714 16 The Principles of Quantum Mechanics. II. The Postulates of Quantum Mechanics
To calculateσpx, we require〈p^2 x〉, which is obtained in the same way as〈px〉except that the
square of the momentum operator is used. The result is〈p^2 x〉h ̄^2 κ^2 (16.5-7)The square of the standard deviation is:σpx^2 〈p^2 x〉−〈px〉^2 h ̄^2 κ^2 −h ̄^2 κ^2 0 (16.5-8)The uncertainty in one of a pair of conjugate variables can vanish only if the uncertainty in
the other variable is infinite, so in this case the uncertainty inxis infinite. If the momentum
is known exactly, we cannot say where the particle is with any finite uncertainty.EXAMPLE16.23
Write the integral that representsσxfor the free particle of Example 16.22 and argue that its
value is infinite.
Solution
We write the integrals prior to taking the limit thatL→∞.〈x〉D∗D∫L−Le−iκxxeiκxdxD∗D∫L−Le−iκxeiκxdx∫L−Lxdx
∫L−Ldx
L^2 / 2 −L^2 / 2
2 L
0〈x^2 〉D∗D∫L−Le−iκxx^2 eiκxdxD∗D∫L−Le−iκxeiκxdx∫L−Lx^2 dx
∫L−LdxL^3 / 3 +L^3 / 3
2 L
L^2
3sx(
L^2
3
− 0) 1 / 2
(
L^2
3) 1 / 2When the limitL→∞is taken, this becomes infinite.From the expectation value of the momentum of a free particle we can now justify
the apparently arbitrary choice of sign that we made in Eq. (16.3-8). It appeared at that
time that eitherihd/dx ̄ or−ihd/dx ̄ could have been chosen as the operator forpx. The
free-particle wave functionDeiκxcorresponds to a positive value of〈px〉, as shown in
Example 16.22. If we combine this coordinate wave function with the appropriate time
factor, exp(−iEt/h ̄), we obtain the time-dependent wave function
ΨDexpi(κx−Et/ ̄h) (16.5-9)
which represents a traveling wave moving to the right (with positive value ofpx). If
ihd/dx ̄ had been chosen for thepxoperator, a negative value for〈px〉would have
resulted, indicating motion in the wrong direction.Exercise 16.13
Show that taking the opposite sign for the momentum operator leads to a negative value of〈px〉
for the wave function of Eq. (15.3-28) ifF0 is taken.