Page 50 - Science at the nanoscale

9:2
RPS: PSP0007 - Science-at-Nanoscale
June 9, 2009
Brief Review of Quantum Mechanics
40
What about the other boundary condition that ψ(x = L) = 0?
Where does it lead us? We have the following equation
ψ(L) = C sin(kL) = 0
(3.22)
Since C cannot be zero (otherwise we will have no wavefunc-
tion), therefore sin(kL) = 0 and this implies kL = nπ where n is
an integer. Substituting this equation back to Eq. (3.18), we have
2 2
n h
(3.23)
E =
2
8mL
Equa-
The number n is known as the Quantum Number.
tion (3.23) shows that the energy of the particle trapped in the
potential box is discrete and cannot take any arbitrary energy. This
situation whereby only certain energy values are allowed is not
peculiar to the particle in a box system. It generally holds in any
bound physical system, i.e. when a particle is in a bound potential
that confines it within a limited region. Such quantization of en-
ergy is a common characteristic of nano-physical systems. For
a long time, the particle in a box problem remained a quantum
mechanics textbook problem. Nowadays, one can readily realise
such a potential in a box situation in an artificial quantum well
where electrons are confined in a narrow region.
In general, the wavefunction for a particle in a 1D potential box
can be expressed as
(3.24)
ψ(x) = C sin(nπx/L)
In order to determine the expression completely, we make use
of the normalisation condition, which requires the probability of ch03
finding the particle everywhere to be equal to 1.
∞ 2
Z
|ψ(x)| dx = 1 (3.25)
−∞
Since the wavefunction is equal to zero everywhere outside the
box, we have
Z L Z L nπx
2
2
|ψ(x)| dx = 1 and so C sin 2 dx = 1 (3.26)
0 0 L
q
2
i.e. C = and hence for a particle in a box, the wavefunction is
L