Download Lecture notes lecture 13 (quantum physics)

Quantum Physics
When we consider the motion of objects on the atomic level, we find
that our classical approach does not work very well.
For example, quantum physics describes how electrons surround the
nucleus of the atom and other subatomic actions.
Therefore, for understanding motion on the microscopic scale we
must use Quantum Mechanics.
"I think I can safely say that nobody understands quantum mechanics."
- Richard P. Feynman
"I don't like it, and I'm sorry I ever had anything to do with it"
- Erwin Schrödinger
"Anyone who is not shocked by the quantum theory has not understood it."
- Niels Bohr
We know some small particles like electrons, protons, and neutrons.
Quantum physics even describes the particles which make these particles!
(The model of an atom that you were taught in high-school is a
approximation). The electrons don't orbit like planets; they form blurred
clouds of probabilities around the nucleus. Protons and neutrons? They're
each made of three quarks, each with its own 'flavor' and one of three
'colors'. Lets not forget the gluons, the even smaller particles that hold this
mess together when they collect and form glueballs. The quantum model of
the atom is much more complex than the traditional model. The world of
subatomic particles is a very bizarre one, filled with quantum probabilities
and organized chaos. For example, the exact position and velocity of an
electron is very hard to find because attempts to "see" it involve bouncing
other particles off of it. By doing this, you've just changed the electron's
velocity, so your data is useless. What quantum physics does is give us the
statistical probability of the electron's location at any one moment.
Blackbody Radiation
One of the earliest indications that classical physics was
incomplete came from attempts to describe blackbody
radiation.
A black body is a theoretical object
that absorbs 100% of the radiation
that hits it. Therefore it reflects
no radiation and appears perfectly
black.
Blackbody radiation is the emission
of electromagnetic waves from the surface of an object.
The distribution of blackbody radiation depends only the
temperature of the object and is independent of the
material.
The Blackbody Distribution
The intensity spectrum emitted
from a blackbody has a
characteristic shape.
The maximum of the intensity is
found to occur at a wavelength given
by Wien’s Displacement Law:
fpeak = (5.88 × 1010 s-1·K-1)T
T = temperature of blackbody (K)
Notice the plot is versus frequency on these
plots, but since it is related to wavelength
via, f λ = c
λmaxT = 2.90 × 10 m i K
−3
Example: The solar radiation curve approximates closely
a blackbody at 5900K. What region of the spectrum is
the sun most intense?
2.90 × 10−3 mi K
−7
λmax =
5900 K
= 4.915 × 10 m
= 491.5nm
Consequences: What part of the spectrum is this?
2.90 × 10−3 mi K
T=300K
= 9.667 × 10−6 m
λmax =
300 K
= 9.667 microns
What portion of the spectrum is this?
2.90 × 10−3 mi K
T=3 million K
= 9.667 × 10−10 m
λmax =
6
3. × 10 K
=9.667Å = 0.97nm
What portion of the spectrum is this?
2.90 × 10−3 mi K
T=3K
= 9.667 × 10−3 m
λmax =
3K
=9.667mm
What portion of the spectrum is this?
The Ultraviolet Catastrophe
Classical physics can
describe the shape of the
blackbody spectrum only at
long wavelengths. At short
wavelengths there is
complete disagreement.
This disagreement between
observations and the
classical theory is known as
the ultraviolet catastrophe.
Planck’s Solution
In 1900, the German physicist Max Planck was able to
explain the observed blackbody spectrum by assuming
that it originated from oscillators on the surface of the
object and that the energies associated with the
oscillators were discrete or quantized:
En = nhf
n = 0, 1, 2, 3…
n is an integer called the quantum number
h is Planck’s constant: 6.63 × 10-34 J·s
= 4.14 10-15 eV·s
f is the frequency
Planck’s idea of quanta was the beginning of quantum physics. The
smallest packet of energy a electron can give or absorb. (Like you
can’t take an elevator to the 32nd ½ floor). With electrons any
energy is a whole number of quantas.
