Theoretical Chemistry Quantum Chemistry. Content 1.Introduction 2.Many-electron systems 3.Two basic approximations 4.Hartree-Fock metod 5.Roothaan approximation.

Theoretical Chemistry Quantum Chemistry

Content 1.Introduction 2.Many-electron systems 3.Two basic approximations 4.Hartree-Fock metod 5.Roothaan approximation and basis sets 6.Electron correlation 7.Post Hartree-Fock methods 8.Geometry optimization 9.Simulation of the IR, NMR and AFM experiments

Introduction Quantum chemistry is essentially quantum mechanics applied to chemical systems Quantum chemistry is very simple and easy science because there is only one equation to be solved. It is the Schrödinger equation of the general form of: (1)

where Ĥ=T+V is the total energy operator called Hamiltonian and φ= φ(q, t) denotes a wave function, a function of a position q and time t. This is the most important quantity in quantum mechanics. It gives so-called the probability density ρ (q)= | Φ(q,t) | 2 = Φ * (q,t) Φ(q,t) = | Ψ(q) | 2 (2)

Very often Hamiltonian of a system does not depend on time. As a consequence the total energy E, an eigenvalue of the Hamiltonian operator Ĥ(q) φ(q, t) = E φ(q, t)(3) is also time independent. This is so called the stationary state.

A stationary state is very unique because of the specific form of its wave function Φ(q,t) = Ψ(q) exp(-i E t/ħ)(4) Thus, in spite of a wave function being time dependent, probability density ρ (q)= | Φ(q,t) | 2 = Φ * (q,t) Φ(q,t) = | Ψ(q) | 2 is constant.

The spatial part of the total wave function (4) for a stationary state is a solution of the time independent Schrödinger equation Ĥ(q) Ψ(q) = E Ψ(q) (5) From hereafter we will consider only stationary states.

The Schrödinger equation (5) for a stationary state has, in general, an infinitive number of solutions and should be rewritten as Ĥ(q) Ψ n (q) = E n Ψ n (q) (6) The above equation has to be solved for a given system. In quantum chemistry the systems we are interested in are molecules (and atoms). How difficult is to solve equation (6) for such systems ?

To answer this question we have to look at the Hamiltonian operator.

Many-electron systems For a molecule containing n electrons and N nuclei we may write Ĥ = T e + T n +V e-n + V e-e + V n-n (7) where the first two terms denote the kinetic energy of electrons and nuclei, respectively, and the last three terms correspond to the potential energy of interaction between electrons and nuclei, among electrons and among nuclei. These terms are given as

T e = Σ i n (-ħ/2m Δ i ) and T n = Σ A N (-ħ/2M A Δ A )(8a) V e-n = Σ i n Σ A N [ - Z A e 2 / |r i – R A |](8b) V e-e = Σ i n Σ j n [ e 2 / | r i – r j | ](8c) V n-n = Σ A N Σ B N [Z A Z B e 2 / |R A – R B | ](8d) here r i =(x i, y i, z i ) and R A =(X A, Y A, Z A ) denote coordinates of i-electron and A-nucleus with masses m, M A, respectively, and Δ i, Δ A are their Laplace operators. |

Now we have to solve the Schrödinger equation (6) with this Hamiltonian Ĥ(r,R) Ψ n (r,R) = E n Ψ n (r,R ) (9) where r=(r 1, r 2, …, r n ) and R=(R 1, R 2,…, R N ). In general such equation can be solved only if the operator (Hamiltonian) is a sum of one-particle terms (electrons and nuclei). In the Hamiltonian (8) only kinetic energy terms (8a) for electrons and nuclei satisfy this condition.

The potential energy terms (8b), (8c) and (8d) representing interactions among particles are defined by pairs and involve distances between two particles 1/ |r i – R A |, 1/ |r i – r j |, 1/ |R A – R B | These terms cannot be rewritten as a sum of separate terms and therefore the Schrödinger equation (9) for a molecule cannot be solved exactly.

It should be also noticed that the size or dimension of the Schrödinger equation (9) may be huge for larger molecules. The number of variables there (x,y and z coordinates for each electron and nucleus) is equal to 3*(n+N). This gives even for rather small benzene (C 6 H 6 ) with 12 nuclei and 42 electrons quite large dimension of 162. Thus, the Schrödinger equation(9) has to be somehow simplified which means that some approximations have to be introduced.

Two basic approximations 1.The first approximation results from the fact that nuclei are much heavier than electrons and therefore their motion is much slower than electrons. So we may go one step further and say that nuclei do not move at all. Thus, in a molecule electrons will move in the field of non- moving nuclei. This assumption is related to so- called Born-Oppenheimer approximation. What do we gain from this approximation ?

