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Lecture 1
Physics 486, Spring ‘
Lecture 1
Course Description;Course Description;
Quantization of Light:Quantization of Light:
Blackbody Radiation,Blackbody Radiation,
Photoelectric Effect, etc.Photoelectric Effect, etc.
Physics 486:Physics 486: Course DescriptionCourse Description
Physics 486 is the first of a two-semester sequence in intermediate quantum mechanics. In this course, we’ll cover the following topics (roughly Griffiths chapters 1-4):
l Quantization of light, matter waves, blackbody radiation
l Schrödinger’s equation (SEQ)
l Bound state solutions to various 1D SEQ
l Expectation values and probabilities
l Particle scattering and tunneling in 1D potentials
l Introduction to approximation methods: perturbation theory and variational methods
l Solutions to 2D and 3D SEQ: Degeneracy and Angular Momentum
l Solutions to the hydrogen atom problem
l Spin angular momentum, addition of angular momentum
Lecture 1
Physics 486Physics 486 Course DescriptionCourse Description
In this course, I’ll provide some of the necessary mathematical and physics background, but it would be helpful if you have working knowledge of:
- Newton’s laws and equations at the level of Phys. 211
- The fundamentals of linear algebra, matrices and matrix algebra (Math 415)
- The properties of complex quantities
- Multivariable calculus and elementary differential equations
Physics 486Physics 486 Course DescriptionCourse Description
Physics 486 is composed of several components:
l Lectures – discussion of concepts, illustration of example problems, some demonstrations when possible
l Discussion – You’ll have a one-hour session on Tuesday or Wednesday evenings, during which you’ll solve relevant problems under the supervision of a TA. Participation constitutes 10% of your total grade.
l Homework – You’ll get an assignment almost every Tuesday, which will be due the following Tuesday. Late homework will be accepted the following week for ½ credit, and no Homework will be accepted more than 2 weeks after it is handed out. HW is an important part of the course, which is why it is worth 45% of your grade!
l Exams – There will be one in-class midterm exam (15%) and a final exam (30%).
l Reading – I’ll provide suggested reading assignments from Griffiths, and occasionally other sources such as Shankar.
Lecture 1
Historical Background (cont.)Historical Background (cont.)
l In 1801, Thomas Young definitively established that light
was a wave by demonstrating interference of light and by
explaining Newton’s rings – although it took time for people
to be convinced of Young’s work, mainly because Isaac
Newton had promoted the corpuscular description of light
(for an interesting book about Young, see The Last Man
Who Knew Everything, by Andrew Robinson).
l In 1865, James Maxwell put the wave theory of light on
firm theoretical footing when he developed a set of
equations describing the generation of, and relationship
between, the electric and magnetic fields comprising light.
l However, around 1900, several phenomena exposed flaws
in the wave description of light:
l Blackbody radiation l Photoelectric effect l Compton scattering l Bremsstrahlung radiation
λmax ~ 1/T
Power ~ T^4
Planck’s distribution:
l The spectral distribution of the radiation is governed by the radiating object’s temperature. Consequently, an object’s temperature can be determined from the distribution of this radiation!
- Early potters estimated the temperatures of their kilns by noting the color of the fire
- Steelmakers estimate the temperature of molten steel by noting its color
Incandescent bulb, heated to T ~ 3000K
very inefficient!
l When an object is heated, it emits radiation consisting
of electromagnetic waves with a broad range of
frequencies.
Blackbody RadiationBlackbody Radiation
Lecture 1
Examples of blackbody radiationExamples of blackbody radiation
l In the late 1800’s and early 1900’s, many scientists, including Max Planck, recognized the fundamental importance of the “blackbody” radiation spectrum, because it was universally observed in a variety of systems:
Penzias and Wilson won the Nobel Prize for “discovering” this!
Solar blackbody spectrum, λmax ~ 0.5 μm
Microwave ‘background’ radiation of Universe, λmax ~ 2 mm remnant radiation from Big Bang!