The correct mathematical work to Planck’s work was later done by
Erwin Schroedinger, an Austrian scientist.
Quantum Mechanics
The essence of quantum mechanics is that
certain physical properties of a system (like
the energy) are not allowed to be just any
value, but instead must be only certain
discrete values.
Quantization of Light
Einstein proposed that light itself
comes in chunks of energy, called
photons. Light is a wave, but also a
particle. The energy of one photon
is
E = hf
where f is the frequency of the light
and h is Planck’s constant.
Useful energy unit: 1 eV = 1.6 × 10-19 J
Example
(a) Find the energy of 1 (red) 650 nm photon.
(b) Find the energy of 100 (red) 650 nm
photons.
E = hf = 6.63 × 10-34 J·s 650 10-9 m
Example A 200 W infrared laser emits photons with a
wavelength of 2.0×10-6 m while a 200 W ultraviolet laser
emits photons with a wavelength of
7.0×10-8 m. (a) Which has greater energy, a single infrared
photon or a single ultraviolet photon?
E = hf =
hc
λ
The UV photon has the greater
energy; its wavelength is smallest.
(b) What is the energy of a single infrared photon and
the energy of a single ultraviolet photon?
EUV =
EIR =
hc
λUV
hc
λIR
= 2.8 × 10 −18 J = 18 eV
= 9.9 ×10 − 20 J = 0.62 eV
(c) How many photons of each kind are emitted per
second?
energy emitted/sec
number of photons emitted per second =
energy/photon
For both lasers the energy emitted per second is 200 J.
The UV laser emits 7.0×1019 photons/sec and the IR
laser emits 2.0 ×1021 photons per second.
The PhotoElectric Effect
When light is incident on
a surface (usually a
metal), electrons can be
ejected. This is known
as the photoelectric
effect.
Around the turn of the century, observations of the
photoelectric effect were in disagreement with the
predictions of classical wave theory.
Experiments show:
1. Brighter light causes more electrons to be ejected, but
not with more kinetic energy.
2. The maximum KE of ejected electrons depends on the
frequency of the incident light.
3. The frequency of the incident light must exceed a certain
threshold, otherwise no electrons are ejected.
4. Electrons are ejected with no observed time delay
regardless of the intensity of the incident light.
1. Brighter light causes more electrons to be
ejected, but not with more kinetic energy.
Wave theory predicts a more intense beam of light, having
more energy, should cause more electrons to be emitted
and they should have more kinetic energy.
Particle theory predicts a more intense beam of light to have
more photons so more electrons should be emitted, but
since the energy of a photon does not change with beam
intensity, the kinetic energy of the ejected electrons should
not change.
2. The maximum KE of ejected electrons
depends on the frequency of the incident
light.
Wave theory cannot explain the frequency dependence of
the maximum kinetic energy.
Particle theory predicts the maximum kinetic energy of the
ejected electrons to show a dependence on the frequency
of the incident light. Each electron in the metal absorbs a
whole photon: some of the energy is used to eject the
electron and the rest goes into the KE of the electron.
3. The frequency of the incident light must
exceed a certain threshold, otherwise no
electrons are ejected.
Wave theory can offer no explanation.
Particle theory predicts a threshold frequency is needed.
Only the incident photons with f>fthreshold will have enough
energy to free the electron from the metal.
4. Electrons are ejected with no observed
time delay regardless of the intensity of
the incident light.
Wave theory predicts that if the intensity of the light is low,
then it will take some time before an electron absorbs
enough energy to be ejected from the metal.
Particle theory predicts a low intensity light beam will just
have a low number of photons, but as long as f>fthreshold an
electron that absorbs a whole photon will be ejected; no
time delay should be observed.
First Photoelectric Experiment:
Photoelectrons stopped by stopping
voltage Vstop. The kinetic energy of
the most energetic photoelectrons is
K max = eVstop
Kmax does not depend on the
intensity of the light!
