Download Lecture_22 - Quantum Mechanics (read Chap 40.2)

Still have a few registered iclickers (3 or 4 ?) that need to be mapped
to names
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Review: Continuous spectra and blackbody radiation
•
A blackbody is an idealized case of
a hot, dense object. The continuous
spectrum produced by a blackbody
at different temperatures is shown
on the right (the sun is another
example)
Note: Usually a heated solid
or liquid produces a
continuous blackbody
frequency spectrum
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Continuous spectra and blackbody radiation
I = sT
4
Stefan-Boltzmann Law for
blackbody radiation (PHYS170);
Here σ is the Stefan-Boltzmann
constant.
W
s = 5.67 ´10
m2 · K 4
-8
Questions: What are the units
of intensity ?
lmT = 2.9 ´10 m · K
-3
The surface of the Sun with a
sunspot. T(sunspot)=4000K,
T(sun)=5800 K; ratio of I’s is
(4000/5800)4 =0.23dark
This is the Wien displacement law (PHYS170).
where λm is the peak wavelength and T is the
temperature
Copyright © 2012 Pearson Education Inc.
Continuous spectra and blackbody radiation
•
A blackbody is an idealized case
of a hot, dense object. The
continuous spectrum produced by
a blackbody at different
temperatures is shown on the right.
A classical Physics calculation by
Lord Rayleigh gives,
I(l ) =
2p ckT
l4
Agrees quite well at large values of
the wavelength λ but breaks down
at small values.
This is the “ultraviolet catastrophe”
of classical physics.
Copyright © 2012 Pearson Education Inc.
Question: why the
colorful name ?
Continuous spectra and blackbody radiation
A calculation by Max Planck assuming
that each mode in the blackbody has
E = hf gives, (he says this “was an act of
desperation”)
2p hc 2
I(l ) = 5 hc/l kT
l (e
-1)
Agrees quite well at all values of
wavelength and avoids the
“ultraviolet catastrophe” of classical
physics.
At large wavelengths, Planck’s result
agrees with Rayleigh’s formula.
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The Heisenberg Uncertainty Principle revisited
• The Heisenberg uncertainty principle for momentum and position applies
to electrons and other matter, as does the uncertainty principle for energy
and time. This gives insight into two-slit interference with electrons
A common misconception is that the interference pattern is
due to two electrons whose waves interfere.
We observe the same interference with a single electron !
Copyright © 2012 Pearson Education Inc.
The Heisenberg Uncertainty Principle revisited
• The Heisenberg uncertainty principle for momentum and position applies
to electrons and other matter, as does the uncertainty principle for energy
and time.
Copyright © 2012 Pearson Education Inc.
The Uncertainty Principle and the Bohr model
An electron is confined within a region of width 5.0 x 1011m (the Bohr radius)
a) Estimate the minimum uncertainty in the x-component of
the electron’s momentum
b) What is the kinetic energy of an electron with this
Kinetic energy
magnitude of momentum ?
h comparable to
atomic energy
DxDpx ³ h / 2 Þ Dpx =
2Dx levels !
h
6.626 ´10 -34 J - s
-24
Dpx =
=
=
1.055
´10
J -s/m
-11
2Dx 2(2p )(5 ´10 m)
p 2 (1.055 ´10 -24 kg · m / s)2
-19
K=
=
=
6.11
´10
J = 3.81eV
-31
2m
2(9.11 ´10 kg)
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The Uncertainty Principle and spectral line width
A sodium atom in an excited state remains in that state for
1.6 x 10-8s before making a transition to the ground state,
emitting a photon with wavelength 589.0nm and energy
2.105 eV
a) What is the uncertainty in energy for the excited state ?
b) What is the fractional wavelength spread of the
corresponding spectral line ?
DEDt ³ h / 2
-34
h
6.626 ´10 J - s
DE =
=
-8
2Dt 2(2p )(1.6 ´10 s)
-27
-8
DE = 3.3´10 J = 2.1´10 eV
Copyright © 2012 Pearson Education Inc.
The Heisenberg Uncertainty Principle revisited
A sodium atom in an excited state remains in that state for
1.6 x 10-8s before making a transition to the ground state,
emitting a photon with wavelength 589.0nm and energy
2.105 eV
Question: What is the fractional wavelength spread of the
corresponding spectral line ?
-27
-8
DE = 3.3´10 J = 2.1´10 eV
-8
DE 2.1´10 eV
=
= 1.0 ´10 -8 eV
E
2.105eV
E=
DE
Dl = l
= 589nm(1.0 ´10 -8 eV ) = 0.0000059nm
E
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hc
l
Double Slit Interference
Quantum Interference
“We choose to examine a
phenomenon which is impossible,
absolutely impossible, to explain in
any classical way, and which has in
it the heart of quantum mechanics.
In reality, it contains the only
mystery.”
--Richard P. Feynman
Copyright © 2012 Pearson Education Inc.
QM Chapter 40: Wave functions in classical physics
¶2 y ( x,t ) 1 ¶2 y ( x,t )
= 2
2
¶x
v
¶t 2
•This is the wave equation for “waves on a string” where y(x,t) is the
displacement of the string and x is the direction of propagation.
•Notice this is a partial differential equation with v = velocity of wave
propagation.
•There are similar wave equations for the E and B fields in E+M waves.
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The Schrödinger equation in 1-D
•
In a one-dimensional model, a quantum-mechanical particle is described
by a wave function Ψ(x, t). [QM: remember point particles are waves]
•
The one-dimensional Schrödinger equation for a free particle of mass m
is
•
The presence of i (the square root of –1) in the Schrödinger equation
means that wave functions are always complex functions.
•
The square of the absolute value of the wave function, |Ψ(x, t)|2, is called
the probability distribution function. It tells us about the probability of
finding the particle near position x at time t.
•
Warning: |Ψ(x, t)|2 is not a probability (rather a probability density
function)
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Euler’s formula
e
iJ
Question: How can we express this in terms
of sines and cosines ?
e = cosq + isinq
iq
We will use this many times in QM
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The Schrödinger equation in 1-D: A free particle
•
If a free particle has definite
momentum p and definite
energy E, its wave function
(see the Figure) is
•
Such a particle is not
localized at all: The wave
function extends to infinity.
Question: What is k ?
Ans: It is the wavenumber, remember k=2π/λ
Copyright © 2012 Pearson Education Inc.
The Schrödinger equation in 1-D: A free particle
If a free particle has definite
momentum p and definite
energy E, its wave function
is
E = hf = hw / 2p
h/ 2 k 2
E = hw =
2m
h(2p )
/
p= h/l =
= hk
2pl
N.B. this is non-relativistic
Question: How does this compare with light waves?
w
/
/
hk
hk
h
Þ =
Þv=
=
k 2m
2m 2ml
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