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LECTURE 7
QUANTUM PHYSICS III
Instructor: Shih-Chieh Hsu
Virtual Visit on Sep 10
2
9:30am Sep 10 (Thu) B305



UW Research Dr. Associate Sam Meehan and UW
Physics Graduate Nikola Whallon from CERN ATLAS
Control Room
ATLAS virtual visit Vidyo – 9213753
http://vidyoportal.cern.ch/flex.html?roomdirect.html&key=P5bTn2wuugkb
10:50am Sep 10 (Thu) PAA A112


LHC Master class – search for Higgs from real data
ATLAS Virtual Visit PAB B305
3

9:20am PAB B305
4
Old Quantum Theory
Three Failures of Classical Physics
5

Black Body Radiation.


Photoelectric Effect.
The Hydrogen Atom
Bohr’s semi-classical model (1913)
6
Energy Quantization
In Aton
En energy level
|ΔEn| = hf=hc/λ
7
New Quantum Theory
Wave-particle duality
8

The classical concepts of waves and particles do not
adequately describe the complete behavior of any
phenomenon.
Everything propagates like a wave and exchanges energy
like a particle.

The wavelength and frequency of matter:
h
l=
p
 For
E
f =
h
macroscopic objects, de Broglie wavelength is too
small to be observed.
Uncertainty principle
9



If we use light with  to measure the position of an object, x, its
uncertainty, x, cannot be less than ~ because of diffraction.
If we use photons with p = h/ to measure the momentum of
an object, p, p of the object cannot be less than ~h/ since
the photon changes the momentum of the object upon
scattering.
The Heisenberg uncertainty principle states that:
It is impossible to simultaneously measure both the
position and the momentum of a particle with unlimited
precision.
DxDpx ³
2
, where
º
h
2p
Quantum Mechanics (1923)
10

In quantum mechanics, a particle is described by a wave
function  that obeys a wave equation called the Schrödinger
equation.
¶
Ñ Y ( r ,t ) + U ( r ) Y ( r ,t ) = i
Y ( r ,t )
2m
¶t
2
2
You absolutely do not need to memorize the formula.

The solution of the equation by itself has no physical meaning.
However, the probability to find a particle in a certain spacetime is:
Time-Independent Schrodinger Equation
11

Solution of the Schrödinger equation.
¶
Ñ Y ( r ,t ) + U ( r ) Y ( r ,t ) = i
Y ( r ,t )
2m
¶t
2
2
Wave function
12


The Schrödinger equation describes a single
particle.
The probability density P(x), the probability per unit
volume (or length in 1-D), of finding the particle as
a function of position is given by
()
()
P x =y2 x

The probability is probability times unit volume, i.e.
P(x) Δx
Normalization condition
13

If we have a particle, the probability of finding the
particle somewhere must be 1. Therefore the wave
function must satisfy the normalization condition.
¥
¥
ò P ( x ) dx = ò y ( x ) dx = 1
2
-¥

-¥
For  to satisfy the normalization condition, it must
approach zero as |x| approaches infinity.
Copenhagen Interpretation
14


1927, Bohr, Heisenberg, Pauli had converged to a consensus
based on Bohr's concept of complementarity, which states
that a physical phenomenon may manifest itself in two
different ‘complementary' ways depending on the
experiment set up to investigate it.
Thus light, for example, could
appear sometimes as a wave
and sometimes as a particle.
Although mutually exclusive,
both pictures were necessary
to obtain a full description of
the phenomenon.
Schrödinger’s Cat
15

1935, Austrian physicist Erwin Schrödinger proposed this
thought experiment, often described as a paradox, to
illustrate what he saw as the problem of the Copenhagen
interpretation of quantum mechanics applied to everyday
objects.
Schrödinger’s Poor Cat
16

The material doesn’t
decay. The cat is alive

The material has decayed.
The cat has been killed by
the poison.
Schrödinger’s Cat: Live or Dead
17

According to the Copenhagen interpretation,
the cat is both alive and dead. It exists in a
state of “superposition”
Probability Calculation for a Classical
Particle
18

A classical point particle moves back and forth with
constant speed between two walls at x = 0 and x =
8.0 cm.
Because the probability density is uniform, the probability
of a particle being in some range Δx in the region 0 < x <
8.0 cm is P0Δx.
Probability Calculation for a Classical
Particle
19

What is the probability density P(x)? .
The probability density P(x) is
uniform between the walls and
zero elsewhere:
probability density x Total length = 1 = P0 x 8cm
Probability Calculation for a Classical
Particle
20

What is the probability of finding the particle at the
point where x equals exactly 2 cm?
On the interval 0 < x < 8.0 cm, the
probability of finding the particle in
some range Δx is proportional to
P0Δx = Δx/(8 cm).
The probability of finding the particle
at the point x = 2 cm is zero because
Δx is zero (no range exists).
Alternatively, because an infinite number of points exists
between x = 0 and x = 8 cm, and the particle is equally likely to
be at any point, the chance that the particle will be at any one
particular point must be zero.
Probability Calculation for a Classical
Particle
21

What is the probability of finding the particle
between x = 3.0 cm and x = 3.4 cm?
Because the probability density
is uniform, the probability of a
particle being in some range Δx
in the region 0 < x < 8.0 cm is
P0Δx.
A particle in a box
22

Consider a particle of mass m confined to a onedimensional box of length L.
Classical Mechanics: The particle with any values of energy
and momentum bounces back and forth between the walls
of the box.
 Quantum Mechanics: The particle is described by a wave
function , and 2 describes the probability of finding the
particle in some region.

