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# Particle in a box

### Introduction

The particle in a box (or the square well) is a simple idealized system that can be completely solved within quantum mechanics. It refers to the situation of a particle confined within a finite region of space (the box) by an infinite potential that exists at the walls of the box. The particle experiences no forces while inside the box, but is constrained by the walls to remain in the box. This is similar to the situation of a gas confined in a container. For simplicity we start with the 1-dimensional case, where all motion is constrained to a single dimension. Later we will extend the discussion to the 2 and 3 dimensional cases.

As we shall see, the solution of the Schrödinger equation for the particle in a box problem reveals some decidedly quantum behavior of the particle that agrees with observation but contrasts sharply with the predictions of classical mechanics. This is a particularily useful illustration because this behaviour is not "forced" on the system, it arises naturally from the initial conditions. It nearly demonstrates that quantum behaviour is a natural outcome of any wave-like system, contrary to the common concept of a "quantum leap" where the behavior is almost magical.

The quantum behavior in the box include:

1. Energy Quantization - It is not possible for the particle to have any arbitrary energy. Instead only discrete energy levels are allowed.
2. Zero-point energy - Even at the lowest possible energy level, the energy of the particle will be non-zero. This lowest possible energy is called the zero-point energy.
3. Nodes - In contrast to classical mechanics, which predicts that a particle of a given energy could be located anywhere in the box, the Schrödinger equation predicts that for some energy levels there are nodes, or positions at which the particle can never be found.

### The Particle in a 1-dimensional Box

For the 1-dimensional case in the $x$ direction, the time-independent Schrödinger equation can be written as:

$-\frac{\hbar^2}{2 m} \frac{d^2 \psi}{d x^2} + V(x) \psi = E \psi \quad (1)$,

where

$\hbar = \frac{h}{2 \pi}$

where h is Planck's constant, m is the mass of the particle, ψ is the (complex valued) wavefunction that we want to find, V(x) is a function describing the potential at each point x and E is the energy, a real number. For the case of the particle in a 1-dimensional box of length L, the potential is zero inside the box, but rises abruptly to infinity at x = 0 and x = L. Thus for the region inside the box V(x) = 0 and Equation 1 reduces to:

$-\frac{\hbar^2}{2 m} \frac{d^2 \psi}{d x^2} = E \psi \quad (2)$

This is a well studied eigenvalue problem with a general solution of:

$\psi = A \sin(kx) + B \cos(kx)$
$E = \frac{k^2 \hbar^2}{2m} \quad (3)$

Here, A and B can be any complex numbers, and k can be any real number (k must be real because E is real).

Now in order find the specific solution for the problem at hand, we must specify the appropriate boundary conditions and find the values for A and B that satisfy those conditions. In this case the boundary conditions are that &psi; is zero at x = 0 and x = L. A handwaving argument for why these boundary conditions are appropriate, is that the particle is unlikely to be found at a location with a high potential (the potential repulses the particle), thus the probability of finding the particle, |ψ|2, must be small in these regions and decreases with increasing potential. For the case of an infinite potential, |ψ|2 must infinitesimally small or 0, thus ψ must also be zero in this region. In summary,

$\psi = 0 \qquad \mathrm{at} \qquad x = 0 \qquad \mathrm{and} \qquad x=L \quad (4)$

Substituting the general solution from Equation 3 into Equation 2 and evaluating at x = 0 (ψ = 0), we find that B = 0 (since sin(0) = 0 and cos(0) = 1). It follows that the wavefunction must be of the form:

$\psi = A \sin(kx) \quad (5)$

and at x = L we find:

$\psi = A \sin(kL) = 0 \quad (6)$

One solution for Equation 6 is A = 0, however, this "trivial solution" would imply that ψ = 0 everywhere (i.e. the particle isn't in the box) and can be thrown out. If A ≠ 0 then sin(kL) = 0, which is only true when:

$kL = n \pi \quad \mathrm{where} \quad n = 1,2,\ldots$
$\mathrm{or} \quad k = \frac{n \pi}{L} \quad (7)$

(note that n = 0 is ruled out because then ψ=0 everywhere, corresponding to the case where the particle is not in the box. Negative values of n are also neglected, since they merely change the sign of sin(nx)). Now in order to find A we recognize that the particle must exist somewhere in space, so the integral of |ψ|2 over all x, which corresponds to the probability of finding the particle somewhere, is equal to 1:

$1 = \int_{-\infty}^{\infty} \left| \psi \right|^2 \, dx = \left| A \right|^2 \int_0^L \sin^2 kx \, dx = \left| A \right|^2 \frac{L}{2}$
or
$\left| A \right| = \sqrt{\frac{2}{L}} \quad (8)$

Thus, A may be any complex number with absolute value √(2/L); these different values of A yield the same physical state, so we choose A = √(2/L) to simplify.

