5. First Quantization
J J Thomson “discovered” the electron and determine it’s charge-to-mass ratio: e/m
Thomson tried to measure the charge of the electron using charged water droplets:
Charge up a large collection of water droplets and measure their total mass and charge
But the water evaporated during the measurement! - Big uncertainties
Robert Millikan switched to oil droplets to eliminate the evaporation problem
Millikan also innovated by examining the dynamics of single
individual charged oil droplets with varying charge and in a variable electric field
6. 6
The oil droplets are charged by the atomizer
They fall into under the influence of gravity
into the capacitor
The droplets experience a drag force and fall
at constant (slow) velocity
The charge on the droplets can change by
means of ionizing radiation sent into the
chamber
One can follow individual droplets through
the microscope
The charge on the droplets can be measured
from a balance of forces involved:
gravity
drag
electric
𝑬
Millikan Oil Drop Experiment
7. 7
Outcomes of the Thomson and Millikan Oil Droplet Experiments
After many observations of droplets in different charge states:
The charge of the droplets was found to be Quantized in units of 𝒆 = 𝟏. 𝟔𝟎𝟐 × 𝟏𝟎−𝟏𝟗 Coulombs
The electron is a “universal” particle that is found in all substances / atoms (follows from Thomson’s cathodes)
The mass and charge of the electron are quantized → First Quantization
Every electron is apparently identical to every other electron
9. 9
Blackbody Radiation
All objects at finite temperature 𝑻 emit electromagnetic radiation
The total amount of radiation increases with the temperature of the object
𝑹𝑻𝒐𝒕𝒂𝒍 = 𝝈𝑻𝟒
𝑹𝑻𝒐𝒕𝒂𝒍 Total radiated power per unit area of object (W/m2)
Absolute temperature (K)
Stefan constant
𝑻
𝝈 𝝈 = 𝟓. 𝟔𝟕𝟎𝟑 × 𝟏𝟎−𝟖 𝑾/(𝒎𝟐𝑲𝟒)
10. Blackbody Radiation is Widely Distributed in Wavelength / Frequency
R(l) (W/(m2-nm))
E
n
e
r
g
y
/
m
^
2
/
n
m
/s
e
c
Wavelength l (Å = 10-10 m) 𝑹𝑻𝒐𝒕𝒂𝒍 =
𝟎
∞
𝑹 𝝀 𝒅𝝀
11. 11
This radiation spectrum is universal in the sense that all objects show the same emission spectrum as long as
they are the same temperature. Universality begs explanation …
16. 16
T = 2.7260±0.0013 K
Cosmic Microwave Background Radiation Spectrum
The radiation is isotropic to roughly one part in 100,000:
the root mean square variations are only 18 µK
https://en.wikipedia.org/wiki/Cosmic_microwave_background
𝟏 𝒄𝒎−𝟏
≈ 𝟑𝟎 𝑮𝑯𝒛
17. 17
What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
First do the classical calculation
𝝆 𝝀 = 𝒌𝑩𝑻 𝒏 𝝀
Energy per mode
Number of modes per unit volume
at wavelength l
Assume Equipartition
of Energy
Every mode acquires 𝒌𝑩𝑻
of energy, on average, for
a system in thermal equilibrium
What is the mode density of electromagnetic fields in the box of wavelength l, 𝒏 𝝀 ?
18. 18
What is the Mode Density 𝒏(𝝀)?
Imagine all the resonant oscillation modes of a string of length L
+ many more modes!
The wavelengths of standing waves on the string are given by
𝑳 = 𝑵
𝝀
𝟐
, where 𝑵 is an integer, 𝑵 = 𝟏, 𝟐, 𝟑, 𝟒, …
We can ask the question: For a given wavelength 𝝀, what mode number
does it correspond to? The answer is 𝑵 = 𝟐𝑳/𝝀.
Doing this in 3 dimensions for a cube of size 𝑳 × 𝑳 × 𝑳 gives:
𝑵 = 𝟒𝝅𝑳𝟑
/𝟑𝝀𝟑
.