→ single photon ejects each electron
Second Photoelectric
Experiment:
Photoelectric effect does not
occur if the frequency is below
the cutoff frequency fo, no
matter how bright the light!
→ single photon with energy
greater than work function Φ
ejects each electron
slope =
ab
bc
The Photoelectric Effect
Photoelectric Equation
The previous two experiments can be summarized by the
following equation, which also expresses energy conservation
hf = K max + Φ
Using
(photoelectric equation)
K max = eVstop → Vstop
Φ
⎛h⎞
=⎜ ⎟ f −
e
⎝e⎠
equation for a straight with slope h/e and intercept –Φ/e
h ab
2.35 V − 0.72 V
-15
slope = =
4.1
10
V ⋅s
=
=
×
14
14
e bc (11.2 × 10 − 7.2 × 10 ) Hz
Multiplying this result by e
h = ( 4.1× 10-15 V ⋅ s )(1.6 × 10-19 C ) = 6.6 × 10-34 J ⋅ s
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The Photoelectric Effect Explained
(Einstein 1905, Nobel Prize 1921)
The photoelectric effect can be understood as follows:
• Electrons are emitted by absorbing a single photon.
• A certain amount of energy , called the work function, Φ,
is required to remove the electron from the material.
• The maximum observed kinetic energy is the difference
between the photon energy and the work function.
Kmax = E - Φ
E = photon energy
"just shows emission" E = Φ means Kmax is zero
Cutoff Frequency f0 = Φ/h
The work function is a property of the individual metal
Element
Work Function
Element
Work Function
(eV)
(eV)
Aluminum
4.3
Nickel
5.1
Carbon
5.0
Silicon
4.8
Copper
4.7
Silver
4.3
Gold
5.1
Sodium
2.7
Application: Photocells, Solar Cells
(5eV = 248nm)
If the energy of a photon is high
enough it can break bonds in
molecules/materials
Zinc and cadmium have photoelectric work
functions given by ΦZn = 4.33 eV and ΦCd = 4.22
eV, respectively. (a) If both metals are illuminated
by UV radiation of the same wavelength, which one
gives off photoelectrons with the greater maximum
kinetic energy? Explain. (b) Calculate the
maximum kinetic energy of photoelectrons from
each surface if λ = 275 nm.
Kmax = E – Φ = h c / λ – 4.33 eV 1.6 10-19 J/eV
= 3.05 10-22 J
Planck’s constant: 6.63 × 10-34 J·s
= 4.14 10-15 eV·s
Example The photoelectric threshold frequency of silver is
1.04×1015 Hz. What is the minimum energy required to
remove an electron from silver?
KEmax = hf threshold − φ = 0
φ = hf threshold
(
)(
= 6.626 ×10 −34 Js 1.04 ×1015 Hz
= 6.89 ×10
−19
J = 4.30 eV
)
Example Two different monochromatic light sources, one
yellow (580 nm) and one violet (425 nm), are used in a
photoelectric effect experiment. The metal surface has a
photoelectric threshold frequency of
6.20×1014 Hz. (a) Are both sources able to eject
photoelectrons from the metal? Explain.
The frequency of each source is
f yellow =
f violet =
c
λyellow
c
λviolet
= 5.17 ×1014 Hz
= 7.06 ×1014 Hz.
Only the violet light is above the threshold frequency.
Example continued:
(b) How much energy is required to eject an electron
from the metal?
KEmax = hf threshold − φ = 0
φ = hf threshold
(
)(
= 6.626 ×10 −34 Js 6.20 ×1014 Hz
= 4.11×10
−19
J = 2.56 eV
)
X-ray Production
When high energy electrons impact a target x-ray photons
can be emitted as the electrons are slowed. This process is
called Bremsstrahlung (German for breaking radiation).
There is a continuous spectrum of radiation emitted up to a
cutoff frequency.
The spikes in the
spectrum are called
characteristic x-rays.
These peaks depend
on the target material.
The Mass and Momentum of a Photon
Photons have momentum, but no mass.
