130_notes.dvi

Putting in the formula forκin terms of the energy.

− 2 mE

̄h^2

=

m^2 a^2 V 02

̄h^4

E=−

ma^2 V 02

2 ̄h^2

There is only one energy for which we can satisfy the boundary conditions. There is onlyone
bound statein an attractive delta function potential.

9.4 The Delta Function Model of a Molecule*.

The use of two delta functions allows us to see, to some extent, howatoms bind into molecules.
Our potential is
V(x) =−aV 0 (δ(x+d) +δ(x−d))

with attractive delta functions atx=±d. This is a parity symmetric potential, so we can assume
that our solutions will be parity eigenstates.

For even parity, our solution in the three regionsis

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.

A little calculation gives (See section 9.7.5)

2 maV 0

κ ̄h^2

= 1 + tanh(κd)

This is a transcendental equation, but we can limit the energy.

2 maV 0

κ ̄h^2

< 2

κ >

maV 0

̄h^2

Sinceκ=maVh ̄ 20 for the single delta function, thisκ=

√

− 2 mE
̄h^2 is larger than the one for the single
delta function. This means thatEis more negative and there ismore binding energy.

Emolecule< Eatom

Basically, the electron doesn’t have to be a localized with two atoms asit does with just one. This
allows the kinetic energy to be lower.

The figure below shows the two solutions plotted on the same graph as the potential.