, the total wave function isantisymmetric. Let P be an operator which interchanges two particles. Due tothe symmetries we have ascribed to our Hamiltonian, this operator commuteswith the total Hamiltonian,

[H, P ] = 0,

meaning that Ψλ(x1, x2, . . . , xA) is an eigenfunction of P as well, that is

PijΨλ(x1, x2, . . . , xi, . . . , xj , . . . , xA) = βΨλ(x1, x2, . . . , xi, . . . , xj , . . . , xA),

where β is the eigenvalue of P . We have introduced the suffix ij in order toindicate that we permute particles i and j. The Pauli principle tells us that thetotal wave function for a system of fermions has to be antisymmetric, resultingin the eigenvalue β = −1.

In our case we assume that we can approximate the exact eigenfunction witha Slater determinant

Φ(x1, x2, . . . , xA, α, β, . . . , σ) = 1√A!

∣∣∣∣∣∣∣∣∣∣ψα(x1) ψα(x2) . . . . . . ψα(xA)ψβ(x1) ψβ(x2) . . . . . . ψβ(xA). . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

ψσ(x1) ψσ(x2) . . . . . . ψσ(xA)

∣∣∣∣∣∣∣∣∣∣,

(3)where xi stand for the coordinates and spin values of a particle i and α, β, . . . , γare quantum numbers needed to describe remaining quantum numbers.

The single-particle function ψα(xi) are eigenfunctions of the onebody Hamil-tonian hi, that is

h0(xi) = t(xi) + uext(xi),

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with eigenvalues

h0(xi)ψα(xi) =(t(xi) + uext(xi)

)ψα(xi) = εαψα(xi).

The energies εα are the so-called non-interacting single-particle energies, orunperturbed energies. The total energy is in this case the sum over all single-particle energies, if no two-body or more complicated many-body interactionsare present.

Let us denote the ground state energy by E0. According to the variationalprinciple we have

E0 ≤ E[Φ] =∫

Φ∗HΦdτ

where Φ is a trial function which we assume to be normalized∫Φ∗Φdτ = 1,

where we have used the shorthand dτ = dx1dr2 . . . drA.In the Hartree-Fock method the trial function is the Slater determinant of

Eq. (3) which can be rewritten as

Φ(x1, x2, . . . , xA, α, β, . . . , ν) = 1√A!

∑P

(−)P Pψα(x1)ψβ(x2) . . . ψν(xA) =√A!AΦH ,

where we have introduced the antisymmetrization operator A defined by thesummation over all possible permutations of two particles.

It is defined asA = 1

A!∑p

(−)pP , (4)

with p standing for the number of permutations. We have introduced for lateruse the so-called Hartree-function, defined by the simple product of all possiblesingle-particle functions

ΦH(x1, x2, . . . , xA, α, β, . . . , ν) = ψα(x1)ψβ(x2) . . . ψν(xA).

Both H0 and HI are invariant under all possible permutations of any twoparticles and hence commute with A

[H0, A] = [HI , A] = 0. (5)

Furthermore, A satisfiesA2 = A, (6)

since every permutation of the Slater determinant reproduces it.The expectation value of H0∫

Φ∗H0Φdτ = A!∫

Φ∗HAH0AΦHdτ

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is readily reduced to ∫Φ∗H0Φdτ = A!

∫Φ∗HH0AΦHdτ,

where we have used Eqs. (5) and (6). The next step is to replace the anti-symmetrization operator by its definition and to replace H0 with the sum ofone-body operators∫

Φ∗H0Φdτ =A∑i=1

∑p

(−)p∫

Φ∗H h0PΦHdτ.

The integral vanishes if two or more particles are permuted in only one ofthe Hartree-functions ΦH because the individual single-particle wave functionsare orthogonal. We obtain then∫

Φ∗H0Φdτ =A∑i=1

∫Φ∗H h0ΦHdτ.

Orthogonality of the single-particle functions allows us to further simplify theintegral, and we arrive at the following expression for the expectation values ofthe sum of one-body Hamiltonians∫

Φ∗H0Φdτ =A∑µ=1

∫ψ∗µ(x)h0ψµ(x)dxdr. (7)

We introduce the following shorthand for the above integral

〈µ|h0|µ〉 =∫ψ∗µ(x)h0ψµ(x)dx,

and rewrite Eq. (7) as ∫Φ∗H0Φdτ =

A∑µ=1〈µ|h0|µ〉. (8)

The expectation value of the two-body part of the Hamiltonian is obtainedin a similar manner. We have∫

Φ∗HIΦdτ = A!∫

Φ∗HAHIAΦHdτ,

which reduces to∫Φ∗HIΦdτ =

A∑i≤j=1

∑p

(−)p∫

Φ∗H v(rij)PΦHdτ,

by following the same arguments as for the one-body Hamiltonian.

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Because of the dependence on the inter-particle distance rij , permutations ofany two particles no longer vanish, and we get∫

Φ∗HIΦdτ =A∑

i<j=1

∫Φ∗H v(rij)(1− Pij)ΦHdτ.

where Pij is the permutation operator that interchanges particle i and particlej. Again we use the assumption that the single-particle wave functions areorthogonal.

We obtain∫Φ∗HIΦdτ = 1

2

A∑µ=1

A∑ν=1

[∫ψ∗µ(xi)ψ∗ν(xj)v(rij)ψµ(xi)ψν(xj)dxidxj (9)

−∫ψ∗µ(xi)ψ∗ν(xj)v(rij)ψν(xi)ψµ(xj)dxidxj

]. (10)

The first term is the so-called direct term. It is frequently also called the Hartreeterm, while the second is due to the Pauli principle and is called the exchangeterm or just the Fock term. The factor 1/2 is introduced because we now runover all pairs twice.

The last equation allows us to introduce some further definitions. The single-particle wave functions ψµ(x), defined by the quantum numbers µ and x aredefined as the overlap

ψα(x) = 〈x|α〉.

We introduce the following shorthands for the above two integrals

〈µν|v|µν〉 =∫ψ∗µ(xi)ψ∗ν(xj)v(rij)ψµ(xi)ψν(xj)dxidxj ,

and〈µν|v|νµ〉 =

∫ψ∗µ(xi)ψ∗ν(xj)v(rij)ψν(xi)ψµ(xj)dxidxj .