First of all the positions R=(R 1, R 2,…, R N ) of non- moving nuclei are fixed so they are not variables any more. Hence the dimension of the Schrödinger equation(9) is reduced from 3*(n+N) to 3*n with only electron coordinates r=(r 1, r 2,…,r n ) being variables (for benzene the dimension is reduced from 162 to 126).

Secondly, the Hamiltonian (7) is drastically simplified: the nuclear kinetic energy term vanishes and the nuclear repulsion term (8d) become constant and can be omitted (instead a constant value will be added to the energy) Ĥ = T e + T n +V e-n + V e-e + V n-n zero constant (omitted)

Such simplified operator is called the electronic Hamiltonian and it is given explicitly as follows Ĥ el = Σ i n (-ħ/2m Δ i ) + Σ i n Σ A N [ - Z A e 2 /(r i – R A ) ] + Σ i n > Σ j n [ e 2 /(r i – r j ) ] (10) = Σ i n h(i) + Σ i n > Σ j n [ e 2 /(r i – r j ) ] with h(i) = (-ħ/2m Δ i ) + Σ A N [ - Z A e 2 /(r i – R A ) ]

The Schrödinger equation with this energy operator Ĥ el (r;R) Ψ n (r;R) = E n (R) Ψ n (r;R )(11) is much simpler than original equation (9) with the full Hamiltonian (7,8) Can we solve the Schrödinger equation with this Electronic Hamiltonian? The answer is NO because of the electron-electron repulsion term in (10) Σ i n > Σ j n [ e 2 / |r i – r j |]

Thus, we have to make further simplification 2.The second approximation tries to get the picture of a many-electron system using one-electron information. This is so-called one-electron approximation which says: each electron in a many-electron system is treated independently and its state will be described by spinorbital, a one-electron wavefunction φ(r) = ψ(r) σ with spinfunction σ=α or β

Thus for the 2n-electron system with the total spin equal zero (closed-shell system) we need 2n spinorbitals {φ 1 (r), φ 2 (r), …,φ 2n (r)} made out of n obitals {ψ 1 (r), ψ 2 (r), …,ψ n (r)}. These spinorbitals are then used to form the total many-electron wavefunction. For the closed-shell system with 2n electrons such a wavefunction takes a single determinant form

φ 1 (1) φ 1 (2) …. φ 1 (2n) φ 2 (1) φ 2 (2) …. φ 2 (2n) Ψ(r 1, r 2, …, r 2n )= N …………………….(12) φ k (1) φ k (2) …. φ k (2n) ……………………. φ 2n (1) φ 2n (2) …. φ 2n (2n) with the normalization factor N=[(2n)!] -1/2. NOTE: these spinorbitals are not known yet and our goal is to determine them. How can we do it?

The best spinorbials will be found using a variational method in which the determinantal wavefunction is used as a trail wavefunction According to the variational method we need to minimize the total energy (expectation value of the Hamiltonian) of a molecular system E ave = ∫ Ψ * (r 1, r 2, …, r 2n ) Ĥ el Ψ(r 1, r 2, …, r 2n ) dv

Substituting the determinantal wavefunction (12) and the electronic Hamiltonian (10) into the expectation energy expression one gets the following explicit form of E ave E ave = 2∑ i h ii + ∑ i ∑ ij [ 2J ij – K ij ](13) where summation runs over spatial orbitals {ψ i } and h ii, J ij, K ij denote one- and two-electron integrals

h ii =∫ ψ i * (1) h(1) ψ i (1) dv 1 J ij =∫∫ ψ i * (1) ψ i (1) 1/r 12 ψ j *(2) ψ j (2) dv 1 dv 2 K ij =∫∫ ψ i * (1) ψ j (1) 1/r 12 ψ j *(2) ψ i (2) dv 1 dv 2 the last two are called the Coulomb and exchange integrals, respectively.

The energy expectation value (13) is now expressed in terms of unknown spatial orbitals {ψ i }. Minimizing energy (13) with respect to these orbitals leads to the following set of one-electron equations f(1) ψ i (1) = ε i ψ i (1) for i=1,n (14) with the orthonormal constrains ∫ ψ i * (1) ψ j (1) dv 1 = δ ij (15)

where so-called the Fock operator is given as f(1)= h(1) + ∑ j [2J j (1) – K j (1)](16) with h(1) = (-ħ/2m Δ 1 ) + Σ A N [ - Z A e 2 /(r 1 – R A ) ] and J j (1) ψ i (1) = [∫1/r 12 ψ j *(2) ψ j (2) dv 2 ] ψ i (1) K j (1) ψ i (1) = [∫1/r 12 ψ j *(2) ψ i (2) dv 2 ] ψ j (1)

Equations (14) and (15) have been derived by Hartree and Fock and therefore this method is known as the Hartree-Fock (HF) method. The Hartree-Fock method plays a central role in quantum chemistry. It usually constitutes the first step towards more accurate approximations

By solving the Hartree-Fock equations we obtain the set of optimal orbitals {ψ i } (molecular or atomic). The essence of the Hartree-Fock method is to replace a very complicated many-electron problem by a one-electron problem in which electron-electron repulsion is treated in an average way.

