130_notes.dvi

We have the equation.
− ̄h^2
2 m

∂^2 ψ(x)

∂x^2

+V 0 ψ(x) =Eψ(x)

∂^2 ψ(x)

∂x^2

=−

2 m

̄h^2

(E−V 0 )ψ(x)

Remember thatxis an independent variable in the above equation whilepandEare constants to
be determined in the solution.

ForE > V 0 , there are solutions
eikx

and
e−ikx

if we definekby the equation ̄hk= +

√

2 m(E−V 0 ). These are waves traveling in opposite directions
with the same energy (and magnitude of momentum).

We could also use the linear combinations of the above two solutions

sin(kx)

and
cos(kx).

There are only two linearly independent solutions. We need to chooseeither the exponentials or the
trig functions, not both. The sin and cos solutions represent states of definite energy but contain
particles moving to the left and to the right. They are not definite momentum states. They will be
useful to us for some solutions.

The solutions are also technically correct forE < V 0 butkbecomes imaginary. Lets write the
solutions in terms of ̄hκ=i ̄hk=

√

2 m(V 0 −E) The solutions are

eκx

and
e−κx.

These are not waves at all, but real exponentials. Note that theseare solutions for regions where
the particle is not allowed classically, due to energy conservation; the total energy is less than the
potential energy. We will use these solutions in Quantum Mechanics.

7.7 Sample Test Problems

1.

2.