Lowe / Peterson | Quantum Chemistry | E-Book | sack.de
E-Book

E-Book, Englisch, 728 Seiten

Lowe / Peterson Quantum Chemistry


3. Auflage 2011
ISBN: 978-0-08-047078-8
Verlag: Elsevier Science & Techn.
Format: EPUB
Kopierschutz: 6 - ePub Watermark

E-Book, Englisch, 728 Seiten

ISBN: 978-0-08-047078-8
Verlag: Elsevier Science & Techn.
Format: EPUB
Kopierschutz: 6 - ePub Watermark



Lowe's new edition assumes little mathematical or physical sophistication and emphasizes an understanding of the techniques and results of quantum chemistry. It can serve as a primary text in quantum chemistry courses, and enables students and researchers to comprehend the current literature. This third edition has been thoroughly updated and includes numerous new exercises to facilitate self-study and solutions to selected exercises. Assumes little initial mathematical or physical sophistication, developing insights and abilities in the context of actual problems Provides thorough treatment of the simple systems basic to this subject Emphasizes UNDERSTANDING of the techniques and results of modern quantum chemistry Treats MO theory from simple Huckel through ab intio methods in current use Develops perturbation theory through the topics of orbital interaction as well as spectroscopic selection rules Presents group theory in a context of MO applications Includes qualitative MO theory of molecular structure, Walsh rules, Woodward-Hoffmann rules, frontier orbitals, and organic reactions Develops MO theory of periodic systems, with applications to organic polymers.

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Chapter 2 Quantum Mechanics of Some Simple Systems 2-1 The Particle in a One-Dimensional “Box”
Imagine that a particle of mass m is free to move along the x axis between x = 0 and x = L, with no change in potential (set V = 0 for 0      (2-1)
Within the box, ? must satisfy the equation
     (2-2)
It should be realized that E must take on the same values for both of these equations; the eigenvalue E pertains to the entire range of the particle and is not influenced by divisions we make for mathematical convenience. Let us examine Eq. (2-1) first. Suppose that, at some point within the infinite barrier, say x = L + dx, ? is finite. Then the second term on the left-hand side of Eq. (2-1) will be infinite. If the first term on the left-hand side is finite or zero, it follows immediately that Eis infinite at the point L + dx (and hence everywhere in the system). Is it possible that a solution exists such that E is finite? One possibility is that ? = 0 at all points where V = 8. The other possibility is that the first term on the left-hand side of Eq. (2-1) can be made to cancel the infinite second term. This might happen if the second derivative of the wavefunction is infinite at all points where V = 8 and ? ? 0. For the second derivative to be infinite, the first derivative must be discontinuous, and so ? itself must be nonsmooth (i.e., it must have a sharp corner; see Fig. 2-2). Thus, we see that it may be possible to obtain a finite value for both E and ? at x = L + dx, provided that ? is nonsmooth there. But what about the next point, x = L + 2 dx, and all the other points outside the box? If we try to use the same device, we end up with the requirement that ? be nonsmooth at every point where V = 8. A function that is continuous but which has a point-wise discontinuous first derivative is a contradiction in terms (i.e., a continuous f cannot be 100% corners. To have recognizable corners, we must have some (continuous) edges. We say that the first derivative of ? must be piecewise continuous.) Hence, if V = 8 at a single point, we might find a solution ? which is finite at that point, with finite energy. If V = 8 over a finite range of connected points, however, either E for the system is infinite, and ? is finite over this region or E is not infinite (but is indeterminate) and ? is zero over this region. Figure 2-2 As the function f (x) approaches being nonsmooth, d capproaches zero (the width of one point) and n approaches infinity. We are not concerned with particles of infinite energy, and so we will say that the solution to Eq. (2-1) is ? = 0.1 Turning now to Eq. (2-2) we ask what solutions ? exist in the box having associated eigenvalues E that are finite and positive. Any function that, when twice differentiated, yields a negative constant times the selfsame function is a possible candidate for ?. Such functions are sin(kx), cos(±kx), and exp(±ikx). But these functions are not all independent since, as we noted in Chapter 1,
     (2-3)
We thus are free to express ? in terms of exp(±ikx) or else in terms of sin(kx) and cos(kx). We choose the latter because of their greater familiarity, although the final answer must be independent of this choice. The most general form for the solution is
     (2-4)
where A, B, and k remain to be determined. As we have already shown, ? is zero at x = 0, x=L and so we have as boundary conditions
     (2-5)

     (2-6)
Mathematically, this is precisely the same problem we have already solved in Chapter 1 for the standing waves in a clamped string. The solutions are
     (2-7)
One difference between Eq. (2-7) and the string solutions is that we have rejected the n = 0 solution in Eq. (2-7). For the string, this solution was for no vibration at all—a physically realizable circumstance. For the particle-in-a-box problem, this solution is rejected because it is not square-integrable. (It gives ? = 0, which means no particle on the x axis, contradicting our starting premise. One could also reject this solution for the classical case since it means no energy in the string, which might contradict a starting premise depending on how the problem is worded.) Let us check to be sure these functions satisfy Schrödinger’s equation
     (2-8)
This shows that the functions (2-7) are indeed eigenfunctions of H. We note in passing that these functions are acceptable in the sense of Chapter 1. The only remaining parameter is the constant A. We set this to make the probability of finding the particle in the well equal to unity:
     (2-9)
This leads to (Problem 2-2)
     (2-10)
which completes the solving of Schrödinger’s time-independent equation for the problem. Our results are the normalized eigenfunctions
     (2-11)
and the corresponding eigenvalues, from Eq. (2-8),
     (2-12)
Each different value of n corresponds to a different stationary state of this system. 2-2 Detailed Examination of Particle-in-a-Box Solutions
Having solved the Schrödinger equation for the particle in the infinitely deep squarewell potential, we now examine the results in more detail. 2-2.A Energies
The most obvious feature of the energies is that, as we move through the allowed states (n = 1, 2, 3, …), E skips from one discrete, well-separated value to another (1, 4, 9 in units of h2/8m L2). Thus, the particle can have only certain discrete energies—the energy is quantized. This situation is normally indicated by sketching the allowed energy levels as horizontal lines superimposed on the potential energy sketch, as in Fig. 2-3a. The fact that each energy level is a horizontal line emphasizes the fact that E is a constant and is the same regardless of the x coordinate of the particle. For this reason, E is called a constant of motion. The dependence of E on n2 is displayed in the increased spacing between levels with increasing n in Fig. 2-3a. The number n is called a quantum number. Figure 2-3 Allowed energies for a particle in various one-dimensional potentials. (a) “box” with infinite walls. (b) quadratic potential, .(c) V = -1/|x| Tendency for higher levels in (b) and (c) not to diverge as in (a) is due to larger “effective box size” for higher energies in (b) and (c). We note also that E is proportional to L-2. This means that the more tightly a particle is confined, the greater is the spacing between the allowed energy levels. Alternatively, as the box is made wider, the separation between energies decreases and, in the limit of an infinitely wide box, disappears entirely. Thus, we associate quantized energies with spatial confinement. For some systems, the degree of confinement of a particle depends on its total energy. For...



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