The second part of the Fock operator (16) v HF (1) = ∑ j [2J j (1) – K j (1)](17) often called the Hartree-Fock potential represents an average potential experienced by one electron due to the presence of other electrons. v HF depends on the orbitals of the other electrons, so the Fock operator depends on its eigenfunctions. Hence the Hartree-Fck equations are nonlinear and have to be solved iteratively.

The procedure for solving the Hartree-Fock equations is called the Self-Consistent-Field (SCF) method. The basic idea of the SCF method is simple. By making an initial guess at the orbitals {ψ i } one can calculate the average filed (i.e. v HF ) seen by each electron and then solve the eigenvalue equation(14) This gives a new set of orbitals which are used to calculate new filed v HF and the whole procedure is repeated until self-consistency is reached.

The solution of the Hartree-Fock eigenvalue problem (14) yields a set of orthonormal Hartree-Fock orbitals {ψ i } with orbital energies {ε i }. In principle, there are an infinitive number of solutions. The n orbitals with the lowest energies are called the occupied orbitals. The Slater determinat formed from these orbitals is the Hartee-Fock ground state wave function and is the best variational approximation to the ground state of a system. The remaining orbitals are called virtual obitals.

Using the one-electron approximation with a single determinant trail wave function within the variational method we have replaced the many- electron Schrödinger equation with the electronic Hamiltonian (10) by a set of one-electron Hartree- Fock equations (14, 15) f(1) ψ i (1) = ε i ψ i (1) for i=1,n ∫ ψ i * (1) ψ j (1) dv 1 = δ ij

At the first look it seems to be much easier now to solve these HF equations than the original many- electron Schrodinger (electronic) equation. However, it is not so simple. The Hartree-Fock equations are very complicated; they are integro-differential in nature through the form of the Fock operator

f(1)= h(1) + ∑ j [2J j (1) – K j (1)] h(1) = (-ħ/2m Δ 1 ) + Σ A N [ - Z A e 2 /(r 1 – R A ) ] J j (1) ψ i (1) = [∫1/r 12 ψ j *(2) ψ j (2) dv 2 ] ψ i (1) K j (1) ψ i (1) = [∫1/r 12 ψ j *(2) ψ i (2) dv 2 ] ψ j (1) The Hartree-Fock equations CANNOT be solved for molecules. An additional approximation is necessary.

The Roothaan analytical approximation to the Hartree-Fock orbitals: ψ i (1) = ∑ μ c μi λ μ (1) (18) where μ=1,m numbers a set of known one-electron basis functions {λ μ }. If this set of the basis functions was complete the above expansion would be exact. Unfortunately, we are always restricted to the finite set of m functions. Nevertheless, the problem of finding HF orbitals is reduced to calculate {c μi }.

The Roothaan analytical approximation replaces the integro-differential Hartree-Fock equations by much simpler Hartree-Fock-Roothaan algebraic equations which can be written in the compact matrix form F C = SCε(19) with the orthonormal conditions C T S C = 1

Here F and S denote the Fock and the overlap matrix with the matrix elements F μν = ∫λ μ (1) f(1) λ v (1) dv 1 = (µ| f |ν) (20) S μν = ∫λ μ (1) λ v (1) dv 1 = (µ|ν) Substituting (18) into f(1) gives finally F μν = h μν + ∑ λσ D λσ [(μν| λσ) – 1/2(μσ| λν)]

where the density matrix elements are given as D λσ = 2 ∑ i c λi c σi and two-electron integrals over basis functions (μν| λσ)=∫∫ μ(1)ν(1)1/r 12 λ(2)σ(2) dv 1 dv 2 The electronic energy (13) is now given as E ave =1/2∑ µν D µν (h µν + F µν )

The total molecular energy, a sum of electronic and nuclear-nuclear repulsion energies E tot = E ave + ∑ A ∑ B Z A Z B /R AB is a very important quantity, particularly in structure determination. It is so because the predicted equilibrium molecular geometry occurs at the minimum of E tot (with respect to the nuclear positions)

