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Ab initio methods: Post-Hartree–Fock Alston J. Misquitta Centre for Condensed Matter and Materials Physics Queen Mary, University of London March 6, 2012

Ab initio methods: Post-Hartree–Fock...Ab initio methods: Post-Hartree–Fock Alston J. Misquitta Centre for Condensed Matter and Materials Physics Queen Mary, University of London

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  • Ab initio methods: Post-Hartree–Fock

    Alston J. Misquitta

    Centre for Condensed Matter and Materials Physics

    Queen Mary, University of London

    March 6, 2012

  • Basis sets CI CC MBPT Properties

    Hartree–Fock equations I

    We have introduced basis sets in solving the Hartree–Fockequations, but what exactly are they?Brief summary of Hartree–Fock theory:

    Born–Oppenheimer approximation: fixed nuclei.

    Ground-state N-electron wavefunction: single Slaterdeterminant:

    0

    (x1

    , x2

    , · · · , xN) = |�1�2 · · ·�Ni (1)

    Variational Principle:

    E

    0

    EHF = minh 0

    |H| 0

    i (2)

    subject to the conditions that the spin-orbitals areorthonormal.

  • Basis sets CI CC MBPT Properties

    Hartree–Fock equations II

    Fock equations:f (i)�(xi ) = ✏�(xi ) (3)

    where f (i) is an e↵ective operator called the Fock operator

    f (i) = �12r2i �

    X

    Z↵

    ri↵+ vHF(i) (4)

    where vHF(i) is the Hartree–Fock e↵ective potential.

    Self-consistent solution (iterations needed).

    Infinity of solutions to the Fock equations.

    Introduce a basis for the spatial part of spin-orbitals:

    �i (r) =X

    m

    Cim�m(r) (5)

  • Basis sets CI CC MBPT Properties

    Hartree–Fock equations III

    Leads to linear equations:

    FC = ✏SC (6)

    Q: How do we choose the basis?

  • Basis sets CI CC MBPT Properties

    Slater-type orbitals I

    A reasonable choice for basis sets for finite systems would be whatare called Slater-type orbitals: these are very like solutions of the1-electron Hamiltonian. They di↵er in two ways: (1) the radialpart is simpler and (2) the exponent is not integral but can bevaried to account for screening e↵ects.

    � = Rnl(r)Ylm(✓,�) (7)

    where Ylm is a (real) spherical harmonic and the radial part isgiven by

    Rnl(r) =(2⇣)n+1/2

    [(2n)!]1/2r

    n�1e

    �⇣r (8)

  • Basis sets CI CC MBPT Properties

    Slater-type orbitals II

    2 · 10�2 4 · 10�2 6 · 10�2 8 · 10�2 0.1

    0.1

    0.2

    0.3

    0.4

    0.5⇣ = 1.0, n = 1⇣ = 1.0, n = 2

  • Basis sets CI CC MBPT Properties

    Slater-type orbitals III

    Comments on Slater-type orbitals:

    GOOD Nuclear cusp condition satisfied.

    @

    @rh⇢(r)isph

    ����r=0

    = �2Z h⇢(0)isph

    GOOD Exact wavefunction has the long-range form of aSlater orbital.If we pull one electron out of an N-electron molecule thewavefunction behaves like

    (N) ! (N � 1)⇥ e�p2⇤I r

    where I is the first (vertical) ionization energy.

    BAD Integrals very di�cult for multi-atom systems.

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) I

    In 1950 S. F. Boys pointed out that the problem of computingintegrals could be resolved by using not Slater-type orbitals, butrather Gaussian-type orbitals (GTOs):

    Rnl ⇠ r le�↵(r�A)2

    (9)

    where A is the centre of the GTO. The main reason for thee�cacy of GTOs is that the product of two GTOs is a third GTO,centred at a point in between:

    exp(�↵(r�A)2) exp(��(r�B)2) = exp(��(A�B)2) exp(�µ(r�P)2)

    where µ = ↵+ �, � = ↵�/µ and P = (↵A+ �B)/µ.

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) II

    2 · 10�2 4 · 10�2 6 · 10�2 8 · 10�2 0.1

    0.2

    0.4

    0.6

    0.8

    1

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) III

    GOOD GTOs makes the integrals that appear in the HFenergy expression much simpler.

    BAD Nuclear cusp condition violated: zero derivative at origin.

    BAD Wrong long-range form: dies o↵ too fast with distance.

    The two negative points can, to some extent, be remedied by usingnot single GTOs, but linear combinations of GTOs. These groupsof GTOs are termed contractions.Basis sets consist of groups of contractions together with someun-contracted GTOs. The better the basis, the more of these therewill be and the more GTOs in a contraction.

