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Quantum Mechanics: Particle in a box
Energy of a particle in a box
Consider a particle of mass 푚, which is trapped inside a one-dimensional box of length
퐿. Inside the box, the particle is free, but the two walls of the box are rigid, and the box
can neither penetrate them, nor go out of the box. Thus the potential experienced by the
particle is zero inside the box and infinite at the two walls and beyond. The Hamiltonian is
given by
2
which can be written in the position representation as
−\
2
2
2
(2)
Notice that since there is only one variable 푥, we use total derivatives instead of partial
derivatives. The time-independent Schrödinger equation can be written as
−\
2
2 휓(푥)
2
This is a 2nd order differential equation, and can be easily solved inside the box, because
the potential energy term is zero there. Inside the box the Schr ¨dinger equation is
2
휓(푥)
2
\
2
which may be written as
\
\
The solution of the equation
푑
푑푥
√
2 푚퐸
\
휓(푥) = 0 will also be a solution of the above
equation. We could have also written the above equation as
\
\
In that case, the solution of the equation
푑
푑푥
√
2 푚퐸
\
휓(푥) = 0 will also be a solution of
the above equation. A second order equation can have only two independent solutions.
These two solutions can easily be gotten as
푖푘푥
, 휓 2 (푥) = 푒
−푖푘푥
,
where 푘 =
2 푚퐸/. The general solition will be a linear combination of these two solutions,
namely
푖푘푥
−푖푘푥
,
where 푐 1
2
are undetermined constants. We have gotten the eigenfunction, but we still
don’t have the energy of the particle. So, what is missing? We have solved the Schrödinger
equation inside the box, but have left out the boundary, and the region of space outside the
box. Instead of putting in infinite potential, which can potentially create problems, we can
simply use the physical condition that the particle cannot penetrate the boundary, and thus
the probability of finding it inside the walls should be zero. From Born’s interpretation of
the wave-function we have learnt that the |휓(푥)|
2
푑푥 is the probability of finding the particle
between 푥 and 푥 + 푑푥. The probability density of finding the particle at a position 푥 is
2
. Since particle cannot penetrate the walls, we have, |휓( 0 )|
2 = 0 and |휓(퐿)|
2 = 0.
This in turn means 휓( 0 ) = 0 and 휓(퐿) = 0. From the first condition we get 푐 2
1
, which
implies
1
sin(푘푥).
The second condition yields
1
sin(푘퐿) = 0 ,
which means 푘퐿 = 푛휋, 푛 = 1 , 2 , 3.... We cannot have 푛 = 0 because that will
make 푘 zero, and the wave-function zero everywhere. The wave-function cannot be zero
everywhere, because that would imply that the probability of finding the particle anywhere
is zero, i.e., the particle does not exist. Thus
and there are many eigenfunctions corresponding to those, which we label by 푛
푛
1
sin(푛휋푥/퐿).
Recalling the relation of 푘 to energy, we get
푛
\
or
2
ℎ
2
2
We arrive at a very interesting result which says that the particle which is trapped inside a
box, cannot just take any value of energy. There are only fixed, quantized values it can
take. The energy of the particle is quantized. This is something that never happens in
classical mechanics. We shall see later that this quantization of energy is not a special
case here, but happens whenver a particle is confined to a small region.
Normalization of the eigenfunctions
Since |휓(푥)|
2 푑푥 is the probability of finding the particle in a small region around 푥, if we
sum it over all 푥, it should give us the probability of finding the particle in all of space. Since
the particle does exist, the probability of finding it anywhere in all of space should be 1:
∞
−∞
∗
(푥)휓(푥)푑푥 = 1. (4)
This is called the normalization condition, and must be satisfied by all wave-functions
representing a physical system. Using the normalization of all eigenfunctions 휓 푛
(푥), we
can find the unknown constant 푐 1 , and get the final normalized eigenfunctions as
푛
2
퐿
sin(푛휋푥/퐿), 푛 = 1 , 2 , 3... (5)