We cannot use the formula p = mv to find
the momentum of the photon. Instead:
hf h
p=
=
c λ
Photons Have Momentum, Compton Shift
hf h
p=
=
c λ
(photon momentum)
Photons Have Momentum, Compton shift
Conservation of energy hf = hf '+ K
Since electron may recoil at speed approaching c
we must use the relativistic expression for K
K = mc 2 (γ − 1) where γ is the Lorentz factor
1
γ =
2
1− (v c)
Substituting K in the energy conservation equation
hf = hf '+ mc 2 (γ − 1)
p = γ mv
Conservation of momentum along x
Conservation of momentum along y
f =c λ
h
→
λ
h
h
λ
=
=
h
λ'
+ mc (γ − 1)
cos φ + γ mv cos θ
λ'
h
0=
sin φ − γ mv sin θ
λ'
Photons Have Momentum, Compton shift cont’d
Want to find wavelength shift
Δλ = λ '− λ
Conservation of energy and momentum provide 3 equations for 5 unknowns
(λ, λ’, v, φ, and θ ), which allows us to eliminate 2 unknowns v and θ.
h
Δλ =
(1-cosφ )
mc
(Compton shift)
λ, λ’, and φ, can be readily measured in the Compton experiment
h
is the Compton wavelength and depends on 1/m of the scattering particle
mc
Loose end: Compton effect can be due to scattering from electrons bound
loosely to atoms (m=me→ peak at θ ≠ 0) or electrons bound tightly to atoms
(m ≈ matom >>me → peak at θ ≈ 0)
Example A photon is incident on an electron at rest. The
scattered photon has a wavelength of 2.81 pm and moves at
an angle of 29.5° with respect to the direction of the incident
photon. (a) What is the wavelength of the incident photon?
The Compton shift is
Δλ = λc (1 − cos θ )
= (2.43 pm )(1 − cos 29.5°)
= 0.315 pm.
The incident wavelength is
λ0 = λ1 − Δλ
= 2.81 pm − 0.314 pm = 2.50 pm.
Example continued:
(b) What is the final kinetic energy of the electron?
The final kinetic energy of the electron is equal
to the change in the photon’s energy.
K = c( p0 − p1 )
⎛ 1 1⎞
⎛ h h⎞
= c⎜⎜ − ⎟⎟ = hc⎜⎜ − ⎟⎟
⎝ λ0 λ1 ⎠
⎝ λ0 λ1 ⎠
−15
= 8.77 × 10 J = 55 keV
The Wave Nature of Particles
We have seen that light is described sometimes
as a wave and sometimes as a particle.
In 1924, Louis deBroglie proposed that particles
also display this dual nature and can be
described by waves too!
The deBroglie wavelength of a particle is related
to its momentum:
λ= h/p
(Use p = γm0v if the velocity is large.)
That is why you can make an atom laser!
Electrons and Matter Waves
If electromagnetic waves (light) can behave like particles (photons), can
particles behave like waves?
λ=
h
p
(de Broglie wavelength) where p is the momentum of the particle
Electrons
θ
38-
We can investigate wave properties of
particles through interference/diffraction
Light as a Probability Wave
How can light act both as a wave and as a particle (photon)?
Standard Version: photons sent
through double slit. Photons detected
(1 click at a time) more often where the
classical intensity:
2
Erms
I=
cμ0
Fig. 38-6
is maximum.
The probability per unit time interval
that a photon will be detected in
any small volume centered on a
given point is proportional to E2 at
that point.
Light is not only an electromagnetic wave but also a probability wave for
detecting photons
Light as a Probability Wave, cont'd
Single Photon Version: photons sent through double slit one at a time. First
experiment by Taylor in 1909.
1. We cannot predict where the photon will arrive on the screen.
2. Unless we place detectors at the slits, which changes the experiment (and
the results), we cannot say which slit(s) the photon went through.
3. We can predict the probability of the photon hitting different parts of the
screen. This probability pattern is just the two slit interference pattern that we
discussed in Ch. 35.
