Understanding Molecular Orbital Theory Numerical Aspects

Lecture 26 Molecular orbital theory II

Numerical aspects of MO theory We learn and carry out a mathematical procedure to determine the MO coefficients, which is based on variational theorem, Lagrange s undetermined multiplier method, and matrix eigenvalue equation. These are mathematical concepts of fundamental importance and their use and benefit go far beyond quantum chemistry.

What is the best wave function? We have not discussed how to determine the expansion coefficients of LCAO MO s (except when they are determined by symmetry). Y = cAA+cBB To find the best LCAO MO s with optimized coefficients, we must have a mathematical criterion by which to identify the best approximate wave function for a given Hamiltonian.

Variational theorem A general mathematical technique applicable to many problems of chemistry, physics, and mathematics (an important tool for many- body problems complementary to perturbation theory). We learn its application to quantum mechanics. Question: we have N normalized wave functions that may or may not be the eigenfunctions of the Hamiltonian H. What is the best wave function for the ground state?

Variational theorem Answer: evaluate the expectation values of energy: EX= YX A * HYXdt B and pick the one with the lowest expectation value. That is the best wave function. Why? C D

Variational theorem Any wave function can be expanded (exactly) as a linear combination of eigenfunctions of H (completeness). YX= c0F0+c1F1+c2F2+ True ground state WF Ground state energy Orthonormality 2E0+ c1 2E1+ c2 2E2+ * HYXdt EX= YX = c0 2E0+ c1 2E0+ c2 2E0+ = E0 Normalization c0

Variational theorem The closer the wave function is to the true ground-state one (which is measured by how close c0 is to unity), the lower the energy gets. Equality holds when and only when the wave function is the true wave function. EX= YX 2E0+ c1 2E1+ c2 2E2+ E0 * HYXdt = c0 We can vary (hence the name variational theorem) a wave function s shape (while maintaining normalization) to minimize its energy expectation value to systematically improve the approximation.

Variational determination of MOs The MO is a linear combination of two AO s: Y = cAA+cBB We find coefficients that minimize the energy (assuming real orbitals and coefficients), E = Y* HYdt +2cAcBA HBdt A HAdt B HBdt = cA +cB 2 2 Under the normalization condition. 2dt 2dt +2cAcBABdt 1= Y*Ydt = cA +cB 2 2 A B

Minimization At a minimum of a function, its first derivative (gradient) must be zero (a necessary but not a sufficient condition). +2cAcBA HBdt A HAdt B HBdt Minimize E = cA +cB 2 2 E c E c = = 0; 0 A B

Constrained minimization Question: How can we incorporate this constraint during minimization (without it, E is not bounded from below and a minimum does not exist)? 2dt 2dt +2cAcBABdt 1= Y*Ydt = cA +cB 2 2 A B Answer: Lagrange s undetermined multiplier method.

Lagranges undetermined multiplier method We need to minimize E by varying cA and cB, E c E c = = 0; 0 A B With the constraint that the wave function remains normalized, 1= Y*Ydt 2dt 2dt +2cAcBABdt = cA +cB 2 2 A B Step 1: We write the constraint into a ... = 0 form: Y*Ydt -1= 0

Lagranges undetermined multiplier method Step 2: We define a new quantity L(Lagrangian) to minimize, where we also introduce an additional parameter (undetermined multiplier) L= E-l ( ) Y*Ydt -1 Step 3: Minimize L by varying cA and cB as well as with no constraint: L cA cB = 0; L = 0; L l= 0

Lagranges undetermined multiplier method L cA = 0; L = 0; L l= 0 cB { } ( ) E-l Y*Ydt l -1 =1- Y*Ydt = 0 { } { } ( ) ( ) E-l Y*Ydt cB -1 E-l Y*Ydt cA -1 The 3rd equation reduces to the constraint, which we must satisfy = 0 = 0 Once the 3rd equation (constraint) is satisfied, the 1st and 2nd equations reduce to minimization of E by varying cA and cB. Minimization of E wrt two parameters with one constraint is equivalent to minimization of L wrt three parameters

Lagranges undetermined multiplier as a buy-one-get-one-free Please minimize this function. Wait let me add this constraint. Don t worry I only added zero.