Finally, substituting the results from Equations 7 and 8 into Equation 3 gives the complete solution for the 1-dimensional particle in a box problem:

$E_n = \frac{n^2\hbar^2 \pi ^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}, \quad n = 1,2, \cdots \quad (9)$
$\psi_n = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right) \quad (10)$

Note, that as mentioned previously, only "quantized" energy levels are possible. Also, since n cannot be zero, the lowest energy from Equation 9 is also non-zero. This zero-point energy, as it is called, can be explained in terms of the Uncertainty Principle. Because the particle is constrained within a finite region, its position is not completely indefinite. This means due to the uncertainty principle that the particle's momentum cannot be zero, so the particle must contain some amount of energy that increases as the length of the box, L, decreases.

Also, since ψ consists of sine waves, for any value of n greater than one, there are regions within the box for which ψ and thus ψ2 both equal zero, indicating that for these energy levels, nodes exist in the box where the probability of finding the particle is zero.

### The Particle in a 2-dimensional or 3-dimensional Box

For the 2-dimensional case the particle is confined to a rectangular surface of length Lx in the x-direction and Ly in the y-direction. Again the potential is zero inside the "box" and infinite at the walls. For the region inside the box, where the potential is zero, the two dimensional analogue of Equation 2 applies:

$-\frac{\hbar^2}{2m} \left( \frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2} \right) =E\psi \quad(11)$

In this case ψ is a function of both x and y, so ψ=ψ(x,y). In order to solve Equation 11, we use the method of separation of variables. First, we assume that ψ can be expressed as the product of two independent functions, the first depending only on x and the second depending only on y, i.e.:

$\psi(x,y) = X(x) Y(y) \quad (12)$

Substituting Equation 12 into Equation 11 and evaluating the partial derivatives gives:

$-\frac{\hbar^2}{2m} \left( Y\frac{\partial^2X}{\partial x^2}+X\frac{\partial^2 Y}{\partial y^2} \right) =E X Y \quad(13)$

which upon dividing by XY and rewriting d2X/dx2 as X" and d2Y/dy2 as Y" becomes:

$-\frac{\hbar^2}{2m} \left( \frac{X}{X}+\frac{Y}{Y} \right) \quad (14)$

Now we note that since X"/X is independent of y, varying y can only change the Y"/Y term. However, from Equation 14 we see that changing Y"/Y without varying X"/X, would also change E, but E is a constant, so Y"/Y must also be a constant, independent of y. The same argument can be applied to show that X"/X is independent of x. Since X"/X and Y"/Y are constants, we can write:

$-\frac{\hbar^2}{2m}\frac{X}{X} = E_x \quad and \quad -\frac{\hbar^2}{2m}\frac{Y}{Y} = E_y \quad (15)$

where Ex + Ey = E. Expanding X" and Y" in terms of the derivatives and rearranging gives:

$-\frac{\hbar^2}{2m}\frac{\partial^2X}{\partial x^2} = E_x X\quad (16)$
$-\frac{\hbar^2}{2m}\frac{\partial^2Y}{\partial y^2} = E_y Y\quad (17)$

each of which are of the same form as the 1-dimensional Schrödinger equation (Equation 2) we solved in the previous section. Thus, adapting the results from the previous section gives:

$X_{n_x}=\sqrt{\frac{2}{L_x}} \sin \left( \frac{n_x \pi x}{L_x} \right) \quad (18)$
$Y_{n_y}=\sqrt{\frac{2}{L_y}} \sin \left( \frac{n_y \pi y}{L_y} \right) \quad (19)$

Finally, since ψ=XY and E = Ex + Ey, we obtain the solutions:

$\psi_{n_x,n_y} = \sqrt{\frac{4}{L_x L_y}} \sin \left( \frac{n_x \pi x}{L_x} \right) \sin \left( \frac{n_y \pi y}{L_y} \right) \quad (20)$

$E_{n_x,n_y} = \frac{h^2}{8m} \left[ \left( \frac{n_x}{L_x} \right)^2 + \left( \frac{n_y}{L_y} \right)^2 \right] \quad (21)$

The same separation of variables technique can be applied to the three dimensional case to give the solution:

$\psi_{n_x,n_y,n_z} = \sqrt{\frac{8}{L_x L_y L_z}} \sin \left( \frac{n_x \pi x}{L_x} \right) \sin \left( \frac{n_y \pi y}{L_y} \right) \sin \left( \frac{n_z \pi z}{L_z} \right) \quad (22)$

$E_{n_x,n_y,n_z} = \frac{h^2}{8m} \left[ \left( \frac{n_x}{L_x} \right)^2 + \left( \frac{n_y}{L_y} \right)^2 + \left( \frac{n_z}{L_z} \right)^2 \right] \quad (23)$

An interesting feature of the above solutions is that when two or more of the lengths are the same (eg. Lx = Ly), there are multiple wavefunctions corresponding to the same total energy. For example the wavefunction with nx = 2, ny = 1 has the same energy as the wavefunction with nx = 1, ny = 2. This situation is called degeneracy and for the case where exactly two degenerate wavefunctions have the same energy that energy level is said to be doubly degenerate. Degenaracy results from symmetry in the system. For the above case two of the lengths are equal so the system is symmetric with respect to a 90° rotation.

### Finite Potential Case

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