However, the most important quantity overall is the molecular wave function (Salter determinant). Once we have it we might evaluate a number of molecular properties such as multiple moments (dipole, quadrupole etc.), field gradient at a nucleus, diamagnetic parts of susceptibility and magnetic shielding. Such properties are described by sums of one-electron operators of the general form O1 = ∑ i o(i)(21)

where o(i) is any operator depending on the coordinates of a single electron. Molecular properties represented by (21) can be calculated as expectation values of such operators with the Slater determinant type of a wave function = (22)

In a given basis set this expression will have always the same general form = ∑ i (ψ i (1)|o(1)| ψ i (1)) = ∑ µν D µν (µ | O(1)| ν )(23) Thus in order to calculate one-electron expectation values we need, in addition to the density matrix D, only a set of one-electron integrals (µ | O(1)| ν ).

Let consider calculations of the dipole moment. The classical definition of the dipole moment of a set of charges q i at the positions r i is μ = ∑ i q i r i For a molecule with n electrons and N nuclei it becomes μ = - ∑ i n e r i + ∑ A N Z A e R A (24)

The quantum mechanical operator representing (24) has the same form (because it contains coordinates only). Its expectation value is given as = (25) (Ψ denotes a Slater determinant). In a given basis set according to (23) it becomes (with e=1) = -∑ µ ∑ ν D v (μ|r|ν) + ∑ A N Z A R A (26)

The first term in (26) represents the electronic part of the total dipole moment and it has a quantum mechanical nature while the second term is purely classical and describes contributions from nuclei. The total dipole moment (26) is, of course, a vector. Its, for example, x-component is given as = -∑ µ ∑ ν D v (μ|x|ν) + ∑ A N Z A X A

Another important quantity obtained from the Hartree-Fock obitals which a deterninantal wave function is consisted of is the total charge density. If there is in a molecule an electron described by the spatial Hartree-Fock orbital ψ i (1) then the probability of finding that electron in a volume dr at a point r is | ψ i (r)| 2 dr

In a closed-shell molecule described by a single determinant wave function with n double occupied orbitals {ψ i (1)} the total charge density ρ(r) = 2 ∑ i n | ψ i (r)| 2 (27) determines the probability of finding any single electron in a volume dr at r. Integration of this total charge density gives the total number of electrons 2n in a system. ∫| ψ i (r)| 2 dr = 2∑ i n ∫ | ψ i (r)| 2 dr=2∑ i n 1= 2n(28)

In a given basis set {λ μ } where the Hartree-Fock orbitals are approximated by (18) ψ i (r) = ∑ μ c μi λ μ (r) the total charge density (27) is expressed as follows ρ(r) = 2 ∑ i n | ψ i (r)| 2 = ∑ μ ∑ ν D μ ν λ μ λ v

Integration of the latter equation together with (28) gives a nice relation 2n = ∑ μ ∑ ν D μν ∫ λ μ (r)λ v (r) dr = ∑ μ ∑ ν D μν S μν Note that both matrices above are symmetric so D μν = D νμ and S μν = S νμ and therefore we can write 2n = ∑ μ ∑ ν D μν S νμ = ∑ μ (DS) μμ

The above equation may be written as 2n = Tr (DS)(29) which says that the total number of electrons is a trace of the matrix being a product of the density and overlap matrices. The relation (29) is very useful in making distribution of an electronic charge in a molecule in different ways.

The diagonal element of the DS matrix, (DS) μ μ can be interpreted as a ”number” of electrons (electronic charge) associated with one basis function λ μ (r). Basis functions are usually centered on atoms (nuclei) in a molecule. By summing over all basis functions on a given atom A the corresponding ”number” of electrons associated with that atom in a molecule can be obtained. This is in general rather a real than natural number q A el

q A el = ∑ μ є A (DS) μμ (30) From this we can get the net charge q A for a given atom A (atomic number Z A ) in a molecule q A = Z A - ∑ μ є A (DS) μμ (31) The procedure described above i.e. distribution of the electronic charge over atoms in a molecule is called population analysis (Mulliken).

Unfortunately, the population analysis defined by (29), (30) and (31) is by no means unique. Consider the following property of the trace of a product of matrices A, B, C Tr ABC = Tr CAB = Tr BCA Using above, equation (29) can be formally written as 2n = Tr (DS) = Tr (D S a S 1-a ) = Tr (S 1-a D S a )

Using above relations, equation (29) can be formally written as 2n = Tr (DS) = Tr (D S a S 1-a ) = Tr (S 1-a D S a ) which is true for any value of a, regardless of the meaning of S a and S 1-a. These matrices can be easily calculated for a=1/2. In such a case we have 2n = Tr (S 1/2 D S 1/2 )

The electronic charge on an atom A resulting from the latter is q A el = ∑ μ є A (S 1/2 D S 1/2 ) μμ (32) which is different than that obtained before according to (30) q A el = ∑ μ є A (DS) μμ (30) These two population analysis have been proposed by Mulliken (30) and Lowdin (32).

