Download Physics IV - Script of the Lecture Prof. Simon Lilly Notes from:

Physics IV - Script of the Lecture
Prof. Simon Lilly
October 21, 2006
Notes from:
Raphael Honegger
$Id: physics4.tex 1507 2006-10-21 15:02:55Z charon $
i
Contents
0 Introduction
1
1 The Limits of Classical Physics & Wave-Particle Duality
1.1 Classical Concepts . . . . . . . . . . . . . . . . . . . . . . .
1.2 Empirical Problems with classical Physics . . . . . . . . . .
1.2.1 Atomic structure & atomic spectra . . . . . . . . . .
1.2.2 Photo-electric effect . . . . . . . . . . . . . . . . . .
1.2.3 Waves behaving as particles . . . . . . . . . . . . . .
1.2.4 Particles acting as waves . . . . . . . . . . . . . . . .
1.2.5 Summay - new ideas . . . . . . . . . . . . . . . . . .
1.3 Reminder Interference & Diffraction . . . . . . . . . . . . .
1.3.1 Young’s slits interference . . . . . . . . . . . . . . .
1.3.2 Diffraction . . . . . . . . . . . . . . . . . . . . . . .
1.3.3 Fourier Optics . . . . . . . . . . . . . . . . . . . . .
1.4 Thermal radiation and Planck’s constant h . . . . . . . . .
1.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . .
1.4.2 Stefan-Bolzmann Law . . . . . . . . . . . . . . . . .
1.4.3 Wien’s Displacement Law . . . . . . . . . . . . . . .
1.4.4 Oscillators . . . . . . . . . . . . . . . . . . . . . . . .
1.4.5 Planck’s hypothesis (1900) . . . . . . . . . . . . . .
1.5 De Broglie waves & Plank’s constant . . . . . . . . . . . . .
1.6 Measurements & Plank’s constant . . . . . . . . . . . . . .
1.7 Wave paricle duality & quantum reality . . . . . . . . . . .
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2
2
2
2
3
3
3
3
3
3
4
5
5
5
6
7
7
8
9
9
10
2 The Schrödinger Equation
2.1 Review of waves . . . . . . . . . . . . . .
2.1.1 Sinusoided waves . . . . . . . . . .
2.1.2 Superposition of waves . . . . . . .
2.1.3 Dispersion relations . . . . . . . .
2.2 Particle wave equation . . . . . . . . . . .
2.3 A Particle in a potential energy field V (r)
2.4 The meaning of Ψ(x, t) . . . . . . . . . . .
2.5 Expectation values and operators . . . . .
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11
11
11
11
11
12
13
13
14
3 Solutions to Schrödinger’s Equation
3.1 Separable solutions of definite energy . . . .
3.2 Example: Particle in a box . . . . . . . . .
3.3 States of uncertain E . . . . . . . . . . . . .
3.4 Finite potential wells . . . . . . . . . . . . .
3.5 Barrier penetration = “quantum tunelling”
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17
17
18
19
20
23
4 The Harmonic Oscillator
4.1 Classical case . . . . . . . . . . . . . . . . . . . .
4.2 Quantum oscillator . . . . . . . . . . . . . . . . .
4.2.1 Starionary states . . . . . . . . . . . . . .
4.2.2 Non-stationary states of uncertain energy
4.2.3 3-dimensional oscillater (degenacy) . . . .
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26
26
26
26
28
29
5 Observables & Operators - Heisenberg Uncertainty Principle 30
5.1 Operators & eigenfunctions . . . . . . . . . . . . . . . . . . . . . 30
5.2 Eigenfunctions for position and momentum . . . . . . . . . . . . 30
5.3 Compatible observables . . . . . . . . . . . . . . . . . . . . . . . 31
5.4 Heisenberg uncertainty principle . . . . . . . . . . . . . . . . . . 32
5.5 Compatibility with Ĥ . . . . . . . . . . . . . . . . . . . . . . . . 34
5.6 Orbital angular momentum . . . . . . . . . . . . . . . . . . . . . 35
5.7 Angular momentum eigenfunctions . . . . . . . . . . . . . . . . . 35
5.8 Stern-Gerlach experiment . . . . . . . . . . . . . . . . . . . . . . 37
5.9 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
ii
6 The
6.1
6.2
6.3
6.4
6.5
6.6
Hydrogen Atom
Motion in “central potentials” . . . . . . . . . . . . . . .
Solutions of Schrödingers equation in a central potential
Hydrogen atom . . . . . . . . . . . . . . . . . . . . . . .
Zeeman effect - Perturbing the Hamiltonian Ĥ . . . . .
Time-dependent perurbations - Radiative transitions . .
Some (small) complicating effects . . . . . . . . . . . . .
6.6.1 The reduced mass effect . . . . . . . . . . . . . .
6.6.2 Spin-orbit coupling . . . . . . . . . . . . . . . . .
6.6.3 Relativist effects . . . . . . . . . . . . . . . . . .
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38
38
38
39
41
42
44
44
44
45
7 Quantum mechanics of identical particles
46
7.1 Wave functions for identical particles . . . . . . . . . . . . . . . . 46
7.2 Exchange symmetry with spin . . . . . . . . . . . . . . . . . . . . 47
8 More Complicated Atoms
49
Stichwortverzeichnis
50
iii
0
Introduction
0
05.04.2006
Introduction
Webpage:
http://www.exp-astro.phys.ethz.ch/PhysikIV
Topics:
1. Failures of Classical Physics; particles and waves
2. The Schrödinger Equation
3. Position and momentum
4. Energy and Time
5. Particles in potentials
6. Harmonic oscillators
7. Observables and Operators
8. Angular Momentum
9. The Hydrogen Atom
10. Bosons and Fermions
11. Atoms
Books:
• Tipler & Mosca: Physik/Physics for Scientists and Engineers
• Phillips: Introduction to Quantum Mechanics
This will be the book, the professor follows more or less.
• Cohen-Tannoudji, Diu, Laloë: Quantum Mechanics vol 1,2 (E/D/F)
A classic book
• Messiah: Quantum Mechanics vol 1,2 (E/D)
A classic book
• Schwabl: Quantenmechanik
1
1
The Limits of Classical Physics & Wave-Particle Duality
1
05.04.2006
The Limits of Classical Physics & Wave-Particle Duality
1.1
Classical Concepts
• We have particles
– A particle is a discrete entity
– It has a precise and well defined position and momentum
– It’s obeying Newton’s Laws of Motion
– In principle, the Physics of Particles is completely deterministic
• We also have the electromagnetic fields and waves
– The electromagnetic fields pervade all space
– They’re governed by Maxwell’s equations
– We have wavelike disturbances which propagate through space
The fields and the particles interact via the Lorentz forces
F
= q(E + v × B)
and we’ve got the equivalent for gravity.
In 1900-1930 there were two revolutions in Physics: Relativity and Quantum
Mechanics. The upshot of this was, that Classical Physics is only an approximation and it isn’t valid on everyday scales (Quantum Mechanics) und speeds
(Relativity).
1.2
1.2.1
Empirical Problems with classical Physics
Atomic structure & atomic spectra
Atoms consist of positive and neutral nuclei and negative electrons (that can be
removed). In Classical Mechanics we assume, that the electrons do orbit like in
the solar system
£
Fg = G
M ¯ mp
r2
1 q + e−
4πε0 r2
Fe =
The orbits are ellipses (sun at a focus) with
r =
p
1 − e cos θ
E =
P 2 = a3
1
Mm
Mm
mv 2 − G
= −G
2
r
2a
4π 2
G(M + m)
where P is the period. As we know, we can get any energy and any period.
Now, we’ve got two problems:
• Atoms gain or loose energy by absorbing or emitting light at particular
frequencies
• The electrons are continuously accelerated, therefore they should loose
energy through electromagnetic radiation
dE
dt
=
2 q 2 a2
3 4πε0 c3
=
q 2 a2
6πε0 c3
This means, the orbits should decay and the electrons should spiral into
the nucleus, which doesn’t happen.
2
1
The Limits of Classical Physics & Wave-Particle Duality
1.2.2
10.04.2006
Photo-electric effect
If we have shining light on a metal surface, there are electrons ejected
£
The incident power per unit area is
2
P = ε0 E c
2
The Ejection of the electrons should not depend on ν, but only on E . What
we see is different. For example, Mg requires ν > 8.9 · 101 4Hz.
1.2.3
Waves behaving as particles
Consider an electromagnetic radiation with very high frequency ν like by X-rays
and wavelength λ
£
The change of K · (1 − cos θ) in λ and ν of an X-ray radiation is a signature of
particle-like behaviour.
1.2.4
Particles acting as waves
Interference (constructive or destructive adding of waves) is fundamentally a
wave phenomena
£
In about 1927, it was seen, that particles also show diffraction patterns. It was
shown, that 54eV acted like a wave with λ = 0.17nm and 40keV like one with
λ = 0.006nm. In 1999 it was even shown, that the same effect occurs with C 60
molecules.
1.2.5
Summay - new ideas
• Consider waves as particles, with specific energy and momentum
• Consider particles as waves, with specific wavelength and frequency
• The enrgy loss/gain is restricted, so ∆E is not continuous
• We’ve got a propagation as waves and an exchange of energy as particles
1.3
Reminder Interference & Diffraction
The Huygens-Fresnel Principle says: “Every point on the wavefront acts as the
source of a secondary spherical wavefront (wavelet), with same frequence and
speed. The amplitude of the field at any point is the superposition of their
wavelets taking into account amplitude and phase”.
