1 / 17

Jack Simons Henry Eyring Scientist and Professor Chemistry Department University of Utah

Electronic Structure Theory Session 1. Jack Simons Henry Eyring Scientist and Professor Chemistry Department University of Utah. The Schrödinger equation for N electrons and M nuclei of a molecule:. H( r,R)  ( r,R, t ) = i  ∂  ( r,R ,t)/∂t, or if H is t-independent,

derek-mccormick
Download Presentation

Jack Simons Henry Eyring Scientist and Professor Chemistry Department University of Utah

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. The Schrödinger equation for N electrons and M nuclei of a molecule: H(r,R) (r,R,t) = i  ∂(r,R,t)/∂t, or if H is t-independent, H(r,R) (r,R) = E (r,R) |(r,R)|2gives probability density for finding electrons at r = r1r2 r3 ... rNand nuclei at R1 R2 R3 ...RM. H contains electronic kinetic energy:Te = -2/2 j=1,N me-1 j2 nuclear kinetic energy:TM= -2/2 j=1,M mj-1 j2 electron-nuclei Coulomb potentials: –j=1,MZj k=1,N e2/|rk-Rj| VeM nuclear-nuclear Coulomb repulsions:+j<k=1,M ZjZke2/|Rk-Rj| and electron-electron Coulomb repulsions:+Vee= j<k=1,Ne2/rj,k It can contain more terms if, for example, external electric or magnetic fields are present (e.g.,k=1,N erkE).

  3. In the Born-Oppenheimerapproximation/separation, we (first) ignore the TM motions of the nuclei (pretend the nuclei are fixed at specified locations R) and solve H0(r|R) =EK(R) (r|R), the so-called electronic Schrödinger equation. H0 contains all of H except TM (Te plus all of the potential energy terms). Because H0 is Hermitian, its eigenfunctions form a complete and orthonormal set of functions of r. So,  can be expanded in the K: (r,R) = KK(r,R) K(R). • The K(r,R) and the K(R) depend on R because H0 does through • j=1,MZj k=1,N e2/|rk-Rj| Z+ j<k=1,MZjZke2/|Rk-Rj|

  4. This expansion can then be used in H(r,R) (r,R) = E (r,R) [H0 -2/2 j=1,M mj-1 j2 -E] KK(r,R) K(R) = 0 to produce equations for theK(R): 0 = [EL(R)-2/2j=1,M mj-1 j2 -E] L(R) + K< L(r,R)| -2/2 j=1,M mj-1 j2 K(r,R)> K(R) + K< L(r,R)| -2j=1,M mj-1 jK(r,R)> jK(R) These are called the coupled-channel equations. If we ignore all of the non-adiabatic terms, we obtain a SE for the vib./rot./trans. motion on the Lth energy surface EL(R) 0 = [EL(R)-2/2j=1,M mj-1 j2 -E] L(R)

  5. Each electronic state L has its own set of rot./vib. wave functions and energies: [EL(R)-2/2j=1,M mj-1 j2 -EL,J,M,] L,J,M, (R) = 0 The non-adiabaticcouplings can induce transitions among these states (radiationless transitions).

  6. There are major difficulties in solving the electronic SE: Vee= j<k=1,Ne2/rj,k makes the equation not separable- this meansis not rigorously a product of functions of individual electron coordinates.  e.g., 1s(1) 1s(2) 2s(3) 2s(4) 2p1(5)) Also,  has cusps The factors (-2/me1/rk /rk –Ze2/rk)  and (-22/me1/rk,l /rk,l +e2/rk,l)  will blow up unless so-called cusp conditions are obeyed by : /rk = -meZe2/ 2as rk0); /rk,l = 1/2 mee2/ 2as rk,l0)

  7. /rk = Ze2as rk0) and /rk,l = - e2as rk,l0). The electrons want to pile up near nuclei and they want to avoid one another. This means when we try to approximately solve the electronic SE, we should use “trial functions” that have such cusps. Slater-type orbitals(exp(-rk))have cusps at nuclei, but Gaussians(exp(-rk2))do not. We rarely use functions with e-e cusps, but we should (this is called using explicit e-e correlation).