T = 2.73 K
Thermal emission:Thermal emission: starsstars
l Star color mostly due to ‘blackbody’ radiation, reflects the temperature of the star:
see T.L. Swihart., “Astrophysics...,” (1968)
red stars are “cool” blue stars are “hot”
λλλλ max ~ 1/T
Power ~ T^4
Planck’s distribution:
It’s not all about temperature!: Notice absorption bands due to gases in sun
Lecture 1
l The reason the Rayleigh-Jeans treatment led to the ultraviolet catastrophe is fairly straightforward to understand:
l The Rayleigh-Jeans formula for energy density in a cavity is the product of two terms: (i) the number of modes per unit frequency per unit volume (which is related to the “density of states”) AND (ii) the average energy per mode. We’ll show that the number of modes per frequency per volume is given by:
l To understand why this is, consider electromagnetic waves confined to a cubical cavity (box) of dimension a on a side. Only standing waves that satisfy the boundary conditions of the cavity are “allowed,” because only these have wavelengths that can fit inside the cavity:
RayleighRayleigh--Jeans LawJeans Law
2
3
8 f
c
π
x
y
E
B
n
a
n
RayleighRayleigh--Jeans LawJeans Law
where n = 1,2,3,...
n
c
f n
a
a
f 3 = 3 f 1 , λ 3 = 3a/
f 2 = 2 f 1 , λ 2 = a
f 1 = c/2a , λ 1 = 2a
0
n ( )^ sin
n
x
x
E E
Reminder: λ=c/f f = ω/2π λ = 2π/k
l So, the allowed frequencies of electromagnetic radiation in a cavity of dimension a are given by:
Lecture 1
2
2
n dn
dN
af a
df
c c
= ×
l Note that, for a 3 dimensional cavity, a similar relationship above applies to the x (nx), y (ny), and z (nz) directions, and the allowed values of n can be thought of as points on a grid in a 3-dimensional cartesian coordinate system (to right).
RayleighRayleigh--Jeans LawJeans Law
nx
ny
nz
n
To calculate the number of allowed states in a particular frequency range, n and n+dn, we need to determine the number of states in a spherical shell of volume (4πn^2 dn)/8 (the division by 8 arises because we’re only interested in the octant in which nx, ny, and nz are positive). In terms of frequency, this can be written:
3 2 3
8 a dN f df c
π
2 2 2
n = n x + ny + nz
( ) 0 sin sin sin
nx x n y^ y nz z
x
a a a
π ^ π π
E E
n
c
f n
a
= (still the
same) polarization
l We’re almost done, except that we have to multiply this result by a factor of 2, because there are 2 independent waves with different polarizations corresponding to each value of frequency. The net result is that the number of electromagnetic wave frequencies in the range f and f+df is given by:
( )
2 3
dN 8 V f f df c
π
RayleighRayleigh--Jeans LawJeans Law
nx
ny
nz
n
Where V=a^3 is the volume of the cavity. dN(f)/df is the number of states per unit frequency interval, usually called the “density of states”.
Lecture 1
Planck’s Law: 2 Key PointsPlanck’s Law: 2 Key Points
l There are 2 key features of Planck’s proposal:
(1). Discrete energies – By postulating that the distribution of energies was DISCRETE, rather than continuous, Planck insured that there would be a well defined crossover between the following regimes:
(i) ∆E << kT: In this limit, the discreteness of the energies doesn’t matter, as all energy levels are equally occupied (i.e., equipartition applies), and the classical result for the average energy, =kT, pertains: (^) En = n ε 4 εεεε 3 εεεε (^2) ε εεε εεεε 0
kT^ ∆E=hf
(ii) ∆E >> kT: In this limit, only the E=0 level is appreciably occupied, giving ~0 for the average energy: E n = n ε 4 εεεε 3 εεεε 2 εεε ε εεεε 0
∆E=hf
kT
E
P(E)
0
E
P(E)
0
All energy levels equally occupied: equipartition!
Only E=0 level is appreciably occupied!
Planck’s Law: 2 Key Points (cont.)Planck’s Law: 2 Key Points (cont.)
(2). Energy spacing scales with frequency f, ∆E = hf:
Small wavelength regime: On the other hand, at large frequencies (small wavelengths), the condition ∆E >> kT applies in Planck’s description (because ∆E = hf), causing the average energy to be ~0 in this regime. This insures that the energy density – which is the product of the density of states and the average energy – will go to zero at small wavelengths!
Large wavelength regime: By postulating that the discrete energy separation scales with frequency, i.e., ∆E = hf, Planck ensured that – for a given, fixed temperature T - his description would give the “classical” regime (∆E<
Lecture 1
l If the energy of an oscillator can only be E=mhf (m=0,1,2,3,…), then the probability that a particular energy of the oscillator is occupied is given by the Boltzmann factor:
Planck to the RescuePlanck to the Rescue
l Now that we understand Planck’s ideas intuitively, let’s do the math:
where Em = mhf, m=0,1,2,3,..