1.3.1
Young’s slits interference
Assume that the slits are very narrow and one-dimensional
£
We have constructive interference if
d sin θ = nλ
n∈Z
and destructive interference if
d sin θ = (n + 1/2)λ
3
n∈Z
1
The Limits of Classical Physics & Wave-Particle Duality
10.04.2006
For a general θ, the phase difference for two slits is
δ = 2π
d sin θ
λ
Place the screen far from the slits (i.e. at a distance L À d). Then, the position
on the screen is
y = L tan θ ∼ L sin θ
This approximation is good for small θ and the maxima and minima are equally
spaced
ymax
ymin
=
=
nλ
L
µd
θmax =
1
n+
2
¶
λ
L
d
nλ
d
θmin =
µ
1
n+
2
¶
λ
d
For the wave amplitude we get
E
=
E0 (sin(ωt) + sin (ωt + δ))
µ
¶
µ ¶
δ
δ
sin ωt +
= 2E0 cos
2
2
The intensity is proportional to E 2 ∝ cos2 (δ/2)
£
What about interference from multiple slits? We need to sum wavelets from all
slits, e.g. using a “phasor diagram”
£
The maximum always has the “vectors” aligned, i.e. δ = n2π. We will have
secondary maxima when
mδ = (2n + 1)π
where m is the number of slits and mδ the total angle rotated in the phasor
diagram (in this case, the end of the phasor-sum is at the other side of the
circle).
£
If we have an infinite number of slits, the interference pattern looks like this
£
1.3.2
Diffraction
Consider an interference from a finite sized aperture. Take for example light
passing through an aperture of finite width and ∞ length
£
We can consider this composed of N finite elements for small θ. The phase
difference between adjacent apertures is
δ = 2π
θ a
λ N
The phasor diagram will become a circle as N goes to ∞
£
4
1
The Limits of Classical Physics & Wave-Particle Duality
12.04.2006
For the angle we get
2πaθ
Nδ =
λ
Now we wet φ =
πaθ
λ
⇒
Eθ
E0 λ
=
sin
πaθ
µ
πaθ
λ
¶
to get
E0
sin φ
φ
Eθ =
The intensity we then can write as
I0
sin2 φ
φ2
This leads us to the diffraction pattern
£
For a À λ, we get that θmin → 0 and for a ∼ λ, that θmin → 1.
Now we can construct the actual pattern from 2 finite slits
£
In two dimensions (e.g. by a rectangular aperture), we get something like
£
and the intensity, we can write as
I = I0
1.3.3
sin2 φa sin2 φb
φ2a
φ2b
φa =
πaθa
λ
φb =
πaθb
λ
Fourier Optics
Consider a general aperture
£
Then we have
E(θ, φ) =
Z Z
dxdy · E0 T (x, y)e−i
2π
λ (θx+φy)
where T (x, y) is the transmission function of the aperture. Set u = λ−1 x,
v = λ−1 y to get
Z Z
E(θ, φ) = E0 λ2
dudv · T 0 (u, v)e−2πi(uθ+vφ)
The diffraction/interference pattern is the Fourier Transform of the aperture
scaled by λ and vice versa.
1.4
Thermal radiation and Planck’s constant h
1.4.1
Introduction
All bodies emit electro magnetic radiation because of their temperature. We
could write
dE(λ, Temperature, material)
dλ =
dt
rate of energy loss per unit area
per unit time in the interval λ→λ+dλ
Different materials absorb different fractions of incident electro magnetic radiation, described by the absorption coefficiant a and the emission coefficiant e.
It was found empirically by Ritchie (1833), that the emission properties are the
same as the absorption properties for all materials, i.e.
e
= const
a
It was shown with an experiment like this
5
1
The Limits of Classical Physics & Wave-Particle Duality
12.04.2006
£
where for the emissive power e and die absorptive power a, we have
eα
eβ
eα a β = e β a α ⇒
=
aα
aβ
We’ve got a maximum e, when a = 1, as by a body that absorbs all radiation
falling on it (black body).
Consider a uniform temperature enclosure in equilibrium at a temperature T
£
It turns out, that the energy density u with [u] = Jm−3 depends only on the
temperature T .
To proof this, we imagine a cavity like
£
If we have uA (T ) > uB (T ), then energy flows from A to B, when the shutter is
opened.
⇒ When we close the shutter, TA has gone down and TB has gone, which is a
violation of the second Law of Thermodynamics.
The radiation escaping from a small hole in a cavity is the same as from an
ideal black body. But what is the form of u(T ) and more specifically, what is
the spectrum of thermal radiation u(λ, T )dλ, which is the energy density for
the interval λ → λ + dλ as f (T )?
1.4.2
Stefan-Bolzmann Law
We want to know, what’s the total energy emitted by a Black Body
£
where
Acu(T )
dE
=
dt
4
(see problem sheet 1). The answer is
dE
dt
= σAT 4
which Stefan got empirically and Boltzmann proves by theory. σ we call the
Stefan-Boltzmann constant. Boltzmann did a Thermodynamic argument to
prove it: Imagine a reflecting enclosure, containing some thermal radiation.
It turns out, that the radiation pressure is equal to 31 u, where the 3 comes from
the three dimensions.
£
The work done is equal to
∆v(p(T ) − p(T − dT )) = ∆vdp =
The efficiency is
1
∆v du
3
1 ∆v dm
dT
=
4
3 3 ∆v u
T
1 du
dT
=
→ u ∝ T4
T
4 u
The energy density u = aT 4 .
The emission from the black body is
ac 4
T = σT 4
4
per unit area.
⇒
6
1
The Limits of Classical Physics & Wave-Particle Duality
1.4.3
19.04.2006
Wien’s Displacement Law
What can we say about the spectrum u(λ, T )dλ. Consider again a cavity undergoing a reversible expansion. In this case, the wavelengths λ are proportional
to the cavity size.
`1
`1
dλ1
λ1
=
=
λ2
`2
dλ2
`2
Using exactly the same argument as before, we get
u(λ2 ) 2
u(λ1 )
dλ1 =
dλ
T14
T24
For an adiabatic process, we have
d[V u(λ) dλ] + p(λ) dV
= 0
1
u(λ) dλ
3
p(λ) =
where the first term is the change in energy and the second is the work.
d[3V p(λ)] + p(λ) dV
3V dp(λ) + 4p(λ) dV
= 0
= 0
4/3
=
p(λ2 )V2
→ u(λ1 )`41 dλ1
→ u(λ1 )λ51
=
=
u(λ2 )`42 dλ2
u(λ2 )λ52
→ p(λ1 )V1
Back to Stefan’s Law
4/3
(1.1)
p(λ1 )
p(λ2 )
=
T14
T24
Now we eliminate p(λ), using equation 1.1 to get
4/3
T14 V1
4/3
= T24 V2
We have
⇒
T1 `1 = T 2 `2
⇒
T1 λ1 = T 2 λ2
u(λ1 )λ51
u(λ2 )λ52
=
f (λ1 , T1 )
f (λ2 , T2 )
⇒ The general result for u(λ, T ) is
u(λ, T ) =
A
f (λT )
λ5
We will also be interested in u(ν) dν which is the energy per unit volume with
ν → ν + dν and
µ ¶
T
1
u(ν, T ) = Bν 3 g
νmax ∝ T
λmax ∝
ν
T
which is the general form of Wien’s Displacement Law.
£
1.4.4
Oscillators
Imagina a cavity surrounded by “oscillators”, each with two degrees of freedom
x
=
x0 sin ωt
ẋ = ωx0 cos ωt
ẍ = −ω 2 x0 sin ωt
The energy of it will be constant as it oscillates and equal to
ε =
1
mω 2 x20
2
7
1
The Limits of Classical Physics & Wave-Particle Duality
19.04.2006
The emission of electro magnetic waves from an oscillator is instantaneously
equal to
dE
q 2 a2
=
dt
6πε0 c3
(see Physics III). Therefore, the time average of this is
dE
dt
1 q 2 ω 4 x20
2 6πε0 c3
=
where we just put in a = ẍ from above and
dE
dt
q2 ω2
ε
6πε0 c3 m
=
For the absorption we get the analogous result (without proof)
dE
dt
πq 2 u(ν)
4πε0 3m
=
where ω = 2πν. In equilibrium, the absorption and the emission are equal and
so we can then write
q2 ω2 ε
6πε0 mc3
=
πq 2 u(ν)
4πε0 3m
⇒
u(ν) =
8πν 2
ε
c3
where ε is the energy of the oscillator. If we look at all oscillators on average,
we get
8πν 2
ε
u(ν) =
c3
where ε is the average energy of all oscillators with frequency ν. So we want to
know what’s the average energy as a function of T . We would expect it to be
ε = kT to get
8πν 2
kT
u(ν) =
c3
which is called the Rayleigh Jeans formula. It satisfies
µ ¶
T
u(ν) = ν 3 g
ν
but it can’t be right, because u → ∞ as , ν → ∞. This is known as the
ultraviolet catastrophe. If we look, where ε = kT comes from, we see that the
analysis to get it assumes, that we have an infinite range of possible energies,
weighted by the Boltzmann factor.
R ∞ N −ε/kT
ε
e
dε
N −ε/kT
e
dε ⇒ ε = R0 ∞ NkT −ε/kT
dn(ε) =
= kT
kt
e
dε
0 kT
1.4.5
Planck’s hypothesis (1900)
Plank’s idea was to ask, what happened, if we allowed only certain energies
εj = jε0
j = γ = 0, 1, 2, ...
Exactly as before, we had nj = Ae−εj /kT , but
A + Ae−ε0 /kT + Ae−2ε0 /kT + Ae−3ε0 /kT + ...
∞
X
A
=
Ae−εj /kT =
−ε0 /kT
1
−
e
j=0
N
=
E
=
0 + Aε0 e−ε0 /kT + A2ε0 e−2ε0 /kT + ...
∞
X
Aε0 e−ε0 /kT
= A
εj e−εj /kT =
(1 − e−ε0 /kT )2
j=0
8
1
The Limits of Classical Physics & Wave-Particle Duality
It follows
ε =
and
E
N
= ¡
ε0 e−ε0 /kT
ε0
¢ = ¡
¢
−ε
/kT
ε
/kT
0
0
1−e
e
−1
ε0
8πν 2
¡
¢
3
ε
/kT
0
3c
e
−1
u(ν) dν =
This works of u(ν) = Bν 3 g
¡T ¢
ν
19.04.2006
if ε0 ∝ ν, i.e.
h = 6.625 · 10−34 Js
ε0 = hν
where h is Plank’s constant. This then gives the final answer for black body
radiation density
u(ν) dν =
8πhν 3
1
¡
¢ dν
3
hν/kT
3c
e
−1
which is called Plank’s formula. It reduces to
8πν 2
kT
3c3
for hν ¿ kT . At the end we’ve got rid of the ultraviolet catastrophe by cutting
out the continuity of the energy.
