We now have determined the parameters Band k, and have the follow-
ing solutions:(10.9)
(10.10)
To determine the last parameter, A, we make use of the normalization
condition that requires the total probability to be equal to one (Chapter 9).
Since there is only one particle in the box, the integral of the probability
over the length of the box must be equal to one, allowing us to write:(10.11)
(10.12)
(10.13)
(10.14)
This gives a final solution of:(10.15)
(10.16)
Properties of the solutions
Energy and wavefunction
The solutions of Schrödinger’s equation for the particle in a box are a series
of wavefunctions and energies related by a quantum number, n, that must
be a positive integer. The allowed energy levels are quantized and increase
as n^2 with their separation increasing as the quantum number increasesE
mn
Lnh
mL=
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
Z^2
(^222)
282
π
ψ
π
( )x sin , , ...
L
nx
L
= n
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
2
123
1
2
000
2
(^222)
=⎡⎣()()−−−⎤⎦=→=
A
L
AL
A
L
1
2
2
2
2
=−sin⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
⎡
⎣
⎢
⎢
A
xnx
LL
nπ
π⎤⎤
⎦
⎥
⎥
0L1
1
2
1
2 2
0=−cos⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
∫ ⎥
A
nx
LxL
π
d1
022
0==() () sin⎛
⎝
⎜⎜
⎞
⎠
∫∫ψψ ⎟⎟
π
*dxxx A dnx
LLL
xxE
k
mmn
Lnh
mL==
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
ZZ^222
(^222)
22 82
π
ψ
π
( )xAsin , , ...
nx
L
= n