E m = m ε
4 εεεε 3 εεεε 2 εε εε εεεε 0
ε = hf
/ / /
m kT m kT P m e e
− ε −ε = (^) ∑
Em / kT P m Ce
−
0
/
C
m
e^ m^^ ε kT
∞
=
−
∑
C is a normalization constant, selected to insure that:
0
n
P m
∞
=
∑^ =
Therefore, (^) and
Planck to the RescuePlanck to the Rescue
l For equally spaced levels, the denominator in,
(where x=e-ε/kT)
E m = m ε
(^4) εεεε 3 εεεε 2 ε εεε εεεε 0
ε = hf
/ / /
m kT m kT P m e e
− ε −ε
So: (^) Z = 1 + xZ
Therefore, the probability that the nth energy level of an oscillator will be populated is:
is straightforward to evaluate:
( )
/ 2 3 0 0 2
m kT m m m
e Z x x x x
x x x xZ
ε
∞ ∞ − = =
∑ ∑
/
1 1-e
m kT m
Z e
x
ε ε
∞ − =
⇒ ∑
( )
/
/^1 1
m kT
m (^) kT
e P e
ε
ε
− = − − −
Lecture 1
Planck to the RescuePlanck to the Rescue
l So finally, the average energy can be written:
Finally, if we multiply this average energy per state times the density of states per energy:
So that:
(where x=e-ε/kT)
We get the famous Planck’s radiation law for the energy density of blackbody radiation:
(Remember this? We derived it earlier!)
( )
( )
( )
2 0 / 1 /^1
m m hf kT hf kT
hf nx
hfx x
E
e e
∞
= − −^ − −
∑
( ) (^ )
/ 0
m hf kT m hf kT hf^ kT hf^ kT
hf mx
hfe hf
E
e e e
∞ − =
∑
2 3
8 f
c
π
( )
3 3 /
hf kT 1
hf
u f T
c e
π
(reduces to kT for large T, which is the classical limit of equipartition)
Planck to the RescuePlanck to the Rescue
l So finally, the average energy density can be written:
( )
3 3 /
hf kT 1
hf
u f T
c e
π
u(f)
hf 2.8 kT
Planck Radiation Law
Summarizing the classical and quantum mechanical results:
l How does energy quantization solve the ultraviolet catastrophe? At high frequencies (low wavelengths), even though many modes are possible (i.e., it’s easier to stuff the short waves into the cavity), not many are excited because it costs too much energy to make a high frequency quantum, i.e., E = hf.
classical
quantum
2
3
8 f
c
π
2
3
8 f
c
of modes per unit
frequency per unit volume
average energy per mode
kT
hf / kT 1
hf
e −
3
3 /
hf kT
hf
c e
2
3
8 f
kT
c
average energy density
Lecture 1
Example: Peak in the Planck distributionExample: Peak in the Planck distribution
At what wavelength is the maximum
in the Planck energy distribution?
( )
3 3 /
hf kT 1
hf
u f T
c e
π
u(f)
hf 2.8 kT
Planck distribution
( ) (^5) /
hc kT 1
hc
u T
e^ λ
π λ λ
But f=c/λ, so:
The maximum of u(λ,T) occurs where ∂u(λ,T)/∂λ =
⇒ (^) ( )
/
a a
e
λ λ
−
= − where a = hc/kT
This transcendental equation can be solved graphically or numerically to give:
6 max
hc m K
k T T
λ
−
× ⋅
( )
( ) ( )
/ 6 / /
a a a
u T (^) hc a e e e
λ λ λ
∂ ^
Planck to the RescuePlanck to the Rescue
l Now, if we want to determine the total energy per unit area radiated by our “blackbody” source, we need to integrate the density over all frequencies:
( )
3 3 / 0 0
hf kT 1
h f
u f T df df
c e
∞ ∞
∫ ∫ −
u(f)
hf 2.8 kT
Planck Radiation Law
Now, let x=hf/kT. Then df=(kT/h)dx, and f^3 =(kTx/h)^3
So one can write: (^) ( )
(^4 )
3 0 0
x 1
h kT x
U T u f T df dx
c h e
π
∞ ∞
∫ (^) ∫ −
(^4 ) 4 3
h kT
U T aT
c h
π ^ π
(a =
5 4 3 3
k
h c
Giving the famous T^4 law^ π of radiated energy )
Lecture 1
RadiationRadiation
P = σ Ae ( T 4 − T 04 )
l Note that an object not only radiates energy at a rate
given by Stefan’s Law, but it also absorbs
electromagnetic radiation from the surroundings. So, if
an object is at a temperature T, and its surroundings are
at a temperature T 0 , then the net energy gained or lost
each second by the object as a result of radiation is:
l When an object is in equilibrium with its surroundings, it
radiates and absorbs energy at the same rate, and so its
temperature remains constant. When an object is hotter
than its surroundings, it radiates more energy than it
absorbs, and its temperature decreases.
l An ideal absorber, often called a black body, absorbs all
the energy incident upon it, and has e=1. By contrast, an
ideal reflector absorbs none of the energy incident upon
it, and so has e=0.
