y 1 , T y2 ’ T y 1 , y2 =

0

= ’i[y1 y2 ]1 = 0.

—

(4.45)

0

Once more, the integrated out part vanishes due to the boundary conditions

satis¬ed by y1 and y2 , so T is nicely Hermitian. Unfortunately, T with these

boundary conditions has no eigenfunctions at all ” never mind a complete

set! Any function satisfying T y = »y will be proportional to ei»x , but an ex-

ponential function is never zero, and cannot satisfy the boundary conditions.

It seems clear that the boundary conditions are the problem. We need

a better de¬nition of “adjoint” than the formal one ” one that pays more

attention to boundary conditions. We will then be forced to distinguish

between mere Hermiticity, or symmetry, and true self-adjointness.

Another disconcerting example: Let p = ’i‚x . Show that the following

operator on the in¬nite real line is formally self-adjoint:

H = x3 p + px3 . (4.46)

92 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

Now let

»

ψ» (x) = |x|’3/2 exp ’ , (4.47)

4x2

where » is real and positive. Show that

Hψ» = ’i»ψ» , (4.48)

so ψ» is an eigenfunction with a purely imaginary eigenvalue. Examine the

usual proof that Hermitian operators have real eigenvalues, and identify at

which point it breaks down.

4.2.3 Adjoint Boundary Conditions

The usual de¬nition of the adjoint operator in linear algebra is as follows:

Given the operator T : V ’ V and an inner product , , we look at

u, T v , and ask if there is a w such that w, v = u, T v for all v. If there

is, then u is in the domain of T † , and T † u = w.

For ¬nite-dimensional vector spaces V there always is such a w, and so

the domain of T † is the entire space. In an in¬nite dimensional Hilbert space,

however, not all u, T v can be written as w, v with w a ¬nite-length element

of L2 . In particular δ-functions are not allowed ” but these are exactly what

we would need if we were to express the boundary values appearing in the

integrated out part, Q(u, v), as an inner-product integral. We must therefore

ensure that u is such that Q(u, v) vanishes, but then accept any u with this

property into the domain of T † . What this means in practice is that we look

at the integrated out term Q(u, v) and see what is required of u to make

Q(u, v) zero for any v satisfying the boundary conditions appearing in D(T ).

These conditions on u are the adjoint boundary conditions, and de¬ne the

domain of T † .

Example: Consider

D(T ) = {y, T y ∈ L2 [0, 1] : y(1) = 0}.

T = ’i‚x , (4.49)

Now,

1 1

— — —

dx(’i‚x u)—v

dx u (’i‚x v) = ’i[u (1)v(1) ’ u (0)v(0)] +

0 0

— —

= ’i[u (1)v(1) ’ u (0)v(0)] + w, v , (4.50)

4.2. THE ADJOINT OPERATOR 93

where w = ’i‚x u. Since v(x) is in the domain of T , we have v(1) = 0, and

so the ¬rst term in the integrated out bit vanishes whatever value we take

for u(1). On the other hand, v(0) could be anything, so to be sure that the

second term vanishes we must demand that u(0) = 0. This, then, is the

adjoint boundary condition. It de¬nes the domain of T † :

T † = ’i‚x , D(T †) = {y, T y ∈ L2 [0, 1] : y(0) = 0}. (4.51)

For our problematic operator

D(T ) = {y, T y ∈ L2 [0, 1] : y(0) = y(1) = 0},

T = ’i‚x , (4.52)

we have

1 1

— —

v]1 dx(’i‚x u)— v

dx u (’i‚x v) = ’i[u +

0

0 0

= 0 + w, v , (4.53)

where again w = ’i‚x u. This time no boundary conditions need be imposed

on u to make the integrated out part vanish. Thus

T † = ’i‚x , D(T † ) = {y, T y ∈ L2 [0, 1]}. (4.54)

Although any of these operators “T = ’i‚x ” is formally self-adjoint we

have,

D(T ) = D(T † ), (4.55)

so T and T † are not the same operator and none of them is truly self-adjoint.

4.2.4 Self-adjoint Boundary Conditions

A formally self-adjoint operator T is truly self adjoint only if the domains of

T † and T coincide. From now on, the unquali¬ed phrase “self-adjoint” will

always mean “truly self-adjoint”.

Self-adjointness is often desirable in physics problems. It is therefore

useful to investigate what boundary conditions lead to self-adjoint operators.

For example, what are the most general boundary conditions we can impose

on T = ’i‚x if we require the resultant operator to be self-adjoint? Now,

1 1

—

dx(’i‚x u)— v = ’i u— (1)v(1) ’ u— (0)v(0) .

dx u (’i‚x v) ’ (4.56)

0 0

94 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

Demanding that the right-hand side be zero gives us, after division by u— (0)v(1),

u— (1) v(0)

= . (4.57)

u— (0) v(1)

We require this to be true for any u and v obeying the same boundary

conditions. Since u and v are unrelated, both sides must equal the same

constant κ, and this constant must obey κ— = κ’1 . Thus, the boundary

condition is

u(1) v(1)

= eiθ

= (4.58)

u(0) v(0)

for some real angle θ. The domain is therefore

D(T ) = {y, T y ∈ L2 [0, 1] : y(1) = eiθ y(0)}. (4.59)

These are twisted periodic boundary conditions.

With these generalized periodic boundary conditions, everything we ex-

pect of a self-adjoint operator actually works:

i) The functions un = ei(2πn+θ)x , with n = . . . , ’2, ’1, 0, 1, 2 . . . are eigen-

functions of T with eigenvalues kn ≡ 2πn + θ.

ii) The eigenvalues are real.

iii) The eigenfunctions form a complete orthonormal set.

Because self-adjoint operators possess a complete set of mutually orthogo-

nal eigenfunctions, they are compatible with the interpretational postulates

of quantum mechanics, where the square of the inner product of a state

vector with an eigenstate gives the probability of measuring the associated

eigenvalue. In quantum mechanics, self-adjoint operators are therefore called

observables.

Example: The Sturm-Liouville equation. With

d d

x ∈ [a, b],