The number of modes with wavelengths between 𝝀 and 𝝀 + 𝒅𝝀 is given by:
𝒅𝑵 =
𝟒𝝅𝑳𝟑
𝝀𝟒 𝒅𝝀.
The number of modes per unit volume per unit wavelength is:
𝒏 𝝀 =
𝟖𝝅
𝝀𝟒 (assuming 2 polarizations for each wavelength)
19. 19
What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
𝝆 𝝀 = 𝒌𝑩𝑻 𝒏 𝝀
𝝆 𝝀 =
𝟖𝝅𝒌𝑩𝑻
𝝀𝟒
Rayleigh-Jeans Law
The ultra-violet catastrophe ensues…
Somehow we have to limit the occupation
of the short-wavelength modes …
Still doing the classical calculation
20. 20
What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
The birth of Quantum Mechanics
The atoms in the walls of the box have
electrons, and when these electrons vibrate at
frequency 𝑓 they emit electromagnetic waves
at frequency 𝑓.
Likewise light at frequency 𝑓 can be absorbed
by the atoms in the walls and start oscillating
at frequency 𝑓.
In equilibrium there is an equal energy flux
from the walls to radiation and from radiation
back into the walls.
21. 21
What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
The birth of Quantum Mechanics
Max Planck (1900) made two (largely unjustified and revolutionary) assumptions:
1) The atoms in the walls of the box have discrete energy levels given by 𝑬𝒏 = 𝒏𝜺, where
𝒏 = 𝟎, 𝟏, 𝟐, … Hence the atoms can only interact with light of energy 𝜺, 𝟐𝜺, 𝟑𝜺, etc.
2) The energy of light is directly related to the frequency of oscillation of the EM fields as
𝜺 = 𝒉𝒇, where 𝒉 is a fudge factor with units of 𝑱/𝑯𝒛 or 𝑱 − 𝒔. Surprisingly, this
energy is independent of the intensity of the light.
Why did Planck take 𝜺 = 𝒉𝒇 and not some other dependence?
22. 22
How Does This Get Rid of the Ultraviolet Catastrophe?
The likelihood of finding an atom in the walls of the box being excited to energy state 𝑬 is given by:
𝒈 𝑬 = 𝑨𝒆−𝑬/𝒌𝑩𝑻
(the Boltzmann factor from classical statistical mechanics)
𝒈 𝑬 is the fraction of atoms in the wall of the box that are excited to a state of energy 𝐸
𝑻 is the temperature of the walls of the box (K)
𝑨 is a normalization constant
This shows that the highly energetic atom energy states (𝑬 >> 𝒌𝑩𝑻) will be occupied with near-zero probability
Hence in equilibrium there will be very few highly energetic electromagnetic modes excited in the cavity
In other words, equipartition of energy breaks down for the energetic modes of the box
𝒌𝑩𝑻~𝟎. 𝟏 𝒆𝑽
𝑬~𝟐 𝒆𝑽
𝒆−𝑬/𝒌𝑩𝑻~𝟏𝟎−𝟗
For example:
23. 23
We can find 𝑨 by using the discrete nature of the energy states of the atoms: 𝑬𝒏 = 𝒏𝜺
In this case we have the discrete fractions: 𝒈𝒏 = 𝑨𝒆−𝑬𝒏/𝒌𝑩𝑻 = 𝑨𝒆−𝒏𝜺/𝒌𝑩𝑻
Find 𝑨 by demanding that the total probability of finding the atom in any state is unity:
𝟏 = 𝒏=𝟎
∞
𝒈𝒏 = 𝒏=𝟎
∞
𝑨𝒆−𝒏𝜺/𝒌𝑩𝑻 = 𝑨 𝒏=𝟎
∞
𝒆−𝜺/𝒌𝑩𝑻 𝒏
We recognize the sum as that for
𝟏
𝟏−𝒙
, 𝒇𝒐𝒓 𝒙 < 𝟏 (note that 𝒙 = 𝒆−𝜺/𝒌𝑩𝑻
< 𝟏):
𝟏
𝟏−𝒙
= 𝒏=𝟎
∞
𝒙 𝒏
S𝐨𝐥𝐯𝐢𝐧𝐠 𝐟𝐨𝐫 𝑨, 𝐰𝐞 𝐠𝐞𝐭 𝑨 = 𝟏 − 𝒆−𝜺/𝒌𝑩𝑻
What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
First step: Find 𝑨
24. 24
What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
Second step: Find the average energy of the atoms in equilibrium
𝑬 =
𝒏=𝟎
∞
𝑬𝒏 𝒈𝒏 = 𝑨
𝒏=𝟎
∞
𝒏𝜺 𝒆−𝜺/𝒌𝑩𝑻 𝒏
Trick to simplify the sum: The pre-factor in the sum, 𝒏𝜺, appears in the exponent. Hence take a derivative
with respect to −𝟏/𝒌𝑩𝑻 to bring it down from there. Then move the derivative outside the sum!