Example: Star Light, Star BrightExample: Star Light, Star Bright
If the measured radiation emitted by a
star has its maximum intensity at
λmax=446 nm, what is the star’s (a) surface
temperature and (b) total power emitted
per unit area?
(a).
( )
4 8 2 4 4 Φ = σ T = 5.67 × 10 −^ Wm −^ K − × 6500 K
⇒
6 9
m K
T K
m
− −
× ⋅
×
(b).
Φ ≅ 101.4 × 106 Wm −^2
Lecture 1
Example:Example: Tungsten FilamentTungsten Filament
If the temperature of a tungsten filament
in an incandescent bulb can be raised to
T=3300K, what is the (a) wavelength of
the peak radiation intensity, and (b) what
is the intensity radiated by the bulb
(a).
( )
4 8 2 4 4 Φ = σ T = 5.67 × 10 −^ Wm −^ K − × 3300 K
⇒
6 max
m K
nm
K
λ
−
× −
(b).
6 2
6.7 10 Wm
−
Φ ≅ ×
Example:Example:
Thermal Radiation from a SolidThermal Radiation from a Solid
Calculate the power radiated by a 1-
cm-diameter sphere of aluminum
at room temperature (20oC).
(Assume it is a perfect radiator.)
4 -8 -2 -4 4 2 2 -2 2 -4 2 2 -4 2
(power radiated per area) (5.670 10 W m K )(293K) 418 W/m Area: A 4 r 4 (0.5 10 ) 3.14 10 Power radiated A (418 W/m )(3.14 10 )
SBT
m m m
σ
π π
= ×
= = × = ×
= Φ × = ×
= 0.13 W
Lecture 1
Einstein to the rescue!Einstein to the rescue!
V
stop (v)
f (x10^14 Hz)
0
1
2
3
0 5 10 15
f 0
h/e
1
Ephoton = hf = hc/λ
K. E .max= e ⋅ Vstop = hf − Φ^ h Φ^ (slope) is Planck’s constant; is the “work function”
l As he described in his now-famous 1905 paper, Einstein realized that the puzzling observations in the photoelectric effect could be understood if one assumes that the incident light consists of quantized “packets” of energy – called photons – which have an energy:
l In this picture, a light quantum delivers all its energy to the electron in the metal. The electrons will lose a certain amount of energy traveling through the material to reach the surface. Furthermore, all the electrons must perform a certain amount of work, Φ, to overcome the potential energy barrier at the surface of the metal.
The resulting simple relationship between the incident light frequency f and the electron kinetic energy KE is consequently:
I won the Nobel Prize in 1921 for explaining this!
When light of wavelength λ = 400 nm shines on a piece of lithium, the stopping voltage of the electrons is Vstop = 0.21 V. What is the work function of lithium?
φ = hf -eVstop
= hc/λ - eVstop
= 4.63× 10 -19^ J
= 2.9 eV
This is from page 2 of your printed notes. f = c/λ.
If V is in Volts, eV is in Joules. e is the charge of the electron, e = 1.60× 10 -19^ Coulomb = 1.60× 10 -19^ J/V.
The definition is: 1 eV ≡ 1.60× 10 -19^ J.
Photoelectric Effect: Example
Solution:
What is the maximum wavelength that can cause the photoelectric effect in lithium?
λmax = hc/φ = 429 nm
To find the maximum wavelength, set Vstop = 0.
Lecture 1
Example: Estimating Planck’s ConstantExample: Estimating Planck’s Constant
l In a photoelectric effect measurement of lead (Pb), you observe that two ultraviolet beams having wavelengths of λ 1 =280nm and λ 2 =490 nm induce photoelectrons with maximum energies 8.57 eV and 6.67 eV, respectively. Obtain an estimate of Planck’s constant from this simple observation.
Vstop
(v)
f (x10^14 Hz)
0
1
2
3
0 5 10 15
f 0
h/e
1
K 1 (^) = hf 1 (^) − Φ = hc / λ 1
K (^) 2 = hf 2 (^) − Φ = hc / λ 2
Taking the difference between these expressions:
K (^) 2 − K (^) 2 = hc ( λ 2 − λ 1 ) /(λ λ 1 2 )^1 2 1 2 1
K K
h
c
( ) ( )( )
19 9 9
8 9 9
8.57 6.67 1.6 10 /^280 10 490
eV J eV^ m^ m
h
m s m m
−^ −^ −
− −
− × ×^ ×^ ×
× × − ×
h ≅ 6.62 × 10 −^34 Js Quite accurate!
