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Theoretical Chemistry
Quantum Chemistry
Many-electron systems
Two basic approximations
Hartree-Fock metod
Roothaan approximation and basis sets
Electron correlation
Post Hartree-Fock methods
Geometry optimization
Simulation of the IR, NMR and AFM experiments
• 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:
 
 H
i t
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
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)
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)
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)
From hereafter we will consider only stationary
The Schrödinger equation (5) for a stationary state
has, in general, an infinitive number of solutions
and should be rewritten as
Ĥ(q) Ψn(q) = En Ψn(q)
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 = Te + Tn +Ve-n + Ve-e + Vn-n
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
Te = Σin (-ħ/2m Δi ) and Tn = ΣAN (-ħ/2MA ΔA ) (8a)
Ve-n = Σin ΣAN [ - ZA e2/ |ri – RA|]
Ve-e = Σin Σjn [ e2/ | ri – rj| ]
Vn-n = ΣAN ΣBN [ZA ZB e2/ |R
| A – RB| ]
here ri=(xi , yi , zi ) and RA=(XA , YA , ZA ) denote
coordinates of i-electron and A-nucleus with
masses m, MA, respectively, and Δi , ΔA are their
Laplace operators.
Now we have to solve the Schrödinger equation (6)
with this Hamiltonian
Ĥ(r,R) Ψn(r,R) = En Ψn(r,R )
where r=(r1, r2, …, rn) and R=(R1, R2,…, RN).
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/ |ri – RA|, 1/ |ri – rj|, 1/ |RA – RB|
These terms cannot be rewritten as a sum of
separate terms and therefore the Schrödinger
equation (9) for a molecule cannot be solved
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 (C6H6) 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 nonmoving nuclei. This assumption is related to socalled Born-Oppenheimer approximation.
What do we gain from this approximation ?
First of all the positions R=(R1, R2,…, RN) of nonmoving 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=(r1, r2,…,rn) being variables
(for benzene the dimension is reduced from 162 to
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 = Te + Tn +Ve-n + Ve-e + Vn-n
Such simplified operator is called the electronic
Hamiltonian and it is given explicitly as follows
Ĥel = Σin (-ħ/2m Δi ) + Σin ΣAN [ - ZA e2/(ri – RA) ]
+ Σin>Σjn [ e2/(ri – rj) ]
= Σin h(i) + Σin>Σjn [ e2/(ri – rj) ]
with h(i) = (-ħ/2m Δi ) + ΣAN [ - ZA e2/(ri – RA) ]
The Schrödinger equation with this energy operator
Ĥel (r;R) Ψn(r;R) = En(R) Ψn(r;R )
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)
Σin>Σjn [ e2/ |ri – rj|]
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
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 {φ1(r), φ2 (r), …,φ2n(r)} made out
of n obitals {ψ1 (r), ψ2 (r), …,ψn (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
φ1(1) φ1(2) …. φ1(2n)
φ2(1) φ2(2) …. φ2(2n)
Ψ(r1, r2 , …, r2n)= N …………………….
φk(1) φk(2) …. φk(2n)
φ2n(1) φ2n(2) …. φ2n(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
Eave = ∫ Ψ*(r1, r2 , …, r2n) Ĥel Ψ(r1, r2 , …, r2n) dv
Substituting the determinantal wavefunction (12)
and the electronic Hamiltonian (10) into the
expectation energy expression one gets the
following explicit form of Eave
Eave= 2∑i hii + ∑i∑ij [ 2Jij – Kij ]
where summation runs over spatial orbitals {ψi}
and hii , Jij , Kij denote one- and two-electron
hii =∫ ψi*(1) h(1) ψi(1) dv1
Jij =∫∫ ψi*(1) ψi(1) 1/r12 ψj*(2) ψj(2) dv1dv2
Kij =∫∫ ψi*(1) ψj(1) 1/r12 ψj*(2) ψi(2) dv1dv2
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 {ψi}.
