CALCULUS OF VARIATIONS
operatorHis called the Hamiltonian and for a particle of massmmoving in a
one-dimensional harmonic oscillator potential is given by
H=−
^2
2 m
d^2
dx^2
+
kx^2
2
, (22.32)
whereis Planck’s constant divided by 2π.
Estimate the ground-state energy of a quantum harmonic oscillator.
Using (22.32) inHψ=Eψ, the Schr ̈odinger equation is
−
^2
2 m
d^2 ψ
dx^2
+
kx^2
2
ψ=Eψ, −∞<x<∞. (22.33)
The boundary conditions are thatψshould vanish asx→±∞. Equation (22.33) is a form
of the Sturm–Liouville equation in whichp=^2 /(2m),q=−kx^2 /2,ρ=1andλ=E;it
can be solved by the methods developed previously, e.g. by writing the eigenfunctionψas
a power series inx.
However, our purpose here is to illustrate variational methods and so we take as a trial
wavefunctionψ=exp(−αx^2 ), whereαis a positive parameter whose value we will choose
later. This function certainly→0asx→±∞and is convenient for calculations. Whether
it approximates the true wave function is unknown, but if it does not our estimate will
still be valid, although the upper bound will be a poor one.
Withy=exp(−αx^2 ) and thereforey′=− 2 αxexp(−αx^2 ), the required estimate is
E=λ=
∫∞
−∞[(
(^2) / 2 m)4α (^2) x (^2) +(k/2)x (^2) ]e− 2 αx^2 dx
∫∞
−∞e
− 2 αx^2 dx =
^2 α
2 m
+
k
8 α
. (22.34)
This evaluation is easily carried out using the reduction formula
In=
n− 1
4 α
In− 2 , for integrals of the form In=
∫∞
−∞
xne−^2 αx
2
dx.
(22.35)
So, we have obtained the estimate (22.34), involving the parameterα, for the oscillator’s
ground-state energy, i.e. the lowest eigenvalue ofH. In line with our previous discussion
we now minimiseλwith respect toα. Puttingdλ/dα= 0 (clearly a minimum), yields
α=(km)^1 /^2 /(2), which in turn gives as the minimum value forλ
E=
2
(
k
m
) 1 / 2
=
ω
2
, (22.36)
where we have put (k/m)^1 /^2 equal to the classical angular frequencyω.
The method thus leads to the conclusion that the ground-state energyE 0 is≤^12 ω.
In fact, as is well known, the equality sign holds,^12 ωbeing just the zero-point energy
of a quantum mechanical oscillator. Our estimate gives the exact value becauseψ(x)=
exp(−αx^2 ) is the correct functional form for the ground state wavefunction and the
particular value ofαthat we have found is that needed to makeψan eigenfunction ofH
with eigenvalue^12 ω.
An alternative but equivalent approach to this is developed in the exercises
that follow, as is an extension of this particular problem to estimating the second-
lowest eigenvalue (see exercise 22.25).