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```22.8 ADJUSTMENT OF PARAMETERS
It is easily veriﬁed that functions (b), (c) and (d) all satisfy (22.30) but, so far as mimicking
the correct solution is concerned, we would expect from the ﬁgure that (b) would be
superior to the other two. The three evaluations are straightforward, using (22.22) and
(22.23):
1
(2 − 2x)2 dx
4/3
= 2.50
=
λb = 10
2 )2 dx
8/15
(2x
−
x
0
1 2
(3x − 6x + 3)2 dx
9/5
λc = 10
=
= 2.80
3 − 3x2 + 3x)2 dx
9/14
(x
0
1 2
(π /4) sin2 (πx) dx
π 2 /8
= 3.29.
=
λ d = 0 1 4
3/8
sin (πx/2) dx
0
We expected all evaluations to yield estimates greater than the lowest eigenvalue, 2.47,
and this is indeed so. From these trials alone we are able to say (only) that λ0 ≤ 2.50.
As expected, the best approximation (b) to the true eigenfunction yields the lowest, and
therefore the best, upper bound on λ0 . We may generalise the work of this section to other diﬀerential equations of
the form Ly = λρy, where L = L† . In particular, one ﬁnds
λmin ≤
where I and J are now given by
b
y ∗ (Ly) dx
I=
I
≤ λmax ,
J
and
J=
a
b
ρy ∗ y dx.
(22.31)
a
It is straightforward to show that, for the special case of the Sturm–Liouville
equation, for which
Ly = −(py ) − qy,
the expression for I in (22.31) leads to (22.22).
Instead of trying to estimate λ0 by selecting a large number of diﬀerent trial
functions, we may also use trial functions that include one or more parameters
which themselves may be adjusted to give the lowest value to λ = I/J and
hence the best estimate of λ0 . The justiﬁcation for this method comes from the
knowledge that no matter what form of function is chosen, nor what values are
assigned to the parameters, provided the boundary conditions are satisﬁed λ can
never be less than the required λ0 .
To illustrate this method an example from quantum mechanics will be used.
The time-independent Schrödinger equation is formally written as the eigenvalue
equation Hψ = Eψ, where H is a linear operator, ψ the wavefunction describing
a quantum mechanical system and E the energy of the system. The energy
795
CALCULUS OF VARIATIONS
operator H is called the Hamiltonian and for a particle of mass m moving in a
one-dimensional harmonic oscillator potential is given by
H =−
kx2
2 d2
,
+
2
2m dx
2
(22.32)
where is Planck’s constant divided by 2π.
Estimate the ground-state energy of a quantum harmonic oscillator.
Using (22.32) in Hψ = Eψ, the Schrödinger equation is
−
kx2
2 d2 ψ
+
ψ = Eψ,
2m dx2
2
−∞ < x < ∞.
(22.33)
The boundary conditions are that ψ should vanish as x → ±∞. Equation (22.33) is a form
of the Sturm–Liouville equation in which p = 2 /(2m), q = −kx2 /2, ρ = 1 and λ = E; it
can be solved by the methods developed previously, e.g. by writing the eigenfunction ψ as
a power series in x.
However, our purpose here is to illustrate variational methods and so we take as a trial
wavefunction ψ = exp(−αx2 ), where α is a positive parameter whose value we will choose
later. This function certainly → 0 as x → ±∞ 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.
With y = exp(−αx2 ) and therefore y = −2αx exp(−αx2 ), the required estimate is
∞
2
[(2 /2m)4α2 x2 + (k/2)x2 ]e−2αx dx
k
2 α
∞
E = λ = −∞
+ .
(22.34)
=
2
−2αx
2m
8α
e
dx
−∞
This evaluation is easily carried out using the reduction formula
∞
n−1
2
xn e−2αx dx.
In =
In−2 , for integrals of the form In =
4α
−∞
(22.35)
So, we have obtained the estimate (22.34), involving the parameter α, for the oscillator’s
ground-state energy, i.e. the lowest eigenvalue of H. In line with our previous discussion
we now minimise λ with respect to α. Putting dλ/dα = 0 (clearly a minimum), yields
α = (km)1/2 /(2), which in turn gives as the minimum value for λ
1/2
ω
k
=
E=
,
(22.36)
2 m
2
where we have put (k/m)1/2 equal to the classical angular frequency ω.
The method thus leads to the conclusion that the ground-state energy E0 is ≤ 12 ω.
In fact, as is well known, the equality sign holds, 12 ω being just the zero-point energy
of a quantum mechanical oscillator. Our estimate gives the exact value because ψ(x) =
exp(−αx2 ) is the correct functional form for the ground state wavefunction and the
particular value of α that we have found is that needed to make ψ an eigenfunction of H
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 secondlowest eigenvalue (see exercise 22.25).
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