Derivation of Hartree-Fock equations in coordinate spaceLet us denote the ground state energy by E0. According to the variationalprinciple we have

E0 ≤ E[Φ] =∫

Φ∗HΦdτ

where Φ is a trial function which we assume to be normalized∫Φ∗Φdτ = 1,

where we have used the shorthand dτ = dx1dx2 . . . dxA.

7

In the Hartree-Fock method the trial function is a Slater determinant whichcan be rewritten as

Ψ(x1, x2, . . . , xA, α, β, . . . , ν) = 1√A!

∑P

(−)PPψα(x1)ψβ(x2) . . . ψν(xA) =√A!AΦH ,

where we have introduced the anti-symmetrization operator A defined by thesummation over all possible permutations p of two fermions. It is defined as

A = 1A!∑p

(−)pP ,

with the the Hartree-function given by the simple product of all possible single-particle function

ΦH(x1, x2, . . . , xA, α, β, . . . , ν) = ψα(x1)ψβ(x2) . . . ψν(xA).

Our functional is written as

E[Φ] =A∑µ=1

∫ψ∗µ(xi)h0(xi)ψµ(xi)dxi+

12

A∑µ=1

A∑ν=1

[∫ψ∗µ(xi)ψ∗ν(xj)v(rij)ψµ(xi)ψν(xj)dxidxj −

∫ψ∗µ(xi)ψ∗ν(xj)v(rij)ψν(xi)ψµ(xj)dxidxj

]The more compact version reads

E[Φ] =A∑µ

〈µ|h0|µ〉+ 12

A∑µν

[〈µν|v|µν〉 − 〈νµ|v|µν〉] .

Since the interaction is invariant under the interchange of two particles itmeans for example that we have

〈µν|v|µν〉 = 〈νµ|v|νµ〉,

or in the more general case

〈µν|v|στ〉 = 〈νµ|v|τσ〉.

The direct and exchange matrix elements can be brought together if we definethe antisymmetrized matrix element

〈µν|v|µν〉AS = 〈µν|v|µν〉 − 〈µν|v|νµ〉,

or for a general matrix element

〈µν|v|στ〉AS = 〈µν|v|στ〉 − 〈µν|v|τσ〉.

It has the symmetry property

〈µν|v|στ〉AS = −〈µν|v|τσ〉AS = −〈νµ|v|στ〉AS .

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The antisymmetric matrix element is also hermitian, implying

〈µν|v|στ〉AS = 〈στ |v|µν〉AS .

With these notations we rewrite the Hartree-Fock functional as∫Φ∗HIΦdτ = 1

2

A∑µ=1

A∑ν=1〈µν|v|µν〉AS . (11)

Adding the contribution from the one-body operator H0 to (11) we obtainthe energy functional

E[Φ] =A∑µ=1〈µ|h|µ〉+ 1

2

A∑µ=1

A∑ν=1〈µν|v|µν〉AS . (12)

In our coordinate space derivations below we will spell out the Hartree-Fockequations in terms of their integrals.

If we generalize the Euler-Lagrange equations to more variables and introduceN2 Lagrange multipliers which we denote by εµν , we can write the variationalequation for the functional of E

δE −A∑µν

εµνδ

∫ψ∗µψν = 0.

For the orthogonal wave functions ψi this reduces to

δE −A∑µ=1

εµδ

∫ψ∗µψµ = 0.

Variation with respect to the single-particle wave functions ψµ yields then

A∑µ=1

∫δψ∗µh0(xi)ψµdxi+

12

A∑µ=1

A∑ν=1

[∫δψ∗µψ

∗ν v(rij)ψµψνdxidxj −

∫δψ∗µψ

∗ν v(rij)ψνψµdxidxj

]+

A∑µ=1

∫ψ∗µh0(xi)δψµdxi+

12

A∑µ=1

A∑ν=1

[∫ψ∗µψ

∗ν v(rij)δψµψνdxidxj −

∫ψ∗µψ

∗ν v(rij)ψνδψµdxidxj

]−

A∑µ=1

Eµ

∫δψ∗µψµdxi−

A∑µ=1

Eµ

∫ψ∗µδψµdxi = 0.

Although the variations δψ and δψ∗ are not independent, they may in fact betreated as such, so that the terms dependent on either δψ and δψ∗ individuallymay be set equal to zero. To see this, simply replace the arbitrary variation δψby iδψ, so that δψ∗ is replaced by −iδψ∗, and combine the two equations. Wethus arrive at the Hartree-Fock equations[−1

2∇2i +

A∑ν=1

∫ψ∗ν(xj)v(rij)ψν(xj)dxj

]ψµ(xi)−

[A∑ν=1

∫ψ∗ν(xj)v(rij)ψµ(xj)dxj

]ψν(xi) = εµψµ(xi).

(13)

9

Notice that the integration∫dxj implies an integration over the spatial coordi-

nates rj and a summation over the spin-coordinate of fermion j. We note thatthe factor of 1/2 in front of the sum involving the two-body interaction, hasbeen removed. This is due to the fact that we need to vary both δψ∗µ and δψ∗ν .Using the symmetry properties of the two-body interaction and interchanging µand ν as summation indices, we obtain two identical terms.

The two first terms in the last equation are the one-body kinetic energy andthe electron-nucleus potential. The third or direct term is the averaged electronicrepulsion of the other electrons. As written, the term includes the self-interactionof electrons when µ = ν. The self-interaction is cancelled in the fourth term, orthe exchange term. The exchange term results from our inclusion of the Pauliprinciple and the assumed determinantal form of the wave-function. Equation(13), in addition to the kinetic energy and the attraction from the atomic nucleusthat confines the motion of a single electron, represents now the motion of asingle-particle modified by the two-body interaction. The additional contributionto the Schroedinger equation due to the two-body interaction, represents a meanfield set up by all the other bystanding electrons, the latter given by the sumover all single-particle states occupied by N electrons.

The Hartree-Fock equation is an example of an integro-differential equation.These equations involve repeated calculations of integrals, in addition to thesolution of a set of coupled differential equations. The Hartree-Fock equationscan also be rewritten in terms of an eigenvalue problem. The solution of aneigenvalue problem represents often a more practical algorithm and the solutionof coupled integro-differential equations. This alternative derivation of theHartree-Fock equations is given below.