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) IV

    2 · 10�2 4 · 10�2 6 · 10�2 8 · 10�2 0.1

    0.1

    0.2

    0.3

    0.4

    0.5⇣ = 1.0, n = 1

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) V

    x

    � ⇣ = 1.0, n = 1STO-1G

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) VI

    x

    4⇡|�|2 ⇣ = 1.0, n = 1STO-1G

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) VII

    2 · 10�2 4 · 10�2 6 · 10�2 8 · 10�2 0.1

    0.1

    0.2

    0.3

    0.4

    0.5⇣ = 1.0, n = 1

    STO-3G

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) VIII

    2 · 10�2 4 · 10�2 6 · 10�2 8 · 10�2 0.1

    0.1

    0.2

    0.3

    0.4

    0.5 ⇣ = 1.0, n = 1STO-3G

  • Basis sets CI CC MBPT Properties

    Gaussian-type orbitals (GTOs) IX

    Basis set recommendations:

    GOOD Complete basis set (CBS) limit

    Geometry optimization: moderate size basis sets. Double-⇣.

    Energies: At least triple-⇣ quality.

    Properties: Triple-⇣ or more.

    We will have another look at basis sets after discussion correlatedmethods.

  • Basis sets CI CC MBPT Properties

    Correlation I

    Q: How to we improve on the complete basis set HF results? I.e.,how do we get beyond the HF limit?

    Variational Principle: More flexibility. Leads to ConfigurationInteraction (CI).

    Perturbation Theory

    Coupled-cluster methods

    Density functional theory

  • Basis sets CI CC MBPT Properties

    Configuration Interaction I

    CI: Increase the flexibility in the wavefunction by including inaddition the the HF ground state, excited states.Q: What are excited states and how to we form them?

  • Basis sets CI CC MBPT Properties

    Configuration Interaction II

    1

    2

    n

    n + 1

    n +m

    Figure: Left: HF ground state configuration. Right: An example of anexcited state configuration. If there are n occupied levels (2 electronseach, so N = 2n) and m virtual (un-occupied) levels, in how many wayscan we form excited states? Each of these states will correspond to aSlater determinant.

  • Basis sets CI CC MBPT Properties

    Configuration Interaction III

    The Full CI (FCI) wavefunction:

    | i = | 0

    i+X

    ar

    c

    ra | rai+

    X

    abrs

    c

    rsab | rsabi+ · · ·

    = | 0

    i+ cS |Si+ cD |Di+ · · ·

    where electrons are excited from the occupied orbitals a, b, c , · · ·to the virtual orbitals r , s, t, · · · .

    GOOD This expansion will lead to the exact energy within thebasis set used.

    BAD There are too many determinants!

    (2(n +m))!

    (2n)!(2m)!

  • Basis sets CI CC MBPT Properties

    Configuration Interaction IV

    One solution to the problem is to use only some of the manydeterminants. For example we could use only double excitations.This leads to the CID method.

    �� CID↵= |

    0

    i+X

    abrs

    c

    rsab | rsabi

    = | 0

    i+ cD |Di

    BAD This theory, like all truncated CI methods, is not sizeextensive.Size-extensivity: If E (N) is the energy of N non-interactingidentical systems then a method is size-extensive ifE (N) = N ⇥ E (1).

  • Basis sets CI CC MBPT Properties

    Configuration Interaction V

    Q: Is CID size-extensive?If T̂

    2

    is an operator that creates all double excitations, then we canwrite the CID wavefunction as

    �� CID↵= |

    0

    i+X

    abrs

    c

    rsab | rsabi

    = | 0

    i+ cD |Di= (1 + T̂

    2

    ) | 0

    i

  • Basis sets CI CC MBPT Properties

    Configuration Interaction VI

    The CID wavefunction for each of two identical non-interactingsystems will be of that form, so the combined wavefunction will be

    �� CIDA↵ �� CIDB

    ↵= (1 + T̂

    2

    (A))��� A

    0

    E(1 + T̂

    2

    (B))��� B

    0

    E

    =(1 + T̂2

    (A) + T̂2

    (B) + T̂2

    (A)T̂2

    (B))��� A

    0

    E ��� B0

    E

    The last excitation term is a quadruple excitation so it will not bepresent in the CID wavefunction for the combined A and Bsystems. Therefore

    E

    CID(AB) 6= ECID(A) + ECID(B).

  • Basis sets CI CC MBPT Properties

    Configuration Interaction VII

    CI is not size-extensive: H2

    �g

    �u

    Figure: An energy level diagram for H2

    with a minimal basis. The twoMOs are as shown. We can form three determinants from them. Left isthe ground state. Middle is a singly excited determinant. Right is adoubly excited determinant. For reasons of symmetry the middleconfiguration does not contribute to the CI expansion.

  • Basis sets CI CC MBPT Properties

    Coupled-cluster Theory I

    The problem of truncated CI methods is severe enough that usingthem is very problematic. A resolution to the problem is the classof coupled-cluster theories. In these the wavefunction is defined as:

    �� CC↵= exp(T̂) |

    0

    i

    where T̂ is an appropriate excitation operator.

