130_notes.dvi

9.7.5 1D Model of a Molecule Derivation*.

ψ(x) =







eκx x <−d

A(eκx+e−κx) −d < x < d

e−κx x > d

κ=

√

− 2 mE
̄h^2
Since the solution is designed to be symmetric aboutx= 0, the boundary conditions at−dare the
same as atd. The boundary conditions determine the constantAand constrainκ.

Continuity ofψgives.

e−κd=A

(

eκd+e−κd

)

A=

e−κd

eκd+e−κd

The discontinuity in the first derivative ofψatx=dis

−κe−κd−Aκ

(

eκd−e−κd

)

=−

2 maV 0

̄h^2

e−κd

− 1 −A

eκd−e−κd

e−κd

=−

2 maV 0

κ ̄h^2

− 1 −

eκd−e−κd

eκd+e−κd

=−

2 maV 0

κ ̄h^2

2 maV 0

κ ̄h^2

= 1 +

eκd−e−κd

eκd+e−κd

2 maV 0

κ ̄h^2

= 1 + tanh(κd)

We’ll need to study this transcendental equation to see what the allowed energies are.

9.7.6 1D Model of a Crystal Derivation*

We are working with the periodic potential

V(x) =aV 0

∑∞

n=−∞

δ(x−na).

Our states have positive energy. This potential has the symmetrythat a translation by the lattice
spacingaleaves the problem unchanged. The probability distributions must therefore have this
symmetry
|ψ(x+a)|^2 =|ψ(x)|^2 ,

which means that the wave function differs by a phase at most.

ψ(x+a) =eiφψ(x)