Simple Nature - Light and Matter

5 In problem 23 on p. 944, you showed that a wavefunction of

the form

Ψ 0 (x) =e−x

(^2) / 2
was a solution of the Schr ̈odinger equation for the quantum har-
monic oscillator in one dimensions. (We ignore units, and the factor
of 1/2 in the exponent is just a convention.) It represents the ground
state. The wavefunction of the first excited state is
Ψ 1 (x) =xe−x
(^2) / 2
,
with the same value ofb.
(a) Show that these states are orthogonal in the sense defined on
p. 982.
(b) What is an observable that would distinguish them?
6 (a) When an excited state in a nucleus undergoes gamma de-
cay, the half-life depends on a variety of factors, but a fairly typical
value would be about 1 ns. Find the uncertainty in energy imposed
by the energy-time uncertainty relation, and compare with a typical
excitation energy of 1 MeV.
(b) Some very neutron-rich nuclei are unstable with respect to emis-
sion of a neutron, and in these cases the half-life is typically on the
order of 10−^21 s. Carry out an estimate as in part a.
7 As you might have guessed from the equations given in problem
5, themth excited state of the one-dimensional quantum harmonic
oscillator has a wavefunction of the form
Ψm(x) =Hm(x)e−x
(^2) / 2
.
HereHmis a polynomial of orderm, andHmis an even function if
mis even, odd ifmis odd. Given these assumptions, it is possible
to find Ψ 2 simply from the requirement that it be orthogonal to Ψ 0
and Ψ 1 , without having to solve the Schr ̈odinger equation. FindH 2
by this method. (Don’t worry about normalization or phase.) Hint:
Near the end of the calculation, you will encounter integrals of the
form

∫∞

−∞x

me−x^2 dx. This can be done using software, or you can

use integration by parts to relate this integral to the corresponding

integral form−2.

√

8 In example 14 on p. 989, we defined a very naughty energy

operator

HˆΨ =iΨ.

Show that it is not hermitian, by directly using the definition on

p. 982.

Problems 1009