variational principle and Schrödinger equation

Transcription

1 variational principle and Schrödinger equation normalized wavefunctions energy functional E[ ]= H variation E[ + ] = H + H + H + H = E[ ]+ E[ ]+O 2 ( ) functional derivative E[ ] = E[ + ] E[ ] ensure normalization via Lagrange multiplier L[ ]=E[ ] ( ) = L[ ] = H variational principle + H gives time-independent Schrödinger equation

2 variational principle and finite basis set trial function = X n n n (orthogonal basis) E[ ] = X n,m L[ ] = X n,m n m n H m X nh nm m 2n = L k = X n nh nk + X m variational principle H km m 2 k = X! nh nk k + X n m H km m k! H H 2 H N H 2 H 22 H 2N 2 = 2 H N H N2 H NN N N

3 non-orthogonal basis set trial function = X n n n (arbitrary basis) overlap matrix S nm = n m L[ ] = X nh nm m X ns nm m variation results in generalized eigenvalue problem H H 2 H N H 2 H 22 H 2N 2 = S S 2 S N S 2 S 22 S 2N 2 H N H N2 H NN N S N S N2 S NN N

4 tight-binding H = 2 + X V n superposition of well separated potentials Vn use solutions of individual potentials H n = 2 + V n H n n = n n as basis set tight-binding approximation for matrix elements: n H n = n + X n V m n m6=n n H m = ( n + m )/2 n m + n V n + V m m /2+ X k6=n,m for n and m nearest neighbors n V k m

5 Hellmann-Feynman theorem & forces molecular dynamics: H( ) ( ) = E( ) ( ) force on nucleus α F = E( )= ( ) H ( ) E( ) = H + H + H = ( ) H( ) ( ) + E( ) ( ) {z ( ) } = {z } = for non-degenerate state

6 Hellmann-Feynman theorem degenerate perturbation theory in Phys Rev 23! PHYSIL REVIEW 68, Hellmann-Feynman theorem at degeneracies Ofir E lon and L S ederbaum Theoretische hemie, Physikalisch-hemisches Institut, Universität Heidelberg, Im Neuenheimer Feld 229, D-692 Heidelberg, Germany Received 2 ugust 22; revised manuscript received 8 March 23; published 3 July 23 The Hellmann-Feynman theorem is extended to account for degenerate states Given a point in parameter space where the energy E( ) is n-fold degenerate, it is shown that the corresponding n forces slopes are obtained by diagonalizing the derivative of the Hamiltonian, H( )/, in the subspace of degenerate eigenstates Such a rotation within the subspace of degenerate eigenfunctions is easy and simple to apply in practical calculations and should be performed separately for each independent direction in parameter space for which the forces are to be calculated DOI: 3/PhysRev68335 PS number s : 35Md, 365 w, 75 m Hellmann-Feynman theorem HFT,2 is widely used in practice to calculate forces slopes in atoms, molecules and crystals, but the way it appears in Refs and 2 and in quantum mechanics textbooks see, eg, Ref 3 is only correct for nondegenerate states see below It should be mentioned that there exists a more general form of HFT, which is defined also for off-diagonal matrix elements see, eg, Refs 4 9 However, the prescribed expressions do not provide an depend also on the direction in parameter space for which one calculates the forces Practically, a set of degenerate eigenfunctions obtained in a numerical calculation has to be appropriately rotated to allow for the correct calculation of forces as expectation values of H( )/ see Eq at degeneracy These detailed and relevant points, all related to the freedom to chose the linear combinations of eigenstates at degeneracy, have been overlooked previously

7 Hellmann-Feynman theorem II a surprising failure of reason PHYSIL REVIEW 66, reakdown of the Hellmann-Feynman theorem: Degeneracy is the key G P Zhang Department of Physics, Indiana State University, Terre Haute, Indiana 4789 Thomas F George Office of the hancellor/departments of hemistry and Physics & stronomy, University of Wisconsin-Stevens Point, Stevens Point, Wisconsin Received 7 May 22; published 29 July 22 The Hellmann-Feynman theorem is a powerful and popular method to efficiently calculate forces in a variety of dynamical processes, but its validity has rarely been addressed Here a surprising failure of this theorem is reported The forces calculated by the theorem can be more than fifty times smaller than the forces calculated by the finite differential method Numerical evidence shows that the energy-level degeneracy is the main reason n analytical proof reveals that although eigenvalues do not depend on a linear combination of degenerate wave functions, forces do sensitively depend on it, which leads to ill-defined forces scheme is proposed to overcome this difficulty DOI: 3/PhysRev6633 PS number s : 75 m The Hellmann-Feynman theorem is powerful and has been widely used in many fields such as dynamical processes, 2 molecular dynamics, 3 chemical reactions, and merical examples for 6 explicitly show that forces calculated by the Hellmann-Feynman theorem can be fifty times smaller than forces calculated by the finite differential

8 Hellmann-Feynman theorem III an elegant method PHYSIL REVIEW 69, Extended Hellmann-Feynman theorem for degenerate eigenstates G P Zhang Department of Physics, Indiana State University, Terre Haute, Indiana 4789, US Thomas F George Office of the hancellor/departments of hemistry & iochemistry and Physics & stronomy, University of Missouri-St Louis, St Louis, Missouri 632, US Received 2 ugust 23; revised manuscript received 26 January 24; published 27 pril 24 In a previous paper, we reported a failure of the traditional Hellmann-Feynman theorem HFT for degenerate eigenstates This has generated enormous interest among different groups In four independent papers by Fernandez, by alawender, Hola, and March, by Vatsya, and by lon and ederbaum, an elegant method to solve the problem was devised The main idea is that one has to construct and diagonalize the force matrix for the degenerate case, and only the eigenforces are well defined We believe this is an important extension to HFT Using our previous example for an energy level of fivefold degeneracy, we find that those eigenforces correctly reflect the symmetry of the molecule DOI: 3/PhysRev69672 PS number s : 75 m The Hellmann-Feynman theorem HFT Ref is an efficient method, but when eigenstates are degenerate, an extension is necessary The reason is that any good linear comculated from two different configurations of 6, the pristine and excited 6 The fivefold degenerate case is from the pristine 6 We present the force matrix elements in Table I