1905 Einstein quantised the radiation in a cavity and called this wave packets
photons with
E = hν
This explained for example the photo-electric effect.
1.5
De Broglie waves & Plank’s constant
The wavelike properties of matter were proposed by de Broglie in 1923 with
λ =
h
p
where p is the momentum. In terms of energy we can rewrite this with
ε2 = m20 c4 + p2 c2
⇒
λ = √
hc
√
ε − m 0 c2 ε + m 0 c2
In a highly relativistic case, i.e. ε À m0 c2 , we can write
λ =
hc
ε
which is identical to photons (ε = hν). In the non-relativistic case, i.e for
p2
is the kinetic energy, we get
ε = m0 c2 + E, where E = 2m
0
λ = √
h
2m0 E
For example, an energy of 1.5eV is a λ = 1nm and an energy of 15keV is a
λ = 0.01nm.
1.6
Measurements & Plank’s constant
This section is still largely classical and shouldn’t be translated into Quantum
Mechanical concepts.
We go through the “thought experiment” of Heisenbergs microscope
£
9
1
The Limits of Classical Physics & Wave-Particle Duality
26.04.2006
We know, that the defraction from the aperture d is of the order ∼
uncertainty ∆x is going to be given by
∆x ∼
λ
d.
The
λ
λ
z ∼
d
α
The scattered photon from the electron must have entered the microscope and
therefore, the photon has a momentum with
|px | <
hα
h
sin α <
λ
λ
For the electron, the uncertainty ∆p in the momentum of the electron must be
of order ∼ hα
λ . We get to the interesting fact
∆x∆p ≥ h
The correct Quantum Mechanical formulation of this is
∆x∆p ≥
1.7
h
4π
Wave paricle duality & quantum reality
Lets do a little thought experiment
£
The electrons hit the detector with a statistical distribution, so we observe
a diffraction pattern in the locations of the detected electrons. This implies
wave properties through the slits. We could ask, whether we can tell which
slit the electron passed through and indeed we can quite easily, but then, the
diffraction pattern dissapears! For example, we could measure the momentum
of the electron by the recoil of the screen. If you think about it, near to the
center of the screen, we would expect
∆pscreen ∼ pe
h d
d
∼
D
λe D
From the previous section we know that however we try to measure the momentum, we have an uncertainty in position that is
∆x ∼
Dλe
h
∼
∆p
d
which is exactly the separation of the interference fringes. So, when we know
which slit the electron passed through, we no longer observe wavelike properties.
When there is no possibility of knowing this, we observe wavelike behavior. We
could say: “The wave is covert”. “The act of measurement brings into existence
a property”.
10
2
The Schrödinger Equation
2
26.04.2006
The Schrödinger Equation
2.1
2.1.1
Review of waves
Sinusoided waves
The simplest wave with definite wavelength λ, wavenumber k = 2π
λ and definite
1
and
frequency
ν
=
are
sinusoided
waves
period τ , angular frequency ω = 2π
τ
τ
as
Ψ(x, t) = A cos (kx − ωt)
The maxima move at speed
ω
k.
The most general sinusoided wave is
Ψ(x, t) = A cos (kx − ωt) + B sin (kx − ωt) = Aei(kx−ωt)
Note, that in Classical Physics we normally take the real part of the complex
wave. In Quantum Mechanics you always consider the complex function.
2.1.2
Superposition of waves
Standing wave: A standing wave we get from a superposition like
Ψ(x, t)
=
=
A cos (kx − ωt) + A cos (kx + ωt)
2A cos (kx) cos (ωt)
So we can separate the spacial and the temporal parts.
Wave packets: A wave packet is a superposition with similar k
Ψ(x, t)
k+∆k
Z
A cos (k 0 x − ω 0 t) dk 0
=
k−∆k
At t = 0, this simplifies to
Ψ(x, 0)
=
k+∆k
Z
A cos (k 0 x) dk 0
k−∆k
=
cos (kx) S(x)
where
S(x) = 2A∆k
sin (∆kx)
∆kx
£
2.1.3
Dispersion relations
Non-dispersive waves are governed by the classical wave-equation
∂2Ψ
1 ∂2Ψ
−
= 0
∂x2
c2 ∂t2
In three dimensions, this is
∇2 Ψ −
1 ∂2Ψ
= 0
c2 ∂t2
The solutions of this equation are
Ψ(x, t) = Aei(kx−ωt)
where ω 2 = c2 k 2 and where vphase = ωk = c is the speed of a maximum in
the wave. We call it phase velocity. Because of linearity, the superposition of
solutions is also a solution.
11
2
The Schrödinger Equation
03.05.2006
Generally, waves are dispersive, that means they have a more complicated ω, k
relation as ωk 6= const. This is due to a more complicated wave equation. The
phase velocity is equal to the speed of individual crests ωk and the group velocity
is the speed of the point of the maximum constrictive interference in a wave
packet vG = dω
dk
£
To see this, consider 2 sinusoided waves with ω1 , ω2 , k1 , k2 . We have constructive
interference, when the phases are the same
(k1 xmax − ω1 t) = (k2 xmax − ω2 t)
⇒
xmax =
ω1 − ω 2
∆ω
t =
t
k1 − k 2
∆k
e.g. for long wavelength water waves in deep water, we have
r
r
p
g
ω
dω
1 g
1
gk
vp =
ω =
=
vG =
=
=
vp
k
k
dk
2 k
2
2.2
Particle wave equation
We want a wave equation for particles by looking at the dispersion relation
(don’t worry yet what the wave is). With de Broglie, we had
λ =
h
p
2π
λ
k =
⇒
p =
h
k =: ~k
2π
A particle with uncertain p is associated with a wave packet with a range of k.
∆p ∼ ~∆k
The “length” of the wave packet
∆x ∼
2π
∆k
Lets set the group velocity vG equal to the velocity of the particle
⇒
dω
dk
=
p
~k
=
m
m
We integrate to find
~k 2
+ (const)
2m
One could ask, what wave equation this dispersion relation gives
ω =
i~
∂Ψ
~2 ∂ 2 Ψ
= −
∂t
2m ∂x2
or in three dimensions
∂Ψ
~2 2
= −
∇ Ψ
∂t
2m
This is the Schrödinger Equation for a “free particle”. Lets take a wave-function
Ψ(x, t) = Aei(kx−ωt) and check
i~
i~(−iω) = −
~2
− k2
2m
⇒
~ω =
~2 k 2
2m
Note:
• Such wave equations have always complex wave solutions.
• The Schrödinger-equation is linear, that means that any superposition of
solutions is also a solution.
12
2
The Schrödinger Equation
03.05.2006
The most general solution then is
Ψ(x, t) =
Z∞
0
A(k 0 )ei(k x−ωt) dk 0
−∞
provided we set
~2 k 0 2
2m
A narrow range of k leads to a wave packet moving with speed
~ω 0 =
dω
~k
=
dk
m
The uncertainty in p and in x (∆p and ∆x) are related by
∆p ∼ ~∆k
2.3
∆x ∼
2π
∆k
A Particle in a potential energy field V (r)
If a particle has a potential energy V (r) as well as a kinetic energy, we modify
the Schrödinger-Equation like
µ
¶
dΨ
~2 2
i~
=
−
∇ + V (r) Ψ
dt
2m
We could try a solution of the form Ψ(x, t) = Aei(kx−ωt) with V (x) = V0 , to get
~2 k 2
+ V0
2m
E = ~ω =
2.4
The meaning of Ψ(x, t)
We’ll first do a digression on probabilities and probability distributions.
Consider n discrete outcomes of an experiment xn , each of which have a probability of occuring pn . We know
X
pn = 1
We define the expectation value of x as the average value, if the experiment is
repeated many times
X
hxi =
x n pn
n
With the standard deviation in x, ∆x, we get the variance (∆x)2 and
X
­ ®
2
2
2
(xn − hxi) pn = x2 − hxi
(∆x) =
n
Consider a continuous variable x, then p(x)dx is the probability of x between x
and x + dx. Like above, we then have
Z
p(x) dx = 1
all x
and futhermore
hxi =
Z
­
xp(x) dx
allx
x
2
®
=
Z
allx
Now, recall classical waves in the two-slit experiment
13
x2 p(x) dx
2
The Schrödinger Equation
03.05.2006
£
By superposition, we have Ψ = ΨA + ΨB and the intensity is proportional to
Ψ2 . By a Quantum wave function, we also have Ψ = ΨA + ΨB and by analogy,
we construct |Ψ|2 = Ψ∗ Ψ. But what does this quantity mean?
£
Born’s interpretation of Ψ(x, t) was the following
Ψ∗ (r, t)Ψ(r, t) dV
Ψ∗ (x, t)Ψ(x, t) dx
or
is the probability of finding a particle in the volume dV (or within the distance
dx) at the time t, when we do some suitable experiment.
As we go on, we can sometimes state some rules about how we have to work
with our wave-functions. Here’s the first one.
Rule 1:
We must always normalize Ψ(R, t), such that
Z
Ψ∗ Ψ dV
= 1
all space
¤
If we have
Ψ(x, t) =
X
Aj ei(kj x−ωj t)
pj = ~kj
j
then we get a probability of measuring the momentum pi which is proportional
to |Ai |2 . We can gereralize this with a continuous distribution
Z∞
Ψ(x, t) =
A(k)ei(kx−ωt) dk
−∞
which effectively is a Fourier Transform. We can also construct a Fourier transform pair as follows
Ψ(x, t)
We know
e t)
Ψ(p,
Z∞
=
=
1
√
2π~
√
1
2π~
∗
Ψ Ψ dx = 1
Z∞
−∞
Z∞
e t)e ipx
~ dp
Ψ(p,
Ψ(x, t)e−
ipx
~
dx
−∞
⇔
−∞
Z∞
−∞
e ∗Ψ
e dp = 1
Ψ
e ∗Ψ
e the probability amplitude
where Ψ∗ Ψ is the probability amplitude for x and Ψ
for p.