𝑬 = 𝑨
𝒅
𝒅(−𝟏/𝒌𝑩𝑻)
𝒏=𝟎
∞
𝒆−𝜺/𝒌𝑩𝑻 𝒏
Now the sum can be easily performed!
𝑬 = 𝑨
𝒅
𝒅(−𝟏/𝒌𝑩𝑻)
𝟏
𝟏 − 𝒆−𝜺/𝒌𝑩𝑻
Finally we get
𝑬 =
𝜺
𝒆𝜺/𝒌𝑩𝑻 − 𝟏
25. 25
𝑬 =
𝜺
𝒆𝜺/𝒌𝑩𝑻 − 𝟏
What is the Energy Density of Electromagnetic Modes in the Box, r(l)?
This is the average energy of the atoms making up the walls of the box
They are in equilibrium with the electromagnetic modes.
Therefore we use Planck’s assumption that 𝜺 = 𝒉𝒇 to find the energy of the light
𝑬 =
𝒉𝒇
𝒆𝒉𝒇/𝒌𝑩𝑻 − 𝟏
=
𝒉𝒄/𝝀
𝒆𝒉𝒄/𝝀𝒌𝑩𝑻 − 𝟏
using the fact that 𝝀𝒇 = 𝒄 for light
This is the average energy in a mode of wavelength 𝝀.
Find the Total energy density of the light by multiplying by the mode density calculated above:
𝝆 𝝀 = 𝑬 𝒏 𝝀 = 𝑬
𝟖𝝅
𝝀𝟒
𝝆 𝝀 =
𝟖𝝅𝒉𝒄/𝝀𝟓
𝒆𝒉𝒄/𝝀𝒌𝑩𝑻 − 𝟏
Planck Blackbody Radiation Formula
26. 26
𝝆 𝝀 =
𝟖𝝅𝒉𝒄/𝝀𝟓
𝒆𝒉𝒄/𝝀𝒌𝑩𝑻 − 𝟏
Planck Blackbody Radiation Formula
This equation fit the data very well! T = 2.7260±0.0013 K
The prediction remains finite as 𝝀 → 𝟎
𝝆 𝝀 ≈
𝟖𝝅𝒉𝒄
𝝀𝟓
𝒆−𝒉𝒄/𝝀𝒌𝑩𝑻
This predicted exponential suppression
at short wavelength is a characteristic
property of blackbody radiation data
27. 27
Thanks for Listening!
Planck has overcome the Ultraviolet Catastrophe, but at what price?
The value of Planck’s constant
𝒉 = 𝟔. 𝟔𝟐𝟔 × 𝟏𝟎−𝟑𝟒
𝑱 − 𝒔
= 𝟒. 𝟏𝟑𝟔 × 𝟏𝟎−𝟏𝟓
𝒆𝑽 − 𝒔
The reduced Planck constant: ℏ = 𝒉/𝟐𝝅
Planck was obsessed with justifying these two assumptions in the 10 years or so that followed
He was a “reluctant revolutionary”
He was relieved when a young Albert Einstein picked up these ideas about light and showed that
they could be used to understand the Photo-electric Effect (a totally different phenomenon), and
determine the value of Planck’s constant with high precision!
Editor's Notes
Get the mass of the droplet from watching it fall under gravity and drag and measuring the radius. Get the charge by balancing the upward electric force with the weight.
