It follows that
L/2
dx |X(x)|
2 = 1,
(3.63)
−L/2
thus satisfying the normalization condition, required for probability.
Repeating this for the y- and z-equations, we obtain
u k (x) = X(x) Y (y) Z(z)
1 i (kx x+ky y+kz z)
= √ e
L 3
1 i k·x
= √ e ,
(3.64)
V
where V = L
3 is the volume of the cube. We have adopted the
vector notation k = (k x , k y , k z ). The components k x , k y , and k z
take on the discrete values
2πn x
2πn y
2πn z
k x =
,
k y =
,
k z =
, (3.65)
L
L
L
where n j = 0, ±1, ±2, . . . . The vector k is called the wave vector.
It is straightforward to show that
d
3 x u ¯ k (x) u k � (x) = δ kk � ,
(3.66)
V
where the integral is performed over the cubic volume V . The
eigenfunctions u k (x) are plane waves, each with a unique wave
vector k.
Each set of n x , n y , and n z represents a distinct state with energy
given by
h
2
h
2
¯
2
2
2
H k =
k x
2 + k y
2 + k z
2 =
n x + n y + n z .
(3.67)
2m
2mL 2
The interval L can be chosen to be arbitrarily large. As L is increased, the energy values become more closely spaced. In the limit
L → ∞, the energy levels approach a continuum. It does not follow that the energy eigenvales become small, since the integers n x ,
3.1. Quantum mechanical description of particle motion
141