Let us consider one more issue concerning the Hartree-Fock method. Until now this method has been viewed as an approximation in which the Hamiltonian operator is exact (electronic) but the wave function is approximated by a single Slater determinant. Now we will concentrate on the Hamiltonian.

Remember that we did not solve the exact electronic Schrödinger equation (11) for a ground state Ĥ el Φ 0 = E 0 Φ 0 but we have rather found an approximation Ψ 0 to Φ 0 using the variational principle i.e. minimizing an expectation energy Ē 0 Ē 0 = ∫ Ψ 0 * Ĥ el Ψ 0 dv

Now the question: is the approximated Ψ 0 we have found an exact eigenfunction of any many-electron Hamiltonian ? Did we solve any N-electron Schrödinger equation exactly ?

The answer is YES !!! It is easy to check that our Ψ 0, a Slater determinant, is an eigenfunction of the following operator Ĥ 0 = ∑ i f(i)(33) where f(i) is the Fock operator (16) for the ith electron f(i) = h(i) + ∑ j [2J j (i) – K j (i)]

Thus we have solved the following many-electron Schrödinger equation Ĥ 0 Ψ 0 = Є 0 Ψ 0 (34) where Є 0 is NOT our approximated Ē 0 (expectation energy) but it is a sum of the Hartree-Fock orbital energies Є 0 = ∑ i ε i (35)

In fact every single determinant build from a set of the Hartree-Fock orbitals (excited states) is an eigen function of Ĥ 0 and satisfies (34) i.e. Ĥ 0 Ψ n = Є n Ψ n with Є n = ∑ i ε i where the sum runs over i-orbitals included in the determinant Ψ n. This could be a basis for a perturbational treatment.

The perturbation theory can be applied here as follows. The equation to be solved is Ĥ el Φ n = E n Φ n Spilt the Hamiltonian into Ĥ el = Ĥ 0 + V where Ĥ 0 Ψ n = Є n Ψ n is known and V = Ĥ el - Ĥ 0

Write the perturbation expansion for the ground state energy and wave function E 0 = Є 0 (0) + Є 0 (1) + Є 0 (2) + … Φ 0 = Ψ 0 (0) + Ψ 0 (1) + Ψ 0 (2) + … Let us begin with the Hartree-Fock energy Ē 0 = ∫ Ψ 0 * Ĥ el Ψ 0 dv =∫Ψ 0 *( Ĥ 0 + V )Ψ 0 dv = ∫Ψ 0 * Ĥ 0 Ψ 0 dv + ∫Ψ 0 * V Ψ 0 dv = Є 0 (0) + Є 0 (1) The H-F energy is exact to the I-st order in perturbation theory.

Electron correlation The Hartree-Fock (and Hartree-Fock-Roothaan) method has a very significant weakness. It suffers from lack of so-called electron correlation. Let’s consider a two-electron system. According to the Hartree-Fock method, its state will be described by the Slater determinant with two spinorbitals Φ 1 and φ 2 φ 1 (1) φ 1 (2) Ψ(r 1, r 2 )= 1/√2 φ 2 (1) φ 2 (2)

These spinorbitals Φ 1 and φ 2,may have in general, different spatial parts (orbitals) ψ 1 and ψ 2. Let consider two cases with two electrons having parallel and opposite spin (1) Φ 1 =ψ 1 α and Φ 2 =ψ 2 α (S=1 → 2S+1=3 triplet) (2) Φ 1 =ψ 1 α and Φ 2 =ψ 2 β (S=0 → 2S+1=1 singlet) We will try to examine possibility of finding two electrons at the same place in space at once.

In the first case (the triplet state) the Slater determinant is: ψ 1 (r 1 )α(1) ψ 1 (r 2 )α(2) Ψ(r 1, r 2 )= 1/√2 ψ 2 (r 1 )α(1) ψ 2 (r 2 )α(2) which gives after expansion Ψ(r 1, r 2 )= 1/√2[ ψ 1 (r 1 )α(1)*ψ 2 (r 2 )α(2) - ψ 2 (r 1 )α(1)*ψ 1 (r 2 )α(2) ]

If we place now both electrons at the same position r 1 = r 2 = r then ψ 1 (r 1 )*ψ 2 (r 2 )= ψ 2 (r 1 )*ψ 1 (r 2 ) = ψ 1 (r)*ψ 2 (r) and the wave function vanishes Ψ(r 1, r 2 )= 1/√2* ψ 1 (r)*ψ 2 (r) [α(1)*α(2) - α(1)*α(2)]=ZERO Thus, as expected, two electrons CANNOT be at the same place at the same time (if they have parallel spins).

