Wave Propagation in Linear Media | re-examined

4 One-dimensional quantum tunnelling
whereas behind the barrier, only the transmitted part of the wave will remain,
3 = C 5 e ,j!t e j 1x : (4.37)
To obtain the coe cients, we set C 1 = 1 and establish the continuity conditions of the wave
functions and their spatial derivatives at the interfaces x = 0 and x = d, respectively,
C 1 + C 2 = C 3 + C 4
1C 1 , 1C 2 = 2C 3 , 2C 4
C 3 e j 2d + C4 e ,j 2d = C5 e j 1d
2C 3 e j 2d , 2C 4 e ,j 2d = 1C 5 e j 1d :
Solving this set of equations, we nd the coe cients
j( 2
C2 =
2 , 1 2 ) sin 2d
2 1 2 cos 2d , j( 1 2 + 2 2 ) sin 2d
( 1 2 + 1 2 )e ,j 2d
C3 =
2 1 2 cos 2d , j( 1 2 + 2 2 ) sin 2d
( 1 2 , 1
C4 =
2 )ej 2d
2 1 2 cos 2d , j( 1 2 + 2 2 ) sin 2d
C 5 =
2 1 2e ,j 1d
2 1 2 cos 2d , j( 1 2 + 2 2 ) sin 2d :
(4.38)
(4.39)
If we furthermore express the propagation constant inside the barrier, 2, bythe one outside as
we did in (4.9) and use our normalised variables (4.12) and (4.13) together with the normalised
barrier thickness D = !p
d, the coe c cients become
C 2 = ,
C 3 =
C 4 =
C 5 =
j sin D p 2 , 1
2 p 2 , 1 cos D p 2 , 1 , j(2 2 , 1) sin D p 2 , 1
( p p
2 2 ,jD 2,1
, 1+ )e
2 p 2 , 1 cos D p 2 , 1 , j(2 2 , 1) sin D p 2 , 1
( p p
2 2 jD 2,1
, 1 , )e
2 p 2 , 1 cos D p 2 , 1 , j(2 2 , 1) sin D p 2 , 1
2 p 2 , 1e ,j D
2 p 2 , 1 cos D p 2 , 1 , j(2 2 , 1) sin D p 2 , 1 :
94
(4.40)