40.4.1 Zero-Point Motion

The lowest energy state in Example 40.2, the ground state, has $E_{1} = 0.38 eV$ . There is no stationary state having $E = 0$ . Unlike a classical particle, a quantum particle in a box cannot be at rest! No matter how much its energy is reduced, such as by cooling it toward absolute zero, it cannot have energy less than $E_{1}$ .

The particle motion associated with energy $E_{1}$ , called the zero-point motion, is a consequence of Heisenberg’s uncertainty principle. Because the particle is somewhere in the box, its position uncertainty is $Δ x = L$ . If the particle were at rest in the box, we would know that its velocity and momentum are exactly zero with no uncertainty: $Δ p_{x} = 0$ . But then $Δ x Δ p_{x} = 0$ would violate the Heisenberg uncertainty principle. One of the conclusions that follow from the uncertainty principle is that a confined particle cannot be at rest.

Although the particle’s position and velocity are uncertain, the particle’s energy in each state can be calculated with a high degree of precision. This distinction between a precise energy and uncertain position and velocity seems strange, but it is just our old friend the standing wave. In order to have a stationary state at all, the de Broglie waves have to form standing waves. Only for very precise frequencies, and thus precise energies, can the standing-wave pattern appear.

EXAMPLE 40.3 Nuclear energies

Protons and neutrons are tightly bound within the nucleus of an atom. If we use a one-dimensional model of a nucleus, what are the first three energy levels of a neutron in a 10-fm-diameter nucleus $(1 fm = 10^{- 15} m) ?$

MODEL Model the nucleus as a one-dimensional box of length $L = 10 fm$ . The neutron is confined within the box.

SOLVE The energy levels, with $L = 10 fm$ and $m = m_{n} = 1.67 \times 10^{- 27} kg,$ are

\begin{array}{l} E_{1} & = & \frac{h^{2}}{8 m L^{2}} = 3.29 \times 10^{- 13} J = 2.06 MeV \\ E_{2} & = & 4 E_{1} = 8.24 MeV \\ E_{3} & = & 9 E_{1} = 18.54 MeV \end{array}

REVIEW You’ve seen that an electron confined in an atom-size space has energies of a few eV. A neutron confined in a nucleus-size space has energies of a few million eV.

EXAMPLE 40.4 The probabilities of locating the particle

A particle in a rigid box of length $L$ is in its ground state.

Where is the particle most likely to be found?
What are the probabilities of finding the particle in an interval of width 0.01 $L$ at $x = 0.00 L, 0.25 L$ , and $0.50 L ?$
What is the probability of finding the particle between $L / 4$ and $3 L / 4$ ?

MODEL The wave functions for a particle in a rigid box have been determined.

VISUALIZE FIGURE 40.10 shows the probability density $P_{1} (x) = {| ψ_{1} (x) |}^{2}$ in the ground state.

The figure illustrates a graphical representation of the probability density for a particle in the ground state.

FIGURE 40.10 Full Alternative Text

SOLVE

The particle is most likely to be found at the point where the probability density $P (x)$ is a maximum. You can see from Figure 40.10 that the point of maximum probability for $n = 1$ is $x = L / 2$ .
For a small width $δ x$ , the probability of finding the particle in $δ x$ at position $x$ is

$Prob(in δ x at x) = P_{1} (x) δ x = | ψ_{1} (x) |^{2} δ x = \frac{2}{L} {sin}^{2} (\frac{π x}{L}) δ x$

The interval $δ x = 0.01 L$ is sufficiently small for this to be valid. The probabilities of finding the particle are

$\begin{array}{l} Prob (in 0.01 L at x = 0.00 L) = 0.000 = 0.0 % \\ Prob (in 0.01 L at x = 0.25 L) = 0.010 = 1.0 % \\ Prob (in 0.01 L at x = 0.50 L) = 0.020 = 2.0 % \end{array}$
You learned in Chapter 39 that the probability of being in an interval is the area under the probability-density curve. We calculate this by integrating:

$\begin{array}{l} Prob (in interval \frac{1}{4} L to \frac{3}{4} L) & = & \int_{L / 4}^{3 L / 4} P_{1} (x) d x \\ = & \frac{2}{L} \int_{L / 4}^{3 L / 4} {sin}^{2} (\frac{π x}{L}) d x \\ = & {[\frac{x}{L} - \frac{1}{π} sin (\frac{π x}{L}) cos (\frac{π x}{L})]}_{L / 4}^{3 L / 4} \\ = & \frac{1}{2} + \frac{1}{π} = 0.818 \end{array}$

The integral of ${sin}^{2}$ was taken from the table of integrals in Appendix A.

REVIEW If a particle in a box is in the $n = 1$ ground state, there is an 81.8% chance of finding it in the center half of the box. The probability is greater than 50% because, as you can see in Figure 40.10, the probability density $P_{1} (x)$ is larger near the center of the box than near the boundaries.

This has been a lengthy presentation of the particle-in-a-box problem. However, it was important that we explore the method of solution completely. Future examples will now go more quickly because many of the issues discussed here will not need to be repeated.