The first equation { } ( ) E-l Y*Ydt cA -1 = 0 E = Y* HYdt +2cAcBA HBdt A HAdt B HBdt = cA +cB 2 2 2dt 2dt 1= Y*Ydt = cA +cB +2cAcBABdt 2 2 A B ( ) 2dt 0= 2cAA HAdt +2cBA HBdt +2cBABdt -l 2cA A

The second equation { } ( ) E-l Y*Ydt cB -1 = 0 E = Y* HYdt +2cAcBA HBdt A HAdt B HBdt = cA +cB 2 2 2dt 2dt 1= Y*Ydt = cA +cB +2cAcBABdt 2 2 A B ( ) 2dt 0= 2cBB HBdt +2cAA HBdt +2cAABdt -l 2cB B

Matrix eigenvalue equation ( ( ) ) L cA = 0 2dt cAA HAdt +cBA HBdt +cBABdt = l cA A 2dt cBB HBdt +cAA HBdt = l cB +cAABdt L cB B = 0 2dt A HAdt A HBdt = l ABdt A cA cB cA cB A HBdt B HBdt 2dt ABdt B or = l Overlap integral aA b b aB Coulomb integral cA cB cA cB 1 S S 1 Resonance integral (negative) or Hc = lSc

Matrix eigenvalue equation Use approximation S = 0 to simplify 1 S 1 0 1 0 c c c c c c S A A A = = 1 B B B This matrix corresponds to 1 in arithmetic: a unit matrix c c c c A A A = B B B Matrix acts on a vector and returns the same vector, apart from a constant factor

Matrix eigenvalue equation Operator eigenvalue equation Hamiltonian operator or matrix H = E eigenfunction eigenvector eigenvalue c c c c A A A = B B B Matrix eigenvalue equation

What is ? This correspondence suggests that (undetermined) actually represents energy! The left-hand side becomes cA ( ( ( ) ) c c 2dt cAA HAdt +cBA HBdt = l cA +cB ABdt A A 2dt +) cBB HBdt +cAA HBdt = l cB +cAABdt B B +2cAcBA HBdt A HAdt B HBdt +cB = E 2 2 The right-hand side becomes )= l 2dt 2dt +2cAcBABdt l cA +cB 2 2 A B We find that in fact = E !

Lagranges undetermined multiplier Many chemistry and physics problems are recast into constrained minimization or maximization. Lagrange s method can convert them into unconstrained minimization or maximization with slightly increased dimension. The undetermined multiplier ends up in representing an important physical quantity.

Matrix eigenvalue equation How do we solve this equation? 1 S c c c c S A A A = E 1 B B B Observation: two equations for three unknowns (cA, cB, and E) indeterminate? The indeterminacy is removed by the normalization constraint. 2dt 2dt +2cAcBABdt 1= Y*Ydt = cA +cB 2 2 A B

Matrix eigenvalue equation 1 S c c c c S A A A = E 1 B B B 0 0 E ES E c c A A = ES B B If this matrix has an inverse, we invariably obtain a trivial, nonphysical solution: 1 0 0 c c E ES E A A = ES B B

Inverse matrix + + + a c b d e g f h ae bg ce dg + af cf bh dh = 1 1 0 0 1 a c b d a c b d Definition of inverse = Unit matrix 1 a c b d d b 1 = c a ad bc We have verified that this is indeed the inverse

Inverse matrix 1 a c b d d b 1 = c a ad bc determinant If the determinant is zero, the inverse does not exist

Inverse matrix 0 0 E ES E c c A A = ES B B Determinant is nonzero trivial nonphysical solution -1 aA- E b - ES b - ES aB- E = 0 0 0 0 = cA cB or physical solution (aA- E)(aB- E)-(b - ES)(b - ES) = 0 Determinant is zero

Summary Variational theorem gives us a way to identify the best approximate wave function for the ground state; it is the one with the lowest energy expectation value. Variational theorem therefore leads to minimization of expectation value with the constraint that the wave function is normalized. Constrained minimization can be converted to unconstrained minimization with slightly higher dimension by Lagrange s undetermined multiplier method. With these, optimization of LCAO MO s becomes a matrix eigenvalue equation.