Minimizing energy (13) with respect to these orbitals
leads to the following set of one-electron equations
f(1) ψi(1) = εi ψi(1)
for i=1,n
with the orthonormal constrains
∫ ψi*(1) ψj(1) dv1= δij
where so-called the Fock operator is given as
f(1)= h(1) + ∑j[2Jj(1) – Kj(1)]
h(1) = (-ħ/2m Δ1 ) + ΣAN [ - ZA e2/(r1 – RA) ]
Jj (1) ψi(1) = [∫1/r12 ψj*(2) ψj(2) dv2] ψi(1)
Kj (1) ψi(1) = [∫1/r12 ψj*(2) ψi(2) dv2] ψj(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 {ψi} (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
The second part of the Fock operator (16)
vHF(1) = ∑j[2Jj(1) – Kj(1)]
often called the Hartree-Fock potential represents
an average potential experienced by one electron
due to the presence of other electrons.
vHF 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)
The basic idea of the SCF method is simple. By
making an initial guess at the orbitals {ψi} one can
calculate the average filed (i.e. vHF ) 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 vHF 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 {ψi} with orbital energies {εi}.
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
Using the one-electron approximation with a
single determinant trail wave function within the
variational method we have replaced the manyelectron Schrödinger equation with the electronic
Hamiltonian (10) by a set of one-electron HartreeFock equations (14, 15)
f(1) ψi(1) = εi ψi(1)
∫ ψi*(1) ψj(1) dv1= δij
for i=1,n
At the first look it seems to be much easier now to
solve these HF equations than the original manyelectron 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[2Jj(1) – Kj(1)]
h(1) = (-ħ/2m Δ1 ) + ΣAN [ - ZA e2/(r1 – RA) ]
Jj (1) ψi(1) = [∫1/r12 ψj*(2) ψj(2) dv2] ψi(1)
Kj (1) ψi(1) = [∫1/r12 ψj*(2) ψi(2) dv2] ψj(1)
The Hartree-Fock equations CANNOT be solved for
An additional approximation is necessary.
The Roothaan analytical approximation to the
Hartree-Fock orbitals:
ψi(1) = ∑μ cμi λμ(1)
where μ=1,m numbers a set of known one-electron
basis functions {λμ}. 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 {cμi}.
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
with the orthonormal conditions
CTS C = 1
Here F and S denote the Fock and the overlap
matrix with the matrix elements
Fμν= ∫λμ(1) f(1) λv(1) dv1 = (µ| f |ν)
Sμν= ∫λμ(1) λv(1) dv1
= (µ|ν)
Substituting (18) into f(1) gives finally
Fμν= hμν + ∑λσ D λσ[(μν| λσ) – 1/2(μσ| λν)]
where the density matrix elements are given as
D λσ = 2 ∑i cλi cσi
and two-electron integrals over basis functions
(μν| λσ)=∫∫ μ(1)ν(1)1/r12 λ(2)σ(2) dv1 dv2
The electronic energy (13) is now given as
Eave=1/2∑µν Dµν(hµν + Fµν)
The total molecular energy, a sum of electronic and
nuclear-nuclear repulsion energies
Etot = Eave + ∑A∑B ZA ZB/RAB
is a very important quantity, particularly in structure
determination. It is so because the predicted
equilibrium molecular geometry occurs at the
minimum of Etot (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)
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
<O1> = < Ψ(r1, r2 , …, r2n) | O1 | Ψ(r1, r2 , …, r2n) >
In a given basis set this expression will have always
the same general form
<O1>= ∑i (ψi(1)|o(1)| ψi(1)) = ∑µν Dµν (µ | O(1)| ν )
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 qi at the positions ri is
μ = ∑i qi ri
For a molecule with n electrons and N nuclei it
μ = - ∑in e ri + ∑AN ZAe RA
The quantum mechanical operator representing (24)
has the same form (because it contains coordinates
only). Its expectation value is given as
<μ> = < Ψ |- ∑in e ri + ∑AN ZAe RA |Ψ >
(Ψ denotes a Slater determinant). In a given basis
set according to (23) it becomes (with e=1)
<μ> = -∑µ ∑ν Dv (μ|r|ν) + ∑AN ZA RA (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
<μx> = -∑µ ∑ν Dv (μ|x|ν) + ∑AN ZA XA
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 ψi(1) then the probability
of finding that electron in a volume dr at a point r is
| ψi(r)|2dr
In a closed-shell molecule described by a single
determinant wave function with n double occupied
orbitals {ψi(1)} the total charge density
ρ(r) = 2 ∑in | ψi(r)|2
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.
