1. 0 Jack SimonsJack Simons, Henry Eyring Scientist and Professor Chemistry Department University of Utah Electronic Structure Theory Session 12

2. 1 analytical derivatives of the energy The use of analytical derivatives of the energy with respect to atomic positions has made evaluation of vibrational frequencies and the mapping out of reaction paths much easier. The first derivative with respect to a Cartesian coordinate (X K ) of an atom is called the gradient g K =  E/  X K. These numbers form the gradient vector. The second derivatives  2 E/  X K  X L form the Hessian matrix H K,L In the old days, g K and H K,L were evaluated by “ finite difference ”. Today, we have analytical expressions for g K and H K,L.

3. 2 How does one use g K and H K,L ? Assume you have g K available at some starting geometry X 0 = {X 1, X z, … X 3N }. One can attempt to move downhill toward a local-minimum by taking small displacements  X K proportional to, but in opposition to, the gradient g K along that direction  X K = - a g K. The energy E is then expected to change by  E = - a ∑ K (g K ) 2. This is the most simple algorithm for “ stepping ” downhill toward a minimum. The parameter a can be used to keep the length of the step small. A series of such “ steps ” from X 0 to X 0 +  X can often lead to a minimum (at which all 3N g K values vanish).

4. 3 One problem with this approach is that if one reaches a point where all 3N g K vanish, one can not be certain it is a minimum; maybe it is a first-, second-, or higher-order saddle point. Minimum: all 3N g K vanish and 3N-6 eigenvalues of the H K,L matrix are positive. First-order saddle (transition state TS): all 3N g K vanish and 3N-7 eigenvalues of the H K,L matrix are positive; one is negative.

5. 4 So, one is usually forced to form H K,L and find its 3N eigenvalues a and eigenvectors V k a  L H K,L V L a = a V k a. 3 of the a have to vanish and the 3 corresponding V k a describe translations of the molecule. 3 more (only 2 for linear molecules) of the a have to vanish and the corresponding V k a describe rotations of the molecule. The remaining 3N-6 (or 3N-5) a and V k a contain the information one needs to characterize the vibrations and reaction paths of the molecule.

6. 5 If one has the gradient vector and Hessian matrix available at some geometry,  E =  K g K  X K + ½  K,L H K,L  X K  X L Because the Hessian is symmetric, its eigenvectors are orthogonal  K V K a V K b =  a,b and they form a complete set  a V K a V L a =  K,L. This allows one to express the atomic Cartesian displacements  X K in terms of displacements  V a along the “ eigenmodes ”  X K =  L  K,L  X L =  a V K a (  L V L a  X L ) =  a V K a  V a.

7. 6 Inserting  X K =  a V K a  V a. into  E =  K g K  X K + ½  K,L H K,L  X K  X L gives  a {g a  V a + ½ a (  V a ) 2 } where g a =  L V L a g L This way of writing  allows us to consider independently maximizing or minimizing along each of the 3N-6 eigenmodes.

8. 7 Setting the derivative of {g a  V a + ½ a (  V a ) 2 } with respect to the  V a displacements equal to zero gives as a suggested “ step ”  V a = - g a / a Inserting these displacements into  a {g a  V a + ½ a (  V a ) 2 } gives  a  {- g a 2 / a + ½ a (-g a / a ) 2 } = -1/2  a g a 2 / a. So the energy will go “ downhill ” along an eigenmode if that mode ’ s eigenvalue a is positive; it will go uphill along modes with negative a values. Once you have a value for  V a, you can compute the Cartesian displacements from  X K =  a V K a  V a

9. 8 If one wants to find a minimum, one can a.Take a displacement  V a = - g a / a along any mode whose a is positive. b.Take a displacement that is small and of opposite sign than - g a / a for modes with negative a values. The energy will then decrease along all 3N-6 modes. What about finding transition states?