2.5
Expectation values and operators
Using the equations from above, we can write down in one dimension
hxi
hpi
=
=
Z
Z
∗
x Ψ Ψ dx =
e ∗Ψ
e dp =
pΨ
14
Z
Z
Ψ∗ xΨ dx
e ∗ pΨ dp
Ψ
2
The Schrödinger Equation
03.05.2006
We can also calculate hpi in the following way
hpi =
Z
Ψ∗ (x, t) (−i~)
∂
(Ψ(x, t)) dx
∂x
allx
which we prove with
Ψ(x, t) = √
1
2π~
We get
∂
−i~
Ψ(x, t)
∂x
=
1
√
2π~
Z
and therefore it follows
=
=
=
=
So, we can write
hxi
hpi
e t)e ipx
~ dp
Ψ(p,


Z
ipx
∂  1
e t)e ~ dp
√
−i~
Ψ(p,
∂x
2π~
=
hpi
Z
e t)e
pΨ(p,
ipx
~
dp
µ
¶
∂
Ψ (x, t) −i~
Ψ(x, t) dx
∂x


Z
Z
ipx
1
e t)e ~ dp dx
pΨ(p,
Ψ∗ (x, t)  √
2π~


Z
Z
ipx
1
e t) dp
√
Ψ∗ (x, t)e ~ dx pΨ(p,
2π~
Z
e ∗ pΨ
e dp
Ψ
Z
∗
=
Z
=
Z
Ψ∗ (x, t)xΨ(x, t) dx
µ
∂
Ψ (x, t) −i~
∂x
∗
¶
(Ψ(x, t)) dx
Now we introduce the concept of operators. The operator for x is x
b = x and
∂
the operator for p is pb = −i~ ∂x
. It turns out, that any obervable quantity can
be represented by an operator. Furthermore, the operators for x2 and p2 are
given by
∂2
2
2
c2 = x2 = (b
p)
x
x)
pb2 = −~2 2 = (b
∂x
In three dimensions, we have
Rule 2:
b = r
r
b = −i~∇
p
We can apply operators to get any expectation value
hξi =
Z∞
−∞
or with dV in three dimension.
b dx
Ψ∗ ξΨ
¤
15
2
The Schrödinger Equation
08.05.2006
Lets look at the operator for the energy E. Classically, we can write it as
E =
p2
+ V (r)
2m
and we would expect the energy operator as
b =
E
pb2
~2 2
b
+ V (r) = −
∇ + V (r) =: H
2m
2m
b is called the Hamiltonian operator. Now
where H
hEi =
Z
b dV
Ψ∗ HΨ
Now going back to Schrödingers-Equation, we see that
µ
¶
∂Ψ
~2 2
b
i~
=
−
∇ + V (r) Ψ = HΨ
∂t
2m
16
3
Solutions to Schrödinger’s Equation
3
3.1
08.05.2006
Solutions to Schrödinger’s Equation
Separable solutions of definite energy
We had the Schrödinger-Equation for a particle in a potential
µ
¶
~2 2
∂Ψ
=
−
∇ + V (r) Ψ
i~
∂t
2m
Now, we’re looking for solutions
Ψ(r, t) = ψ(r)T (t)
We put it in
dT
i~ψ(r)
dt
µ
=
¶
~2 2
−
∇ ψ + V (r)ψ T
2m
and separate the variables
i~ dT
T dt
µ
1
ψ
=
−
~2 2
∇ ψ + V (r)ψ
2m
¶
This equation is true for all t and for all r, so the terms have to be equal to a
constant K. It follows
dT
dt
i
= − KT
~
⇒
T (t) = Ae−
iKt
~
and on the other hand
µ
−
¶
~2 2
∇ + V (r) ψ = Kψ
2m
b associated with the eigenvalues
So, we’re looking for the eigenfunctions ψn of H
Kn , i.e.
b n = K n ψn
Hψ
With Ψn (r, t) = ψn (r)Ae−
hEi =
Z
iKt
~
we then get
b n
Ψ∗n HΨ
3
d r =
Z
b n d3 r = K n
ψn∗ Hψ
So, Kn is equal to the energy of the system and we can write
b
Hψ(r)
= Eψ(r)
which is called the time-independent Schrödinger equation. We also get
­
E
2
®
=
Z
b 2 Ψ d3 r = E 2
Ψ∗ H
and for the energy uncertainty, we get from before
q
2
∆E =
hE 2 i − hEi = 0
b the energy of the
So, if the wave function of a system is an eigenfunction of H,
system is precisely determined and is equal to to eigenvalue En associated to
the eigenfunction. If we make a measurement to determine the energy E, Ψ becomes the eigenfunction associated with the actual value measured. This sort of
strange idea is called the collapse of the wave function during the measurement.
Note: The time part T is
T (t) = Ae−
iEn t
~
En = hνn = ~ωn
17
3
Solutions to Schrödinger’s Equation
3.2
08.05.2006
Example: Particle in a box
Consider the potential
V (x) =
½
0,
0<x<a
∞, elsewhere
a
0
x
We had the 1-dimensional Schrödinger-Equation
µ
¶
~2 d 2
∂Ψ
=
−
+
V
(r)
Ψ
i~
∂t
2m dx2
Look for separable solutions of it as
iEn t
~
Ψ(x, t) = ψ(x)e−
so we get
µ
¶
~2 d 2
−
+ V (x) ψn (x) = En ψ(x)
2m dx2
~2 2
2m kn
Inside the box, V = 0 and with En =
∂2ψ
∂x2
we get the equation
= −kn2 ψ
The boundary conditions are ψ(0) = 0, ψ(a) = 0, so that ψ is continuous. Then
the solutions are
nπ
ψn = A sin kn x
kn =
a
Now, for the n-th energy-state we get
En =
n 2 π 2 ~2
a2 2m
n = 1, 2, 3, ...
and we can write
Ψn = A sin
We can draw out these energy states
³ nπx ´
a
e−
iEn t
~
n=4
n=3
E
n=2
n=1
Note: The distance En+1 − En goes up as we increase n, but
En+1 − En
E
18
≈
2
n
3
Solutions to Schrödinger’s Equation
10.05.2006
goes down as n and therefore the energy E gets bigger. The minimum E 6= 0
we have for n = 1, so
~2 π 2 1
E1 =
2m a2
For 3-dimensional potential wells
c
b
a
Its easy to show, that
ψ(x, y, z) = A sin
³ n πx ´
³ n πy ´
³ n πz ´
x
y
z
sin
sin
a
b
c
and of course then the energy of this system is
Ã
!
n2y
n2z
~2 π 2 n2x
+ 2 + 2
E(nx , ny , nz ) =
2m
a2
b
c
We can see, that we have more complex energy-states, which lead to something
like
£
In dependency of a, b, c we can find equal energy-states E for different nx , ny , nz .
We call this degeneracy.
3.3
States of uncertain E
We represent more general states by a sum over the eigenstates
Ψ(r, t) =
∞
X
cn ψn (r)e−
iEn t
~
n=1
b form a complex orthonormal set of
This is, because the eigenfunctions of H
basis functions. That is
½
Z
1, m = n
3
∗
ψm ψn d r =
0, m 6= n
It is again easy to prove that
cn =
Z
Ψ∗ (r, 0)ψn (r) d3 r =
and then
hEi =
X
n
Z
ψn∗ (r)Ψ(r, 0) d3 r
X
­ 2®
E
=
|cn |2 En2
|cn |2 En
n
If the Quantum particle is represented by this general Ψ, that satisfies the
conditions we’ve set, then the result of an experiment to measure E will be
on En with probability |cn |2 . Remember, that the eigenfunctions represent
quantum states, with
• precicely defined energy E
• observable properties that are time independent, like the position probability of the n-th quantum state
Ψ∗n Ψn = ψn∗ e
iEn t
~
19
ψ n e−
iEn t
~
6= f (t)
3
Solutions to Schrödinger’s Equation
Example 3.1:
10.05.2006
We want to go through the example of a simple superposition
iE1 t
iE2 t
1
1
Ψ(r, t) = √ ψ1 e− ~ + √ ψ2 e− ~
2
2
Then we get
hEi =
and therefore
X
q
∆E =
­
1
(E1 + E2 )
2
|cn |2 En =
hE 2 i − hEi
2
=
E2
®
=
¢
1¡ 2
E1 + E22
2
1
|E1 − E2 |
2
Look at Ψ∗ Ψ as function of t
µ
¶
i(E1 −E2 )t
i(E1 −E2 )t
1 ∗
1
1
1
~
~
Ψ∗ Ψ =
+ ψ2∗ ψ1 e−
ψ1 ψ1 + ψ2∗ ψ2 + ψ1∗ ψ2 e
2
2
2
2
Therefore Ψ∗ Ψ oscillates with frequency
E1 − E 2
h
ν =
or
ω =
E1 − E 2
~
This of course is a non stationary state. The time scale for change is δt ∝
where δt∆E ∼ ~.
