Physical Chemistry Third Edition

16.5 The Uncertainty Principle of Heisenberg 711

c.Calculate the probability that a quantum-mechanical

oscillator in thev1 state is farther from its

equilibrium position than the turning-point value. You

can use the identity

∫x

0

t^2 e−at

2

dt

√

π

4 a^3 /^2

erf(

√

ax)−

x

2 a

e−ax

2

16.30a.Calculate the probability that a particle in a
one-dimensional box will be found within one
standard deviation of its mean position for then 2
stationary state. Compare your result with the value
for then1 state from the previous example.
b.Calculate the probability that a particle in a
one-dimensional box will be found within one
standard deviation of its mean position for then 3
stationary state.

16.31a.Show that application of the definition of the standard

deviation in Eq. (16.4-26) to the Gaussian probability

distribution in Eq. (16.4-28) gives the same standard

deviation as specified in that equation.

b.By comparison with the Gaussian distribution, find an

expression for the standard deviation of the position of

the harmonic oscillator in thev0 state.

c.For a Gaussian distribution, show that the probability

that the variables lies betweenμ−σandμ+σis

equal to 0.683.

16.32For a particle in a one-dimensional box, find〈E〉andσE

for the coordinate wave function

ψ

√

1

3

ψ 1 +

√

2

3

ψ 2

whereψ 1 andψ 2 are the first two energy eigenfunctions.

16.5 The Uncertainty Principle of Heisenberg

If the statistical case applies we want to be able to predict the uncertainty in a proposed

measurement. If a single measurement is to be made, there is roughly a two-thirds prob-

ability that the result will lie within one standard deviation of the expectation value. We

will use the standard deviation as a prediction of the uncertainty. In Example 16.16 we

determined that the standard deviation of the position of a particle in a one-dimensional

box of lengthais equal to 0.180756afor then1 state. We now find the standard

deviation of the momentum for this state.

EXAMPLE16.21

Find〈px〉,〈p^2 x〉, andσpxfor then1 state of a particle in a one-dimensional box.

Solution

〈px〉

2

π

∫a

0

sin

(

πx

a

)

h ̄

i

d

dx

sin

(

πx

a

)

dx

2 h ̄

πi

∫a

0

sin

(

πx

a

)

cos

(

πx

a

)

dx 0

〈p^2 x〉

2

a

∫a

0

sin

(

πx

a

)(

−h ̄^2

)d 2

dx^2

sin

(

πx

a

)

dx



2

a ̄

h^2

(

π

a

) 2 ∫a

0

sin

(

πx

a

)

sin

(

πx

a

)

dx



2

a

h ̄^2

(

π

a

) 2

a

2

 ̄

h^2 π^2

a^2

σpx

[

〈p^2 x〉−〈px〉^2

] 1 / 2

〈p^2 x〉^1 /^2  ̄

hπ

a



h

2 a

There is roughly a two-thirds probability that the momentum lies between−h/ 2 aandh/ 2 a.

The statistical case applies, as it did with the position. We can predict the mean and the

standard deviation of a set of many measurements of the momentum, but it is not possible to

predict the outcome of a single measurement.