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Young/Freeman
University Physics 11e
Ch 40 Quantum
Mechanics
© 2005 Pearson Education
Wave functions with a precise energy
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For a given wave function, we cannot predict
where a particle will be found. We are faced with
probabilities.
For the energy it is different. Special wave
functions can be calculated such that the energy
is well defined. These wave functions are found
by solving the time independent Schrödinger
equation.
Several examples will be described.
Big surprise:
‘energy quantization’
Classical localised systems have quantised
energy values.
This fact explains emission and absorption
events by atoms and molecules.
A classical localized particle
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If the potential energy has
the shape shown in the
figure a particle with energy
E will be confined to a
region [x1,x2].
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We say that the particle is
localized.
Classical physics Quantum
physics
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When a particle with such a potential energy is
treated by quantum mechanics we find that all
energy values are not allowed, energy is
quantized .
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We will consider three examples of this
statement 1) a particle in a box, 2) a finite box 3)
a harmonic oscillator
Energy E calculation in QM.
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Energy values and related wave functions of
a quantum system are calculated by solving
the time independent Schrodinger equation.
d  ( x)

 U ( x) ( x)  E ( x)
2
2m dx
2

2
The problem is to find a function Y(x) and E
which satisfy this equation and where Y(x) is
normalizable.
Moreover…
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Moreover the function Y(x) and its derivative
Y‘(x) should be continuous everywhere.
This is required because a derivative Y‘(x) does
not exist when the function Y(x) is not
continuous. Also a second derivative Y”(x) exists
only when the derivative Y‘(x) is continuous.
( the second derivative Y”(x) occurs in the SE.)
40.1 Particle in a box
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We have a very simple situation
for a localized particle when we
take U(x)=0 inside an interval
[0,L] and U(x) is infinitely high
outside that interval.
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The latter implies that a particle
cannot come there or Y(x) =0
for x<0 and x>L.
General solution of the SE inside the box
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We check by substitution that the general
solution is of the form
Y(x) = Asin(kx) + Bcos(kx)
with A and B are (complex) constants and k is
related to the energy by 2mE =(ћk)2.
Energy values
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Because of continuity: Y(x) =0 at x=0 and
Y(x) =0 at x=L.
It follows from the first requirement that B=0
and from the second that kL = np with n an
integer n=1,2,3...
Therefore we find the quantized energy values
En =n2E1 with E1 = h2/(8mL2) n=1,2,3,…
Final results
For a particle in a box there are wave functions
Yn(x) with a well defined energy En
The solutions are normalized when
C = sqrt(2/L)
© 2005 Pearson Education
n
energy levels, particle
in a box
© 2005 Pearson Education
Probability and normalization
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Normalization
condition



2
 ( x) dx  1
Particle in a box
 n ( x) 
2
npx
sin
L
L
The measurement in QM
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QM says: every measurement influences
the particle and thus the wave function.
Rule:
‘‘Immediately after the measurement the
wave function is restricted to what has
been measured. A new wave function
normalization is required”.
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A position measurement.
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Say we measure the
particle described by
Y(x) to be in the interval
[a,b].
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Immediately after the
measurement the new
wave function is the
previous wave function
limited to the interval
[a,b].
An energy measurement
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Suppose before the energy measurement the
wave function is
Y(x) = 1/sqrt(5)(Y1(x) +2iY2(x))
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If we measure the energy we can only find a
quantized value. In this case the wave function is
a sum of Y1(x) and Y2(x). Following QM we
may find E1 with probability 1/5 and E2 with
probability 4/5.
What do you think?
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Say we measure the energy and find E1. What is
the normalized wave function after the
measurement? (hint: consider what happened
after a position measurement)
Y(x) = ?
Understanding emission and
absorption processes.
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Absorption. When a
particle is in the state with
n=1 it can absorp a
photon when this has a
correct energy.
Similarly a particle in an
excited state (n>1) can fall
back to a lower state by
emitting a photon
The same photon
frequencies are observed.
40.2 Potential Wells
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Another simple example
is a box with a finite high
potential wall Uo.
Now we must distinguish
between energies lower
than Uo and energies
higher than Uo.
The result: energies
below Uo are quantized,
those higher than Uo are
not.
Solutions for the square well potential
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For E<Uo the wave function
extends outside the well but far
away goes to zero and the energy
is quantized.
The number of such quantized
energies is limited and depends on
Uo.
These functions look very much
like the solutions for a infinite box
potential
Probabilities
40.4 An atomic spring
d 2 ( x)