Analysis of Hartree-Fock equations in coordinate spaceA theoretically convenient form of the Hartree-Fock equation is to regard thedirect and exchange operator defined through

V dµ (xi) =∫ψ∗µ(xj)v(rij)ψµ(xj)dxj

andV exµ (xi)g(xi) =

(∫ψ∗µ(xj)v(rij)g(xj)dxj

)ψµ(xi),

respectively.The function g(xi) is an arbitrary function, and by the substitution g(xi) =

ψν(xi) we get

V exµ (xi)ψν(xi) =(∫

ψ∗µ(xj)v(rij)ψν(xj)dxj)ψµ(xi).

We may then rewrite the Hartree-Fock equations as

hHF (xi)ψν(xi) = ενψν(xi),

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with

hHF (xi) = h0(xi) +A∑µ=1

V dµ (xi)−A∑µ=1

V exµ (xi),

and where h0(i) is the one-body part. The latter is normally chosen as a partwhich yields solutions in closed form. The harmonic oscilltor is a classicalproblem thereof. We normally rewrite the last equation as

hHF (xi) = h0(xi) + uHF (xi).

Hartree-Fock by varying the coefficients of a wave functionexpansionAnother possibility is to expand the single-particle functions in a known basisand vary the coefficients, that is, the new single-particle wave function is writtenas a linear expansion in terms of a fixed chosen orthogonal basis (for examplethe well-known harmonic oscillator functions or the hydrogen-like functions etc).We define our new Hartree-Fock single-particle basis by performing a unitarytransformation on our previous basis (labelled with greek indices) as

ψHFp =∑λ

Cpλφλ. (14)

In this case we vary the coefficients Cpλ. If the basis has infinitely many solutions,we need to truncate the above sum. We assume that the basis φλ is orthogonal.A unitary transformation keeps the orthogonality, as discussed in exercise 1below.

It is normal to choose a single-particle basis defined as the eigenfunctions ofparts of the full Hamiltonian. The typical situation consists of the solutions ofthe one-body part of the Hamiltonian, that is we have

h0φλ = ελφλ.

The single-particle wave functions φλ(r), defined by the quantum numbers λand r are defined as the overlap

φλ(r) = 〈r|λ〉.

In our discussions hereafter we will use our definitions of single-particle statesabove and below the Fermi (F ) level given by the labels ijkl · · · ≤ F for so-calledsingle-hole states and abcd · · · > F for so-called particle states. For generalsingle-particle states we employ the labels pqrs . . . .

In Eq. (12), restated here

E[Φ] =A∑µ=1〈µ|h|µ〉+ 1

2

A∑µ=1

A∑ν=1〈µν|v|µν〉AS ,

11

we found the expression for the energy functional in terms of the basis functionφλ(r). We then varied the above energy functional with respect to the basisfunctions |µ〉. Now we are interested in defining a new basis defined in terms ofa chosen basis as defined in Eq. (14). We can then rewrite the energy functionalas

E[ΦHF ] =A∑i=1〈i|h|i〉+ 1

2

A∑ij=1〈ij|v|ij〉AS , (15)

where ΦHF is the new Slater determinant defined by the new basis of Eq. (14).Using Eq. (14) we can rewrite Eq. (15) as

E[Ψ] =A∑i=1

∑αβ

C∗iαCiβ〈α|h|β〉+ 12

A∑ij=1

∑αβγδ

C∗iαC∗jβCiγCjδ〈αβ|v|γδ〉AS . (16)

We wish now to minimize the above functional. We introduce again a setof Lagrange multipliers, noting that since 〈i|j〉 = δi,j and 〈α|β〉 = δα,β , thecoefficients Ciγ obey the relation

〈i|j〉 = δi,j =∑αβ

C∗iαCiβ〈α|β〉 =∑α

C∗iαCiα,

which allows us to define a functional to be minimized that reads

F [ΦHF ] = E[ΦHF ]−A∑i=1

εi∑α

C∗iαCiα. (17)

Minimizing with respect to C∗iα, remembering that the equations for C∗iα andCiα can be written as two independent equations, we obtain

d

dC∗iα

E[ΦHF ]−∑j

εj∑α

C∗jαCjα

= 0,

which yields for every single-particle state i and index α (recalling that thecoefficients Ciα are matrix elements of a unitary (or orthogonal for a realsymmetric matrix) matrix) the following Hartree-Fock equations

∑β

Ciβ〈α|h|β〉+A∑j=1

∑βγδ

C∗jβCjδCiγ〈αβ|v|γδ〉AS = εHFi Ciα.

We can rewrite this equation as (changing dummy variables)

∑β

〈α|h|β〉+A∑j

∑γδ

C∗jγCjδ〈αγ|v|βδ〉AS

Ciβ = εHFi Ciα.

Note that the sums over greek indices run over the number of basis set functions(in principle an infinite number).

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Defining

hHFαβ = 〈α|h|β〉+A∑j=1

∑γδ

C∗jγCjδ〈αγ|v|βδ〉AS ,

we can rewrite the new equations as∑γ

hHFαβ Ciβ = εHFi Ciα. (18)

The latter is nothing but a standard eigenvalue problem. Compared with Eq. (13),we see that we do not need to compute any integrals in an iterative procedurefor solving the equations. It suffices to tabulate the matrix elements 〈α|h|β〉 and〈αγ|v|βδ〉AS once and for all. Successive iterations require thus only a look-upin tables over one-body and two-body matrix elements. These details will bediscussed below when we solve the Hartree-Fock equations numerical.

Hartree-Fock algorithmOur Hartree-Fock matrix is thus

hHFαβ = 〈α|h0|β〉+A∑j=1

∑γδ

C∗jγCjδ〈αγ|v|βδ〉AS .

The Hartree-Fock equations are solved in an iterative waym starting with aguess for the coefficients Cjγ = δj,γ and solving the equations by diagonalizationtill the new single-particle energies εHF

i do not change anymore by a prefixedquantity.

Normally we assume that the single-particle basis |β〉 forms an eigenbasis forthe operator h0, meaning that the Hartree-Fock matrix becomes

hHFαβ = εαδα,β +A∑j=1

∑γδ

C∗jγCjδ〈αγ|v|βδ〉AS .

