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 <x <L). At x = 0 and L and at all points beyond these limits the particle encounters an infinitely repulsive barrier (V = ∞ 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

     (2-1)

Within the box, ψ 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, ψ 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 ψ = 0 at all points where V = ∞. 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 = ∞ and ψ ≠ 0. For the second derivative to be infinite, the first derivative must be discontinuous, and so ψ 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 ψ at x = L + dx, provided that ψ 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 ψ be nonsmooth at every point where V = ∞. 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 ψ must be piecewise continuous.) Hence, if V = ∞ at a single point, we might find a solution ψ which is finite at that point, with finite energy. If V = ∞ over a finite range of connected points, however, either E for the system is infinite, and ψ is finite over this region or E is not infinite (but is indeterminate) and ψ is zero over this region.

We are not concerned with particles of infinite energy, and so we will say that the solution to Eq. (2-1) is ψ = 0.1

Turning now to Eq. (2-2) we ask what solutions ψ 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 ψ. 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 ψ 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, ψ 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 ψ = 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.