130_notes.dvi

Theeigenfunctions are orthogonal.

What if two of the eigenfunctions have thesame eigenvalue? Then, our proof doesn’t work.
Assume〈ψ 2 |ψ 1 〉is real, since we can always adjust a phase to make it so. Since any linear combination
ofψ 1 andψ 2 has the same eigenvalue, we can use any linear combination. Our aim willbe tochoose
two linear combinations which are orthogonal. Lets try

ψ+ =

1

√

2

(ψ 1 +ψ 2 )

ψ− =

1

√

2

(ψ 1 −ψ 2 )

so

〈ψ+|ψ−〉 =

1

2

(1−1 + (〈ψ 1 |ψ 2 〉−〈ψ 2 |ψ 1 〉))

=

1

2

(〈ψ 1 |ψ 2 〉−〈ψ 2 |ψ 1 〉) = 0.

This is zero under the assumption that the dot product is real.

We have thus found anorthogonal set of eigenfunctions even in the case that some of the
eigenvalues are equal(degenerate). From now on we will just assume that we are workingwith
an orthogonal set of eigenfunctions.

8.7.2 Continuity of Wavefunctions and Derivatives

We can use the Schr ̈odinger Equation to show that thefirst derivative of the wave function
should be continuous, unless the potential is infinite at the boundary.

− ̄h^2

2 m

d^2 ψ

dx^2

= (E−V(x))ψ

d^2 ψ

dx^2

=

2 m

̄h^2

(V(x)−E)ψ

Integrate both sides from just below a boundary (assumed to be atx= 0) to just above.

∫+ǫ

−ǫ

d^2 ψ

dx^2

dx=

2 m

̄h^2

∫+ǫ

−ǫ

(V(x)−E)ψdx→ 0

Letǫgo to zero and the right hand side must go to zero for finite potentials.

dψ

dx

∣

∣

∣

∣

+ǫ

−

dψ

dx

∣

∣

∣

∣

−ǫ

→ 0

Infinite potentials are unphysical but often handy. The delta function potential is very handy, so we
will derive a special continuity equation for it. AssumeV(x) =V 0 δ(x). Integrating the Schr ̈odinger
Equation, we get
∫+ǫ

−ǫ

d^2 ψ

dx^2

dx=

2 m

̄h^2

∫+ǫ

−ǫ

(V 0 δ(x)−E)ψdx