The Hartree-Fock eigenvalue problem∑β

hHFαβ Ciβ = εHFi Ciα,

can be written out in a more compact form as

hHF C = εHFC.

The Hartree-Fock equations are, in their simplest form, solved in an iterativeway, starting with a guess for the coefficients Ciα. We label the coefficients asC

(n)iα , where the subscript n stands for iteration n. To set up the algorithm we

can proceed as follows:

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• We start with a guess C(0)iα = δi,α. Alternatively, we could have used

random starting values as long as the vectors are normalized. Anotherpossibility is to give states below the Fermi level a larger weight.

• The Hartree-Fock matrix simplifies then to (assuming that the coefficientsCiα are real)

hHFαβ = εαδα,β +A∑j=1

∑γδ

C(0)jγ C

(0)jδ 〈αγ|v|βδ〉AS .

Solving the Hartree-Fock eigenvalue problem yields then new eigenvectorsC

(1)iα and eigenvalues εHF (1)

i .

• With the new eigenvalues we can set up a new Hartree-Fock potentialA∑j=1

∑γδ

C(1)jγ C

(1)jδ 〈αγ|v|βδ〉AS .

The diagonalization with the new Hartree-Fock potential yields new eigenvectorsand eigenvalues. This process is continued till for example∑

p |ε(n)i − ε(n−1)

i |m

≤ λ,

where λ is a user prefixed quantity (λ ∼ 10−8 or smaller) and p runs over allcalculated single-particle energies and m is the number of single-particle states.

• TODO: add code with Hartree-Fock for nuclear system

Analysis of Hartree-Fock equations and Koopman’s theo-remWe can rewrite the ground state energy by adding and subtracting uHF (xi)

EHF0 = 〈Φ0|H|Φ0〉 =A∑i≤F

〈i|h0+uHF |j〉+12

A∑i≤F

A∑j≤F

[〈ij|v|ij〉 − 〈ij|v|ji〉]−A∑i≤F

〈i|uHF |i〉,

which results in

EHF0 =A∑i≤F

εHFi + 12

A∑i≤F

A∑j≤F

[〈ij|v|ij〉 − 〈ij|v|ji〉]−A∑i≤F

〈i|uHF |i〉.

Our single-particle states ijk . . . are now single-particle states obtained from thesolution of the Hartree-Fock equations.

14

Using our definition of the Hartree-Fock single-particle energies we obtainthen the following expression for the total ground-state energy

EHF0 =A∑i≤F

εi −12

A∑i≤F

A∑j≤F

[〈ij|v|ij〉 − 〈ij|v|ji〉] .

This form will be used in our discussion of Koopman’s theorem.In the atomic physics case we have

E[ΦHF(N)] =H∑i=1〈i|h0|i〉+ 1

2

N∑ij=1〈ij|v|ij〉AS ,

where ΦHF(N) is the new Slater determinant defined by the new basis of Eq. (14)for N electrons (same Z). If we assume that the single-particle wave functionsin the new basis do not change when we remove one electron or add one electron,we can then define the corresponding energy for the N − 1 systems as

E[ΦHF(N − 1)] =N∑

i=1;i6=k〈i|h0|i〉+ 1

2

N∑ij=1;i,j 6=k

〈ij|v|ij〉AS ,

where we have removed a single-particle state k ≤ F , that is a state below theFermi level.

Calculating the difference

E[ΦHF(N)]−E[ΦHF(N−1)] = 〈k|h0|k〉+12

N∑i=1;i6=k

〈ik|v|ik〉AS+12

N∑j=1;j 6=k

〈kj|v|kj〉AS ,

we obtain

E[ΦHF(N)]− E[ΦHF(N − 1)] = 〈k|h0|k〉+N∑j=1〈kj|v|kj〉AS

which is just our definition of the Hartree-Fock single-particle energy

E[ΦHF(N)]− E[ΦHF(N − 1)] = εHFk

Similarly, we can now compute the difference (we label the single-particlestates above the Fermi level as abcd > F )

E[ΦHF(N + 1)]− E[ΦHF(N)] = εHFa .

These two equations can thus be used to the electron affinity or ionization energies,respectively. Koopman’s theorem states that for example the ionization energy ofa closed-shell system is given by the energy of the highest occupied single-particlestate. If we assume that changing the number of electrons from N to N + 1 doesnot change the Hartree-Fock single-particle energies and eigenfunctions, then

15

Koopman’s theorem simply states that the ionization energy of an atom is givenby the single-particle energy of the last bound state. In a similar way, we canalso define the electron affinities.

As an example, consider a simple model for atomic sodium, Na. Neutralsodium has eleven electrons, with the weakest bound one being confined the 3ssingle-particle quantum numbers. The energy needed to remove an electron fromneutral sodium is rather small, 5.1391 eV, a feature which pertains to all alkalimetals. Having performed a Hartree-Fock calculation for neutral sodium wouldthen allows us to compute the ionization energy by using the single-particleenergy for the 3s states, namely εHF

3s .From these considerations, we see that Hartree-Fock theory allows us to make

a connection between experimental observables (here ionization and affinityenergies) and the underlying interactions between particles. In this sense, we arenow linking the dynamics and structure of a many-body system with the laws ofmotion which govern the system. Our approach is a reductionistic one, meaningthat we want to understand the laws of motion in terms of the particles or degreesof freedom which we believe are the fundamental ones. Our Slater determinant,being constructed as the product of various single-particle functions, follows thisphilosophy.

With similar arguments as in atomic physics, we can now use Hartree-Focktheory to make a link between nuclear forces and separation energies. Changingto nuclear system, we define

E[ΦHF(A)] =A∑i=1〈i|h0|i〉+ 1

2

A∑ij=1〈ij|v|ij〉AS ,

where ΦHF(A) is the new Slater determinant defined by the new basis of Eq. (14)for A nucleons, where A = N + Z, with N now being the number of neutronsand Z th enumber of protons. If we assume again that the single-particle wavefunctions in the new basis do not change from a nucleus with A nucleons to anucleus with A− 1 nucleons, we can then define the corresponding energy forthe A− 1 systems as

E[ΦHF(A− 1)] =A∑

i=1;i 6=k〈i|h0|i〉+ 1

2

A∑ij=1;i,j 6=k

〈ij|v|ij〉AS ,

where we have removed a single-particle state k ≤ F , that is a state below theFermi level.

