Chapter 4 Linear Differential Operators

120 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS
iii) the normalized eigenfunctions ϕn(x) = √ 2 sin nπx are complete: any
function in L 2 [0, 1] has an (L 2 ) convergent expansion as
where
y(x) =
an =
1
0
∞
n=1
√
an 2 sin nπx (4.43)
y(x) √ 2 sin nπx dx. (4.44)
This all looks very good — exactly the properties we expect for finite Hermitian
matrices. Can we carry over all the results of finite matrix theory to
these Hermitian operators? The answer sadly is no! Here is a counterexample:
Let
Again
T = −i∂x, D(T ) = {y, T y ∈ L 2 [0, 1] : y(0) = y(1) = 0}. (4.45)
〈y1, T y2〉 − 〈T y1, y2〉 =
1
0
dx {y ∗ 1 (−i∂xy2) − (−i∂xy1) ∗ y2}
= −i[y ∗ 1y2] 1 0 = 0. (4.46)
Once more, the integrated out part vanishes due to the boundary conditions
satisfied 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 = λy will be proportional to e iλx , but an exponential
function is never zero, and cannot satisfy the boundary conditions.
It seems clear that the boundary conditions are the problem. We need
a better definition 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.
Exercise 4.3: Another disconcerting example. Let p = −i∂x. Show that the
following operator on the infinite real line is formally self-adjoint:
Now let
H = x 3 p + px 3 . (4.47)
ψλ(x) = |x| −3/2
exp − λ
4x2
, (4.48)