The wave traveling from the source to the screen is a probability wave, which
produces a pattern of "probability fringes" at the screen.
38-
Light as a Probability Wave, cont'd
Conclusions from the previous versions/experiments:
1. Light generated at source as photons.
2. Light absorbed at detector as photons.
3. Light travels between source and detector as a probability wave.
38-
Schrödinger’s Equation
For light E(x, y, z, t) characterizes its wavelike nature, for matter the wave
function Ψ(x, y, z, t) characterizes its wavelike nature.
Like any wave, Ψ(x, y, z, t) has an amplitude and a phase (it can be shifted in
time and or position), which can be conveniently represented using a
complex number a+ib where a and b are real numbers and i2 = -1.
On the situations that we will discuss, the space and time variables can be
grouped separately:
Ψ ( x, y, z , t ) = ψ ( x, y, z , t ) e − iω t
where ω=2πf is the angular frequency of the matter wave.
Schrödinger’s Equation, cont’d
What does the wave function mean? If the matter wave reaches a particle
detector that is small, the probability that that a particle will be detected there
in a specified period of time is proportional to Iψ|2, where Iψ| is the absolute
value (amplitude) of the wave function at the detector’s location.
The probability per unit time interval
of detecting a particle in a small
volume centered on a given point in
a matter wave is proportional to the
value Iψ|2 at that point.
Since ψ is typically complex, we obtain Iψ|2 by multiplying ψ by its complex
conjugate ψ* . To find ψ* we replace the imaginary number i in ψ with –i
wherever it occurs.
ψ = ψψ * = ( a + ib )( a − ib )
2
Schrödinger’s Equation, cont’d
How do we find (calculate) the wave function? Matter waves are described
by Schrödinger’s equation. Light waves a described Maxwell’s equations,
matter waves are described by Schrödinger’s equation.
For a particle traveling in the x direction through a region in which forces on
the particle cause it to have a potential energy U(x), Schrödinger’s equation
reduces to:
d 2ψ 8π 2 m
+ 2 ⎡⎣ E - U ( x ) ⎤⎦ψ =0 (Schrodinger's eq. in 1D)
2
dx
h
where E is the total energy of the particle.
If U(x) = 0, this equation describes a free particle. In that case the total
energy of the particle is simple its kinetic energy (1/2)mv2 and the equation
becomes:
2
d ψ 8π m ⎛ mv 2 ⎞
d 2ψ ⎛
p⎞
+ 2 ⎜
⎟ψ =0 → 2 + ⎜ 2π ⎟ ψ =0
2
dx
h ⎝ 2 ⎠
dx ⎝
h⎠
2
2
Schrödinger’s Equation, cont’d
In the previous equation, since λ=h/p, we can replace the p/h with 1/λ, which
in turn is related to the angular wave number k=2π/λ.
d 2ψ
2
+
k
ψ =0 (Schrodinger's eq., free particle)
2
dx
The most general solution is:
ψ ( x ) =Aeikx + Be − ikx
Leading to:
Ψ ( x, t ) = ψ ( x ) e − iω t = ( Aeikx + Be − ikx ) e − iω t
= Ae
i ( kx −ω t )
+ Be
− i ( kx +ω t )
38-
Finding the Probability Density Iψ|2
eiθ = cos θ + i sin θ
and e − iθ = cos θ − i sin θ
Choose arbitrary constant B=0 and let A= ψ0
ψ ( x ) =ψ 0 eikx
2
ψ = ψ 0e
e
2
ikx 2
= (ψ 0 ) e
2
ikx 2
= ( eikx )( eikx ) * = eikx e − ikx = eikx −ikx = 1
ψ = ψ 0e
Fig. 38-12
ikx 2
ikx 2
= (ψ 0 ) 1 = ψ 02 (a constant)
2
2
38-
Heisenberg Uncertainty Principle
To describe a particle a physicist would refer four properties, the
position of the electron, its momentum, its energy, and the time.
