1000 Solved Problems in Modern Physics

3.3 Solutions 187

3.39 (a)

[(

−

^2

2 m

)

∇^2 +V

]

ψ(x,y,z)=Eψ(x,y,z)(1)

PutV= 0

(

−

^2

2 m

)[

∂^2

∂x^2

+

∂^2

∂y^2

+

∂^2

∂z^2

]

ψ(x,y,z)=Eψ(x,y,z)

Let

ψ(x,y,z)=X(x)Y(y)Z(x) (2)

YZ

∂^2 X

∂x^2

+ZX

∂^2 Y

∂y^2

+XY

∂Z

∂z^2

=−

(

2 mE

^2

)

XY Z

Dividing throughout byXYZ

1

Y

∂^2 Y

∂y^2

+

1

Z

∂^2 Z

∂z^2

=−

1

X

∂^2 X

∂x^2

−

2 mE

^2

(3)

LHS is a function ofyandzonly while the RHS is a function ofxonly.

The only way (3) can be satisfied is that each side is equal to a constant, say –

α^2.

1

X

d^2 X

dx^2

+

2 mE

^2

−α^2 = 0

∂^2 X

dx^2

+

(

2 mE

^2

−α^2

)

X= 0

Or

∂^2 X

∂x^2

+β^2 X= 0

where

β^2 =

(

2 mE

^2

)

−α^2 (4)

X=Asinβx+Bcosβx

Take the origin at the corner

Boundary condition:X=0 whenx=0. This givesB=0.

X=Asinβx

Further,X=0 whenx=a

Sinβa= 0 →βa=nxπ

Or

β=

nxπ

a

(nx=integer) (5)