Transcript power point

Metals: Electron in a Box
2
E ( x) 
 ( x)  V ( x) ( x)
2
2m  x

2
V(x) = 0 for | x | ≤ L/2
inside
V(x) =  for | x | > L/2
outside


V(x)
solve for (x) within the box
2
E ( x) 
 ( x)
2m  2 x

-L/2
2
or
 0e
ikx
where
use boundary conditions to get A, B, k (or 0)

 /  
L
2
L

/2  0
L/2
E  E (k )
solutions are of the form
 ( x)  A cos kx  B sin kx
0

2mE ½
k

Metals: Electron in a Box

2mE ½
k
 ( x)  A cos kx  B sin kx


 /  
L
2
L

/2  0
 /   0  A cosk /  B sin k / 
A = B = 0 trivial solution
  /   0  A cos k /   B sin  k / 
there are no values of k that
subtract: 2 B sin k /   0
make these both true for
add: 2 A cosk /   0
arbitrary, non-zero A and B

L
L
2
k is quantized, not continuous
L
2
L
2
L
2
L
2
2
L
2
L
2
 2 solution sets
{0, p, 2p, 3p, etc….}


kL
nπ

n
'
π
k

sin k / 2  0 
set 1: A = 0
n = even
n
2
L
n = principle quantum number
 ( x)  B sin kx
kL
nπ
 ( n'  ½ ) π k n 
cos k L / 2  0 
set 2: B = 0
n = odd
2
L
 ( x)  A coskx
{p/2, 3p/2, 5p/2, etc….}

L

Metals: Electron in a Box
nπx

set 1:  ( x)  B sin 
 L 
n = even

2mE ½
k

nπx
n = odd

 L 
to solve for A and B, normalize according to
kn 
nπ
L
set 2:  ( x)  A cos
L/2
 * ( x) ( x)dx  1
L / 2
use
 cos
2
 A B
zdz  ½ z  ¼ sin 2 z
kn
 n ( x) 
2
 n( x) 
2
2
nπx
/ L sin 
 n = even
L


nπx
/ L cos
 n = odd
L


eigenfunctions
/L
general wave equation
 2π x
f ( x)  A cos




2L
n 
n
wave-length
kn 
2π
n
wave-vector