Heisenberg showed that no matter how accurate the
instruments used, quantum mechanics limits the precision
when two properties (let´s built two pairs: momentum-position
or energy-time) are measured at the same time.
"The more precisely
the POSITION is
determined,
the less precisely
the MOMENTUM is known"
WERNER HEISENBERG (1901 - 1976)
Heisenberg’s Uncertainty Principle
In the previous example, the momentum (p or k) in the x-direction was exactly
defined, but the particle’s position along the x-direction was completely
unknown. This is an example of an important principle formulated by
Heisenberg: Measured values cannot be assigned to the position r and the
momentum p of a particle simultaneously with unlimited precision.
= h 2π
Δx ⋅ Δ p x ≥
where
Δy ⋅ Δ p y ≥
(Heisenberg's uncertainty principle)
Δz ⋅ Δp z ≥
Note that if Δpx = Δp y = Δpz = 0
Then Δx → ∞, Δy → ∞, and Δz → ∞
38-
Determinism of Classical
Mechanics
• Suppose the positions and speeds of all particles in
the universe are measured to sufficient accuracy at
a particular instant in time
• It is possible to predict the motions of every particle
at any time in the future (or in the past for that matter)
“An intelligent being knowing, at a given instant of time, all forces
acting in nature, as well as the momentary positions of all things of
which the universe consists, would be able to comprehend the
motions of the largest bodies of the world and those of the smallest
atoms in one single formula, provided it were sufficiently powerful
to subject all the data to analysis; to it, nothing would be uncertain,
both future and past would be present before its eyes.”
Pierre Simon Laplace
Measuring the position
of an electron
• Shine light on electron and detect reflected
light using a microscope
• Minimum uncertainty in position
is given by the wavelength of the
light
• So to determine the position
accurately, it is necessary to use
light with a short wavelength
Measuring the momentum
of an electron
• By Planck’s law E = hc/λ, a photon with a short
wavelength has a large energy
• Thus, it would impart a large ‘kick’ to the electron
• But to determine its momentum accurately,
electron must only be given a small kick
• This means using light of long wavelength!
Fundamental Trade Off …
• Use light with short wavelength:
accurate measurement of position but not momentum
• Use light with long wavelength:
accurate measurement of momentum but not position
h
ΔxΔp ≥
= (h _ bar )
2π
The more accurately you know the position (i.e.,
the smaller Δx is) , the less accurately you
know the momentum (i.e., the larger Δp is);
and vice versa
Heisenberg’s Uncertainty Principle
involving energy and time
h
ΔEΔt ≥
=
2π
• The more accurately we know the energy of a body,
the less accurately we know how long it possessed
that energy
• The energy can be known with perfect precision (ΔE
= 0), only if the measurement is made over an infinite
period of time (Δt = ∞)
Barrier Tunneling
As puck slides up hill, kinetic energy K is converted to gravitational potential
energy U. If the puck reaches the top its potential energy is Ub. The puck can
only pass over the top if its initial mechanical energy E> Ub. Otherwise the puck
eventually stops its climb up left side of hill and slides back top left. For
example, if Ub=20 J and E=10 J, the puck will not make pass over the hill, which
acts as a potential barrier.
Fig. 38-13
38-
Barrier Tunneling, cont’d
What about an electron approaching an
electrostatic potential barrier?
Fig. 38-14
Due to the nature of quantum
mechanics, even if E< Ub there is a
non-zero transmission probability
(transmission coefficient T) that the
electron will get through (tunnel) to the
other side of the electrostatic potential
barrier!
8π m (U b − E )
where b =
h2
2
T ≈e
−2 bL
Fig. 38-15
Fig. 38-16
38-
The Scanning Tunneling Microscope (STM)
As tip is scanned laterally across the surface, the tip is moved up or down to
keep the tunneling current (tip to surface distance L) constant. As a result the
tip maps out the contours of the surface with resolution on the scale of 1 nm
instead of >300 nm for optical microscopes!
http://www.youtube.com/watch?v=8GZdZUouzBY
38-