In the second case (the singlet state) the Slater determinant is: ψ 1 (r 1 )α(1) ψ 1 (r 2 )α(2) Ψ(r 1, r 2 )= 1/√2 ψ 2 (r 1 )β(1) ψ 2 (r 2 )β(2) which gives after expansion Ψ(r 1, r 2 )= 1/√2[ ψ 1 (r 1 )α(1)*ψ 2 (r 2 )β(2) - ψ 2 (r 1 )β (1)*ψ 1 (r 2 )α(2) ]

Now placing both electrons at the same position r 1 = r 2 = r gives as before ψ 1 (r 1 )*ψ 2 (r 2 )= ψ 2 (r 1 )*ψ 1 (r 2 )= ψ 1 (r)*ψ 2 (r) and the wave function becomes Ψ(r 1, r 2 )= 1/√2* ψ 1 (r)*ψ 2 (r) [α(1)*β(2) - β (1)*α(2)]= NON ZERO Thus, two electrons with opposite spins CAN be found at the same place at the same time.

This is physically impossible !!! Electrons will always try to avoid each other. The above considerations show that at the Hartree- Fock level electron motions are NOT properly correlated. This is so-called the electron correlation problem. It brings significant consequences. The fact that electrons are allowed to be too close to each other results in errors in essentially all quantities and properties calculated at the Hartree-Fock level.

This error in the energy is called correlation energy. It is defined as a difference between the exact (nonrelativistic) energy of a system (E 0 ) and the Hartree-Fock energy (E 0 HF ) E corr = E 0 – E 0 HF(41) where for the ground state E 0 HF = ∫ Ψ 0 * Ĥ el Ψ 0 dv = Ĥ el Φ 0 = E 0 Φ 0

Note that because the Hartree-Fock energy is an upper bound to the exact energy, the correlation energy (41) is negative. How big is the correlation energy E corr ? In general it is small, usually less than 1% of the total energy. So, may be it is not really important ?

Unfortunately, it is very important !!! It is so because we are usually interested not in absolute energies but rather in energy differences (e.g. between conformations, isomers, two states etc.) and such differences are in the same range of value as the correlation energy. Electron correlation effects are also very important for accurate results in calculated molecular properties. For instance, bond lengths calculated at the Hartree- Fock level are much too short.

Thus in general, electron correlation effects not only should be but have to be taken into account in accurate quantum chemical calculations. How can we include electron correlation effects? To answer this question we have to find out first why the electron correlation problem occurs. The problem with electron correlation at the Hartree- Fock level comes from one-electron approximation where electrons are treated almost independently.

Thus, to include electron correlation effects one has to go beyond one-electron approximation. For a closed-shell system within the framework of one-electron approximation a many-electron wave function is a single Slater determinant Ψ 0. Thus, if we approximate an exact wave function Φ 0 not by just one Slater determinant but by a linear combination of at least two Slater determinants then we are already beyond one-electron approximation.

Let’s consider the beryllium atom Be. Its ground state electron configuration is 1s 2 2s 2. Within one-electron approximation we need here four spinorbitals φ 1 =(1sα), φ 2 =(1sβ), φ 3 =(2sα), φ 4 =(2sβ)(42a) and we can write the ground state Slater determinant symbolically as Ψ 0 = det| φ 1 φ 2 φ 3 φ 4 |(42b)

But we can also use an excited state with, let say, 1s 2 2p 2 electron configuration. This would required the following spinorbitals φ 1 =(1sα), φ 2 =(1sβ), φ 5 =(2pα), φ 6 =(2pβ)(43a) and the corresponding Slater determinant for this excited state would be Ψ 1 = det| φ 1 φ 2 φ 5 φ 6 |(43b)

In the Hatree-Fock metod (one-electron approximation) the exact, ground state wave function Φ 0 is approximated by the single determinant Ψ 0 (42) Φ 0 = Ψ 0 (44) Now we can use also excited configuration Ψ 1 (43) to describe the ground state of beryllium Φ 0 = A 0 Ψ 0 + A 1 Ψ 1 (45)

The coefficients A 0 and A 1 can be found variationally. The ratio A 0 / A 1 is about 3.04 which means that the ground state of beryllium does not have simple 1s 2 2s 2 electron configuration but it contains also about 25% of the excited 1s 2 2p 2 configuration !!! The energies obtained with Φ 0 =Ψ 0 (Hartree-Fock) and Φ 0 =A 0 Ψ 0 +A 1 Ψ 1 (two configuration interaction) where -14.573 au and -14.617 au, respectively, while the experimental value is -14.667 au.

