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 F = 0 for 0 x L). At x = 0 and L and at all points heyond these limits the particle encounters an infinitely repulsive barrier (V = oo for X 0, X L). The situation is illustrated in Fig. 2-1. Because of the shape of this potential, this problem is often referred to as a particle in a square well or a particle in a box problem. It is well to bear in mind, however, that the situation is really like that of a particle confined to movement along a finite length of wire. 

When the potential is discontinuous, as it is here, it is convenient to write a wave equation for each region. For the two regions beyond the ends of the box 

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, xfr 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 E is 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 = oo. 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 wavefimction is infinite at aU points where V = oo and 0. 

Eor the second derivative to be infinite, the first derivative must be discontinuous, and so xir itself must be nonsmooth (i.e., it must have a sharp comer see Fig. 2-2). Thus, we see that it may be possible to obtain a finite value for both E and atx — L+ dx, provided that xfr is nonsmooth there. But what about the next point, x = L + 2dx, and all the other points outside the box If we try to use the same device, we end up with the requirement that xjr be nonsmooth at every point where V = oo. A function that is 

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To begin a more general approach to molecular orbital theory, we shall describe a variational solution of the prototypical problem found in most elementary physical chemistry textbooks the ground-state energy of a particle in a box (McQuanie, 1983) The particle in a one-dimensional box has an exact solution... 

Figure 3.3 (a) The potential energy function assumed in the particle-in-a-one-dimensional-box model, (b) A wave function satisfying the boundary conditions, (c) An unacceptable wave function. (Reproduced with permission from P. A. Cox, Introduction to Quantum Theory and Atomic Structure, 1996, Oxford University Press, Oxford, Figure 2.6.)... 

The results of the particle in a one-dimensional box problem can be used to describe the delocalized n electrons in (linear) conjugated polyenes. Such an approximation is called the free-electron model. Take the butadiene molecule CH2=CH-CH=CH2 as an example. The four n electrons of this system would fill up the [Pg.16]

Now that we know the allowed values of k and A, we can specify the wave function for the particle in a one-dimensional box as... 

Unlike the particle in a one-dimensional box, the electron of the hydrogen atom moves in three dimensions and has potential energy, because of its attraction to the positive nucleus at the atom s center. These differences can be easily accounted for by including the second derivatives with respect to all... 

Less obviously the trigonometric functions are important in quantum mechanics. The simple system of the particle in a one-dimensional box has been met previously in Chapters 13, 17 and 18 here the particle is able to move in the x-direction from x = 0 to x = a, a being the length of the box. 

A simple example of the wave equation, the particle in a one-dimensional box, shows how these conditions are used. We will give an outline of the method details are available elsewhere. The box is shown in Figure 2.3. The potential energy F(x) inside the box, between jc = 0 and jc = a, is defined to be zero. Outside the box, the potential energy is infinite. This means that the particle is completely trapped in the box and would require an infinite amount of energy to leave the box. However, there are no forces acting on it within the box. 

FIGURE 2.2 Lowest four energy levels for the particle in a one-dimensional box. 

For the particle in a one-dimensional box of length /, we could have put the coordinate origin at the center of the box. Find the wave functions and energy levels for this choice of... 

For each of the following systems, give the expression for dimensional harmonic oscillator, (c) A one-particle, three-dimensional system where Cartesian coordinates are used, (d) The hydrogen atom, using spherical coordinates. 

EXAMPLE Devise a trial variation function for the particle in a one-dimensional box of length L The wave function is zero outside the box and the boundary conditions require that — 0 at X = 0 and at x = /. Hie variation function must meet these boundary conditions of being zero at the ends of the box. As noted after Eq. (4.59), the ground-state has no nodes interior to the boundary points, so it is desirable that have no interior nodes. A simple function that has these properties is the parabolic function... 

For a particle in a three-dimensional box with sides of length a, b, c, write down the variation function that is the three-dimensional extension of the variation function = x(l — x) used in Section 8.1 for the particle in a one-dimensional box. Use the integrals in the equations following (8.11) to evaluate the variational integral for the three-dimensional case find the percent error in the ground-state energy. 

Consider, for example, the particle in a one-dimensional box (Section 2.2). The transition dipole moment is where Q is the particle s charge and x is its... 

The wavefunctions, probability densities [given by iA (x)], and energies for the first four energy levels for the particle in a one-dimensional box are plotted in Eigure 1.24. 

Sketch the probability densitieA Qf O thRDF energy levels of the particle in a one-dimensional box. Without doing any calculations, determine the average value of the position of the particle (x) corresponding... 

Calculate the frequency of light required to promote a particle from the n = 2 to the n = 3 levels of the particle in a one-dimensional box, assuming that L = 5.00 A and that the mass is equal to that of an electron. 

Molecules with delocalized molecular orbitals are generally more stable than those containing molecular orbitals locahzed on only two atoms. The benzene molecule, for example, which contains delocalized molecular orbitals, is chanically less reactive (and hence more stable) than molecules containing localized C=C bonds, such as ethylene. Benzene is so stable because the energy of the pi electrons is lower when the electrons are delocalized over the entire molecule than when they are localized in individual bonds, much as the energy of the particle in a one-dimensional box is lowered when the length of the box is increased (see Section 1.3). 

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