1000 Solved Problems in Modern Physics

158 3 Quantum Mechanics – II

This is the continuity equation where the probability currentJ= 2 im^1 (φ∗∇φ−

φ∇φ∗)

And probability density

ρ=

i

2 m

(

φ∗

∂φ

∂t

−φ

∂φ∗

∂t

)

For a force free particle the solution of the Klein – Gordan equation isφ=

Aei(p.x−Et)

The probability density is

ρ=

i

2 m

[(

A∗e−i(p.x−Et)

)

(iAE)e−i(p.x−Et)−

(

Ae−i(p.x−Et)

)(

iA∗E

)

e−i(p.x−Et)

]

=

i

2 m

[

A∗A(−iE)−AA∗(iE)

]

=

|A|^2

2 m

[E+E]=E

|A|^2

m

AsEcan have positive and negative values, the probability density could

then be negative

3.6 (a) Class I: Refer to Problem 3.25

ψ 1 =Aeβx(−∞<x<−a)

ψ 2 =Dcosαx(−a<x<+a)

ψ 3 =Ae−βx(a<x<∞)

Normalization implies that

∫−a

−∞

|ψ 1 |^2 dx+

∫a

−a

|ψ 2 |^2 dx+

∫∞

a

|ψ 3 |^2 dx= 1

∫−a

−∞

A^2 e^2 βxdx+

∫a

−a

D^2 cos^2 αxdx+

∫∞

a

A^2 e−^2 βxdx= 1

A^2 e−^2 βa/ 2 β+D^2 [a+sin(2αa)/ 2 α]+A^2 e−^2 βa/ 2 β= 1

Or

A^2 e−^2 βa/β+D^2 (a+sin(2αa)/ 2 α)=1(1)

Boundary condition atx=agives

Dcosαa=ae−βa (2)

Combining (1) and (2) gives

D=

(

a+

1

β

)− 1

A=eβacosαa

(

a+

1

β

)− 1