The above consideration concerning beryllium can be generalized. We can go beyond one-electron approximation using a number of Slater determinants. Thus, the exact wave function for the ground state and excited states of any N-electron system can be written as a linear combination of all possible N-electron Slater determinants form from the complete set of Hartree-Fock spinorbitals (infinitive):

Φ 0 = c 0 Ψ 0 + ∑ c a r Ψ a r + ∑ c ab rs Ψ ab rs (46) + ∑ c abc rst Ψ abc rst ……. Ψ 0 is the Hartree-Fock ground state Salter determinant and all others are described by reference to Ψ 0 :

Ψ a r is a determinant obtained from Ψ 0 by promoting one electron from the occupied (in Ψ 0 ) a-spinorbital to the virtual r-spinorbital (single excitation) Ψ ab rs is a determinant obtained from Ψ 0 by promoting two electrons from the occupied (in Ψ 0 ) a and b spinorbitals to the virtual spinorbitals r,s (double excitations) The whole (infinitive) set of all Slater determinants { Ψ 0, Ψ a r, Ψ ab rs, Ψ abc rst etc ….} forms a complete N-electron basis set for an exact wave function.

In practice we deal with a finite set of Slater determinants obtained using finite one-electron basis set (Roothaan). If we used K basis function then we generate K Hartree-Fock spatial orbitals and 2K spinorbitals for a system with N electrons. This gives us 2K N possible Slater determinants that in principle can be used to approximate an exact wave function. This can be a huge number !!!

Note that in our approach where the exact wave function is taken in the form of (46) Φ 0 = c 0 Ψ 0 + ∑ c a r Ψ a r + ∑ c ab rs Ψ ab rs + ∑ c abc rst Ψ abc rst ……. each Slater determinant corresponds to a given electron configuration. Therefore this metod is called Configuration Interaction (CI). This is conceptually the simplest metod to include electron correlation effects.

The basic idea in the CI metod is to form the N- electron Hamiltonian matrix in the N-electon basis set and then to diagonalize it. In other words, we represent the exact wave function as a linear combination of N-electron trial functions and use linear variational metod. The CI method is variational i.e. gives an energy which is an upper bound to the exact energy but it is size consistent only if all possible excitations are Included (full CI).

Electron correlation effects can be also taken account through the perturbational treatment. Such a method was developed already in 1934, however, its first practical implementation appeared in late sixtieth. This method is known as the Möller-Plesset perturbation theory and depending on the order it is called MP2, MP3 etc. It should be emphasized that, unlike CI, the MP n metod is size consistent at every order of pertubation.

Alike CI, the MP-n method uses the Hatree-Fock solutions. The correlation energy is calculated in terms of n-order corrections to the Hartree-Fock energy. The Möller-Plesset method is based on the ordinary Rayleigh-Schrödinger perturbation theory with a specific partitioning of the Hamiltonian.

As usually, we are interested in solving the electronic Schrödinger equation for n-electron system Ĥ el Φ n = E n Φ n (51) with Ĥ el = Σ i n h(i) + Σ i n > Σ j n 1/r ij (52) Since this equation cannot be solved exactly, we would like to apply a perturbation theory to find approximated solutions.

In order to use a perturbation theory, the total Hamiltonian has to be first split into two parts Ĥ el = Ĥ (0) + λ V(53) with a formal parameter λ, which will be later set to 1. The partitioning of the total Hamiltonian has to satisfy two conditions (1)the solutions of Ĥ (0) Ψ n (0) = Є n (0) Ψ n (0) have to be known (2)the perturbation λ V has to be small

Do we have any n-electron Hamiltonian that could be use as Ĥ (0) ? Yes, we do. It is the Hartree-Fock n-electron Hamiltonian (33), a sum of one-electron Fock operators Ĥ (0) = ∑ i f(i)(33) where f(i) is the Fock operator (16) for the ith electron f(i) = h(i) + ∑ j [2J j (i) – K j (i)] = h(i) + v HF (i)

Solving the Hartree-Fock equations one gets a set of Hartree-Fock orbitals f(1) ψ i (1) = ε i ψ i (1) which multiplied by alpha and/or beta spin functions generate a set of spinorbitals {φ i }. For n-electron systemany set of n such spinorbitals forms a Slater determinant that is an eigenfunction of Ĥ (0) Ĥ (0) Ψ n (0) = Є n (0) Ψ n (0) with Є n (0) = ∑ i ε i