10. 9 If one is already at a geometry where one a is negative and the 3N-7 other a values are positive, one should a.Visualize the eigenvector V k a belonging to the negative a to make sure this displacement “ makes sense ” (i.e., looks reasonable for motion away from the desired transition state). b.If the mode having negative eigenvalue makes sense, one then takes  V a = - g a / a for all modes. This choice will cause  a  {- g a 2 / a + ½ a (-g a / a ) 2 } = -1/2  a g a 2 / a to go downhill along 3N-7 modes and uphill along the one mode having negative a. Following a series of such steps may allow one to locate the TS at which all g a vanish, 3N-7 a are positive and one a is negative.

11. 10 At a minimum or TS, one can evaluate harmonic vibrational frequencies using the Hessian. The classical dynamics Hamiltonian for displacements  X K is H =  K,L ½ H K,L  X K  X L + ½  K m K (d  X K /dt) 2 Introducing the mass-weighted Cartesian coordinates  MWX K = (m K ) 1/2  X K allows the Hamiltonian to become H =  K,L ½ MWH K,L  MWX K  MWX L + ½  K (d  MWX K /dt) 2

12. 11 Expressing the Cartesian displacements in terms of the eigenmode displacements  X K =  a V K a  V a allows H to become H =  a { ½ a (  V a ) 2 + ½ (dV a /dt) 2 }. This is the Hamiltonian for 3N-6 uncoupled harmonic oscillators having force constants a and having unit masses for all coordinates. Thus, the harmonic vibrational frequencies are given by  a = ( a ) 1/2 so the eigenvalues of the mass-weighted Hessian provide the harmonic vibrational frequencies. At a TS, one of the a will be negative.

13. 12 To trace out a reaction path starting at a transition state, one first finds the Hessian eigenvector {V K 1 } belonging to the negative eigenvalue. One takes a very small step along this direction. Next, one re-computes the Hessian and gradient (n.b., the gradient vanishes at the transition state but not once begins to move along the reaction path) at the new geometry X K +  X K where one finds the eigenvalues and eigenvectors of the mass-weighted Hessian and uses the local quadratic approximation  a {g a  V a + ½ a (  V a ) 2 } to guide one downhill. Along the eigenmode corresponding to the negative eigenvalue 1, the gradient g 1 will be non-zero while the components of the gradient along the other eigenmodes will be small (if one has taken a small initial step). One is attempting to move down a streambed whose direction of flow initially lies along V K 1 and perpendicular to which there are harmonic sidewalls ½ a (  V a ) 2.

14. 13 One performs a series of displacements by a)moving (in small steps) downhill along the eigenmode that begins at V K 1 and that has a significant gradient component g a, b)while minimizing the energy (to remain in the streambed ’ s bottom) along the 3N-7 other eigenmodes (by taking steps  V a = - g a / a that minimize each {g a  V a + ½ a (  V a ) 2 }. As one evolves along this reaction path, one reaches a point where 1 changes sign from negative to positive. This signals that one is approaching a minimum. Continuing onward, one reaches a point where the gradient ’ s component along the step displacement vanishes and along all other directions vanishes. This is the local minimum that connects to the transition state at which the reaction path started. One needs to also begin at the transition state and follow the other branch of the reaction path to be able to connect reactants, transition state, and products.

15. 14 So, how does one evaluate the gradient and Hessian analytically? For methods such as SCF, CI, and MCSCF that compute the energy E as E = /, one makes use of the chain rule to write  E/  X K =  I  E/  C I  C I /  X K +  i    E/  C i   C i  /  X K + /. For MCSCF,  E/  C I and  E/  C i  are zero. For SCF   E/  C i  are zero and  E/  C I does not exist. For CI,  E/  C I are zero, but   E/  C i  are not. So, for some methods, one needs to solve “ response equations ” for  E/  C i  

