Quantum physics
“Anyone who is not shocked by the quantum theory has not
understood it.” – Niels Bohr, Nobel Price in 1922 (1885-1962)
“I can safely say that nobody understand quantum
physics” – Richard Feynman Nobel Lecture, 1966, (1918-1988)
PHY232
Remco Zegers
zegers@nscl.msu.edu
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
quiz (extra credit)
 a distant star moves at a velocity of 0.5 times the speed of
light away from us. It emits light that can be detected in
earth-based telescopes. What is the speed of the light of
the radiation we receive from these quasars? Ignore the
fact that the telescope is in air and not in vacuum.
 a) it is less than the speed of light in vacuum (3x108 m/s)
by a factor 1/=(1-v2/c2)=(1-(0.5c)2/c2)=0.87
 b) it is larger than the speed of light in vacuum (3x108 m/s)
by a factor =1/(1-v2/c2)=1/(1-(0.5c)2/c2)=1.15
 c) it is equal to the speed of light in vacuum (3x108 m/s)
answer: c the speed of light is independent of the motion
of the source or observer
PHY232 - Remco Zegers
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so far…
 we have treated light as being waves and used that
formalism to treat optics and interference
 we have seen that under extreme conditions (very high
velocities) the Newtonian description of mechanics
breaks down and the relativistic treatment designed by
Einstein must be used.
 Now, we will see that the description of light in terms of
waves breaks down when looking at very small scales. In
addition, we will see that objects that we usually refer to
as particles (like electrons) exhibit wave-phenomena.
PHY232 - Remco Zegers
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photoelectric effect
 when light hits a metal, electrons are released.
By providing a voltage difference between the
metal and a collector, these electrons are
collected and produce a current.
 if light is described in terms of waves one would
expect that (classical description):
 independent of the frequency of the light,
electrons should be emitted if one waits long
enough for sufficient energy to be absorbed
by the metal
 the maximum kinetic energy depends on the
intensity (more energy absorbed)
 the kinetic energy of the electrons is
independent on the frequency (wavelength
of the light) and only depends on the
intensity
 electrons take a little time to be released
since sufficient energy needs to be absorbed
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however…
one observes that:
• if the frequency of the light is too low, no electrons
are emitted
•the maximum kinetic energy of the electrons is
independent of the intensity.
• the maximum kinetic energy increases linearly
with frequency
•the electrons are emitted almost instantaneously,
even at very low light intensities
These observations contradict the classical
description. It suggest that energy is delivered to the
electrons in the metal in terms of well-localized
packets of energy. The photons in the light beam
are thus seen as ‘particles’ that deliver packets of
energy (so-called energy quanta) to the electron it
strikes.
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photo-electric effect
The energy carried by a photon: E=hf
h: planck’s constant (h=6.63x10-34 Js)
f: frequency with c=f
The energy is localized in the photon-particle
The maximum kinetic energy of a released
electron: KEmax=hf-
with: : the workfunction (binding of electron to
the metal)
So only if hf> will electrons be released from
the metal
fc=/h : fc is the cut-off frequency
c=c/fc=(hc)/: the cut-off wavelength
see table 27.1 for work functions for various metals
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example
light with a wavelength of 400 nm is projected on a sodium metal
surface (=2.46 eV).
a) what is the energy carried by a single photon?
b) what is the maximum kinetic energy of the released electrons?
c) what is the cut-off wave length for sodium?
d) what happens if light with a wavelength of 600 nm is used?
a) E=hf=hc/=6.63x10-34Js x 3x108/(400x10-9)=4.97x10-19 J
in eV (1 eV=1.6x10-19 J)
=3.11 eV
b) KEmax=hf-=3.11-2.46 eV = 0.65 eV
c) c=c/fc=(hc)/=6.63x10-34x3x108/(2.46x1.6x10-19J)=505x10-9 m
the cut-off wavelength is 505 nm.
d) if light with a wavelength of 600 nm is used: no electrons are emitted
Note: if f<fc no electrons emitted
if > c no electrons emitted
PHY232 - Remco Zegers
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particle-wave dualism
 So, is light a wave or particle phenomenon?
experiment
can be described by can be described by
light as waves
light as particles
reflection
X
X
refraction
X
X
interference
X
diffraction
X
polarization
X
X
photo-electric effect
 answer: it depends!
