Introductory квантовая химия, Introductory quantum chemistry(T)(649s)_Ch

Introductory квантовая химия

Introductory quantum chemistry(T)(649s)_Ch_.pdf

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Date Dec 18, 2003

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2. The Section Simple Molecular Orbital Theory deals with atomic and molecular orbitals in a qualitative manner, including their symmetries, shapes, sizes, and energies. It introduces bonding, non-bonding, and antibonding orbitals, delocalized, hybrid, and Rydberg orbitals, and introduces Hückel-level models for the calculation of molecular orbitals as linear combinations of atomic orbitals (a more extensive treatment of...

10. Introduction to Quantum Mechanics, L. Pauling and E. B. Wilson, Dover Publications, Inc., New York, N. Y. (1963)- Pauling and Wilson. 11. Modern Quantum Chemistry, A. Szabo and N. S. Ostlund, Mc Graw-Hill, New York (1989)- Szabo and Ostlund. 12. Quantum Chemistry, I. N. Levine, Prentice Hall, Englewood Cliffs, N. J. (1991)Levine. 13. Energetic Principles of Chemical Reactions, J. Simons, Jones and Bartlett, Portola Valley, Calif. (1983),...

where Ψ(qj,t) is the unknown wavefunction and H is the operator corresponding to the total energy physical property of the system. This operator is called the Hamiltonian and is formed, as stated above, by first writing down the classical mechanical expression for the total energy (kinetic plus potential) in cartesian coordinates and momenta and then replacing all classical momenta pj by their quantum mechanical operators pj = - ih∂/∂qj . For the H2O example used above, the classical mechanical energy of all thirteen particles is E = Σ i { pi2/2me + 1/2 Σ j e2/ri,j - Σ a Zae2/ri,a } + Σ a {pa2/2ma + 1/2 Σ b ZaZbe2/ra,b }, where the indices i and j are used to label the ten electrons whose thirty cartesian coordinates are {qi} and a and b label the three nuclei whose charges are denoted {Za}, and whose nine cartesian coordinates are {qa}. The electron and nuclear masses are denoted me and {ma}, respectively. The corresponding Hamiltonian operator is H = Σ i { - (h2/2me) ∂2/∂qi2 + 1/2 Σ j e2/ri,j - Σ a Zae2/ri,a } + Σ a { - (h2/2ma) ∂2/∂qa2+ 1/2 Σ b ZaZbe2/ra,b }. Notice that H is a second order differential operator in the space of the thirty-nine cartesian coordinates that describe the positions of the ten electrons and three nuclei. It is a second order operator because the momenta appear in the kinetic energy as pj2 and pa2, and the quantum mechanical operator for each momentum p = -ih ∂/∂q is of first order. The Schrödinger equation for the H2O example at hand then reads...

The number of dimensions depends on the number of particles and the number of spatial (and other) dimensions needed to characterize the position and motion of each particle 1. The Schrödinger Equation Consider an electron of mass m and charge e moving on a two-dimensional surface that defines the x,y plane (perhaps the electron is constrained to the surface of a solid by a potential that binds it tightly to a narrow region in the z-direction), and assume that the electron experiences a constant potential V0 at all points in this plane (on any real atomic or molecular surface, the electron would experience a potential that varies with position in a manner that reflects the periodic structure of the surface). The pertinent time independent Schrödinger equation is: - h2/2m (∂2/∂x2 +∂2/∂y2)ψ(x,y) +V 0ψ(x,y) = E ψ(x,y). Because there are no terms in this equation that couple motion in the x and y directions (e.g., no terms of the form xayb or ∂/∂x ∂/∂y or x∂/∂y), separation of variables can be used to write ψ as a product ψ(x,y)=A(x)B(y). Substitution of this form into the Schrödinger equation, followed by collecting together all x-dependent and all y-dependent terms, gives; - h2/2m A-1∂2A/∂x2 - h2/2m B-1∂2B/∂y2 =E-V0. Since the first term contains no y-dependence and the second contains no x-dependence, both must actually be constant (these two constants are denoted Ex and Ey, respectively), which allows two separate Schrödinger equations to be written: - h2/2m A-1∂2A/∂x2 =Ex, and - h2/2m B-1∂2B/∂y2 =Ey. The total energy E can then be expressed in terms of these separate energies Ex and Ey as Ex + Ey =E-V0. Solutions to the x- and y- Schrödinger equations are easily seen to be: A(x) = exp(ix(2mEx/h2)1/2) and exp(-ix(2mEx/h2)1/2) ,...

The two (1/2L)1/2 factors are included to guarantee that ψ is normalized: ∫ |ψ(x,y)| 2 dx dy = 1. Normalization allows |ψ(x,y)| 2 to be properly identified as a probability density for finding the electron at a point x, y. 4. Quantized Action Can Also be Used to Derive Energy Levels There is another approach that can be used to find energy levels and is especially straightforward to use for systems whose Schrödinger equations are separable. The socalled classical action (denoted S) of a particle moving with momentum p along a path leading from initial coordinate qi at initial time ti to a final coordinate qf at time tf is defined by: qf;tf S = ⌠ p•dq . ⌡ qi;ti Here, the momentum vector p contains the momenta along all coordinates of the system, and the coordinate vector q likewise contains the coordinates along all such degrees of freedom. For example, in the two-dimensional particle in a box problem considered above, q = (x, y) has two components as does p = (P x, p y) , and the action integral is: x f;yf;tf S = ⌠ (p x d x + p y dy) . ⌡ xi;yi;ti In computing such actions, it is essential to keep in mind the sign of the momentum as the particle moves from its initial to its final positions. An example will help clarify these matters. For systems such as the above particle in a box example for which the Hamiltonian is separable, the action integral decomposed into a sum of such integrals, one for each degree of freedom. In this two-dimensional example, the additivity of H:...

This equation is not separable in cartesian coordinates (x,y,z) because of the way x,y, and z appear together in the square root. However, it is separable in spherical coordinates − h2  ∂  2µr2  ∂r ∂ ∂ψ  2 ∂ψ  1 r  + 2  Sinθ  ∂θ   ∂r   r Sinθ ∂θ ...

∂ ∂z ∂ ∂ = = - Sinθ . ∂θ ∂θ ∂z ∂z The range of values for θ was 0 ≤ θ < π , so the range for z is -1 < z < 1. The equation for Θ , when expressed in terms of P and z, becomes d dP m2P (1-z2) dz  + λP = 0. dz   1-z2 Now we can look for polynomial solutions for P, because z is restricted to be less than unity in magnitude. If m = 0, we first let...