A particle in a box: conditions for 
23

The particle is somewhere in the box.
L
2
y
ò ( x ) dx = 1
0

The particle is not outside the box.
y ( x ) = 0 for x £ 0 and x ³ L

 is continuous everywhere.
y ( 0) = 0 and y ( L) = 0

This is the same boundary condition as the condition for standing waves
on a string fixed at x = 0 and x = L and satisfies following equation
A particle in a box: allowed wavelengths
24


The boundary condition
restricts the allowed
wavelengths for a particle
in a box.
The box length L equals an
integral number of half
wavelengths.
L=n
ln
; n = 1,2,3,
2
Standing wave condition
A particle in a box: allowed energies
25
p
mv 2
=
E=
2
2m
2

h2
=
2ml 2
L=n
ln
2
; n = 1,2,3,
The standing wave condition yields
the allowed energies.
2
h
2
En = n 2
=
n
E1
2
8mL


The lowest allowed energy, E1, is
called ground state energy.
Note that E1 is not zero, and
depends on the size of the box.
Electron bound to an atom
26



If an electron is constrained to be within an
atom, the electron is confined in one of the
allowed energy states.
The electron can make a transition to and from
one energy state, Ei, to another, Ef, by the
emission of a photon (if Ei > Ef).
The frequency and wavelength of the emitted
photon are:
Ei - Ef
f =
h
c
hc
l= =
f Ei - Ef
Quantum number
27




The number “n” is called a quantum number.
It characterizes n for a particular state and for the
energy of that state, En.
For a particle in a 1-D box, a quantum number arises
from the boundary condition on :
(0) = 0 and (L) = 0.
For a particle in a 3-D box, three quantum numbers
arise, one associated with a boundary condition in
each dimension.
Standing wave functions: probability densities
28



The probability per unit length of finding the particle as a
function of position is n2(x).
The particle is most likely to be found near the maxima. The
particle cannot be found where 2 = 0.
For very large values of n, the maxima and minima are so
closely spaced that 2 cannot be distinguished from its
average value. The particle is equally likely to be found
anywhere in the box, the same as in the classical result.
Large quantum number
29



The fractional energy difference of adjacent states
becomes very small as the quantum number
increases.
For a very large n, energy quantization is not
important.
Bohr’s correspondence principle states:
In the limit of very large quantum numbers, the
classical calculation and the quantum calculation
must yield the same results.
Clicker Question 19-2

There are three 1-D boxes, B1, B2, and B3, with length
L, 2L, and 3L, respectively. Each box contains an
electron in the state for n = 10. Rank the boxes
according the number of maxima for the probability
density of the electron, greatest first.




B1, B2, B3
B3, B2, B1
B2, B3, B1
They are all tie.
Example 1
31

An electron is in the initial state ni = 3 of an 1-D
box of length 100 pm. If it is to make a quantum
jump to the state nf = 6 by absorbing a photon,
what must be the energy and wavelength of the
photon?
2
h
2
En = n 2
=
n
E1
2
8mL
E6 - E3
hc
2
(hc)
h2
2
2
= 27
= (6 - 3 )
2
8(mc 2 )L2
8mL
(1240eV ·nm)2
= 27
8(5.11´105 eV )(0.1nm)2
=1015.12eV
hc
E = ®l =
l
E
1240ev ·nm
=
= 1.22nm
1015.12eV
Hydrogen Atom
32
an electron is bound to a proton
by the electrostatic force of
attraction
Timeline
33
34
Backup
Expectation Value
35
Example
36
A particle in a one-dimensional box of length L is in the
ground state. Find the probability of finding the particle (a) in
the region that has a length Δx = 0.01L and is centered at x =
L and (b) in the region 0 < x < L.
(a) 0
(b) 1
Example
37
A particle in a one-dimensional box of length L is in the
ground state. Find the probability of finding the particle (a) in
the region that has a length Δx = 0.01L and is centered at x =
L/2 and (b) in the region 0 < x < L/4.
Example 3
38

The photons in a monochromatic beam are scattered
by electrons. The wavelength of the photons that are
scattered at an angle of 135° with the direction of
the incident photon beam is 2.3 percent more than
the wavelength of the incident photons.
a)
b)
What is the wavelength of the incident photons?
What is the kinetic energy of the electron?
Harmonic oscillator potential well
39


Consider a particle with mass, m, on a spring with force
constant, k.
Potential energy function for a harmonic oscillator is parabolic.
()
U x = 12 kx 2 = 12 mw 02 x 2
where w 0 = k m is the natural

frequency of the oscillator.
Classically, the object oscillates between
±A, and its total energy, E, can have
any nonnegative value, including zero.
1
Eclassic = mw 02 A2
2
Parabolic well
Harmonic oscillator: allowed energies
40

Normalizable n(x) occur only for discrete values of
the energy En given by
(
)
En = n + 12 hf0
n = 0, 1, 2,
Note that the ground
state energy is not 0.
Equally spaced levels: hf0
Example 3
41

An electron in a harmonic oscillator is initially in the
n = 4 state. It drops to n = 2 state and emits a
photon with wavelength 500 nm. What is the
ground state energy of this harmonic oscillator?
(
)
En = n + 12 hf 0
E4 - E2
= 2hf 0
1240ev ·nm
=
= 4.96eV
500nm
hc = 1240 eV·nm = 1.988
 10-25 J·m
(
)
E0 = 0 + 12 hf 0 = 4.96eV / 4 =1.24eV