130_notes.dvi

We now have two pairs of equations for then+ 1 coefficients in terms of thencoefficients.

An+1 =

2 maV 0

̄h^2 k

Bncos(ka)−Bnsin(ka) +Ancos(ka)

Bn+1 =

2 maV 0

̄h^2 k

Bnsin(ka) +Bncos(ka) +Ansin(ka)

An+1 = eiφAn

Bn+1 = eiφBn

Using the second pair of equations to eliminate then+ 1 coefficients, we have

(eiφ−cos(ka))An=

(

2 maV 0

̄h^2 k

cos(ka)−sin(ka)

)

Bn

(

eiφ−cos(ka)−

2 maV 0

̄h^2 k

sin(ka)

)

Bn= sin(ka)An.

Now we can eliminate all the coefficients.

(eiφ−cos(ka))(eiφ−cos(ka)−

2 maV 0

̄h^2 k

sin(ka))

=

(

2 maV 0

̄h^2 k

cos(ka)−sin(ka)

)

sin(ka)

e^2 iφ−eiφ

(

2 maV 0

̄h^2 k

sin(ka) + cos(ka) + cos(ka)

)

+

2 maV 0

̄h^2 k

sin(ka) cos(ka) + cos^2 (ka)

=

2 maV 0

̄h^2 k

sin(ka) cos(ka)−sin^2 (ka)

e^2 iφ−eiφ

(

2 maV 0

̄h^2 k

sin(ka) + 2 cos(ka)

)

+ 1 = 0

Multiply bye−iφ.

eiφ+e−iφ−

(

2 maV 0

̄h^2 k

sin(ka) + 2 cos(ka)

)

= 0

cos(φ) = cos(ka) +

maV 0

̄h^2 k

sin(ka)

This relation puts constraints onk, like the constraints that give us quantized energies for bound
states. Since cos(φ) can only take on values between -1 and 1, there are allowed bands ofkand gaps
between those bands.

9.8 Examples

This whole section is examples.