1000 Solved Problems in Modern Physics

180 3 Quantum Mechanics – II

Region 3: (x>a)V= 0

Solution:ψ 3 =Dexp(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)→Bexp(k 2 a)+Cexp(−k 2 a)=Dexp(ik 1 a)(3)

dψ 2

dx

∣

∣

∣

∣

x=a

=

dψ 3

dx

∣

∣

∣

∣

x=a

→k 2 (Bexp(k 2 a)−Ck 2 exp(−k 2 a))

=ik 1 Dexp(ik 1 a)

(4)

EliminateAbetween (1) and (2) to get

B(k 2 +ik 1 )−C(k 2 −ik 1 )= 2 ik 1 (5)

EliminateDbetween (3) and (4) to get

k 2 (Bexp(k 2 a)−Ck 2 exp(−k 2 a))=ik 1 (Bexp(k 2 a)+Cexp(−k 2 a)) (6)

Solve (5) and (6) to get

B=

2 ik 1 (k 2 +ik 1 )

[

(k 2 +ik 1 )^2 −exp(2k 2 a)(k 2 −ik 1 )^2

] (7)

C=

2 ik 1 (k 2 −ik 1 )e^2 k^2 a

[

(k 2 +ik 1 )^2 −e^2 k^2 a(k 2 −ik 1 )^2

] (8)

Using the values ofBandCin (3),

τ=D=

4 ik 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)Ftrans=τ∗τ=|D|^2 =^16

k^21 k^22

(k^21 +k^22 )^2 (e^2 k^2 a+e−^2 k^2 a)−2(k 24 − 6 k 22 k 12 +k^41 )

This expression simplifies to

Ftrans=T=

4 k 12 k^22

(k^21 +k^22 )^2 sinh^2 (k 2 a)+ 4 k^21 k^22

(10)

usek^21 = 2 mE/^2 andk^22 = 2 m(Vb−E)/^2

The reflection coefficientRis obtained by substituting (7) and (8) in (1)

to find the value ofA. After similar algebraic manipulations we find

R=|A|^2 =

(k^21 +k 22 )^2 sinh^2 (k 2 a)

(k 12 +k^22 )^2 sinh^2 (k 2 a)+ 4 k^21 k^22

(11)

Note thatR+T= 1

(b) WhenE>Vb,k 2 becomes imaginary and

sinh(k 2 a)=isin(k 2 a) (12)