∫| ψi(r)|2dr = 2∑in ∫ | ψi(r)|2dr=2∑in 1= 2n
In a given basis set {λμ} where the Hartree-Fock
orbitals are approximated by (18)
ψi(r) = ∑μ cμi λμ(r)
the total charge density (27) is expressed as follows
ρ(r) = 2 ∑in | ψi(r)|2 = ∑μ ∑ν Dμ ν λμ λv
Integration of the latter equation together with (28)
gives a nice relation
2n = ∑μ ∑ν Dμν ∫ λμ(r)λv (r) dr = ∑μ ∑ν Dμν Sμν
Note that both matrices above are symmetric so
Dμν = Dνμ and Sμν = Sνμ and therefore we can write
2n = ∑μ ∑ν Dμν Sνμ = ∑μ (DS)μμ
The above equation may be written as
2n = Tr (DS)
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
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 λμ(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 qAel
qAel = ∑μ є A (DS)μμ
From this we can get the net charge qA for a given
atom A (atomic number ZA) in a molecule
qA = ZA - ∑μ є A (DS)μμ
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 Sa S1-a) = Tr (S1-a D Sa)
Using above relations, equation (29) can be formally
written as
2n = Tr (DS) = Tr (D Sa S1-a) = Tr (S1-a D Sa)
which is true for any value of a, regardless of the
meaning of Sa and S1-a . These matrices can be
easily calculated for a=1/2. In such a case we have
2n = Tr (S1/2 D S1/2 )
The electronic charge on an atom A resulting from
the latter is
qAel = ∑μ є A (S1/2 D S1/2 )μμ
which is different than that obtained before according
to (30)
qAel = ∑μ є A (DS)μμ
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
Remember that we did not solve the exact electronic
Schrödinger equation (11) for a ground state
Ĥel Φ0 = E0 Φ0
but we have rather found an approximation Ψ0 to Φ0
using the variational principle i.e. minimizing an
expectation energy Ē0
Ē0 = ∫ Ψ0*Ĥel Ψ0 dv
Now the question:
is the approximated Ψ0 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 Ψ0 , a Slater determinant,
is an eigenfunction of the following operator
Ĥ0 = ∑i f(i)
where f(i) is the Fock operator (16) for the ith electron
f(i) = h(i) + ∑j[2Jj(i) – Kj(i)]
Thus we have solved the following many-electron
Schrödinger equation
Ĥ0 Ψ0 = Є0 Ψ0
where Є0 is NOT our approximated Ē0 (expectation
energy) but it is a sum of the Hartree-Fock orbital
Є0 = ∑i εi
In fact every single determinant build from a set of the
Hartree-Fock orbitals (excited states) is an eigen
function of Ĥ0 and satisfies (34) i.e.
Ĥ0 Ψn = Єn Ψn
with Єn = ∑i εi where the sum runs over i-orbitals
included in the determinant Ψn . This could be a basis
for a perturbational treatment.
The perturbation theory can be applied here as
follows. The equation to be solved is
Ĥel Φn = En Φn
Spilt the Hamiltonian into
Ĥel = Ĥ0 + V
Ĥ0 Ψn = Єn Ψn
is known
V = Ĥel - Ĥ0
Write the perturbation expansion for the ground state
energy and wave function
E0 = Є0(0) + Є0(1) + Є0(2) + …
Φ0 = Ψ0(0) + Ψ0(1) + Ψ0(2) + …
Let us begin with the Hartree-Fock energy
Ē0 = ∫ Ψ0*Ĥel Ψ0 dv =∫Ψ0*(Ĥ0 + V )Ψ0 dv
= ∫Ψ0*Ĥ0 Ψ0 dv + ∫Ψ0*V Ψ0 dv
= Є0(0) + Є0(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
Φ1 and φ2
φ1(1) φ1(2)
Ψ(r1, r2 )= 1/√2
φ2(1) φ2(2)
These spinorbitals Φ1 and φ2,may have in general,
different spatial parts (orbitals) ψ1 and ψ2.
