Demanding that

ψ1 , Hψ2 = Hψ1 , ψ2 , (4.73)

we ¬nd

1 1

— —

— —

ψ1L (cψ2R + dψ2R ) ’ ψ 1L (aψ2R + bψ2R ) = ψ1R ψ2R ’ ψ 1R ψ2R ,

2mL 2mR

(4.74)

and this must hold for arbitrary ψ2R , ψ2R , so, picking o¬ the coe¬cients of

these expressions and complex conjugating, we ¬nd

a— ’b—

mR

ψ1R ψ1L

= . (4.75)

’c— d—

ψ1R ψ1L

mL

Because we wish the domain of H † to coincide with that of H, these must

be same conditions that we imposed on ψ2 . Thus we must have

’1

a— ’b—

mR

ab

= . (4.76)

’c— d—

cd mL

Since

’1

1 ’b

ab a

= , (4.77)

’c

cd d

ad ’ bc

we see that this requires

mL

ab AB

= eiφ , (4.78)

cd CD

mR

where φ, A, B, C, D are real, and AD’BC = 1. Demanding self-adjointness

has therefore cut the original eight real parameters down to four. These

can be determined either by experiment or by performing the microscopic

calculation3 . Note that 4 = 22 , a perfect square, as required by the Weyl-

Von Neumann theory.

3

T. Ando, S. Mori, Surface Science 113 (1982) 124.

4.3. COMPLETENESS OF EIGENFUNCTIONS 99

4.3 Completeness of Eigenfunctions

Now that we have a clear understanding of what it means to be self-adjoint,

we can reiterate the basic claim: an operator, self-adjoint with respect to

an L2 inner product, possesses a complete set of mutually orthogonal eigen-

functions. The proof that the eigenfunctions are orthogonal is identical to

that for ¬nite matrices. We will provide a proof of completeness in the next

section.

The set of eigenvalues is, with some mathematical cavils, called the spec-

trum of the operator. It is usually denoted by σ(L). An eigenvalue is said to

belong to the point spectrum when its associated eigenfunction is normaliz-

able i.e is a bona-¬de member of L2 having a ¬nite length. Usually (but not

always) the eigenvalues of the point spectrum form a discrete set. When the

operator acts on functions on an in¬nite interval, the eigenfunctions may fail

to be normalizable. The associated eigenvalues are then said to belong to

the continuous spectrum. Sometimes, e.g. the hydrogen atom, the spectrum

is partly discrete and partly continuous. There is also something called the

residual spectrum, but this does not occur for self-adjoint operators.

4.3.1 Discrete Spectrum

The simplest problems have a purely discrete spectrum. We have eigenfunc-

tions φn (x) such that

Lφn (x) = »n φn (x), (4.79)

where n is an integer. After multiplication by suitable constants, the φn are

orthonormal,

φ— (x)φm (x ) dx = δnm , (4.80)

n

and complete. We can express the completeness condition as the statement

that

φn (x)φ— (x ) = δ(x ’ x ). (4.81)

n

n

If we take this representation of the delta function and multiply it by f (x )

and integrate over x , we ¬nd

φ— (x )f (x ) dx .

f (x) = φn (x) (4.82)

n

n

100 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

So,

f (x) = an φn (x) (4.83)

n

with

φ— (x )f (x ) dx .

an = (4.84)

n

This means that if we can expand a delta function in terms of the φn (x), we

can expand any (square integrable) function.

Note: The convergence of the series n φn (x)φ— (x ) to δ(x ’ x ) is neither

n

2

pointwise nor in the L sense. The sum tends to a limit only in the sense

of a distribution ” meaning that we must multiply the partial sums by a

smooth test function and integrate over x before we have something that

actually converges in any meaningful manner. As an illustration consider

√

our favourite orthonormal set: φn (x) = 2 sin(nπx) on the interval [0, 1]. A

plot of the ¬rst m terms in the sum

∞√ √

2 sin(nπx) 2 sin(nπx ) = δ(x ’ x )

n=1

will show “wiggles” away from x = x whose amplitude does not decrease as

m becomes large ” although they become of higher and higher frequency.

When multiplied by a smooth function and integrated, the contributions

from adjacent positive and negative wiggle regions tend to cancel, and it is

only after this integration that the sum tends to zero away from the spike at

x=x.

60

40

20

0.2 0.4 0.6 0.8 1

70

The sum 2 sin(nπx) sin(nπx ) for x = 0.4.

n=1

4.3. COMPLETENESS OF EIGENFUNCTIONS 101

Rayleigh-Ritz and Completeness

For the Schr¨dinger eigenvalue problem

o

Ly = ’y + q(x)y = »y, x ∈ [a, b], (4.85)