# Particle in a Box (PIB). Simple case of infinite square well potential in one dimension.

## Introduction

This is one of the simplest quantum mechanical systems. Its Schroedinger equation is easily solved using elementary calculus. This solution is commonly used in elementary courses to illustrate several fundamental concepts of quantum theory.

1. Energy quantization, stationary states, eigenvalues and eigenfunctions.
2. Wave functions, nodes, parity, orthogonality, normalization.
3. Probability distributions.
4. Tunneling when the PIB is extended to the finite square well.
5. Momentum wave functions, momentum distributions.
6. Symmetry operations and symmetry groups, accidental degeneracy.
7. Quasibound states above the "continuum" limit.
8. Relativistic quantum theory of "free" particles.

(Topics 5 through 8 are appropriate to advanced courses.) Besides these important pedagogical uses, the particle in a box is also a model for several important physical appplications.

In addition to the pedagogical uses listed above, the particle in a box problem is also a model for several important applications.

1. Absorption of light by long conjugated dye molecules.
2. The free electron model of electrical conductors (metals).
3. Translational energy levels and statistical thermodynamics of perfect gases
4. Quantum electronics and quantum computers.

## Objectives

The following sections are devoted to solving the PIB problem, and interpreting the solution. Following this, several related PIB problems are stated, some are solved, and others are exercises for the reader.

## References

1. B. D. Anderson, J. Chem. Ed. 74, 985 (1997) [Conjugated dyes and PIB models]
2. A. Das and A.C. Melissinos, Quantum Mechanics [Gordon and Breach Science Publishers, 1986], p. 43 ff.
3. C. E. Dykstra, Introduction to Quantum Chemistry [Prentice Hall, 1994], p. 47 ff.
4. P. Atkins, Physical Chemistry [W.H. Freeman and Company, 1998], p. 314 ff.

## Statement of the Problem

The PIB is an idealized physical system consisting of a particle constrained to move in one dimension between infinite potential walls. Thus, the particle is subject to no forces between the walls but infinite force at the walls to prevent its escape from the box. The total energy of the particle is the sum of kinetic and potential terms as represented by the Hamiltonian operator of quantum mechanics: . Here , and h=6.626 10-34 joule sec is Planck's constant, the fundamental constant of quantum mechanics. Two other constants are required to specify the system: m, the particle mass, and L, the length of the box. Then the object is to solve Schroedinger's time-independent equation, Hy = Ey, for the allowed energy levels, E, and states y(x) (energy eigenvalues and eigenfunctions) of the PIB.

## Dimensionless Form of the Schroedinger Equation and "Natural Units"

The solution is made easier if the coordinate is transformed to x=x/L and at the same time the energy is divided by . With these changes, the Schroedinger equation becomes . Now, wave functions are required to be "well behaved"; usually that means that (a) y is finite, differentiable (at least through first order), and continuous, (b) also must be continuous (except possibly at a finite number of points).

## Analytical Solution of the PIB problem

Elementary calculus is enough to find the solutions as shown here. Briefly, the eigenvalues (in dimensionless energy units) are En=n2 for n = 1, 2, ... , and the corresponding wave functions are . Here is a plot to show the energy levels with wave functions superimposed.

## Figure 1. PIB E, y Plot

The vertical scale is energy (in natural energy units, Enatural) and the energy levels are shown as horizontal lines. The horizontal scale is position, x/L, in the box of length L.Wave functions, y(x), are drawn using the corresponding energy level as the zero of y. Wave functions are normalized, but their vertical scale on the graph is arbitrary. This graph belongs to a system consisting of a mass, m = 4 10-30 kg in a box of length L = 10 Angstrom.

The lowest 4 levels are shown: n = 1, 2, 3, 4. Not counting the nodes at the walls of the box, n is the number of interior nodes in the wave function. The parity of states alternates with n; thus, the lowest state is even, y1(-x) = y1(x) , the first excited state is odd, y2(-x) = -y2(x), and so forth.

Analytical solutions of PIB are not always possible when the potential inside the box is modified. In such cases numerical solutions can be found. For comparison, a numerical solution of the simple PIB problem is given in an rtf file and an interactive mathcad  solution file.

## Problems

The simple PIB system discussed above can be "decorated" in various ways to model important physical systems. In the following problems the potential is V(x) = U(x)(|x|<L) + ¥(|x|>L), that is the potential is infinite outside the range -L<x<L and varies according to the function U(x) inside the range. If U(x) = 0, we have the elementary PIB described above. The solution to each of these problems can be simplified by adopting natural units of length and energy. Complete analytical expressions should be sought but if these are not found, then a numerical solution for the lowest few (say 4 or 5) energy states is acceptable.

1. Solve the PIB system in which a central potential well is added to the potential. This problem is a primitive model for an electron in a quantum dot.
In this case, U(x) = -U0(|x|<a/L) . That is, the potential inside the box is -U0 (U0 is positive) in a centered region where -a/L<x<a/L . You may assume a = 1/3 and U0 = 100Enatural. An analytical solution is possible for this case; see the rtf file. A numerical method of solution is also possible; see the mathcad file. The numerical approach can be applied even when anaytical solutions are not possible. [Hint, the ground state energy is -86.998 Enatural and the ground state wave function looks like this:

Note that the well is from x=-1/3 to x=1/3. The wave function penetrates beyond the well into regions where the potential energy is larger than the total particle energy - an illustration of tunneling.]

2. Solve the PIB with a central potential barrier. Model for an electron in a metal-oxide-metal junction.
Here, instead of a central well, the particle sees a barrier of height U0 in the box. Use U(x) = +U0(|x|<a/L) in place of the potential of problem 1.
3. Solve the PIB with a linear varying potental in the box. Models an electron in a gap having an electric field.
Let U(x) = g x for -L<x<L; that is, there is a constant force on the particle tending to push it to the left (g>0).
4. Solve the PIB with a symmetric quadratic potential in the box. Let U(x) = kx2 for -L<x<L. Models a particle on a spring between two rigid stops.

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Created or up-dated 08/03/99   by R.D. Poshusta