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Chem 120A
Probability Current and Free Particles
02/17/06
Spring 2006
Lecture 14
READING:
Engel (QCS): Chapter 4 (you should finish both Ch. 4 and 5 before the end of next week)
The finite squarewell
E
<
0 solutions
Last time we used parity considerations to find the solutions for the allowed energies of the particle in a
finite squarewell potential (Figure ) with
E
<
0. For solutions with even parity, we applied the boundary
conditions to find the condition for the allowed energies,
k
tan
±
ka
2
²
=
κ
.
(1)
The allowed energies can be found graphically by plotting both sides of Eq.1 as a function of
E
and finding
the energies where they are equal. Similarly, for the odd parity solutions we find
k
cot
±
2
²
=
−
.
(2)
V
0
x
V(x)
Figure 1: Potential for a quantum particle in a 1D well of
length a.
V(x) = V
0
,
a/2 < x < a/2
V(x) = 0,
x
r
a/2
a/2
a/2
II
I
III
V=0
This is the general procedure we have used to solve the Schr¨odinger equation for the infinite and finite
squarewell potentials:
1. Divide into regions (i.e. I, II, and III) and examine symmetry.
2. Write the Schr¨odinger equation for each region.
Chem 120A, Spring 2006, Lecture 14
1
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View Full Document 3. Apply boundary conditions and get either implicit or explicit solutions.
Comments about finite squarewell solutions
1. We will always have at least one boundstate solution.
2. The deeper the well (larger magnitude of
V
0
) the more possible bound solutions there are.
3. This is truly a 1D problem, meaning that the dimensions are
−
∞
≤
x
≤
∞
. In 3D we use spherical
coordinates and define
r
as a relative coordinate with 0
≤
r
≤
∞
. The potential in Figure3 is one
example of a problem where we use the relative coordinate
r
. In this particular case, we do not always
have a bound state. There is some minimum value of

U
0

that is required to have a bound state.
V(x)
8
U
0
V=0
r
r
0
V = 0,
r>r
0
V = U
0
,
0<r<r
0
V = infinity, r<0
Figure 2: Pseudo1D potential used to treat the deuteron.
0
Example: the deuteron. This is the nucleus of deuterium,
2
H+. The exact nature of the interaction
between the proton and the neutron is not known, but the potential in Figure3 models this system
well. There is only one bound state in this model, with energy
E
1
=
2
.
8 MeV, and radius
r
0
=
2
×
10
−
15
m. For hydrogen,
1
H, the electronproton radius is
r
0
∼
0
.
5
×
10
−
10
m, which is about 5
orders of magnitude larger than the neutronproton distance (i.e. the nucleus is small compared to the
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This note was uploaded on 09/29/2009 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at University of California, Berkeley.
 Spring '07
 Whaley
 Physical chemistry, pH

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