180
3 Quantum Mechanics – II
Region 3: (x > a) V = 0
Solution: ψ 3 = D exp(ik 1 x)
(b) Boundary conditions:
ψ 1 (0) = ψ 2 (0) → 1 + A = B + C
(1)
dψ 1
dx
x=0
=
dψ 2
dx
x=0
→ ik 1 (1 − A) = k 2 (B − C)
( 2 )
ψ 2 (a) = ψ 3 (a) → B exp(k 2 a) + C exp(−k 2 a) = D exp(ik 1 a)
( 3 )
dψ 2
dx
x=a
=
dψ 3
dx
x=a
→ k 2 (B exp(k 2 a) − Ck 2 exp(−k 2 a))
= ik 1 D exp(ik 1 a)
(4)
Eliminate A between (1) and (2) to get
B(k 2 + ik 1 ) − C(k 2 − ik 1 ) = 2ik 1
(5)
Eliminate D between (3) and (4) to get
k 2 (B exp(k 2 a) − Ck 2 exp(−k 2 a)) = ik 1 (B exp(k 2 a) + C exp(−k 2 a)) (6)
Solve (5) and (6) to get
B =
2ik 1 (k 2 + ik 1 )
(k 2 + ik 1 )
2
− exp(2k 2 a)(k 2 − ik 1 ) 2
(7)
C =
2ik 1 (k 2 − ik 1 )e
2k2a
(k 2 + ik 1 )
2
− e 2k 2 a (k 2 − ik 1 )
2
(8)
Using the values of B and C in (3),
τ = D =
4ik 1 k 2 exp(−ik 1 a)
(k 2 + ik 1 ) 2 exp(−k 2 a) − (ik 1 − k 2 ) 2 exp(k 2 a)
(9)
3.31 (a) F trans = τ
∗
τ = |D|
2
= 16
k
2
1 k
2
2
(k
2
1 + k
2
2 ) 2 (e 2k 2 a + e −2k 2 a ) − 2(k
4
2 − 6k
2
2 k
2
1 + k
4
1 )
This expression simplifies to
F trans = T =
4k
2
1 k
2
2
(k
2
1 + k
2
2 ) 2 sinh 2 (k 2 a) + 4k
2
1 k
2
2
(10)
use k
2
1 = 2m E/
2 and k
2
2 = 2m(V b − E)/
2
The reflection coefficient R is obtained by substituting (7) and (8) in (1)
to find the value of A. After similar algebraic manipulations we find
R = |A|
2
=
(k
2
1 + k
2
2 )
2 sinh
2 (k 2 a)
(k
2
1 + k
2
2 ) 2 sinh 2 (k 2 a) + 4k
2
1 k
2
2
(11)
Note that R + T = 1
(b) When E > V b , k 2 becomes imaginary and
sinh (k 2 a) = i sin (k 2 a)
(12)