1000 Solved Problems in Modern Physics

3.3 Solutions 169

Or

d^2 ψ(x)

dx^2

+

(

2 mE

^2

)

ψ(x)=0(2)

Writing

α^2 =

2 mE

^2

(3)

Equation (2) becomes

d^2 ψ

dx^2

+α^2 ψ=0(4)

which has the solution

ψ(x)=Asinαx+Bcosαx (5)

where AandBare constants of integration. Take the origin at the left

corner, Fig 3.5.

Fig. 3.5Square potential
well of infinite depth

Boundary condition:

ψ(0)=0;ψ(a)= 0

The first one givesB=0. We are left with

ψ=Asinαx (6)

The second one gives

αa=nπ,n= 1 , 2 , 3 ... (7)

n=0 is excluded as it would give a trivial solution.

Using the value ofαin (6)

ψn(x)=Asin

(nπx

a

)

(8)

This is an unnormalized solution. The constantAis determined from

normalization condition.