  8. Digression into atomic units. Often, we use so-called atomic units. We let each coordinate be represented in terms of a parameter a0 having units of length and a dimensionless quantity: rja0 rj. The kinetic and potential energies thus contain T = -(2/2m)(1/a0)2j2 and Ven= -Ze2(1/a0) 1/rj, Vee= e2(1/a0) 1/rj,i respectively, where the variables now are the dimensionless ones. Factoring e2/a0 out from both the kinetic and potential energies as T = e2/a0{-(2/2m)(1/e2a0) j2} and V = e2/a0 {-Z/rj +1/rj,i} Choosinga0 = 2/(e2m) = 0.529 Å, when m is the electron mass, allows T and V to be written in terms of a common factor: e2/a0 = 1 Hartree = 27.21 eV: T = -1/2 j2while V = - Z/rj.

  9. Addressing the non-separability problem: If Vee could be replaced (or approximated) by a one-electron additive potential VMF = j=1,N VMF(rj) each of the solutionsLwould be a product (an antisymmetrized product called a Slater determinant) of functions of individual electron coordinates (spin-orbitals) j(r): | r1)(r2) (r3) (r4)  (r5) | = (N!)-1/2 P=1,N! SP P r1)(r2) (r3) (r4)  (r5) where SP is the sign of the permutation P. For example: | r1)(r2) | = (2!)-1/2 [r1)(r2) - 2(r1) 1(r2)]

  10. Before considering how we might find a goodVMF, let’s examine how important permutational antisymmetry is by considering some spin-orbital product (Slater determinant) wave functions for two electrons in and * orbitals. The purposes of this exercise are to remind one how to form singlet and triplet functions, to see how the different spin states have different physical content (i.e., charge distribution) even when the orbital occupancy is the same, and to introduce the idea that it is simply not possible, in certain circumstances, to use a single Slater determinant function as an approximation toK. So, let’s think of how the low-energy states of an olefin with two electrons in its  orbital framework should behave as we twist and break the  bond.

  11. Singlet2| |= 2-1/2 Triplet*||= 2-1/2 [ ||= 2-1/2 [ 2-1/2 [||+|   • Singlet2-1/2 [|| – | • -  Note: single determinant not possible here Singlet *2||= 2-1/2

  12. Now think of  2-1/2(L + R) and 2-1/2 (-L + R) to consider the behavior when -bond cleavage occurs upon rotation. Singlet 2 || = 2-1[|RR|+ |LL| + |RL|+ |LR|] ionic + diradical Triplet  || = 2-1[|LR|- |RL|] = |LR| diradical • Singlet 2 • || = 2-1[|RR|+ |LL| • |RL|- |LR|] • ionic - diradical

  13. Singlet * 2-1/2 [|| - | - |RR|+ |LR|- |RL| – |LL|] – 2-3/2[|RR|+|LR| – |RL|- |LL|] - |RR|+|LR|] ionic

  14. So, the spin state and orbital occupancy plus the antisymmetry have effects on the ionic/radical character of the function. To adequately describe the (singlet)bond breaking, we need to mix the and configuration state functions (CSF). This shows how single configuration functions and single determinants may not be adequate.

  15. To adequately describe the (singlet)bond breaking to give the singlet diradical, we need to mix the and configuration state functions (CSF). • || = 2-1[|RR|+ |LL| • + |RL|+ |LR|] • ionic + diradical • || = 2-1[|RR|+ |LL| • |RL|- |LR|] • ionic - diradical So, one must combine 2-1/2{|| - ||} to obtain a diradical state and 2-1/2{|| + ||} to obtain an ionic state. A single determinant function won’t work!

  16. Analogous “trouble” occurs if one uses a single determinant wave function to describe a bond that one wants to break: H2(2)  H(1sA) + H(1sB) H3C-CH3(2)  H3C + CH3 The |(1)(2)| determinant has diradical and ionic terms at large-R. It is possible to use a trial function of the form |(1)’(2)| and to allow and ’ to evolve from being and  near the equilibrium bond length into L and R at large-R. However, the function |R(1)L(2)| is not a singlet; it is a mixture of singlet and triplet, and, how do you let  and ’ evolve? Nevertheless, these are the approaches used in so-called unrestricted and restricted Hartree-Fock theory (RHF, UHF)

  17. |(1) ’(2)| is neither a singlet nor a triplet, but a mixture: 2-1/2[ |(1) ’ (2)| - |(1) ’ (2)|] +2-1/2[ |(1) ’ (2)| + |(1) ’ (2)|] |(1) ’(2)| dissociates to diradicals, but is neither singlet nor triplet |(1) (2)| is a singlet, but dissociates incorrectly One experiences jerks (not in the energy, but in its derivatives as well as spin impurity when the RHF and UHF curves connect.

More Related