130_notes.dvi

1.12 The Harmonic Oscillator in One Dimension

Next we solve for the energy eigenstates of the harmonic oscillator(See section 9.2) potential
V(x) =^12 kx^2 =^12 mω^2 x^2 , where we have eliminated the spring constantkby using the classical

oscillator frequencyω=

√

k

m. The energy eigenvalues are

En=

(

n+

1

2

)

̄hω.

The energy eigenstates turn out to be a polynomial (inx) of degreentimese−mωx

(^2) / ̄h

. So the ground
state, properly normalized, is just

u 0 (x) =

(mω

π ̄h

)^14

e−mωx

(^2) / ̄h
We will later return the harmonic oscillator to solve the problem by operator methods.

1.13 Delta Function Potentials in One Dimension

The delta function potential (See section 9.3) is a very useful one to make simple models of molecules
and solids. First we solve the problem with one attractive delta functionV(x) =−aV 0 δ(x). Since
the bound state has negative energy, the solutions that are normalizable areCeκxforx <0 and
Ce−κxforx >0. Makingu(x) continuous and its first derivative have a discontinuity computed
from the Schr ̈odinger equation atx= 0, gives us exactly one bound state with

E=−

ma^2 V 02

2 ̄h^2

.

Next we use two delta functions to model a molecule (See section 9.4), V(x) =−aV 0 δ(x+d)−
aV 0 δ(x−d). Solving this problem by matching wave functions at the boundariesat±d, we find
again transcendental equations for two bound state energies. The ground state energy is more
negative than that for one delta function, indicating that the molecule would be bound. A look at
the wavefunction shows that the 2 delta function state can lower the kinetic energy compared to the
state for one delta function, by reducing the curvature of the wavefunction. The excited state has
more curvature than the atomic state so we would not expect molecular binding in that state.

Our final 1D potential, is a model of a solid (See section 9.5).

V(x) =−aV 0

∑∞

n=−∞

δ(x−na)

This has a infinite, periodic array of delta functions, so this might be applicable to a crystal. The
solution to this is a bit tricky but it comes down to

cos(φ) = cos(ka) +

2 maV 0

̄h^2 k

sin(ka).

Since the right hand side of the equation can be bigger than 1.0 (or less than -1), there are regions

ofE= ̄h

(^2) k 2
2 m which do not have solutions. There are also bands of energies with solutions. These
energy bands are seen in crystals (like Si).