 U ( x) ( x)  E ( x)
2
2m dx
2
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Another simple system is a
particle with a potential
energy like a spring. Then
U(x) = ½ kx2 with k the
spring constant.
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The solution of the time
independent Schrodinger
equation in this case
requires higher mathematics
Results
The shape of the wave
functions is similar to a
particle in a square well
potential.
 The energy levels are
equally spaced and of
the form with
n=0,1,2,3…
w2 = k/m
m =particle mass
Non localised systems
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When the particle is not localised in the classical
sense the energies are not quantized.
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We discuss two examples.
1) a free particle 2) a particle with a potential
energy in the shape of a potential barrier. This
represents a situation where a particle is scattered
by an obstacle.
A free particle: U(x) =0 everywhere
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The general solution of the SE is of the form
Y(x)= A cos(kx) + Bsin(kx) where
E = (ћk)2/2m
There is no restriction on k and so every energy E is
possible: no quantization.
Note that Y(x) is periodic: Y(x+l)= Y(x) for l=2p/k.
As the energy is also written as E=p2/2m we find p= ћ k
= h/l which is the relation of the Broglie we mentioned
earlier. This was to be expected.
Quantum scattering
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Classically a particle can
approach an obstacle and
for E<Uo will bounce
back.
In QM this is not so.
To see what happens we
look for a solution of the
SE with an energy
E<Uo.
The tunnel effect for E<Uo
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For x<0 and x>L we have the
situation of a free particle
with a solution cos(kx) or
sin(kx).
Unlike the classical situation
we see that there is a chance
that the particle is found at
x>L.
This remarkable behavior is
called the tunnel effect. It has
several applications.
Application: a scanning tunneling
microscope (Binnig and Rohrer)
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A fine tip moves slightly
above a surface.
A potential difference is
set up between surface
and tip: tip negative and
sample positive.
Because of the tunnel
effect electrons tunnel
through the gap between
tip and surface
Scanning tunneling microscope
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When the tip comes
closer to the surface the
current rises
Keeping the current
constant requires
adaptation of the tip
height. So surface
geometry is recorded.
The tip scans the surface
and represents the
elevations in an image.
A view of a crystal surface
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Crystal surface is
positively charged
with respect to tip.
Because of the tunnel
effect electrons can
jump over the barrier
from tip to the crystal
Individual atoms can
be observed on the
surface
Tunneling application:
Emission of an a particle by a nucleus.
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In a nucleus of an atom two
protons and 2 neutrons form
a very stable group called an
a particle.
The potential of the a
particle is shown in the
figure.
We observe that sometimes
an a particles leaks out of
the nucleus by tunneling. We
speak about a decay.
40.5 Three-Dimensional Problems
three-dimensional Schrödinger equation
 2   2 x, y, z   2 x, y, z   2 x, y, z  

  U x, y, z   x, y, z   E x, y, z 



2
2
2
2m 
x
y
z

It is very difficult to find solutions Y(x) and
energy values E which satisfy this equation.
If more particles are involved it is even much
more difficult. Normally one resorts to
computer calculations.
© 2005 Pearson Education
END
THE END
© 2005 Pearson Education
Differential equations
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The Schrodinger equation is a differential
equation of second order. We mentioned already
that a differential equation has a general
solution.
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For a better understanding some basic
information is given in the next slides.
A first order differential equation
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Consider an unknown function of a variable x, we note y(x).
Say we are given that y(x) satisfies y’(x) =2. This is called a
first order differential equation as only a first derivative
enters.
Then how is the function y(x)?
We see that there are many functions satisfying this
requirement: e.g. y(x) =2x , y(x) =2x+5, y(x) = 2x-1 etc.
In general a solution has the form y(x) = 2x + c with c some
constant.
A first order differential equation
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The solution of a differential equation is thus
not uniquely defined.
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For a unuique solution a further condirion is
required like demanding that y(x=0)=7. There
there is only one solution y(x) = 2x+7.
A second order differential equation
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Say we are given that y(x) satisfies y”(x) =2. This is
called a second order differential equation as a
second derivative enters the equation.
What about this function y(x)?
Also here there are many functions satisfying this
requirement: e.g. y(x) =x2 , y(x) = x2+x+8 ,, y(x) =
x2-5 etc.
In general a solution has the form y(x) = x2+ax+b
with a and b some constant.
A unique solution?
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To have a unique solution again further
conditions have to be imposed. Now two
constants should be fixed, so we need two
conditions.
We might ask a solution satisfying {y(x=0)=3
and y’(x=0)=1} or {y(x=0)=3 and y(x=1)=2}.
The Schrodinger equation
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The SE is a second order differential equation.
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Two conditions for uniqueness are that the
solution goes to zero at plus and minus infinity.
1)Y(x) →0 as x → +∞
2)Y(x) →0 as x → - ∞
Questions
(you may discuss these questions with your friends).
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1. Consider example 40.1 in this ppt. Suppose a photon is
absorped so that the system jumps from the state n=1 to
n=2. Is this photon transition in the visible part of the
spectrum?
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2. A two-atomic molecule may vibrate. In a simple model
we may describe it as a vibrating atomic spring. For HCl
the observed force constant is 482 N m-1. Considering the
transition from n=1 to n=2, is the frequency of this
transition in the visible part of the spectrum? What is the
transition energy in eV. Is this energy larger or smaller
than typical atomic energies. (hint: an atom is roughly like
a ball of diameter L= 10-10m)