130_notes.dvi

〈φ|λ 1 ψ 1 +λ 2 ψ 2 〉=λ 1 〈φ|ψ 1 〉+λ 2 〈φ|ψ 2 〉

〈λ 1 φ 1 +λ 2 φ 2 |ψ〉=λ∗ 1 〈φ 1 |ψ〉+λ∗ 2 〈φ 2 |ψ〉

〈ψ|ψ〉is real and greater than 0. It equals zero iffψ= 0. We may also derive the Schwartz inequality.

〈ψ 1 |ψ 2 〉≤

√

〈ψ 1 |ψ 1 〉〈ψ 2 |ψ 2 〉

Linear operators take vectors in the space into other vectors.

ψ′=Aψ

8.5 The Particle in a 1D Box

As a simple example, we will solve the 1DParticle in a Boxproblem. That is a particle confined
to a region 0< x < a. We can do this with the (unphysical) potential which is zero with in those
limits and +∞outside the limits.

V(x) =

{

0 0< x < a

∞ elsewhere

Because of the infinite potential, this problem has veryunusual boundary conditions. (Normally
we will require continuity of the wave function and its first derivative.) The wave function must be
zero atx= 0 andx=asince it must be continuous and it is zero in the region of infinite potential.
The first derivative does not need to be continuous at the boundary (unlike other problems), because
of the infinite discontinuity in the potential.

The time independentSchr ̈odinger equation(also called the energy eigenvalue equation) is

Huj=Ejuj

with the Hamiltonian (inside the box)

H=−

̄h^2

2 m

d^2

dx^2

Our solutions will have
uj= 0

outside the box.

Thesolution inside the boxcould be written as

uj=eikx

wherekcan be positive or negative. We doneed to choose linear combinations that satisfy
the boundary conditionthatuj(x= 0) =uj(x=a) = 0.

We can do this easily bychoosing
uj=Csin(kx)