Thus we have satisfied one condition with Ĥ (0). What about the second condition concerning small perturbation V ? The only perturbation V we can use here is the difference between the exact electronic Hamiltonian and our chosen Ĥ (0) V = Ĥ el – Ĥ (0) Is it small enough? Yes, it is V = ∑h(i) + ∑∑1/r ij - ∑f(i) = ∑∑1/r ij - ∑v HF (i)

Now we expend the exact eigenfunctions and eigen- values in a Taylor series in λ Φ n = Ψ n (0) + λ Ψ n (1) + λ 2 Ψ n (2) + λ 3 Ψ n (3) + … (54) E n = Є n (0) + λ Є n (1) + λ 2 Є n (2) + λ 3 Є n (3) + … and evaluate n-order corrections to the energy and wave function. These correctionse can be obtained from a set of perturbational equations. Substituting expansions (54) and (51) into the Schrödinger equation (51) and collecting terms with a given pawer of λ we get

Є n (1) = Є n (2) = (55) Є n (3) = and so on. The first- and the second-order corrections to the wave function are the solutions of the first- and second-order pertubational equations ( Ĥ (0) - Є n (0) )Ψ n (1) = - ( V - Є n (1) ) Ψ n (0) (56a) ( Ĥ (0) - Є n (0) )Ψ n (2) = - ( V - Є n (1) ) Ψ n (1) + Є n (2) Ψ n (0) (56b) The first-order wave function can be eliminated from the expression for Є n (3).

Multiplying (56a) and (56b) respectively by Ψ n (2) and Ψ n (1) from left and integrating one gets = and Є n (3) = (57) Thus, knowledge of Ψ n (0) and Ψ n (1) is sufficient to calculate energy corrections up to third-order.

The first-order wave function can be expressed as a linear combination of the zero-order solutions (Slater determinants) Ψ n (1) = ∑ i c in Ψ i (0) (58) From the first-order equation (56a) the coefficients are given as c in = V in /(Є n (0) - Є i (0) ) (59) with V in =

The final expressions for the energy corrections can be written as Є n (1) = Є n (2) = ∑ i V ni V in /(Є n (0) - Є i (0) )(60) Є n (3) = ∑ ij V ni V ij V jn /[(Є n (0) - Є i (0) )(Є n (0) - Є j (0) )] - Є n (1) ∑ i V ni V in /(Є n (0) - Є i (0) ) 2 Remember that the Hartree-Fock energy is the expectation value of the total Hamiltonian and

Ē n HF = ∫ Ψ n (0)* Ĥ el Ψ n (0) dv =∫Ψ n (0)*( Ĥ (0) + V )Ψ n (0) dv = ∫Ψ n (0)* Ĥ (0) Ψ n (0) dv + ∫Ψ n (0)* V Ψ n (0) dv = Є n (0) + Є n (1) it is correct to the first order. Thus the first correction to the Hartree-Fock energy is the second-order correction Є n (2). The only thing we need is to evaluate matrix elements V in = with different Slater determinants What Slater determinants will we use?

For the ground state n=0 we take of course the Hartree-Fock Ψ n (0) = Ψ 0 (0). But waht about Ψ i (0) ? It can be shown that non-zero V i0 = elements exist only with Ψ i (0) being double excitations with respect to the Hartree-Fock Ψ 0 (0). Ψ i (0) =Ψ ab rs where a,b denote occupied (in Ψ 0 (0) ) and r,s virtual spinorbitals.

The final expression for the second-order correlation energy in terms of spinorbitals becomes Є n (2) =∑ a | 2 /( є a + є b - є r + є s )(61) Where two-electron integrals over spinorbitals are = - = ∫∫ a*(1)b*(2)1/r 12 r(1)s(2) dv1dv2 - ∫∫a*(1)b*(2)1/r 12 s(1)r(2) dv1dv2 The Moller-Plesset MP2 perturbation theory is wldely used in electron correlation calculations

Theoretical Chemistry Quantum Chemistry. Content 1.Introduction 2.Many-electron systems 3.Two basic approximations 4.Hartree-Fock metod 5.Roothaan approximation.

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Theoretical Chemistry Quantum Chemistry. Content 1.Introduction 2.Many-electron systems 3.Two basic approximations 4.Hartree-Fock metod 5.Roothaan approximation.

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Presentation on theme: "Theoretical Chemistry Quantum Chemistry. Content 1.Introduction 2.Many-electron systems 3.Two basic approximations 4.Hartree-Fock metod 5.Roothaan approximation."— Presentation transcript:

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