Let consider two cases with two electrons having
parallel and opposite spin
(1) Φ1=ψ1α and Φ2=ψ2α (S=1 → 2S+1=3 triplet)
(2) Φ1=ψ1α and Φ2=ψ2β (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(r1, r2 )= 1/√2
which gives after expansion
Ψ(r1, r2 )= 1/√2[ ψ1(r1)α(1)*ψ2(r2)α(2) - ψ2(r1)α(1)*ψ1(r2)α(2) ]
If we place now both electrons at the same position
r 1= r 2= r
ψ1(r1)*ψ2(r2)= ψ2(r1)*ψ1(r2) = ψ1(r)*ψ2(r)
and the wave function vanishes
Ψ(r1, r2 )= 1/√2* ψ1(r)*ψ2(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
In the second case (the singlet state) the Slater
determinant is:
Ψ(r1, r2 )= 1/√2
which gives after expansion
Ψ(r1, r2 )= 1/√2[ ψ1(r1)α(1)*ψ2(r2)β(2) - ψ2(r1)β (1)*ψ1(r2)α(2) ]
Now placing both electrons at the same position
r 1= r2= r
gives as before
ψ1(r1)*ψ2(r2)= ψ2(r1)*ψ1(r2)= ψ1(r)*ψ2(r)
and the wave function becomes
Ψ(r1, r2 )= 1/√2* ψ1(r)*ψ2(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 HartreeFock 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 (E0) and the
Hartree-Fock energy (E0HF)
Ecorr = E0 – E0HF
where for the ground state
E0HF = ∫ Ψ0* Ĥel Ψ0 dv = < Ψ0 |Ĥel |Ψ0 >
Ĥel Φ0 = E0 Φ0
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 Ecorr ?
In general it is small, usually less than 1% of the total
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 HartreeFock 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 HartreeFock 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 Ψ0 . Thus, if we
approximate an exact wave function Φ0 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 1s22s2. Within one-electron
approximation we need here four spinorbitals
φ1=(1sα) , φ2=(1sβ) , φ3=(2sα) , φ4=(2sβ)
and we can write the ground state Slater determinant
symbolically as
Ψ0 = det| φ1 φ2 φ3 φ4|
But we can also use an excited state with, let say,
1s22p2 electron configuration. This would required
the following spinorbitals
φ1=(1sα) , φ2=(1sβ) , φ5=(2pα) , φ6=(2pβ)
and the corresponding Slater determinant for this
excited state would be
Ψ1 = det| φ1 φ2 φ5 φ6|
In the Hatree-Fock metod (one-electron
approximation) the exact, ground state wave function
Φ0 is approximated by the single determinant Ψ0(42)
Φ0 = Ψ 0
Now we can use also excited configuration Ψ1 (43) to
describe the ground state of beryllium
Φ0 = A0 Ψ0 + A1 Ψ1
The coefficients A0 and A1 can be found variationally.
The ratio A0 / A1 is about 3.04 which means that
the ground state of beryllium does not have simple
1s22s2 electron configuration but it contains also
about 25% of the excited 1s22p2 configuration !!!
The energies obtained with Φ0=Ψ0 (Hartree-Fock) and
Φ0=A0Ψ0+A1Ψ1 (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):
Φ0 = c0 Ψ0 + ∑ car Ψar
+ ∑ cabrs Ψabrs
+ ∑ cabcrst Ψabcrst …….
Ψ0 is the Hartree-Fock ground state Salter
determinant and all others are described by
reference to Ψ0 :
Ψar is a determinant obtained from Ψ0 by promoting
one electron from the occupied (in Ψ0) a-spinorbital to
the virtual r-spinorbital (single excitation)
Ψabrs is a determinant obtained from Ψ0 by promoting
two electrons from the occupied (in Ψ0) a and b
spinorbitals to the virtual spinorbitals r,s (double
The whole (infinitive) set of all Slater determinants
{ Ψ0 , Ψar , Ψabrs , Ψabcrst 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
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)
Φ0 = c0 Ψ0 + ∑ car Ψar
+ ∑ cabrs Ψabrs
+ ∑ cabcrst Ψabcrst …….