Calculating the difference

E[ΦHF(A)]−E[ΦHF(A−1)] = 〈k|h0|k〉+12

A∑i=1;i 6=k

〈ik|v|ik〉AS+12

A∑j=1;j 6=k

〈kj|v|kj〉AS ,

which becomes

E[ΦHF(A)]− E[ΦHF(A− 1)] = 〈k|h0|k〉+A∑j=1〈kj|v|kj〉AS

16

which is just our definition of the Hartree-Fock single-particle energy

E[ΦHF(A)]− E[ΦHF(A− 1)] = εHFk

Similarly, we can now compute the difference (recall that the single-particlestates abcd > F )

E[ΦHF(A+ 1)]− E[ΦHF(A)] = εHFa .

If we then recall that the binding energy differences

BE(A)−BE(A− 1) and BE(A+ 1)−BE(A),

define the separation energies, we see that the Hartree-Fock single-particleenergies can be used to define separation energies. We have thus our first linkbetween nuclear forces (included in the potential energy term) and an observablequantity defined by differences in binding energies.

We have thus the following interpretations (if the single-particle fields do notchange)

BE(A)−BE(A− 1) ≈ E[ΦHF(A)]− E[ΦHF(A− 1)] = εHFk ,

andBE(A+ 1)−BE(A) ≈ E[ΦHF(A+ 1)]− E[ΦHF(A)] = εHF

a .

If we use 16O as our closed-shell nucleus, we could then interpret the separationenergy

BE(16O)−BE(15O) ≈ εHF0pν1/2

,

andBE(16O)−BE(15N) ≈ εHF

0pπ1/2.

Similalry, we could interpret

BE(17O)−BE(16O) ≈ εHF0dν5/2

,

andBE(17F)−BE(16O) ≈ εHF

0dπ5/2.

We can continue like this for all A± 1 nuclei where A is a good closed-shell (orsubshell closure) nucleus. Examples are 22O, 24O, 40Ca, 48Ca, 52Ca, 54Ca, 56Ni,68Ni, 78Ni, 90Zr, 88Sr, 100Sn, 132Sn and 208Pb, to mention some possile cases.

We can thus make our first interpretation of the separation energies in termsof the simplest possible many-body theory. If we also recall that the so-calledenergy gap for neutrons (or protons) is defined as

∆Sn = 2BE(N,Z)−BE(N − 1, Z)−BE(N + 1, Z),

for neutrons and the corresponding gap for protons

∆Sp = 2BE(N,Z)−BE(N,Z − 1)−BE(N,Z + 1),

17

we can define the neutron and proton energy gaps for 16O as

∆Sν = εHF0dν5/2

− εHF0pν1/2

,

and∆Sπ = εHF

0dπ5/2− εHF

0pπ1/2.

Exercise 1: Derivation of Hartree-Fock equationsConsider a Slater determinant built up of single-particle orbitals ψλ, withλ = 1, 2, . . . , N .

The unitary transformation

ψa =∑λ

Caλφλ,

brings us into the new basis. The new basis has quantum numbers a = 1, 2, . . . , N .aragraph!paragraph>paragraph>-0.5em

a) Show that the new basis is orthonormal.aragraph!paragraph>paragraph>-0.5em

b) Show that the new Slater determinant constructed from the new single-particle wave functions can be written as the determinant based on the previousbasis and the determinant of the matrix C.

aragraph!paragraph>paragraph>-0.5em

c) Show that the old and the new Slater determinants are equal up to acomplex constant with absolute value unity.

Hint. Use the fact that C is a unitary matrix.

Exercise 2: Derivation of Hartree-Fock equationsConsider the Slater determinant

Φ0 = 1√n!

∑p

(−)pPn∏i=1

ψαi(xi).

A small variation in this function is given by

δΦ0 = 1√n!

∑p

(−)pPψα1(x1)ψα2(x2) . . . ψαi−1(xi−1)(δψαi(xi))ψαi+1(xi+1) . . . ψαn(xn).

aragraph!paragraph>paragraph>-0.5em

a) Show that

〈δΦ0|n∑i=1{t(xi) + u(xi)}+

12

n∑i 6=j=1

v(xi, xj)|Φ0〉 =n∑i=1〈δψαi |t+u|φαi〉+

n∑i6=j=1

{〈δψαiψαj |v|ψαiψαj 〉 − 〈δψαiψαj |v|ψαjψαi〉

}

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Exercise 3: Hartree-Fock calculations in atomic physicsIn this exercise we will develop two simple models for studying the helium atom(with two electrons) and the beryllium atom with four electrons. The reason whywe chose these systems is simply due to the fact that we can use hydrogen-likesingle-particle state functions. These functions allow for simple evaluations ofthe two-body integrals needed in a Hartree-Fock calculation. In most cases, theseintegrals have a simple closed form solution, as indicated in the table below.

After having introduced the Born-Oppenheimer approximation which ef-fectively freezes out the nucleonic degrees of freedom, the Hamiltonian for Nelectrons takes the following form

H =N∑i=1

t(xi)−N∑i=1

kZe2

ri+

N∑i<j

ke2

rij,

with k = 1.44 eVnm. Througout this work we will use atomic units, this meansthat ~ = c = e = me = 1. The constant k becomes also equal 1. The resultingenergies have to be multiplied by 2 × 13.6 eV in order to obtain energies ineletronvolts.

We can rewrite our Hamiltonians as

H = H0 + HI =N∑i=1

h0(xi) +N∑i<j

1rij,

where we have defined rij = |ri − rj | and h0(xi) = t(xi) − Zri

The variable xcontains both the spatial coordinates and the spin values. The first term of theprevious equation, H0, is the sum of the N one-body Hamiltonians h0. Eachindividual Hamiltonian h0 contains the kinetic energy operator of an electronand its potential energy due to the attraction of the nucleus. The second term,HI , is the sum of the N(N − 1)/2 two-body interactions between each pair ofelectrons. Note that the double sum carries a restriction i < j.

As basis functions for our calculations we will use hydrogen-like single-particlefunctions. This means the onebody operator is diagonal in this basis for statesi, j with quantum numbers nlmlsms with energies

〈i|h0|j〉 = −Z2/2n2δi,j .