3.4
1
∆E ,
¤
Finite potential wells
Consider a V (x) as
V
∼ unbound
0
ε = bound energy
E
a
−V0
0
Now we look for states of definite E solving
−
~2 ∂ 2 ψ
+ V (x)ψ = Eψ
2m ∂x2
where
V (x) :=


∞, x < 0
−V0 0 < x < a

0 x>a
∂ψ
Set ψ(x) and ∂ψ
∂x to be continuous at x = 0, x = a (except ∂x at x = 0). Look
first at the bound states with −V0 < E < 0. For 0 < x < a, we get
∂2ψ
∂x2
with k02 =
2m
~2
= −
2m
(E + V0 ) ψ = −k02 ψ
~2
(E + V0 ). This has the general solution
ψin = C sin (k0 x + γ)
20
3
Solutions to Schrödinger’s Equation
10.05.2006
where γ is an arbitrary phase. But γ = 0 satisfies ψ = 0 at x = 0, so
ψin = C sin (k0 x)
For x > a, we have
∂2ψ
∂x2
= −
2mE
ψ = α2 ψ
~2
with α2 = − 2mE
~2 and we get solutions
ψout = Ae−αx + A0 eαx
We can set A0 = 0, so that ψ doesn’t blow up for x → ∞ (which wouldn’t have
any Physical meaning). Now stitch together the solutions at x = a. To get a
continuous ψ(x), we need
C sin (k0 a) = Ae−αa
and for the first derivative
k0 C cos (k0 a) = −αAe−αa
Dividing these, we get
k0 cot (k0 a) = −α
About the energy, we know that
E =
~2 2
~2 2
k0 − V 0 = −
α
2m
2m
V0 =
~2 2
w
2m 0
where w0 is a measure of depth of the well. Now there are two things required
1) k02 + α2 = w02
2) k0 cot (k0 a) = −α
α
w0 = 3
w0 = 2
w0 = 1
k0
We have only one solution for k0 , α for
and so on.
21
π
2a
< ω0 <
3π
2a ,
two for
3π
2a
< ω0 <
5π
2a
3
Solutions to Schrödinger’s Equation
10.05.2006
Note:
• The number of solutions depends on the well depth and it goes to ∞ as
−V0 → ∞.
• ψ(x) for the bound states has an exponential tail extending into the classically forbidden region. That was essentially because
2mε
~2
We get a long tail, when ε is small and a small tail, when ε is very large.
α2 =
ψout ∝ e−αx
E
−V0
Now, look at the unbound states E > 0. We write the energy for 0 < x < a and
x>a
~2 2
~2 2
k0 − V 0 =
k
E =
2m
2m
Then we write down the solutions
ψin (x)
ψout (x)
=
=
C sin (k0 x)
D sin (kx − δ)
Now ψ(x) and ψ 0 (x) shall be continuous at x = a, so
C sin (k0 a)
=
D sin (ka + δ)
k0 C cos (k0 a)
=
kD cos (ka + δ)
After dividing them, we get
k0 cot (k0 a) = k cot (ka + δ)
and as before
2m
~2
At the end, there’s an infinite number of solutions for k and k0 , all with own δ.
This, if you like, makes sense.
k02 − k 2 = V0
Note: These are solutions for the stationary states. This does make sense for
the unbound particle, because we can view stationary solutions as a standing
wave due to a total reflection of particles at x = 0.
C
Introduce A0 = − 2i
, then
ψin = A0 e−ik0 x − A0 eik0 x
Now we can write
Ψin = A0 e−ik0 x−
iEt
~
− A0 eik0 x−
iEt
~
= ψin (x)e−
iEt
~
where the first term describes a travelling wave in negative x-direction and
the second one a travelling wave in positive x-direction. For Ψout , introduce
D −iδ
A = − 2i
e . Applying algebra, we get
Ψout = Ae−ikx−
iEt
~
− Ae2iδ eikx−
iEt
~
= ψout (x)e−
iEt
~
Here, the first term describes the incident and the second the reflected wave.
22
3
Solutions to Schrödinger’s Equation
15.05.2006
£
Note: The reflected wave has a phase shift of 2δ. We’d like to know where this
comes from
It is due to the time delay in traversing x = a to x = 0 and back and therefore,
we would expect δ to be a function of the energy δ(E). In practice, the actual
particles will have a range of E (like an uncertain E) and then, there’ll be a
wave packet
Ψincident =
Z∞
“
p0 x
E0 t
~ + ~
0
−i
Z∞
c(E 0 )e
c(E )e
”
dE 0
p0 =
√
2mE 0
0
and the reflected wave will be
Ψreflected =
−i
“
”
p0 x
E0 t
~ − ~
0
e2iδ(E ) dE 0
0
3.5
Barrier penetration = “quantum tunelling”
V0
V =0
x=a
x=0
Classically, a particle will be reflected by the barrier. We use the same approach
as before, so we solve the time-independent Schroedinger-Equation to get
Ψ(x, t) = ψE (x)e−
iEt
~
−
~2 d 2 ψ E
+ V (x)ψE = EψE
2m dx2
For x < 0, we have
d 2 ψE
dx2
= −k 2 ψE
E =
~2 k 2
2m
because V (x) = 0. Therefore, we get the general solution
⇒ ψE (x) = AI eikx + AR e−ikx
where the first part is the incident component and the socond part the reflected
one.
For 0 < x < a, we get two cases
a) E > V0 :
~2 2
d 2 ψE
2
ψ
E
=
k + V0
=
−k
E
0
dx2
2m 0
So we get the same solution as above, but with k0 instead of k.
b) E < V0 : This is classically forbidden, but here we get
d 2 ψE
= α 2 ψE
dx2
which has as general solution
E = −
α 2 ~2
+ V0
2m
→ ψE (x) = Be−αx + B 0 eαx
23
3
Solutions to Schrödinger’s Equation
15.05.2006
For x > a, we get the same as for x < 0, means
ψE (x) = AT eikx
where AT can be complex to account for phasing.
Continuity at x = 0 requires
AI + A R = B + B 0
ikAI − ikAR = −αB + αB 0
Continuity at x = a requires
Be−αa + B 0 eαa = AT eika
− αBe−αa + αB 0 eαa = ikAT eika
We set B 0 = 0, assuming that we have a “wide” barrier. So we get at x = 0
2ikAI ≈ B(ik − α)
and at x = a
AT eika ≈
Now eliminate B to get
⇒ AT eika ≈
2α
Be−αa
(α − ik)
4ik
(α − ik)
2
αe−αa AI
The transmission probability then is
¯
¯
¯ A T ¯2
16k 2 α2
−2αa
¯ ≈
¯
T = ¯
2 e
AI ¯
(α2 + k 2 )
where
2mE
~2
Therefore, we can rewrite this as
k2 =
T ≈
α2 =
2m (V0 − E)
~2
16E (V0 − E) −2αa
e
V02
Remember, this is all right for e−2αa ¿ 1. If we also have E ¿ V0 , we can write
T ≈ 16
E −2αa
e
V0
Example 3.2: Fusion in stars
It turns out, that in the center of the sun, we have H and He gas at temperature
T ∼ 107 K. The energy of a proton then is about
Ep = kT ∼ 1keV
As protons approach with E ∼ 1keV, the radius of closest approach is r ∼
10−12 m. To get to r ∼ 10−15 m, which is necessary for fusion, we need an
energy E ∼ 1MeV. The Coulomb-Potential is
V (r) =
Z A Z B e2
q2
=
4πε0 r
4πε0 r
So, fusion should not occur classically at 107 K. We need to solve
µ 2
¶
~
mA mB
Z A Z R e2
− ∇2 +
ψ = Eψ
µ =
2µ
4πε0 r
(mA + mB )
where µ is the reduced mass. ψ(r) for a spherically symmetric wave function
has the form
u(r)
ψ(r) =
r
It turns out, that u(r) satisfies
−~2 d2 u ZA ZB e2
+
u = Eu
2mr dr2
4πε0
which is the same as in the one-dimensional case for x.
24
3
Solutions to Schrödinger’s Equation
22.05.2006
V ∼
rN
1
r
rC
Now, the tunneling probability, we get by
¯
 r
¯2
sµ
¯
¯
¶
ZC
¯
¯
Z A Z B e2
2m
T = ¯¯exp − β dr¯¯
−E
β(r) =
4πε
r
~2
0
¯
¯
rN
This leads us to
³ p
´
T ∼ exp − EG /E
EG =
µ
e2
4πε0 ~c
¶2
2π 2 µc2
where we call EG the Gamow energy. In numbers, this would be
T ∼ e−22 ∼ 3 · 10−10
by a temperature T ∼ 107 K and therefore an E ∼ 1keV. The rate of fusion
dN
dt
= e−βE
−1/2
e−αE n2 σV dE
−1
where e−βE 2 is the tunneling probability and e−αE is the Maxwellian distribution, n is the density
N
Note:
1) There is a sharp peak in energy of particles that can fuse e−αE e−βE
−1/2
2) The total rate is very strong temperature dependent
¤
25
4
The Harmonic Oscillator
4
4.1
22.05.2006
The Harmonic Oscillator
Classical case
We have the restoring force which is proportional to a displacement
F = −kx
and the potential
Z∞
V (x) =
kx0 dx0 =
1 2
kx
2
0
We can write down the differential equation
d2 x
m 2 = −kx
dt
2
↔
ẍ = −ω x
ω =
r
k
m
which has solutions
x = A cos (ωt + α)
The total energy is
Etot =
4.2
4.2.1
1
1
1
mẋ2 + kx2 =
mω 2 A2
2
2
2
Quantum oscillator
Starionary states
Lets write down the Hamiltonian in terms of the kinetic and the potential energy
p̂2
1
~2 d 2
1
+ mω 2 x̂2 = −
+ mω 2 x2
2m 2
2m dx2
2
Ĥ =
and Schrödinger’s Equation
∂Ψ
= ĤΨ
∂t
i~
Now look for stationary states, that is
~2 d 2
1
ψn + mω 2 x2 ψn = Eψn
2
2m dx
2
p
Change variables, i.e. E = ε~ω, x = q ~/mω, to get
µ
¶
d2
2
− 2 + q ψ(q) = 2εψ(q)
dq
−
Remember
¶µ
¶
µ
d
d
q−
f (q)
q+
dq
dq
µ
¶µ
¶
d
d
q−
q+
f (q)
dq
dq
d
d2
df
+
(qf )
q2f − 2 f − q
dq
dq
dq
µ
¶
d2
=
q 2 − 2 + 1 f (q)
dq
µ
¶
d2
2
=
q − 2 − 1 f (q)
dq
=
So we can write
and also
µ
d
q+
dq
¶µ
d
q−
dq
¶
ψ(q) = (2ε + 1) ψ(q)
µ
q−
d
dq
¶µ
q+
d
dq
¶
ψ(q) = (2ε − 1) ψ(q)
26
4
The Harmonic Oscillator
22.05.2006
We can quickly find two solutions
a) ε = −1/2:
µ
¶
1 2
d
Ψ = 0
ψ(q) = Ae 2 q
dq
which does not work, because the energy can’t be negative.
