eventually becomes negligeable compared to »y, and then we can solve the

problem with sines and cosines. We see that there is no upper limit to the

magnitude of the eigenvalues. It can be shown that the eigenvalues of the

Sturm-Liouville problem

Ly = ’(py ) + qy = »y, x ∈ [a, b], (4.86)

are similarly unbounded. We will use this unboundedness of the spectrum to

make an estimate of the rate of convergence of the eigenfunction expansion

for functions in the domain of L, and extend this result to prove that the

eigenfunctions form a complete set.

We know from chapter one that the Sturm-Liouville eigenvalues are the

stationary values of y, Ly when the function y is constrained to have unit

length, y, y = 1. The lowest eigenvalue, »0 , is therefore given by

y, Ly

»0 = inf . (4.87)

y, y

y∈D(L)

As the variational principle, this formula provides a well-known method of

obtaining approximate ground state energies in quantum mechanics. Part of

its e¬ectiveness comes from the stationary nature of y, Ly at the minimum:

a crude approximation to y often gives a tolerably good approximation to »0 .

In the wider world of eigenvalue problems, the variational principle is named

after Rayleigh and Ritz4 .

Suppose we have already found the ¬rst n normalized eigenfunctions

y0 , y1 , . . . , yn’1 . Let the space spanned by these functions be Vn . Then an

obvious extension of the variational principle gives

y, Ly

»n = inf⊥ . (4.88)

y, y

y∈Vn

4

J. W. Strutt (later Lord Rayleigh), In Finding the Correction for the Open End of

¨

an Organ-Pipe. Phil. Trans. 161 (1870) 77; W. Ritz, Uber eine neue Methode zur L¨sung

o

gewisser Variationsprobleme der mathematischen Physik. J. reine angew. Math. 135

(1908)

102 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

We now exploit this variational estimate to show that if we expand an arbi-

trary y in the domain of L in terms of the full set of eigenfunctions ym ,

∞

y= am ym , (4.89)

m=0

where

am = ym , y , (4.90)

then the sum does indeed converge to y.

Let

n’1

hn = y ’ am ym (4.91)

m=0

⊥

be the residual error after the ¬rst n terms. By de¬nition, hn ∈ Vn . Let

us assume that we have adjusted, by adding a constant to q if necessary, L

so that all the »m are positive. This adjustment will not a¬ect the ym . We

expand out

n’1

»m |am |2 ,

hn , Lhn = y, Ly ’ (4.92)

m=0

where we have made use of the orthonormality of the ym . The subtracted

sum is guaranteed positive, so

hn , Lhn ¤ y, Ly . (4.93)

Combining this inequality with Rayleigh-Ritz tells us that

y, Ly hn , Lhn

≥ ≥ »n . (4.94)

hn , hn hn , hn

In other words

n’1

y, Ly

am ym 2 .

≥ y’ (4.95)

»n m=0

Since y, Ly is independent of n, and »n ’ ∞, we have y ’ n’1 am ym 2 ’ 0.

0

Thus the eigenfunction expansion indeed converges to y, and does so faster

than »’1 goes to zero.

n

Our estimate of the rate of convergence applies only to the expansion of

functions y for which y, Ly is de¬ned ” i.e. to functions y ∈ D (L). The

domain D (L) is always a dense subset of the entire Hilbert space L2 [a, b],

4.3. COMPLETENESS OF EIGENFUNCTIONS 103

however, and, since a dense subset of a dense subset is also dense in the larger

space, we have shown that the linear span of the eigenfunctions is a dense

subset of L2 [a, b]. Combining this observation with the alternative de¬nition

of completeness in 2.2.3, we see that the eigenfunctions do indeed form a

complete orthonormal set. Any square integrable function therefore has a

convergent expansion in terms of the ym , but the rate of convergence may

well be slower than that for functions y ∈ D (L).

Operator Methods

Sometimes there are tricks for solving the eigenvalue problem.

Example: Harmonic Oscillator. Consider the operator

H = (’‚x + x)(‚x + x) + 1 = ’‚x + x2 .

2

(4.96)

This is in the form Q† Q + 1, where Q = (‚x + x), and Q† is its formal adjoint.

If we write these in the other order we have

QQ† = (‚x + x)(’‚x + x) = ’‚x + x2 + 1 = H + 1.

2

(4.97)

Now, if ψ is an eigenfunction of Q† Q with non-zero eigenvalue » then Qψ is

eigenfunction of QQ† with the same eigenvalue. This is because

Q† Qψ = »ψ (4.98)

implies that

Q(Q† Qψ) = Qψ, (4.99)

or

QQ† (Qψ) = »(Qψ). (4.100)

The only way that this can go wrong is if Qψ = 0, but this implies that

Q† Qψ = 0 and so the eigenvalue was zero. Conversely, if the eigenvalue is

zero then

0 = ψ, Q† Qψ = Qψ, Qψ , (4.101)

and so Qψ = 0. In this way, we see that the Q† Q and QQ† have exactly the

same spectrum, with the possible exception of any zero eigenvalue.

Now notice that Q† Q does have a zero eigenvalue because

12

ψ0 = e’ 2 x (4.102)

104 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

obeys Qψ0 = 0 and is normalizable. The operator QQ† , considered as an

operator on L2 [’∞, ∞], does not have a zero eigenvalue because this would

require Q† ψ = 0, and so

12

ψ = e+ 2 x , (4.103)

which is not normalizable, and so not an element of L2 [’∞, ∞].

Since

H = Q† Q + 1 = QQ† ’ 1, (4.104)

we see that ψ0 is an eigenfunction of H with eigenvalue 1, and so an eigenfunc-

tion of QQ† with eigenvalue 2. Hence Q† ψ0 is an eigenfunction of Q† Q with

eigenvalue 2 and so an eigenfunction H with eigenfunction 3. Proceeding in

the way we ¬nd that

ψn = (Q† )n ψ0 (4.105)

is an eigenfunction of H with eigenvalue 2n + 1.