The quantum number n refers to the number of nodes of the wave function.Observe that this expectation value is independent of spin.

We will in all calculations here restrict ourselves to only so-called s -waves,that is the orbital momentum l is zero. We will also limit the quantum numbern to n ≤ 3. It means that every ns state can accomodate two electrons due tothe spin degeneracy. In the calculations you will need the Coulomb interactionwith matrix elements involving single-particle wave functions with l = 0 only,the so-called s-waves. We need only the radial part since the spherical harmonicsfor the s-waves are rather simple. We omit single-particle states with l > 0. Our

19

radial part of the wave functions are

Rn0(r) =(

2Zn

)3/2√ (n− 1)!2n× n! L

1n−1(2Zr

n) exp (−Zr

n),

where L1n−1(r) are the so-called Laguerre polynomials. These wave functions

can then be used to compute the direct part of the Coulomb interaction

〈αβ|V |γδ〉 =∫r2

1dr1

∫r2

2dr2R∗nα0(r1)R∗nβ0(r2) 1

|r1 − r2|Rnγ0(r1)Rnδ0(r2)

Observe that this is only the radial integral and that the labels αβγδ refer onlyto the quantum numbers nlml, with ml the projection of the orbital momentum l.A similar expression can be found for the exchange part. Since we have restrictedourselves to only s-waves, these integrals are straightforward but tedious tocalculate. As an addendum to this project we list all closed-form expressionsfor the relevant matrix elements. Note well that these matrix elements do notinclude spin. When setting up the final antisymmetrized matrix elements youneed to consider the spin degrees of freedom as well. Please pay in particularspecial attention to the exchange part and the pertinent spin values of thesingle-particle states.

We will also, for both helium and beryllium assume that the many-particlestates we construct have always the same total spin projection MS = 0. Thismeans that if we excite one or two particles from the ground state, the spins ofthe various single-particle states should always sum up to zero.

aragraph!paragraph>paragraph>-0.5em

a) We start with the helium atom and define our single-particle Hilbert spaceto consist of the single-particle orbits 1s, 2s and 3s, with their correspondingspin degeneracies.

Set up the ansatz for the ground state |c〉 = |Φ0〉 in second quantizationand define a table of single-particle states. Construct thereafter all possibleone-particle-one-hole excitations |Φa

i 〉 where i refer to levels below the Fermilevel (define this level) and a refers to particle states. Define particles and holes.The Slater determinants have to be written in terms of the respective creationand annihilation operators. The states you construct should all have total spinprojection MS = 0. Construct also all possible two-particle-two-hole states |Φabij 〉in a second quantization representation.

aragraph!paragraph>paragraph>-0.5em

b) Define the Hamiltonian in a second-quantized form and use this to computethe expectation value of the ground state (defining the so-called reference energyand later our Hartree-Fock functional) of the helium atom. Show that it is givenby

E[Φ0] = 〈c|H|c〉 =∑i

〈i|h0|i〉+ 12∑ij

[〈ij|1

r|ij〉 − 〈ij|1

r|ji〉].

Define properly the sums keeping in mind that the states ij refer to all quantumnumbers nlmlsms. Use the values for the various matrix elements listed at the

20

end of the project to find the value of E as function of Z and compute E asfunction of Z.

aragraph!paragraph>paragraph>-0.5em

c) Hereafter we will limit ourselves to a system which now contains onlyone-particle-one-hole excitations beyond the chosen state |c〉. Using the possibleSlater determinants from exercise a) for the helium atom, find the expressions(without inserting the explicit values for the matrix elements first) for

〈c|H|Φai 〉,

and〈Φai |H|Φbj〉.

Represent these expressions in a diagrammatic form, both for the onebody partand the two-body part of the Hamiltonian.

Insert then the explicit values for the various matrix elements and set up thefinal Hamiltonian matrix and diagonalize it using for example Octave, Matlab,Python, C++ or Fortran as programming tools.

Compare your results from those of exercise b) and comment your results.The exact energy with our Hamiltonian is −2.9037 atomic units for helium. Thisvalue is also close to the experimental energy.

aragraph!paragraph>paragraph>-0.5em

d) We repeat exercises b) and c) but now for the beryllium atom. Definethe ansatz for |c〉 and limit yourself again to one-particle-one-hole excitations.Compute the reference energy 〈c|H|c〉 as function of Z. Thereafter you will needto set up the appropriate Hamiltonian matrix which involves also one-particle-one-hole excitations. Diagonalize this matrix and compare your eigenvalues with〈c|H|c〉 as function of Z and comment your results. The exact energy with ourHamiltonian is −14.6674 atomic units for beryllium. This value is again close tothe experimental energy.

With a given energy functional, we can perform at least two types of varia-tional strategies

1. Vary the Slater determinant by changing the spatial part of the single-particle

wave functions themselves.

1. Expand the single-particle functions in a known basis and vary the coeffi-cients,

that is, the new function single-particle wave function |p〉 is written as a linearexpansion in terms of a fixed basis φ (harmonic oscillator, Laguerre polynomialsetc)

ψp =∑λ

Cpλφλ,

21

Both cases lead to a new Slater determinant which is related to the previous viaa unitary transformation. Below we will set up the Hartree-Fock equations usingthe second option. We assume that our basis is still formed by the hydrogen-likewave functions. We consider a Slater determinant built up of single-particleorbitals φλ where the indices λ refer to specific single-particle states. As anexample, you could think of the ground state ansatz for the Beryllium atom.

The unitary transformation

ψp =∑λ

Cpλφλ,

brings us into the new basis ψ. The new basis is orthonormal and C is a unitarymatrix.

aragraph!paragraph>paragraph>-0.5em

e) Minimizing with respect to C∗pα, remembering that C∗pα and Cpα (andthat the indices contain all single-particle quantum numbers including spin) areindependent and defining

hHFαγ = 〈α|h|γ〉+∑p

∑βδ

C∗pβCpδ〈αβ|V |γδ〉AS ,

show that you can write the Hartree-Fock equations as∑γ

hHFαγ Cpγ = εHFp Cpα.

