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Schrödinger’s Equation in 1D: Some Examples
Michael Fowler, UVa.
9/24/07
Curvature of Wave Functions
Schrödinger’s equation in the form
2
22
() 2 ( ()
)
()
dx
m
V
x
E
x
dx
ψ
−
=
=
can be interpreted by saying that the lefthand side, the rate of change of slope, is the
curvature
– so the curvature of the function is proportional to
( )
( ) ()
.
Vx E
x
−
This
means that if
E
>
V
(
x
), for
x
positive
( )
x
is curving negatively, for
x
negative
x
is curving positively.
In both cases,
( )
x
is always curving towards the xaxis
—
so, for
E
>
V
(
x
),
x
has a kind of stability: its curvature is always bringing it back
towards the axis, and so generating oscillations.
The simplest example is that of a
constant potential
V
(
x
) =
V
0
<
E
, for which the wave function is
(
)(
sin
xA
k
x
)
δ
=+
with
a constant and
2
0
2/
.
km
E
V
=−
=
On the other hand, for
V
(
x
) >
E
, the curvature is always
away
from the axis.
This means
that
x
tends to diverge to infinity. Only with
exactly
the right initial conditions will
the curvature be just right to bring the wave function to zero as
x
goes to infinity. (This is
possible because as
x
tends to zero, the curvature tends to zero, too.)
For a constant potential
V
0
>
E
, the wave function is
( )
x
x
x
Ae
Be
α
−
, with
2
0
.
mV
E
=
=
−
Of course, this wave function will diverge in at least one
direction!
However, as we shall see below, there are situations with spatially varying
potentials where this wave function is only relevant for positive
x
, and the coefficients
A
,
B
are functions of the energy—for certain energies it turns out that
0
A
=
, and the wave
function converges.
One Dimensional Infinite Depth Square Well
In an earlier lecture, we considered in some detail the allowed wave functions and
energies for a particle trapped in an infinitely deep square well, that is, between infinitely
high walls a distance
L
apart.
For that case, the potential between the walls is identically
zero so the wave function has the form
( ) ( )
sin
.
k
x
The wave
function
x
necessarily goes to zero right at the walls, since it cannot have a
discontinuity, and must be zero just inside the wall.
Even a quantum particle cannot
penetrate an infinite wall!
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An immediate consequence is that the lowest state cannot have zero energy, since
k
= 0
gives a constant
()
x
ψ
.
Rather, the lowest energy state must have the minimal amount of
bending of the wave function necessary for it to be zero at
both
walls but nonzero in
between—this corresponds to half a period of a sine or cosine (depending on the choice
of origin), these functions being the solutions of Schrödinger’s equation in the zero
potential region between the walls.
The allowed wave functions (eigenstates) found as
the energy increases have successively
0, 1, 2, … zeros (nodes) in the well.
Parity of a Wave Function
Notice that the allowed wave eigenfunctions of the Hamiltonian for the infinite well are
symmetrical or antisymmetrical about the center,
( ) ( )
.
x
x
=
±−
We call the operator that reflects a function in the origin the
parity
operator
P
, so these
eigenstates of the Hamiltonian are
also
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This note was uploaded on 12/07/2011 for the course PHYSICS 751 taught by Professor Michaelfowler during the Fall '07 term at UVA.
 Fall '07
 MichaelFowler
 Schrodinger Equation

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