b) ε = 1/2:
µ
q−
d
q+
dq
¶
1
ψ(q) = Ae− 2 q
ψ = 0
2
This is the ground state of the oscillator, so if ε = 1/2, we have E0 = 12 ~ω
and
r
x2
~
− 2a
− 12 q 2
2
a =
⇒ ψ0 (x) = Ae
ψ(q) = Ae
mω
Now return to the eigenvalue equation
µ
¶µ
¶
d
d
q+
q−
ψn (q)
dq
dq
µ
¶µ
¶µ
¶
d
d
d
q−
q+
q−
ψn (q)
dq
dq
dq
|
{z
}
ψm
but we had
µ
q−
d
dq
¶µ
q+
¶
d
ψm (q)
dq
⇒ 2εn + 1
=
(2εn + 1) ψn (q)
¶
µ
d
ψn (q)
= (2εn + 1) q −
| {z }
dq
|
{z
}
2εm −1
ψm
⇒ εn
So the operator
µ
q−
d
dq
=
(2ε − 1) ψm (q)
=
2εm − 1
=
εm − 1
¶
is the operator to raise n by 1, means
¶
µ
d
ψn = ψn+1
q−
dq
Now apply the “raising operator” to ψ0 , ψ1 , ψ2 , ..., so
εn = ε0 + n~ω
and
ψn (q)
=
ψn (x)
=
⇒
En = ε0 + n~ω
¶n
µ
1 2
d
An q −
e− 2 q
dq
r
³x´
x2
~
− 2a
2
A n Hn
e
a =
a
mω
where Hn is a Hermite polynomial of xa . We might write out some of the
normalized functions
µ
¶ 12
x2
1
√
ψ0 (x) =
e− 2a2
a π
µ
¶ 12 ³ ´
1
x − x22
√
ψ1 (x) =
e 2a
2
a
2a π
µ
¶ 12 µ
³ x ´2 ¶
x2
1
√
ψ2 (x) =
2−4
e− 2a2
a
8a π
¶ 21 µ ³ ´
µ
³ x ´3 ¶
x2
x
1
√
12
e− 2a2
−8
ψ3 (x) =
a
a
48a π
..
.
27
4
The Harmonic Oscillator
22.05.2006
Notes:
1) Hn has either odd or even powers of x, as n is odd or even, so
ψn (−x) = ψn (x) (−1)
n
2) The states (of definite E) are time-independent (unlike the classical oscillator)
3) ψ ∗ ψ extends to x = ±∞
4) hxi = 0 for all n, both sides are equally likely and
r
µ
¶
­ 2®
1
1
2
=
n+
x
a
⇒ ∆x = a n +
2
2
5) hpi = 0
It follows
­ 2®
p
=
µ
1
n+
2
¶
~2
a2
⇒
µ
∆x∆p =
~
∆p =
a
n+
1
2
¶
6) For the energies we get
r
n+
1
2
~
2
~ ≥
­ 2®
p
1
hEkin i =
=
En
2m
2
­ ®
1
1
mω 2 x2 =
En
hEpot i =
2
2
To be sure, that ³
ψ0 (x) is´ really the “ground state”, we look at the energy
d
lowering operator q + dq
, that is
µ
¶
¶
µ
d
d
ψn = ψn−1
ψ0 = 0
q+
q+
dq
dq
where ψ0 has the term e−
4.2.2
q2
2
in it.
Non-stationary states of uncertain energy
We write
Ψ(x, t) =
∞
X
cn ψn (x)e−i
En t
~
n=1
The probability density in x, will oscillate in a complicated way
∞ ³
∞ ³
´X
´
X
Em t
En t
Ψ∗ Ψ =
cm ψm e−i ~
c∗n ψn∗ ei ~
=
m=1
n=1
∞
X
c∗n cm ψn∗ ψm e−i
(Em −En )t
~
n,m=1
where
|Em − En |
~
is an integer multiple of ω, but hxi oscillates at only one frequency, ω
Z∞
hxi (t) =
Ψ∗ xΨ dx
ωm,n =
−∞
=
X
c∗m cn e−
m,n=1
i(En −Em )
t
~
Z∞
ψn∗ xψm dx
−∞
which is 0 unless |m − n| ≤ 1. It is called the quasi classical state. In fact, we
can find a set of cn , so that hxi and hpi change with t as in the classical case
nn −n
e
n!
with n À 1. This is called the Poisson distribution.
|cn |2 =
28
4
The Harmonic Oscillator
4.2.3
29.05.2006
3-dimensional oscillater (degenacy)
Now we solve the 3-dimensional Ĥ, which is seperable in x, y, z
Ĥx
=
Ĥy
=
Ĥz
=
~2
2m
~2
−
2m
~2
−
2m
−
1
∂2
+ mωx2 x2
2
∂x
2
1
∂2
+ mωy2 y 2
∂y 2
2
2
1
∂
+ mωz2 z 2
∂z 2
2
Ĥ = Ĥx + Ĥy + Ĥz
From this, we get directly
µ
¶
µ
¶
µ
¶
1
1
1
nx +
En x n y n z =
~ωx + ny +
~ωy + nz +
~ωz
2
2
2
or if ωx = ωy = ωz = ω
En x n y n z =
µ
nx + n y + n z +
∆E
29
3
2
¶
~ω
5
Observables & Operators - Heisenberg Uncertainty Principle
5
5.1
29.05.2006
Observables & Operators - Heisenberg Uncertainty Principle
Operators & eigenfunctions
We had the eigenvalue problem
Ĥψ = Eψ
Having it resolved, we could decompose
X
Ψ(r, t) =
cn (t)ψn (r) +
n
Z
c(E 0 , t)ψE 0 (r) dE 0
where the sum is for the bound and the integral for the continues case. |c n (t)|2
is the probability, that we measure En and |c(E 0 , t)|2 dE 0 the probability, that
we measure an energy E between E 0 and E 0 + dE 0 . We can now generalize this
to any operator  corresponding to an observable A, provided
(1) Â must be a linear operator. So, if we have Âψ1 = φ1 and Âψ2 = φ2 , then
 (c1 ψ1 + c2 ψ2 ) = c1 φ1 + c2 φ2
We need this because it allows the whole concept of superposition of states.
(2) Â must be Hermitian, then
Z
Ψ∗1 ÂΨ2 d3 r =
Z ³
ÂΨ1
´∗
Ψ 2 d3 r
It guarantees that the eigenvalues of  are real and that hAi is real.
(3) The eigenfunctions must form a complete basis set. So any Ψ can be
represented as the sum or the integral of the eigenfunctions, one of which
is “selected” when we make a measurement. We can rewrite generally
Ψ(r, t) =
X
can (t)ψan (r) +
n
Z
c(a0 , t)ψa0 (r) da0
where |can (t)|2 is the probability of measuring a = an and |c(a0 , t)|2 da0 is
the probability of measuring a between a0 and a0 + da0 .
5.2
Eigenfunctions for position and momentum
For the position, we have the equation
x̂ψx0 (x) = x0 ψx0 (x)
where ψx0 (x) is the eigenfunction with eigenvalue x0 . Because x̂ = x, we get
xψx0 (x) = x0 ψx0 (x)
This has the Dirac δ-function as solution
ψx0 (x) = δ (x − x0 )
where by definition
Z∞
f (x)δ (x − x0 ) dx = f (x0 )
−∞
30
5
Observables & Operators - Heisenberg Uncertainty Principle
29.05.2006
The eigenfunctions of position are continuous. Therefore, we can take our general decomposition equation in one dimension
Ψ(x, t)
=
=
Z∞
−∞
Z∞
c (x0 , t) ψx0 (x) dx0
c(x0 , t)δ (x − x0 ) dx0
−∞
=
c(x, t)
where again |c(x, t)|2 dx is the probability of finding a value between x and
x + dx.
For the momentum, we get the eigenequation
p̂ψp0 (x) = p0 ψp0 (x)
which is equivalent to
−i~
∂
ψp0 (x) = p0 ψp0 (x)
∂x
It has the solutions
ψp0 (x) = √
ip0 x
1
e ~
2π~
−1/2
where (2π~)
is an arbitrary constant. We expect, that the function is either
continuous in p0 or discrete, so
Ψ(x, t)
=
Z∞
c(p0 , t)ψp0 (x) dp0
−∞
=
1
√
2π~
Z∞
c(p0 , t)e
ip0 x
~
dp0
−∞
which is the Fourier integral from before with Ψ̃(p, t) = c(p, t).
5.3
Compatible observables
There’s an important idea: We can not neccessarily specify a quantum state
using different observables (c.f. x, px , y, py , z, pz in classical mechanics), but we
specify the position by a Ψ which is a superposition of momentum states and
we specify the momentum by a Ψ which is a superposition of position states.
The observables must be “compatible”, but x, p are incompatible. To explore
this, we introduce the commutator of two operators
h
i
Â, B̂ = ÂB̂ − B̂ Â
If [Â, B̂] = 0, then A, B are comatible. If [Â, B̂] 6= 0, A, B are uncompatible.
We take x̂, p̂
µ
¶
∂
∂
x̂p̂Ψ − p̂x̂Ψ = x −i~
Ψ + i~
(xΨ)
∂x
∂x
∂
∂
= x (−i~)
Ψ + xi~
Ψ + i~Ψ
∂x
∂x
It follows
[x̂, p̂] = i~ 6= 0
It is easy to prove that there are no simultaneous eigenfunctions of x̂ and p̂. If
there were, we had
p̂ψx0 p0 (x) = p0 ψx0 p0 (x)
x̂ψx0 p0 (x) = x0 ψx0 p0 (x)
31
5
Observables & Operators - Heisenberg Uncertainty Principle
29.05.2006
and therefore
[x̂, p̂] ψx0 p0 (x) = (x0 p0 − p0 x0 ) ψx0 p0 (x) = 0
It followed
i~ψx0 p0 (x) = 0
⇒
ψx0 p0 (x) = 0
Therefore, if we have a non-zero commutator, we have non identical eigenfunctions.
