The particle in a box

When studying this section, keep in mind that the simplest and earliest system taught, the particle in a box, is still an excellent model (with various adjustments) illuminating the behaviour of electrons in a metal, nucleons in a nucleus, plasma in a star, amongst other systems.

Considering just a one-dimensional box of length $L$, we see that we have the potential

$$U(x) = \left\{ \begin{array}{ll} \infty & x < 0 \\ 0 & 0 \leq x \leq L \\ \infty & x > L \end{array} \right. .$$ (56)

Figure 11: The box in one dimension : a square potential well with infinitely high walls.

In one dimension, the time-independent Schrödinger equation is

$$- \frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2} \psi(x) + U(x)\psi(x) = E \psi(x)$$ (57)

However, in the case of a particle confined between infinitely high walls, we see that the particle will be restricted to the potential free region $0 \leq x \leq L$ between the walls. This is because the particle cannot have infinitely high potential energy behind the walls. Classically, the particle could have any energy, moving back and forth making elastic collisions with the walls of the box. Quantum mechanically, we must solve for the wave function that satisfies the equation

$$\left [ \frac{\partial^2 }{\partial x^2} + \frac{2m}{\hbar^2}E\right ] \psi(x) = 0$$ (58)

subject to the appropriate boundary conditions. Clearly, on the boundaries, we must require

$$\psi(x) = 0 \mbox{ for } x=0 \mbox{ and } x=L.$$ (59)

The general solution of equation 56 is

$$\psi(x) = A\sin\frac{\sqrt{2mE}}{\hbar}x +B\cos\frac{\sqrt{2mE}}{\hbar}x$$ (60)

as can be verified by back substitution. There are two constants $A$ and $B$. These are effectively the two ``constants of integration'' which result from the solution (by integration) of the 2nd order differential equation. The value of one of the constants, as well as the value of $E$ are determined by application of the boundary conditions. The value of the other constant is determined by the normalisation condition

$$\int _{-\infty} ^{\infty} \vert\psi(x)\vert^2 dx = \int _0^L \vert\psi(x)\vert^2 dx = 1.$$ (61)

The normalisation condition is chosen so that $\vert\psi(x)\vert^2$ represents a probability density.

The boundary conditions clearly require that the coefficient of the cosine term is zero ($B=0$). So we now have

$$\psi(x) = A\sin\frac{\sqrt{2mE}}{\hbar}x$$ (62)

Applying the boundary conditions again to the remaining sine term, we have to insist

$$\frac{\sqrt{2mE}}{\hbar}L = n\pi \qquad \mbox{ n=1,2,3\ldots}.$$ (63)

There are therefore many (discrete) values of $E$ which will satisfy the boundary conditions. Quantisation of energy has therefore arisen in a natural way,

$$E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \qquad \mbox{ n=1,2,3\ldots}.$$ (64)

Note that this is very different to the classical case, where any energy values were possible.

The unnormalised solution is

$$\psi_n(x) = A\sin\frac{\sqrt{2mE_n}}{\hbar}x = A\sin\frac{n\pi x}{L}$$ (65)

Now applying the normalisation requirement

$$\int _{-\infty} ^{\infty} \vert\psi(x)\vert^2 dx = \int _0^L \vert\psi(x)\vert^2 dx$$

$$= A^2 \int _0^L \sin^2\frac{n\pi x}{L} dx$$

$$= A^2 \frac{L}{2}$$

$$= 1.$$ (66)

Therefore

$$A = \sqrt{\frac{2}{L}}.$$ (67)

Exercise 2.8
Verify the integration

Finally, the normalised wave functions for the particle in a one-dimensional box of length $L$ are

$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\frac{n\pi x}{L} \qquad \mbox{ n=1,2,3\ldots}$$ (68)

Figure 12: a) The first three wave functions and b) the correspond probability densities for a quantum particle in a box.

The first three wave functions $\psi_n(x)$ for $n=1,2,3$ are displayed in figure 12 with the corresponding probability densities. Note that the wave function $\psi_n(x)$ may be positive as well as negative. The corresponding probability density $\vert\psi_n(x)\vert^2$ is always positive. The probability density of the quantum particle is dependent on the quantum number $n$. For example, for $n=1$, the particle is most likely to be in the middle of the box. Note also that the quantum number $n$ is always one less than the number of nodes. There are always nodes at the walls of the box. For a given energy state, a quantum particle may not be anywhere in the box with equal probability. In contrast, for any kinetic energy, a classical particle is equally likely to be found anywhere in the box.

The lowest energy that the particle can have is

Exercise 2.9
Find the probability that a particle trapped in a box $L$ wide can be found between 0.45$L$ and 0.55$L$ for the ground and first excited states.

Exercise 2.10
Find the expectation value $\langle x \rangle$ of a particle trapped in a box $L$ wide.