130_notes.dvi

For this problem, both regions haveE > V, so we will use the complex exponential solutions in both
regions. This is essentially a 1D scattering problem. Assume there is abeam of particles with
definite momentumcoming in from the left and assume there is no flux of particles coming from
the right.

Forx <0, the solution is

u(x) =eikx+Re−ikx

k=

√

2 mE

̄h^2

.

Note we have assumed the coefficient of the incident beam is 1. (Multiplying by some number does
not change the physics.) Forx >0 the solution is

u′(x) =Teik

′x

k′=

√

2 m(E−V 0 )

̄h^2

(Note that a beam coming from the right, would have given ae−ik
′x
term forx >0.)

0

V(x)

incident wave

reflected

transmitted

ik’x

V 0

0 x

E

e + Re

Energy

Te

ikx -ikx

There aretwo unknown coefficientsRandTwhich will be determined by matching boundary
conditions. We will not require normalization to one particle, since we have a beam with definite
momentum, which cannot be so normalized. (A more physical problemto solve would use an
incoming wave packet with a spread in momentum.)

Continuity of the wave functionatx= 0 implies

1 +R=T.

The exponentials are all equal to 1 there so the equation is simple.