5.4
Heisenberg uncertainty principle
We apply these ideas to get the uncertainties in x and p. The variance in x is
(∆x)
(∆p)
2
=
2
=
­ 2®
2
x − hxi =
­ 2®
2
p − hpi =
Z∞
−∞
Z∞
−∞
³ ´2
c Ψ dx
Ψ∗ ∆x
³ ´2
c Ψ dx
Ψ∗ ∆p
c = x̂ − hxi and ∆p
c = p̂ − hpi. So
Where ∆x
h
i
c ∆p
c = (x̂ − hxi) (p̂ − hpi) − (p̂ − hpi) (x̂ − hxi) = [x̂, p̂] = i~
∆x,
To go on, we need to write down some general properties for Hermitian operators:
(1) For the expectation value of A2 , where  is Hermitian, we get
­
A
2
®
Z∞
=
∗
Ψ Â(ÂΨ) dx =
Z∞
(ÂΨ)∗ (ÂΨ) dx
−∞
−∞
(2) With two Hermitian operators, we get  and B̂
Z∞
∗
Ψ ÂB̂Ψ dx
=
−∞
=
=
Z∞
−∞
Z∞
−∞
Z∞
(ÂΨ)∗ B̂Ψ dx
(B̂ ÂΨ)∗ Ψ dx
Ψ(B̂ ÂΨ)∗ dx
−∞
=


Z∞
−∞
This implies, that the integral
Z∞
Ψ
∗
−∞
h
Â, B̂
i
+
Ψ dx =
Z∞
−∞
∗
Ψ∗ B̂ ÂΨ dx
´
³
Ψ∗ ÂB̂ + B̂ Â Ψ dx
is real and
Z∞
−∞
Ψ
∗
h
i
Â, B̂ Ψ dx =
Z∞
−∞
is purely imaginary.
32
³
´
Ψ∗ ÂB̂ − B̂ Â Ψ dx
5
Observables & Operators - Heisenberg Uncertainty Principle
29.05.2006
(3) Schwarz’ inequality: Suppose α(x), β(x) are complex functions of x with
finite integrals
Z∞
Z∞
∗
α α dx,
−∞
Z∞
∗
β β dx,
−∞
α∗ β dx
−∞
Construct φ(x) = α + λβ, λ ∈ C
Z∞
Z∞
∗
φ φ dx =
−∞
∗
∗
α α dx + λ λ
−∞
Z∞
β ∗ β dx
−∞
+λ
∗
Z∞
∗
β α dx + λ
−∞
Z∞
α∗ β dx ≥ 0
−∞
This is true for any λ, so true for
λ =
−
R∞
β ∗ α dx
−∞
R∞
β ∗ β dx
−∞
It follows
λ∗
Z∞
β ∗ α dx = −λ∗ λ
−∞
Z∞
β ∗ β dx
−∞
and therefore
Z
φ∗ φ dx
=
Z∞
α∗ α dx −
R∞
α∗ β dx
β ∗ α dx
−∞
−∞
−∞
R∞
R∞
β ∗ β dx
−∞
≥
0
That is
Z
α∗ α dx
Z
β ∗ β dx ≥
Z
α∗ β dx
Z
¯2
¯ ∞
¯
¯Z
¯
¯
∗
∗
¯
α β dx¯¯
β α dx = ¯
¯
¯
−∞
Now, set α = ÂΨ and β = B̂Ψ to get
¯ ∞
¯2
¯Z
¯
(1,3)
¯
¯
­ 2® ­ 2®
∗
¯
A
B
≥ ¯
(ÂΨ) (B̂Ψ) dx¯¯
¯
¯
−∞
¯ ∞
¯2
¯Z
¯
¯
¯
= ¯¯
Ψ∗ ÂB̂Ψ dx¯¯
¯
¯
−∞
¯2
¯ ∞
¯
¯Z
µh
i¶
i
h
¯
¯
1
∗
¯
= ¯
Ψ
Â, B̂ + Â, B̂ Ψ dx¯¯
2
+
¯
¯
−∞
(2)
=
=
¯ ∞
¯2
¯2 ¯
¯ Z
¯
¯
¯ Z
i
i
h
h
¯1
¯
¯
¯
1
∗
∗
¯
¯
¯
Ψ Â, B̂ Ψ dx¯¯
Ψ Â, B̂ Ψ dx¯ + ¯
¯2
2
+
¯
¯
¯
¯
−∞
¯ ∞
¯ ∞
¯2
¯2
¯Z
¯Z
¯
¯
h
h
i
i
¯
¯
¯
¯
1¯
∗
∗
¯ +1¯
¯
Ψ
Ψ
Â,
B̂
Ψ
dx
Â,
B̂
Ψ
dx
¯
¯
¯
¯
4¯
4¯
+
¯
¯
−∞
−∞
33
5
Observables & Operators - Heisenberg Uncertainty Principle
It follows
2
h∆xi h∆pi
31.05.2006
¯2
¯
¯
¯Z
¯
¯
1 2 ¯
≥
~ + ¯ Ψ∗ [x̂, p̂]+ Ψ dx¯¯
4
¯
¯
2
which gives us the Heisenberg uncertainty inequality
∆x∆p ≥
~
2
If we want equality, we also need
Z∞
ψ ∗ [x̂, p̂]+ ψ dx = 0
−∞
Note: In three dimensions, we can “mix” the position and momentum observables to specify a state, like x, y and pz
[x̂, ŷ] = 0
[x̂, p̂z ] = 0
[ŷ, p̂z ] = 0
Then we could write any wave function as superposition of functions
ψx0 y0 p0z = √
ip0z z
1
δ (x − x0 ) δ (y − y 0 ) e− ~
2π~
which are the eigenfunctions of x̂, also ŷ, also p̂z .
5.5
Compatibility with Ĥ
Consider the expectation value for an observable A for a particle with wave
function Ψ(r, t)
Z
Ψ(r, t)∗ ÂΨ(r, t) d3 r
hA(t)i =
R3
Using Schrödingers Equation
i~
∂
Ψ(r, t) = ĤΨ(r, t)
∂t
Z
∂Ψ∗
ÂΨ d3 r +
∂t
we get
d
hA(t)i
dt
=
R3
=
−
Z
Ψ∗ Â
R3
1
i~
Z
(ĤΨ)∗ ÂΨ d3 r +
R3
=
1
i~
Z
R3
=
∂Ψ 3
d r
∂t
1
i~
Z
R3
1
i~
Z
Ψ∗ ÂĤΨ d3 r
R3
´
³
Ψ∗ ÂĤ − Ĥ Â Ψ d3 r
i
h
Ψ∗ Â, Ĥ Ψ d3 r
where by the first step we assume that  6= f (t). If  is compatible with Ĥ,
i.e. [Â, Ĥ] = 0, then
d hAi
= 0
dt
that means, A is a constant of motion.
34
5
Observables & Operators - Heisenberg Uncertainty Principle
5.6
31.05.2006
Orbital angular momentum
Classically, the angular momentum is given by
L = r×p
In Quantum Mechanics, we write
L̂ = r̂ × p̂ = −i~r × ∇
or
µ
¶
∂
∂
= ŷ p̂z − ẑ p̂y = −i~ y
−z
∂z
∂y
¶
µ
∂
∂
= ẑ p̂x − x̂p̂z = −i~ z
−x
∂x
∂z
µ
¶
∂
∂
= x̂p̂y − ŷ p̂x = −i~ x
−y
∂y
∂x
L̂x
L̂y
L̂x
It is easy to show by expanding, that
i
h
L̂x , L̂y
h
i
L̂y , L̂z
h
i
L̂z , L̂x
but that
h
2
L̂ , L̂x
i
=
h
2
=
L̂z [ẑ, p̂z ]
=
L̂x [x̂, p̂x ]
=
L̂y [ŷ, p̂y ]
| {z }
=i~6=0
L̂ , L̂y
We can also verify, that
h
i
L̂i , ∇2 = 0
It follows
for i = x, y, z
i
=
h
h 2
i
L̂ , L̂z = 0
L̂i , V (r)
h
i
L̂i , Ĥ = 0
i
= 0
for all Ĥ associated with a “central potential V (r)”
h 2 i
⇒ L̂ , Ĥ = 0
As expected, L is conserved in a central potential. The most surprising thing
is that we can define (or measure) the total L2 , plus only one component of it.
Conventionally, we take Lz .
5.7
Angular momentum eigenfunctions
We first explore a Cartesian function
ψ(0,0) = R(r)
so a spherically symmetric function. We go on with
z
x + iy
x − iy
ψ(1,0) = R(r)
ψ(1,+1) = R(r)
ψ(1,−1) = R(r)
r
r
r
Now we want to know what happens, if we apply L̂x , L̂y , L̂z and L̂
• ψ(0,0) : We get
L̂x ψ(0,0)
=
=
2
µ
¶
∂R ∂r
∂R ∂r
−i~ y
−z
∂r ∂z
∂r ∂y
y´
∂R ³ z
−i~
y −z
= 0
∂r
r
r
Any spherically symmetric ψ has L2 = Lx = Ly = Lz = 0 and therefore
L̂x ψ(0,0) = 0ψ(0,0)
L̂y ψ(0,0) = 0ψ(0,0)
2
L̂ ψ(0,0) = 0
35
L̂z ψ(0,0) = 0ψ(0,0)
5
Observables & Operators - Heisenberg Uncertainty Principle
31.05.2006
• ψ(1,0) : First we write down, that
µ
¶
¶
µ
R(r)
R(r)
R(r)
R(r)
L̂
+
z
= z L̂
L̂z =
L̂z
r
r
r
r
Furthermore, we get
L̂x ψ(1,0)
=
L̂y ψ(1,0)
=
L̂z ψ(1,0)
=
R(r)
R(r)
L̂x z = −i~
y
r
r
R(r)
x
i~
r
0
ψ(1,0) is not an eigenfunction of L̂x or L̂y , but is of L̂z , with eigenvalue 0.
