130_notes.dvi

Continuity of the derivative of the wavefunctionatx= 0 gives

[ikeikx−ikRe−ikx]x=0= [ik′Teik

′x

]x=0

Evaluate and plug inTfrom the equation above. We can solve the problem.

k(1−R) =k′(1 +R)

(k+k′)R= (k−k′)

The coefficients are

R=

k−k′

k+k′

T= 1 +R=

2 k

k+k′

We now have thefull solution, given our assumption of particles incident from the left.

u(x) =

{

eikx+k−k

′

k+k′e

−ikx x < 0

2 k

k+k′e

ik′x x > 0

Classically, all of the particles would be transmitted, continuing on to infinity.

In Quantum Mechanics, some probability is reflected.

Preflection=|R|^2 =

(

k−k′

k+k′

) 2

(Note that we can simply use the coefficient ofe−ikxbecause the incoming term has a coefficient of
1 and because the reflected particles are moving with the same velocity as the incoming beam.)

If we wish to compute thetransmission probability, the easy way to do it is to say that its

Ptransmission= 1−Preflection=

4 kk′

(k+k′)^2

We’ll get the same answers for the reflection and transmission coefficients using the probability flux
(See section 9.7.1) to solve the problem.

The transmission probability goes to 1 onek =k′ (since there is no step). The transmission
probability goes to 0 fork′= 0 (since the kinetic energy is zero).

9.1.3 The Potential Well withE > 0 *.

With positive energy, this is again a scattering type problem, now withthree regions of the potential,
all withE > V.

V(x) =







0 x <−a

−V 0 −a < x < a

0 x > a