Explain the meaning of the different terms and define the Hartree-Fock operatorin second quantization. Write down its diagrammatic representation as well.The greek letters refer to the wave functions in the original basis (in our casethe hydrogen-like wave functions) while roman letters refer to the new basis.

aragraph!paragraph>paragraph>-0.5em

f) The Hartree-Fock equations with a variation of the coefficients Cpα lead toan eigenvalue problem whose eigenvectors are the coefficients Cpα and eigenvaluesare the new single-particle energies. Use the single-particle orbits 1s− 3s and setup the Hartree-Fock matrix for both the helium atom and the beryllium atom.Find after the first diagonalization the new single-particle energies and the newground state energy. Compare these results with those you obtained under theminimization of the ground states as functions of Z and the full diagonalization.When setting up the Hartree-Fock matrix in the first iteration, our guess for thecoefficients Cpβ etc. is Cpβ = 1 for p = β and zero else. The final stage is toset up an iterative scheme where you use the new wave functions determinedvia the coefficients Cpα to solve iteratively the Hartree-Fock equations till agiven self-consistency is reached. A typical way of doing this is to compare thesingle-particle energies from the previous iteration with those obtained from thenew diagonalization. If the total difference is smaller than a prefixed value, theiterative process is stopped.

22

Compare these results with the those you obtained under the minimizationof the ground states as functions of Z and the full diagonalization. Discuss yourresults.

Solution. The following Python program set ups a simple Hartree-Fock pro-gram for solving the above problem, using as input the integrals listed at theend here.

"""Program for solving Hartree Fock iteratively.1. Import radial integral results2. Set up HF matrix using C = I as initial guess3. Solve eigenvalue problem using numpy.linalg4. Use resulting eigenvectors to assemble new HF matrix5. Repeat steps 3 and 4 until convergence is met.6. Calculate ground state energy from resulting C."""

from __future__ import divisionfrom sympy import *from numpy import *import pickle, sys

Z = Symbol("Z")

class HF:"""Class for solving the Hartree Fock equations iteratively."""

def __init__(self, N, basis, Z_value, first_C=’identity’):"""N is the number of particles in the system and basisis the single-particle basis for the system."""# Read in the radial_integrals from pickled objectwith open("radial_integrals.p", "rb") as infile:

self.radial_integrals = pickle.load(infile)

self.N = N # number of particles in the systemself.basis = basis # single-particle basis for the systemself.Z_value = Z_value # atomic number of the atomself.n = len(basis) # number of single-particle basis statesself.ek = array((0,0,0,0,0,0)) # new single-particle energiesself.E = 0 # energy# Set up the first coefficient matrix to be usedif first_C == ’identity’:

self.C = identity(self.n)elif first_C == ’zero’:

self.C = zeros((self.n, self.n))elif first_C == ’rand’:

self.C = random.rand(self.n, self.n)else:

print "first_C argument not understood"print "Legal values are: ’identity’, ’zero’, ’rand’"sys.exit(1)

self.h_HF = zeros((self.n, self.n))self.assemble_HF_matrix() # set up HF matrix for C = I

23

def h_0(self, p, q):"""Takes the integer values of states and returns theasymmetrized twobody matrix element <pq||rs>."""n1, s1 = self.basis[p]n2, s2 = self.basis[q]

if n1 != n2 or s1 != s2:return 0

else:return -Z**2/(2*n1**2)

def rad(self, n1, n2, n3, n4):"""Returns the radial integral <n1, n2|v|n3, n4>."""return self.radial_integrals[n1-1, n2-1, n3-1, n4-1]

def h_1(self, p, q, r, s):"""Takes the integer values of four basis-states and returnsthe asymmetrized twobody matrix element <pq||rs>."""n1, s1 = self.basis[p]n2, s2 = self.basis[q]n3, s3 = self.basis[r]n4, s4 = self.basis[s]

if s1 == s2 == s3 == s4:return self.rad(n1, n2, n3, n4) - self.rad(n1, n2, n4, n3)

if s1 == s3 and s2 == s4:return self.rad(n1, n2, n3, n4)

if s1 == s4 and s2 == s3:return -self.rad(n1, n2, n4, n3)

else:return 0

def assemble_HF_matrix(self):"""Assemble the HF matric from the coefficient matrix."""n, N = self.n, self.NC = self.C

for a in range(n):for g in range(n):

s = self.h_0(a,g)for p in range(N):

for b in range(n):for d in range(n):

s += C[p,b]*C[p,d]*self.h_1(a,b,g,d)

self.h_HF[a,g] = s.subs(Z, self.Z_value)

def reorder_coefficients(self):ek, C = self.ek, self.C

# Sort eigenvalues and coefficient matrix using numpy.argsortindices = argsort(ek)

24

ek = ek[indices]C = C[:, indices]

self.ek, self.C = ek, C.T

def calc_energy(self):"""Calculates the ground state energy from thecurrent coefficient matric."""n, N = self.n, self.NC = self.C

e = 0for p in range(N):for a in range(n):

for b in range(n):e += C[p,a]*C[p,b]*self.h_0(a,b)for q in range(N):for c in range(n):for d in range(n):e += 0.5*C[p,a]*C[q,b]*C[p,c]*C[q,d]*self.h_1(a,b,c,d)

self.E = e.subs(Z, Z_value).evalf()return self.E

def solve(self, tol=1e-6, max_iters=40):iterations = 0n, N = self.n, self.NEp = 0ekp = array((0,0,0,0,0,0))

while iterations < max_iters:iterations +=1

# Find eigenvalues and eigenvector of HF matrixself.ek, self.C = linalg.eig(self.h_HF)

# Reorder eigenvalues and eigenvectorself.reorder_coefficients()

# Assemble the new HF matrixself.assemble_HF_matrix()

# Test tolerance of lowest eigenvalueprint self.calc_energy()error = sum(abs(ekp - self.ek[0]))if error < tol:

print "Solver converged after %d iterations." % (iterations)return

Ep = self.Eekp = self.ek[0]

print "Solver failed to converge in %d iterations." % (iterations)

N = 4Z_value = 4basis = [(1,1), (1,-1), (2,1), (2,-1), (3,1), (3,-1)]

25

print "Solving with initial guess C=I."solver = HF(N, basis, Z_value, first_C=’identity’)solver.solve(max_iters=100)

print "\n\n\n Solving with initial guess C=0."solver = HF(N, basis, Z_value, first_C=’zero’)solver.solve(max_iters=100)

print "\n\n\n Solving with random initial guess."solver = HF(N, basis, Z_value, first_C=’rand’)solver.solve(max_iters=100)

We conclude by listing the matrix elements for the radial integrals to be usedfor the direct part and the exchange part. Note again that these integrals donot include spin. The nomenclature is 1 = 1s, 2 = 2s, and 3 = 3s, with no spindegrees of freedom.