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question
 light from a far-away star is used to perform a double slit
experiment. Approximately once per 10 minutes will a
single photon from the star arrive at the double slit setup
on earth. Which of the following is true?
 a) since light is a wave-phenomenon, an interference
pattern will be seen on a screen placed behind the
double slits.
 b) since only one photon arrives every 10 minutes,
interference is not possible since one can hardly think of
the light coming in as waves
interference is a pure wave-phenomena; it doesn’t depends on how
many photons are there!
PHY232 - Remco Zegers
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question
 instead of a light source, an electron gun firing electrons at high
speeds is used in a double-slit experiment. which of the following is
true?
 a) since electrons are massive particles, an interference pattern is not
produced
 b) electrons are similar to photons; they exhibit both wave and
particle phenomena. In this case, electrons behave like waves and
an interference pattern is produced.
All particles can be associated with a characteristic wavelength,
with =h/(mv) (the so-called ‘de Broglie’ wavelength and thus exhibit
wave-phenomena under the right conditions
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interference pattern
A
P1
P1
B
P2
P2
If one of the slits in a double slit experiment is closed one sees only
a diffraction pattern from a single slit (P1). If the other slit is opened and
the first one closed, one sees only the diffraction pattern from the other
slit (P2). If both are opened, one does not simply see the sum of P1 and
P2 (like in A), but the double-slit interference pattern (like in B).
The reason is the following:
Remember that the intensity (I) is proportional to the E-field squared:
I~E2=E02cos2. In A, it is assumed that the intensities add: Isum=I1+I2 .
However, one should add the E-fields (which can be positive or
negative) and than squared, like in B: Isum=(E1+E2)2 where E1 and E2 are
treated as vectors.
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and if you think that you’ve seen it all…
A
B
let’s assume I determine through which hole the photon (or electron)
goes by placing a detector before the slits. Would I still observe an
interference pattern like in B?
Answer: no! By measuring the location of the photon, we have ‘turned’
the light-wave into a particle and the interference pattern gets lost.
PHY232 - Remco Zegers
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Schrödinger’s cat
A cat is placed in a closed box. Inside the box a radioactive source
is placed in which on average once per hour a radioactive decay
takes place. If the decay takes place, a bottle of poison breaks, killing
the cat.
In quantum-physical sense, the cat is 50% dead and 50 alive after
half an hour. Since we can’t see it, it is in a superposition of those two
states and there is a certain probability of being in one of either states.
Only when we open the box, do we determine what state the cat is in.
The observation is crucial to determine the state of the cat.
PHY232 - Remco Zegers
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PHY232 - Remco Zegers
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heisenberg’s uncertainty principle
If we want to determine the location and
velocity (momentum) of an electron at a
certain point in time, we can only do that with
limited precision.
Let’s assume we can locate the electron using
a powerful light microscope. Light scatters off
the electron and is detected in the
microscope. However, some of the
momentum is transferred and observing the
electron means we can only determine its
note ħ=h/(2)
velocity (momentum) with limited accuracy.
with x: precision of position measurement
with p: precision of momentum (mv) measurement
this can also be expressed in terms of energy and time measurements
Eth/(4)
with E: precision of energy measurement
with p: precision of time measurement
xph/(4)
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example
The location of an electron is measured with an uncertainty of 1 nm.
One also tries to measure the velocity of the electron. What is the
(minimum) uncertainty in the velocity measurement? The mass of the
electron is 9.11x10-31 kg.
use the uncertainty principle:
xph/(4) with x=1x10-9 m, h=6.63x10-34 Js
so p 6.63x10-34/(4 x 1x10-9)=5.28x10-26 kgm/s
vmin = 5.28x10-26/9.11x10-31=5.79x104 m/s
(use p=mv)
note that the uncertainty principle works for the three
dimensions separately:
xpxh/(4)
ypyh/(4)
PHY232 - Remco Zegers
- quantum physics
zpzh/(4)
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photons as particles and quanta
Some other examples of where the particle nature of
light plays a role:
•Photo-electric effect
•Black-body radiation
•bremmstrahlung
•Compton effect
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black-body radiation
A black body is an object that absorbs all electromagnetic radiation
that falls onto it. They emit radiation, depending on their temperature. If
T<700 K, almost no visible light is produced (hence a
‘black’ body).
The energy emitted from a black body: P=T4 with =5.67x10-8 W/m2K4
The peak in the intensity spectrum
varies with wavelength using the
Wien displacement law:
maxT=0.2898x10-2 mK
(classical)
Until 1900, the intensity distribution,
predicted using classical equations,
predicted a steep rise at small
wavelengths. However, the opposite
was determined experimentally…
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Planck to the rescue
 Max Planck devised a theory for a
simple black body that could describe
the measured spectra.