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 Nelectron 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
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
Ĥel Φn = En Φn
Ĥel = Σin h(i) + Σin>Σjn 1/rij
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
Ĥel = Ĥ(0) + λ V
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) Ψn(0) = Єn(0) Ψn(0) have to be
(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)
where f(i) is the Fock operator (16) for the ith electron
f(i) = h(i) + ∑j[2Jj(i) – Kj(i)] = h(i) + vHF(i)
Solving the Hartree-Fock equations one gets a set of
Hartree-Fock orbitals
f(1) ψi(1) = εi ψi(1)
which multiplied by alpha and/or beta spin functions
generate a set of spinorbitals {φi }. For n-electron
systemany set of n such spinorbitals forms a Slater
determinant that is an eigenfunction of Ĥ(0)
Ĥ(0) Ψn(0) = Єn(0) Ψn(0) with Єn(0) = ∑i εi
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 = Ĥel – Ĥ(0)
Is it small enough? Yes, it is
V = ∑h(i) + ∑∑1/rij - ∑f(i) = ∑∑1/rij - ∑vHF(i)
Now we expend the exact eigenfunctions and eigenvalues in a Taylor series in λ
Φn = Ψn(0) + λ Ψn(1) + λ2 Ψn(2) + λ3 Ψn(3) + … (54)
En = Єn(0) + λ Єn(1) + λ2 Єn(2) + λ3 Єn(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
Єn(1) = < Ψn(0)| V | Ψn(0) >
Єn(2) = < Ψn(0)| V | Ψn(1) >
Єn(3) = < Ψn(0)| V | Ψn(2) >
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) - Єn(0) )Ψn(1) = - ( V - Єn(1) ) Ψn(0)
(Ĥ(0) - Єn(0) )Ψn(2) = - ( V - Єn(1) ) Ψn(1) + Єn(2) Ψn(0) (56b)
The first-order wave function can be eliminated from
the expression for Єn(3) .
Multiplying (56a) and (56b) respectively by Ψn(2) and
Ψn(1) from left and integrating one gets
<Ψn(2)|V - Єn(1)|Ψn(0)>= <Ψn(1)|V - Єn(1)|Ψn(1)>
Єn(3) = <Ψn(1)|V - Єn(1)|Ψn(1)>
Thus, knowledge of Ψn(0) and Ψn(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
Ψn(1) = ∑i cin Ψi(0)
From the first-order equation (56a) the coefficients are
given as
cin = Vin /(Єn(0) - Єi(0))
with Vin= < Ψi(0) |V|Ψn(0) >
The final expressions for the energy corrections can
be written as
Єn(1) = < Ψn(0)| V | Ψn(0) >
Єn(2) = ∑i Vni Vin /(Єn(0) - Єi(0))
Єn(3) = ∑ijVni Vij Vjn /[(Єn(0) - Єi(0))(Єn(0) - Єj(0))]
- Єn(1) ∑i Vni Vin /(Єn(0) - Єi(0))2
Remember that the Hartree-Fock energy is the
expectation value of the total Hamiltonian and
ĒnHF = ∫ Ψn(0)*Ĥel Ψn(0) dv =∫Ψn(0)*(Ĥ(0) + V )Ψn(0) dv
= ∫Ψn(0)*Ĥ(0)Ψn(0)dv + ∫Ψn(0)*V Ψn(0) dv
= Єn(0) + Єn(1)
it is correct to the first order. Thus the first correction
to the Hartree-Fock energy is the second-order
correction Єn(2) .
The only thing we need is to evaluate matrix elements
Vin= < Ψi(0) |V|Ψn(0) > with different Slater determinants
What Slater determinants will we use?
For the ground state n=0 we take of course the
Hartree-Fock Ψn(0)= Ψ0(0). But waht about Ψi(0) ?
It can be shown that non-zero Vi0= < Ψi(0) |V|Ψ0(0)>
elements exist only with Ψi(0) being double excitations
with respect to the Hartree-Fock Ψ0(0).
Ψi(0) =Ψabrs
where a,b denote occupied (in Ψ0(0)) and r,s virtual
The final expression for the second-order correlation
energy in terms of spinorbitals becomes
Єn(2)=∑a<b ∑r<s |<ab||rs>|2 /( єa + єb - єr + єs )
Where two-electron integrals over spinorbitals are
<ab||rs> = <ab|rs> - <ab|sr>
= ∫∫ a*(1)b*(2)1/r12 r(1)s(2) dv1dv2
- ∫∫a*(1)b*(2)1/r12 s(1)r(2) dv1dv2
The Moller-Plesset MP2 perturbation theory is wldely
used in electron correlation calculations