CHE Inorganic, Physical & Solid State Chemistry Advanced Quantum Chemistry: lecture 2 Rob Jackson LJ1.16,

che Advanced QC lecture 2 2 Lecture 2 contents 1.Trial wavefunctions and expectation values 2.The Variation Principle 3.Recap. on Molecular Orbitals: the hydrogen molecule ion 4.Molecular Orbitals for diatomic molecules 5.Secular Determinants

Trial wavefunctions and expectation values – (i) In quantum chemistry, one of the main problems faced is determining the wavefunction of an atom or molecule. If the ground state energy of a molecule is E 0, we expect the following to apply (if we know the wavefunction): H  0 = E 0  0 –Remember that this means that H acts on the wavefunction of the ground state to give the energy of the ground state. che Advanced QC lecture 2 3

Trial wavefunctions and expectation values – (ii) Normally we can’t solve the Schrödinger equation analytically, so what do we do? If we choose a ‘trial wavefunction’, , (a guess at what the wavefunction should be), we can define the energy associated with the wavefunction as:  E  =    H  d  /     d  This is called the Expectation Value of the energy. –notation: * means ‘complex conjugate’, and  means ‘all space’ che Advanced QC lecture 2 4

The Variation Principle If we use a trial wavefunction, , and calculate the expectation value of the energy, how can we judge our choice of wavefunction? The Variation Principle states that: –For any trial wavefunction, the expectation value of the energy can never be less than the true ground state energy. This is expressed as E 0     H  d  /     d  –The expectation value is only E 0 if the trial wavefunction is equal to the true one. che Advanced QC lecture 2 5

Why is the Variation Principle useful? It enables different trial wavefunctions to be tested and evaluated, since the wavefunction giving the lowest energy expectation value should be closest to the true wavefunction. For the hydrogen atom we know the ground state energy so the variation principle can be tested for different wavefunctions. A reference to an example calculation will be given and discussed. che Advanced QC lecture 2 6

Molecular orbitals for H (i) For H 2 + we have 2 1s atomic orbitals, each centred on a H atom. We can form a trial wavefunction by combining the 1s orbitals  A and  B. –Remember molecular orbital diagrams here! This can be written as   = N  (  A   B ) –Here, N  is a normalising constant. The expectation value for the energy will be:  E  = (N  ) 2  (  A   B ) H (  A   B ) d  che Advanced QC lecture 2 7

Molecular orbitals for H (ii) The 2 molecular orbitals  +,  - are the bonding and antibonding orbitals in the H 2 + molecule. In Hayward’s book there is a more detailed derivation, using expressions for the 1s orbitals  A and  B (pp ) The shape of the wavefunction and the potential energy diagram will be shown and discussed in the lecture (see also Hayward figures 8.2, 8.3, pp ). che Advanced QC lecture 2 8

Illustration of formation of H 2 + orbitals ch09_s03 che Advanced QC lecture 2 9

Molecular orbitals for diatomic molecules – (i) We can follow the same procedure for diatomic molecules as demonstrated for H 2 + : If the molecule is AB (i.e. different nuclei) we can suggest a trial wavefunction based on orbitals on atoms A and B:  = c A  A + c B  B (where c A, c B are weighting coefficients) The expectation value for the energy of this trial wavefunction will be (see definition on slide 5) :  E  =    H  d  /     d  che Advanced QC lecture 2 10

Molecular orbitals for diatomic molecules – (ii) We will assume that the wavefunctions are not ‘complex’, so  * =  –Substituting for the wavefunction gives:  E  =  (c A  A + c B  B ) H (c A  A + c B  B ) d  /  (c A  A + c B  B ) 2 d  –On p. 150 of Hayward’s book, the expression is expanded to give:  E  = (c 2 A H AA + 2 c A c B H AB + c 2 B H BB )/ (c 2 A S AA + 2 c A c B S AB + c 2 B S BB ) –Here H AA, H AB, H BB, S AA, S AB and S BB are all integrals which are defined in the book and the next slide: che Advanced QC lecture 2 11

Overlap, Coulomb and Resonance integrals The terms in the 2nd equation, slide 10 are: S AA =   A  A d , S BB =   B  B d  ( =1 for normalised orbitals) S AB =   A  B d  = S BA =   B  A d  (Overlap integrals) –The value depends on the amount of orbital overlap. H AA =   A H  A d , H BB =   B H  B d  –Coulomb integral, the energy an electron would have in a particular orbital. H AB =   A H  B d  = H BA =   B H  A d  –Resonance integrals, = 0 if there is no overlap. che Advanced QC lecture 2 12

So how do we get the energy? We need to evaluate the integrals, and get the values of c A and c B that give the lowest energy (Variation Principle). The general approach to this will be explained (not restricted to diatomics), and applied to specific cases. We start by writing the trial wavefunction as a linear combination of atomic orbitals:  i =  k c ik  k So a trial wavefunction for a triatomic molecule would look like:  i = c i1  1 + c i2  2 + c i3  3 che Advanced QC lecture 2 13

Obtaining the secular determinant – (i) Writing down the Schrödinger equation gives: H  k c ik  k =  i  k c ik  k Rearranging (moving H inside the summation sign):  k c ik H  k =  i  k c ik  k Multiplying both sides by  l :  k c ik  l H  k =  i  k c ik  l  k Finally integrate over all electron positions:  k c ik   l H  k d  =  i  k c ik   l  k d  The integrals in bold we have already seen on slide 11. che Advanced QC lecture 2 14

Obtaining the secular determinant – (ii) We define: H lk =   l H  k d  and S lk =   l  k d  The expression at the bottom of slide 13 becomes:  k c ik H lk =  i  k c ik S lk Rearranging gives:  k c ik (H lk -  i S lk ) = 0 Using a mathematical rule, the above expression implies that: det  H lk -  S lk  = 0 (‘det’ means determinant) –This will be explained for a diatomic molecule che Advanced QC lecture 2 15

The secular determinant for a diatomic molecule (i) We expand our trial wavefunction (molecular orbital) in terms of 2 atomic orbitals:  i = c i1  1 + c i2  2 If there are 2 AOs there will be 2 MOs:  1 = c 11  1 + c 12  2  2 = c 21  1 + c 22  2 Following slide 14 this leads to 2 equations: c 11 (H 11 -  1 S 11 ) + c 12 (H 12 -  1 S 12 ) = 0 c 21 (H 21 -  2 S 21 ) + c 22 (H 22 -  2 S 22 ) = 0 che Advanced QC lecture 2 16

The secular determinant for a diatomic molecule (ii) The energy  for each orbital i is obtained from:  H 11 -  S 11 H 12 -  S 12  = 0  H 21 -  S 21 H 22 -  S 22  This is the secular determinant for a diatomic molecule. This can be expanded to get a quadratic equation which can be solved to give , provided we know H lk and S lk. This method can be used to get the secular determinant for any molecule (which can be solved by a computer !) Energies of orbitals can be obtained in this way. che Advanced QC lecture 2 17

Lecture summary The Expectation Value of a property has been introduced and defined. The Variation Principle has been introduced and explained. Molecular Orbitals for H 2 + and diatomic molecules in general have been calculated. Coulomb, Overlap and Resonance integrals have been defined Secular Determinants have been introduced. che Advanced QC lecture 2 18