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Element Expression as function of charge Z〈11|v|11〉 = (5*Z)/8〈11|v|12〉 = (4096*Sqrt[2]*Z)/64827〈11|v|13〉 = (1269*Sqrt[3]*Z)/50000〈11|v|21〉 = (4096*Sqrt[2]*Z)/64827〈11|v|22〉 = (16*Z)/729〈11|v|23〉 = (110592*Sqrt[6]*Z)/24137569〈11|v|31〉 = (1269*Sqrt[3]*Z)/50000〈11|v|32〉 = (110592*Sqrt[6]*Z)/24137569〈11|v|33〉 = (189*Z)/32768〈12|v|11〉 = (4096*Sqrt[2]*Z)/64827〈12|v|12〉 = (17*Z)/81〈12|v|13〉 = (1555918848*Sqrt[6]*Z)/75429903125〈12|v|21〉 = (16*Z)/729〈12|v|22〉 = (512*Sqrt[2]*Z)/84375〈12|v|23〉 = (2160*Sqrt[3]*Z)/823543〈12|v|31〉 = (110592*Sqrt[6]*Z)/24137569〈12|v|32〉 = (29943*Sqrt[3]*Z)/13176688〈12|v|33〉 = (1216512*Sqrt[2]*Z)/815730721〈13|v|11〉 = (1269*Sqrt[3]*Z)/50000〈13|v|12〉 = (1555918848*Sqrt[6]*Z)/75429903125〈13|v|13〉 = (815*Z)/8192〈13|v|21〉 = (110592*Sqrt[6]*Z)/24137569〈13|v|22〉 = (2160*Sqrt[3]*Z)/823543〈13|v|23〉 = (37826560*Sqrt[2]*Z)/22024729467〈13|v|31〉 = (189*Z)/32768〈13|v|32〉 = (1216512*Sqrt[2]*Z)/815730721〈13|v|33〉 = (617*Z)/(314928*Sqrt[3])〈21|v|11〉 = (4096*Sqrt[2]*Z)/64827〈21|v|12〉 = (16*Z)/729〈21|v|13〉 = (110592*Sqrt[6]*Z)/24137569〈21|v|21〉 = (17*Z)/81〈21|v|22〉 = (512*Sqrt[2]*Z)/84375〈21|v|23〉 = (29943*Sqrt[3]*Z)/13176688〈21|v|31〉 = (1555918848*Sqrt[6]*Z)/75429903125〈21|v|32〉 = (2160*Sqrt[3]*Z)/823543〈21|v|33〉 = (1216512*Sqrt[2]*Z)/815730721〈22|v|11〉 = (16*Z)/729〈22|v|12〉 = (512*Sqrt[2]*Z)/84375〈22|v|13〉 = (2160*Sqrt[3]*Z)/823543〈22|v|21〉 = (512*Sqrt[2]*Z)/84375〈22|v|22〉 = (77*Z)/512〈22|v|23〉 = (5870679552*Sqrt[6]*Z)/669871503125〈22|v|31〉 = (2160*Sqrt[3]*Z)/823543〈22|v|32〉 = (5870679552*Sqrt[6]*Z)/669871503125〈22|v|33〉 = (73008*Z)/9765625〈23|v|11〉 = (110592*Sqrt[6]*Z)/24137569〈23|v|12〉 = (2160*Sqrt[3]*Z)/823543〈23|v|13〉 = (37826560*Sqrt[2]*Z)/22024729467〈23|v|21〉 = (29943*Sqrt[3]*Z)/13176688〈23|v|22〉 = (5870679552*Sqrt[6]*Z)/669871503125〈23|v|23〉 = (32857*Z)/390625〈23|v|31〉 = (1216512*Sqrt[2]*Z)/815730721〈23|v|32〉 = (73008*Z)/9765625〈23|v|33〉 = (6890942464*Sqrt[2/3]*Z)/1210689028125〈31|v|11〉 = (1269*Sqrt[3]*Z)/50000〈31|v|12〉 = (110592*Sqrt[6]*Z)/24137569〈31|v|13〉 = (189*Z)/32768〈31|v|21〉 = (1555918848*Sqrt[6]*Z)/75429903125〈31|v|22〉 = (2160*Sqrt[3]*Z)/823543〈31|v|23〉 = (1216512*Sqrt[2]*Z)/815730721〈31|v|31〉 = (815*Z)/8192〈31|v|32〉 = (37826560*Sqrt[2]*Z)/22024729467〈31|v|33〉 = (617*Z)/(314928*Sqrt[3])〈32|v|11〉 = (110592*Sqrt[6]*Z)/24137569〈32|v|12〉 = (29943*Sqrt[3]*Z)/13176688〈32|v|13〉 = (1216512*Sqrt[2]*Z)/815730721〈32|v|21〉 = (2160*Sqrt[3]*Z)/823543〈32|v|22〉 = (5870679552*Sqrt[6]*Z)/669871503125〈32|v|23〉 = (73008*Z)/9765625〈32|v|31〉 = (37826560*Sqrt[2]*Z)/22024729467〈32|v|32〉 = (32857*Z)/390625〈32|v|33〉 = (6890942464*Sqrt[2/3]*Z)/1210689028125〈33|v|11〉 = (189*Z)/32768〈33|v|12〉 = (1216512*Sqrt[2]*Z)/815730721〈33|v|13〉 = (617*Z)/(314928*Sqrt[3])〈33|v|21〉 = (1216512*Sqrt[2]*Z)/815730721〈33|v|22〉 = (73008*Z)/9765625〈33|v|23〉 = (6890942464*Sqrt[2/3]*Z)/1210689028125〈33|v|31〉 = (617*Z)/(314928*Sqrt[3])〈33|v|32〉 = (6890942464*Sqrt[2/3]*Z)/1210689028125〈33|v|33〉 = (17*Z)/256

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