19.3 EXERCISES
that would involve a number of non-trivial integrals if tackled using explicit
wavefunctions.
Given that the first-order change in the ground-state energy of a quantum system when
it is perturbed by a small additional termH′in the Hamiltonian is〈 0 |H′| 0 〉, find the first-
order change in the energy of a simple harmonic oscillator in the presence of an additional
potentialV′(x)=λx^3 +μx^4.
From the definitions ofAandA†, equation (19.39), we can write
x=
1
√
2 mω
(A+A†) ⇒ H′=
λ
(2mω)^3 /^2
(A+A†)^3 +
μ
(2mω)^2
(A+A†)^4.
We now compute successive values of (A+A†)n| 0 〉forn=1, 2 , 3 ,4, remembering that
A|n〉=
√
n|n− 1 〉 and A†|n〉=
√
(n+1)|n+1〉:
(A+A†)| 0 〉=0+^1 /^2 | 1 〉,
(A+A†)^2 | 0 〉=| 0 〉+
√
2 | 2 〉,
(A+A†)^3 | 0 〉=0+^3 /^2 | 1 〉+2^3 /^2 | 1 〉+
√
6 ^3 /^2 | 3 〉
=3^3 /^2 | 1 〉+
√
6 ^3 /^2 | 3 〉,
(A+A†)^4 | 0 〉=3^2 | 0 〉+
√
18 ^2 | 2 〉+
√
18 ^2 | 2 〉+
√
24 ^2 | 4 〉.
To find the energy shift we need to form the inner product of each of these state vectors
with| 0 〉.But| 0 〉is orthogonal to all|n〉ifn= 0. Consequently, the term〈 0 |(A+A†)^3 | 0 〉
in the expectation value is zero, and in the expression for〈 0 |(A+A†)^4 | 0 〉only the first
term is non-zero; its value is 3^2. The perturbation energy is thus given by
〈 0 |H′| 0 〉=
3 μ^2
(2mω)^2
.
It could have been anticipated on symmetry grounds that the expectation ofλx^3 ,an
odd function ofx, would be zero, but the calculation gives this result automatically. The
contribution of the quadratic term in the perturbation would have been much harder to
anticipate!
19.3 Exercises
19.1 Show that the commutator of two operators that correspond to two physical
observables cannot itself correspond to another physical observable.
19.2 By expressing the operatorLz, corresponding to thez-component of angular
momentum, in spherical polar coordinates (r, θ, φ), show that the angular mo-
mentum of a particle about the polar axis cannot be known at the same time as
its azimuthal position around that axis.
19.3 In quantum mechanics, the time dependence of the state function|ψ〉of a system
is given, as a further postulate, by the equation
i
∂
∂t
|ψ〉=H|ψ〉,
whereHis the Hamiltonian of the system. Use this to find the time dependence
of the expectation value〈A〉of an operatorAthat itself has no explicit time
dependence. Hence show that operators that commute with the Hamiltonian
correspond to the classical ‘constants of the motion’.