Physical Chemistry Third Edition

16.4 Postulate 4 and Expectation Values 703

The right one-third of the box will have the same probability as the left one-third because of

the symmetry of the probability density. The probability of finding the particle in the center

region will be

(Probability) 1 −2(0.402249) 0. 195501

The node in the wave function atxa/2 results in a smaller probability for the middle third

of the box than for the left third or right third of the box.

Figure 16.5 shows the probability densities for the first few energy eigenfunctions

of a harmonic oscillator. The horizontal axis for each graph is placed at a height in the

figure equal to its energy eigenvalue, and the potential energy is also plotted in the figure.

The classical turning point for each state is located where the axis for that state crosses

the potential energy curve, since that is the point where the classical energy equals the

potential energy and the kinetic energy vanishes. Classical mechanics predicts that the

oscillator cannot go beyond the turning point. For the quantum-mechanical harmonic

oscillator the probability density is nonzero in the classically forbidden region. This

penetration into a classically forbidden region is calledtunneling. The name was chosen

because a tunnel into a hillside allows access to a location under an inaccessible location

of high gravitational potential energy.

23

Energy / 0

h^

or wave function squared

1

2

3

4

5

22 210123

ŒWa z

Figure 16.5 The Probability Den-
sity for the First Few Energy Eigen-
states of the Harmonic Oscillator.
EXAMPLE16.14
Calculate the probability that the harmonic oscillator will be found in the classically forbidden
region for thev0 state.
Solution
For a classical energy equal to the quantum mechanical energy forv0, the turning point
is given by

xt^2 

hv

k



h

k

1

2 π

√

k

m

 ̄

h

√

km



1

a

|xt|

√

1

a

The probability that the harmonic oscillator is in the classically permitted region is:

(Probability)

(

a

π

) 1 / 2 ∫√ 1 /a

−√ 1 /a

e−ax

2

dx 2

(

a

π

) 1 / 2 ∫√ 1 /a

0

e−ax

2

dx

where we have used the fact that since the integrand is an even function, the integral from

0to

√

1 /ais equal to half of the integral from−

√

1 /ato

√

1 /a. This integral is related to

theerror function, which is discussed in Appendix C. The error function is defined by the

equation:^3

erf(z)

2

√

π

∫z

0

e−y

2

dy (definition) (16.4-22)

(^3) M. Abramowitz and I. A. Stegun, eds.,Handbook of Mathematical Functions with Formulas, Graphs,
and Mathematical Tables, U.S. Government Printing Office, Washington, DC, 1964. See Appendix C.