 He assumed that the walls consisted of
little radiators that only emitted light at
certain discrete energies: E=nhf
 f the frequency of the light (Hz)
 h: planck’s constant (6.63x10-34 Js)
 n: integer.
 His achievement was really the first
success of quantum theory
 In essence, his theory showed that
because the energy is quantized, it is
hard to emit light of small wavelengths
(high frequency) since a lot of energy is
required
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example
Hot lava can be considered as a black
body emitting radiation at a variety of
temperatures.
If temperature of molten lava is about 1200 0C,
what is the peak wave length of the light
emitted?
maxT=0.2898x10-2 mK
T=1200+273=1473 K
max= 0.2898x10-2/1473
=1.96x10-6 m=1970 nm
The peak is in the infrared region (not
visible by eye), but closest to the colors
red/orange in the visible spectrum
PHY232 - Remco Zegers
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X-rays
 when energetic electrons are shot on a
material, photons with small
wavelengths (~0.1 nm) are produced.
 The spectrum consist of two
components
 broad bremsstrahlung spectrum
 peaks at characteristic wavelengths
depending on the material (see
next chapter)
 the bremsstrahlung (braking radiation)
is due to the deflection of the electron
in the field of the charged nucleus.
 a light quantum is produced when the
electron is deflected. It takes away
energy from the electron
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bremmsstrahlung
 assume electrons are accelerated in a potential of V Volts.
 their kinetic energy is E=eV with e=1.6x10-19 C and V the potential
 If the electron is completely stopped in the material, all its kinetic
energy is converted into the photon with maximum frequency fmax
and hence minimum wavelength min
 if it merely deflected, the frequency f is smaller than fmax and its
wavelength  larger than min.
 so eV=hfmax=hc/min
PHY232 - Remco Zegers
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example
 an X-ray spectrum is analyzed and the
minimum wavelength is found to be
0.35 angstrom (1 angstrom: 10-10 m).
What was the potential over which the
electrons were accelerated before the
interacted with the material?
eV=hfmax=hc/min
V=hc/(mine)
=6.6x10-34x3x108/(0.35x10-10x1.6x10-19)
=3.55x104 V
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Braggs law
X-rays scattered off atoms in e.g. will interfere and the
interference pattern can be used to identify/study
materials.
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question
 X-rays are sometimes used to identify crystal structures of materials.
this is done by looking at the diffraction pattern of X-rays scattered off
the material (see ch 27.4). Why are X-rays used for this and not for
example visible light?
 a) the wavelength of X-rays is close to the spacing between atoms in
a crystal
 b) since the frequency (and thus energy) of X-rays is much larger
than that of visible light, they are easier to detect
 c) X-rays are much easier to produce than visible light
In order to observe structures of a given scale, the probe
must have a wavelength of the same scale.
e.g.: to observe airplanes by radar, radiation of very long
wavelength should be used
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compton effect
 When photons (X-rays) of a certain wavelength are directed towards
a material, they can scatter off the electrons in the material
 If we assume the photon and the electron to be classical particles, we
can describe this as a normal collision in which energy and
momentum conservation must hold.
 after taking into account relativistic effects (see previous chapter) one
finds that:
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compton scattering
 =-0=h/(mec) x (1-cos)
with: : wavelength of photon after collision
0: wavelength of photon before collision
h/(mec): Compton wavelength (2.43x10-3 nm)
me: mass of electron
: angle of outgoing X-ray relative
to incoming direction

0
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=-0=h/(mec) x (1-cos)
example
 A beam of X-rays with 0=10-12 m is used to bombard a material.
 a) What is the maximum shift in wavelength that can be observed due
to Compton scattering?
 b) What is the minimum shift in wavelength that can be observed due
to Compton scattering?
 c) What are the minimum and maximum kinetic energies of the struck
electrons, ignoring binding to the material they are in.
a) maximum shift occurs if cos=-1 (=1800). This is usually referred to as
Compton backscattering. in that case:
=2h/(mec)=2x2.43x10-12=4.86x10-12 m
b) minimum shift occurs if cos=1 (=00) in which case essentially no
collision takes place: =0
c) gain in kinetic energy by electron is loss in energy of x-ray:
case b) no kinetic energy gained by electron
case a) energy of X-ray before collision: hf=hc/0=1.98x10-13 J
energy of X-ray after collision: hf=hc/(0+) =3.38x10-14 J
kinetic energy gained by electron: 1.64x10-13 J=1.0 GeV (giga)
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applications
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