  • Basis sets CI CC MBPT Properties

    Coupled-cluster Theory II

    For example, in CCD theory we use T̂2

    as the excitation operator.This gives:

    �� CCD↵= exp(T̂

    2

    ) | 0

    i

    = (1 + T̂2

    +1

    2!T̂

    2

    2

    + · · · ) | 0

    i

    The first two terms give us CID theory. The rest are needed tomake CCD size-extensive:

    exp(T̂2

    (A))��� A

    0

    E⇥ exp(T̂

    2

    (B))��� B

    0

    E= exp(T̂

    2

    (A) + T̂2

    (B))��� A

    0

    E ��� B0

    E

    ⌘ exp(T̂2

    (AB))��� A

    0

    E ��� B0

    E

  • Basis sets CI CC MBPT Properties

    Coupled-cluster Theory III

    GOOD CC theories can be systematically improved.

    GOOD CCSD(T) is a very accurate theory. Here single anddouble excitations are included as described above and tripleexcitations are included through a perturbative treatment.

    GOOD Size-extensive.

    BAD Computationally very expensive: CCSD(T) scales asO(N7). So double the system size and the calculation costs128 times more.

    BAD (kind of!) These are single-determinant theories asdescribed. If the system is multi-configurational (more thanone state contributing dominantly) the standard CC methodsare not appropriate.

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory I

    Brief recap of Raleigh–Schrödinger perturbation theory:

    Split the Hamiltonian into two parts:

    H = H0

    + �V

    where H0

    is a Hamiltonian which we know how to solve andV contains that troublesome parts. We expect V to be aperturbation so it must be small in some sense.� is a complex number that will be 1 for the physical solution.

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory II

    Let the solutions of H0

    be:

    H0

    (0)i = E(0)

    i (0)

    i

    Here the ‘0’ indicates that these eigenvalues andeigenfunctions are of zeroth-order in the perturbation V. Wewill use the short-form:

    | (0)i i ⌘ |ii

    Express the solutions of H in a power-series:

    i = (0)

    i + � (1)

    i + �2 (2)i + · · · =

    X

    n

    �n (n)i

    Ei = E(0)

    i + �E(1)

    i + �2

    E

    (2)

    i + · · · =X

    n

    �nE (n)i

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory III

    Energies can be calculated by collecting terms at variousorders:

    E

    (0)

    i = h (0)

    i |H0| (0)

    i i = hi |H0|iiE

    (1)

    i = h (0)

    i |V| (0)

    i i = hi |V|iiE

    (2)

    i = h (0)

    i |V| (1)

    i i

    etc. for higher order terms.

    The first-order correction to the wavefunction is given by

    | (1)i i =X

    n 6=0

    |nihn|V|iiE

    (0)

    i � E(0)

    n

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory IV

    So we get

    E

    (2)

    i =X

    n 6=0

    |hn|V|ii|2

    E

    (0)

    i � E(0)

    n

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory V

    Many-body perturbation theory (MBPT) starts from Hartree–Focktheory:

    H0

    =NX

    i=1

    f (i) =nX

    i=1

    �h(i) + vHF(i)

    �(10)

    where h(i) = �12

    r2i �P

    ↵Z↵ri↵

    We can now define the perturbation as

    V =NX

    i=1

    NX

    j>i

    1

    rij�

    NX

    i=1

    v

    HF(i) (11)

    Unlike vHF, the perturbation V is a 2-electron operator.

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory VI

    MBPT energy at �0:

    E

    (0)

    0

    = h0|F|0i =X

    a2occ✏a

    At first-order we get (no proof):

    E

    (1)

    0

    = �12

    NX

    a=1,b=1

    ⇥hab|r�1

    12

    |abi � haa|r�112

    |bbi⇤

    We have not seen this before, but the sum of E (0)0

    and E (1)0

    is justthe Hartree–Fock ground state energy:

    E

    HF = E (0)0

    + E (1)0

    (12)

    This means that we need to get to at least second-order inperturbation theory to go beyond the Hartree–Fock description.

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory VII

    Here is what the second-order MBPT energy expression looks like:

    E

    (2)

    0

    =occX

    a,b>a

    virX

    r ,s>r

    ⇥hab|r�1

    12

    |rsi � hab|r�112

    |sri⇤2

    ✏r + ✏s � ✏a � ✏b(13)

    This expression is termed as MBPT2 or MP2. The latter namecomes from the other name for this kind of perturbation theory:Møller–Plesset perturbation theory.

  • Basis sets CI CC MBPT Properties

    Møller–Plesset Perturbation Theory VIII

    BAD A problem with Møller–Plesset perturbation theory: itdiverges! See Olsen et al. J. Chem. Phys. 112, 9736 (2000)for details. We now rarely go beyond MP2 in practicalcalculations.

    GOOD MP2 contains correlation.

    BAD But not enough correlation. Problems with systems withsmall HOMO-LUMO gaps (band gap — HOMO is highestoccupied MO and LUMO is lowest unoccupied MO).

    GOOD (kind of!) It has a computational cost of O(N5). I.e.,double the system in size and it will cost 32 times morecomputational power.

    GOOD MBPT is size-consistent

  • Basis sets CI CC MBPT Properties

    Expectation Values and Forces I

    Once we’ve got the HF solution, how do we calculate expectationvalues of operators and forces?

    Basis setsCICCMBPTProperties