130_notes.dvi

We can subtract the same equations to most easily solve forR.

2 Reika=

1

2

Teika

[(

k

k′

−

k′

k

)

e−^2 ik

′a

+

(

k′

k

−

k

k′

)

e^2 ik

′a

]

R=

1

4

T

[

− 2 i

k

k′

sin(2k′a) + 2i

k′

k

sin(2k′a)

]

R=

i

2

Tsin(2k′a)

[

k′

k

−

k

k′

]

R=

ikk′e−^2 ikasin(2k′a)

[

k′

k−

k

k′

]

2 kk′cos(2k′a)−i(k^2 +k′^2 ) sin(2k′a)

R=

i

(

k′^2 −k^2

)

sin(2k′a)e−^2 ika

2 kk′cos(2k′a)−i(k^2 +k′^2 ) sin(2k′a)

We have solved the boundary condition equations to find the reflection and transmission amplitudes

R = ie−^2 ika

(k′^2 −k^2 ) sin(2k′a)

2 kk′cos(2k′a)−i(k′^2 +k^2 ) sin(2k′a)

T = e−^2 ika

2 kk′

2 kk′cos(2k′a)−i(k′^2 +k^2 ) sin(2k′a)

The squares of these give the reflection and transmission probability, since the potential is the same
in the two regions.

9.7.3 Bound States of a 1D Potential Well*.

In the two outer regions we have solutions

u 1 (x) =C 1 eκx

u 3 (x) =C 3 e−κx

κ=

√

− 2 mE

̄h^2

.

In the center we have the same solution as before.

u 2 (x) =Acos(kx) +Bsin(kx)

k=

√

2 m(E+V 0 )

̄h^2

(Note that we have switched fromk′tokfor economy.) We will have 4 equations in 4 unknown
coefficients.

At−awe get

C 1 e−κa=Acos(ka)−Bsin(ka)

κC 1 e−κa=kAsin(ka) +kBcos(ka).