40.6.1 The Classically Forbidden Region

The extension of a particle’s wave functions into the classically forbidden region is an important difference between classical and quantum physics. Let’s take a closer look at the wave function in the region x≥L of Figure 40.13a. The potential energy in the classically forbidden region is U0; thus the Schrödinger equation for x≥L is

d2ψdx2=−2mℏ2(E−U0)ψ(x)

We’re assuming a confined particle, with E less than U0, so E−U0 is negative. It will be useful to reverse the order of the two energies and write

d2ψdx2=2mℏ2(U0−E)ψ(x)=1η2ψ(x) (40.34)

where

η2=ℏ22m(U0−E) (40.35)

is a positive constant. As a homework problem, you can show that the units of η are meters.

The Schrödinger equation of Equation 40.34 is one we can solve by guessing. We simply need to think of two functions whose second derivatives are a positive constant times the functions themselves. Two such functions, as you can quickly confirm, are ex/η and e−x/η. Thus, according to Equation 40.8, the general solution of the Schrödinger equation for x≥L is

ψ(x)=Aex/η+Be−x/η for x≥L (40.36)

One requirement of the wave function is that ψ→0 as x→∞. The function ex/η diverges as x→∞, so the only way to satisfy this requirement is to set A=0. Thus

ψ(x)=Be−x/η for x≥L (40.37)

This is an exponentially decaying function. Notice that all the wave functions in Figure 40.14a look like exponential decays for x>L.

The wave function must also be continuous. Suppose the oscillating wave function within the potential well (x≤L) has the value ψedge when it reaches the classical boundary at x=L. To be continuous, the wave function of Equation 40.37 has to match this value at x=L. That is,

ψ(at x=L)=Be−L/η=ψedge (40.38)

This boundary condition at x=L is sufficient to determine that the constant B is

B=ψedgeeL/η (40.39)

If we use the Equation 40.39 result for B in Equation 40.37, we find that the wave function in the classically forbidden region of a finite potential well is

ψ(x)=ψedgee−(x−L)/η  for x≥L (40.40)

In other words, the wave function oscillates until it reaches the classical turning point at x=L, then it decays exponentially within the classically forbidden region. A similar analysis could be done for x≤0.

FIGURE 40.15 shows the wave function in the classically forbidden region. You can see that the wave function at x=L+η has decreased to

ψ(at x=L+η)=e−1ψedge=0.37ψedge

Although an exponential decay does not have a sharp ending point, the parameter η measures “about how far” the wave function extends past the classical turning point before the probability of finding the particle has decreased nearly to zero. This distance is called the penetration distance:

penetration distance η=ℏ2m(U0−E) (40.41)

FIGURE 40.15 The wave function in the classically forbidden region.

The figure illustrates a wave function in a classically forbidden region.
FIGURE 40.15 Full Alternative Text

A classical particle reverses direction at the x=L turning point. But atomic particles are not classical. Because of wave–particle duality, an atomic particle is “fuzzy” with no well-defined edge. Thus an atomic particle can spread a distance of roughly η into the classically forbidden region.

The penetration distance is unimaginably small for any macroscopic mass, but it can be significant for atomic particles. Notice that the penetration distance depends inversely on the quantity U0−E, the distance of the energy level below the top of the potential well. You can see in Figure 40.14a that η is much larger for the n=4 state, near the top of the potential well, than for the n=1 state.

NOTE   In making use of Equation 40.41, you must use SI units of J s for ℏ and J for the energies. The penetration distance η is then in meters.

EXAMPLE 40.7 Penetration distance of an electron

An electron is confined in a 2.0-nm-wide region with a potential-energy depth of 1.00 eV. What are the penetration distances into the classically forbidden region for an electron in the n=1 and n=4 states?

MODEL The electron is in the finite potential well whose energies and wave functions were shown in Figure 40.14a.

SOLVE The ground state has U0−E1=1.000 eV−0.068 eV =0.932 eV. Similarly, U0−E4=0.051 eV in the n=4 state. We can use Equation 40.41 to calculate

η=ℏ2m(U0−E)={0.20 nmn=10.86 nmn=4

REVIEW These values are consistent with Figure 40.14a.