16. 15 What is / ? =  L,J C L C J and each of the Hamiltonian matrix elements is given via Slater-Condon rules in terms of 1- and 2- electron integrals and The only places the nuclear positions X K appear are 1.in the basis functions appearing in  J =     C J,  and 2.in V e,n = -  a Z a e 2 /|r-R A | So, will involve as well as derivatives  /  X K of the   appearing in and in

17. 16  /  X A V e,n = -  a Z a (x-X A ) e 2 /|r-R A | 3 When put back into and into the Slater-Condon formulas, these terms give the Hellmann-Feynman contributions to the gradient. These are not “ difficult ” integrals, but they are new ones that need to be added to the usual 1- electron integrals. The  /  X K derivatives of the  appearing in +  a and in present major new difficulties because they involve new integrals +  a

18. 17 When Cartesian Gaussians  a,b,c (r, ,  ) = N' a,b,c,  x a y b z c exp(-  r 2 ) are used, the derivatives  /  X K  (r) can be done because X K appears in ( x-X K ) a and in r 2 = (x-X K ) 2 + (y-Y K ) 2 + (z-Z K ) 2. These derivatives give functions of one lower (from  /  X K (x-X K ) a ) and one higher (from  /  X K exp(-  r 2 )) angular momentum value. So, the AO integral list must be extended to higher L-values. More troublesome are because there are now 4 times ( the original plus  /  X K,  /  Y K,  /  Z K ) the number of 2-electron integrals.

19. 18 The good news is that the Hellmann-Feynman and integral derivative terms can be evaluated and thus the gradients can be computed as  E/  X K =  I  E/  C I  C I /  X K +  i    E/  C i   C i  /  X K + / = / for SCF or MCSCF wavefunctions. What about CI, MPn, or CC wave functions? What is different?

20. 19  E/  X K =  I  E/  C I  C I /  X K +  i    E/  C i   C i  /  X K + /. For CI, the  E/  C I term still vanishes and the / term is handled as in MCSCF, but the  E/  C i  terms do not vanish For MPn, one does not have C I parameters; E is given in terms of orbital energies  j and 2-electron integrals over the  j. For CC, one has t i,j m,n amplitudes as parameters and E is given in terms of them and integrals over the  j. So, in CI, MPn, and CC one needs to have expressions for  C i  /  X K and for  t i,j m,n /  X K. response equations These are called response equations.

21. 20 The response equations for  C i  /  X K are obtained by taking the  /  X K derivative of the Fock equations that determined the C i,   /  X K    |h e |   > C J,  =  /  X K  J   C J,  This gives   [  |h e |   >-  J ]  /  X K C J,      /  X K [  |h e |   >-  J   ] C J,  Because all the machinery to evaluate the terms in  /  X K [  |h e |   >-  J   ] exists as does the matrix  |h e |   >-  J, one can solve for  /  X K C J, 

22. 21 A similar, but more complicated strategy can be used to derive equations for the  t i,j m,n /  X K that are needed to achieve gradients in CC theory. The bottom line is that for MPn, CI, and CC, one can obtain analytical expressions for g K =  E/  X K. To derive analytical expressions for the Hessian  2 E/  X K  X L is, or course, more difficult. It has been done for HF and MCSCF and CI and may exist (?) for CC theory. As you may expect it involves second derivatives of 2- electron integrals and thus is much more “ expensive ”. There are other kinds of responses that one can seek to treat analytically. What if one added to the Hamiltonian an electric field term such as  k=1,N er k E +  a=1,M e Z a R a E ?

23. 22 So, H is now H +  k=1,N er k E +  a=1,M e Z a R a E. The wavefunction  (E) and energy E(E) will now depend on the electric field E. dE/dE =  I  E/  C I  C I /  E+  i   E/  C i   C i  /  E + /. / =, the dipole moment expectation value. This is the final answer for HF and MCSCF, but not for MPn, CI, CC. For these cases, we also need  C I /  E and  C i  /  E response contributions. So, the expectation value of the dipole moment operator is not always the correct dipole moment!