Going on, we get
L̂2x ψ(1,0) = ~2
and therefore
L̂
2
R(r)
z = ~2 ψ(1,0) = L̂2y ψ(1,0)
r
= L̂2x + L̂2y + L̂2z = 2~2 ψ(1,0)
2
so ψ(1,0) is also an eigenfunction of L̂ with eigenvalue 2~2 .
• ψ(1,1) :
ψ(1,1) = R(r)
x + iy
r
2
ψ(1,1) is an eigenfunction of L̂ with
2
L̂ ψ(1,1) = 2~2 ψ(1,1)
and an eigenfunction of L̂z with
L̂z ψ(1,1) = ~ψ(1,1)
• ψ(1,−1) :
L̂2 ψ(1,−1) = 2~2 ψ(1,−1)
L̂z ψ(1,−1) = −~ψ(1,−1)
This can be illustrated in the following way
√
2~
ψ1,1
ψ1,0
ψ1,−1
2
All this is much easier in spherical coordinates (r, θ, φ), then L̂z and L̂ operate
on spherical harmonic functions, where
2
L̂ (Y`m (θ, φ))
L̂z (Y`m (θ, φ))
=
=
` (` + 1) ~2 Y`m (θ, φ)
m~Y`m (θ, φ)
with m = −`, −` + 1, ..., 0, ..., `.
L
p
l(l + 1)
36
5
Observables & Operators - Heisenberg Uncertainty Principle
07.06.2006
Note that ψ0,0 ∝ Y00 , ψ1,−1 ∝ Y1−1 , ... and that Y`m are orthonormal
Z2πZπ
0
∗
Y`0 m0 sin θ dθ dφ = δmm0 δ``0
Y`m
0
So we have the two important things
(1) L2 and Lz define orbital angular momentum quantum states
(2) The orbital angular momentum is quantised in units of ~
p
L =
` (` + 1) ~
Lz = 0, ±1~, ±2~, ..., ±`~
5.8
Stern-Gerlach experiment
The angular momentum of charged particles lets us expect a magnetic moment
µ. e.g. for the electron
=
µorbital
z
and for the spin
=
µspin
z
e
−e
Lz = −
m` ~
2me
2me
−2e
e
Sz = −
ms ~
2me
me
Recall that µ in an inhomogeneous B field suffers a force
Fz = µ z
∂B
∂z
Silver
Atoms
Precisely
two
beams
The result is, that we get exactly two beams, that is Lz is quantised by ± 21 ~.
5.9
Spin
We have an angular momentum also from the spin of particles, which is not
really meaningful as classical spinning object. There we have eigenvalues
p
s (s + 1) ~
S =
and ms = −s, −s + 1..., s − 1, s, but s can now be a half-integer. The electron
for example has s = 21 and therefore ms = ± 21 or the W -Boson has s = 1 and
therefore ms = −1, 0, 1.
37
6
The Hydrogen Atom
6
07.06.2006
The Hydrogen Atom
6.1
Motion in “central potentials”
The hydrogen atom is the simplest one, because it is a proton p+ and an orpiting
electron e− . We say e− orbits, because mp À me . There we have a central
potential
1 −e2
V (r) =
4πε0 r
which is as known similar to the potential of planetary motions. The central
force acts radially and therefore, the angular momentum L is conserved
F
= −
dV
r̂
dr
L = mr × v = mr ×
dr
dt
N = r×F
= 0
where N is the torque.
v=
dr
dt
r
A = 21 r × dr =
1
2m Ldt
We can split the momentum into a radial and a tangential part
pr = m
dr
dt
pt =
L
r
The kinetic energy then is given by
p2
p2r
L2
=
+
2m
2m 2mr2
and therefore the total energy by
E =
L2
p2r
p2r
+
+ Ve
+
V
(r)
=
2m 2mr2
2m
Ve :=
L2
+ V (r)
2mr2
where we call Ve the effective potential with effective force
Fe = −
6.2
dVe
dr
=
L2
dV
−
3
mr
dr
Solutions of Schrödingers equation in a central potential
For a wave function, describing a quantum state with sharply defined energy E,
we write
iEt
Ψ(r, θ, φ, t) = ψ(r, θ, φ)e− ~
where ψ(r, θ, φ) is an energy eigenfunction satisfying the eigenvalue equation
¸
·
~2 2
∇ + V (r) ψ = Eψ
(6.1)
−
2m
If we also assume, that the quantum state has as well definite angular momentum
properties, we can write
ψ(r, θ, φ) = R(r)Y`,m` (θ, φ)
(6.2)
Note that
∇2 ψ =
1
1 ∂2
(rψ) + 2
r ∂r2
r
µ
∂ 2 ψ cos θ ∂ψ
1 ∂2ψ
+
+
2
∂θ
sin θ ∂θ
sin2 θ ∂φ2
38
¶
6
The Hydrogen Atom
12.06.2006
and
L̂
2
= −~
So we can also write
2
µ
∂2
cos θ ∂
1
∂2
+
+
2
∂θ2
sin θ ∂θ sin θ ∂φ2
¶
2
1 ∂2
1
(rψ) − 2 2 L̂ ψ
r ∂r2
r ~
Substituting into equation (6.1), using equation (6.2), we get
∇2 ψ =
−
(6.3)
~2 2
∇ R(r)Y`,m`
2m
2
~2 ∂ 2
1
= −
(rR(r)Y`,m` (θ, φ)) +
L̂ (R(r)Y`,m` (θ, φ))
2mr ∂r2
2mr2
1
~2 ∂ 2
(rR(r)Y`,m` (θ, φ)) +
R(r)` (` + 1) ~2 Y`,m` (θ, φ)
= −
2mr ∂r2
2mr2
= ER(r)Y`,m` (θ, φ) − V (r)R(r)Y`,m` (θ, φ)
Here we can remove the spherical harmonic Y`,m` (θ, φ) and multiply by r to get
` (` + 1) ~2
~2 ∂ 2
(rR(r)) +
rR(r) + V (r)rR(r)Y`,m` = ErR(r)
2
2m ∂r
2mr2
If we set u(r) = rR(r), we can rewrite
·
¸
~2 ∂ 2
` (` + 1) ~2
−
u(r) +
+ V (r) u(r) = Eu(r)
2m ∂r2
2mr2
−
Here we will get eigenfunctions unr ,` (r) with energy Enr ,` and therefore
unr ,` (r)
Y`m (θ, φ)
r
ψnr ,`,m (r, θ, φ) =
2
which must be an eigenfunction of Ĥ, L̂z , L̂ .
Introduce the parity of a wave function
ψ(−r) = ±ψ(r)
The eigenvalues are +1 (even parity) and −1 (odd parity). In a central potential,
the parity is even if ` is even and it is odd, if ` is odd. The parity is an observable
and the parity operator eigenvalues are ±1. Ĥ and the parity operator commute
if Ĥ(−r) = H(r).
6.3
Hydrogen atom
We would like to solve the radial Schrödinger Equation
−
~2 ∂ 2
` (` + 1) ~2
e2
un ,` = Enr ,` unr ,`
u
+
u
−
n
,`
n
,`
r
r
2me ∂r2
2me r2
4πε0 r r
with u → 0 as r → 0 (because ψ is finite at r = 0) and as r → ∞. We define
the effective potential
Veff (r) :=
` (` + 1) ~2
e2
−
2
2me r
4πε0 r
Veff (r)
r
increasing `
`=0
39
6
The Hydrogen Atom
12.06.2006
The minimum, we get by
dVeff
dr
= −
` (` + 1) ~2
e2
+
me r 3
4πε0 r2
where
r = rmin = ` (` + 1) ~2
4πε0
me e 2
= 0
=: ` (` + 1) a0
and a0 is the Bohr radius
4πε0 ~2
e 2 me
a0 =
= 0.53 · 10−10 m
For the energy at rmin we get
Veff (` (` + 1) a0 ) = −
ER
1 e2
1
= −
2 eπε0 a0 ` (` + 1)
` (` + 1)
where ER is the Rydberg energy
ER :=
e2
8πε0 a0
= 136eV
For high `, we have a high energy and the probability gets further out. Introduce
the fine structure constant
α :=
1
e2
=
4πε0 ~c
137.035...
which is a dimensionless constant. Now
a0 =
1 ~
α me c
ER =
e2
8πε0 a0
=
α2
me c 2
2
and we solve
~2 ∂ 2
−
un ,` +
2me ∂r2 r
µ
` (` + 1) ~2 2
e2
~
−
2me r2
4πε0 r
¶
unr ,` = Enr ,` unr ,`
Setting r = qa0 , E = −γ 2 ER , we can write this as
¶
µ
∂2u
2 ` (` + 1)
u = γ2u
+
−
∂q 2
q
q2
(6.4)
We have the boundary conditions u(q) = 0 at q = 0 and as q → ∞. Then, at
large q, we get the equation
d2 u
= γ2u
dq 2
which has the general solutions
u(q) = Ae−γq + Beγq = Ae−γq
where we set B = 0, because u(q) → 0 as q → ∞. For small q, we get
d2 u ` (` + 1) u
−
= 0
dq 2
q2
which has solutions
u(q) = Aq `+1
or
u(q) = Bq −`
where the second one wouldn’t satisfy u(q) = 0 at q = 0. So we look for
u(q) = f (q)q `+1 e−γq
40
6
The Hydrogen Atom
14.06.2006
which (by substituting into equation (6.4)) leads to
q
df
d2 f
+ 2 [(` + 1) − γq]
+ 2 [1 − γ(` + 1)] f = 0
dq 2
dq
This has solutions
X
f (q) = pnr ,` (q) =
as q s
s
with
as+1 =
·
¸
2γ (s + ` + 1) − 2
as
s(s + 1) + 2(s + 1)(` + 1)
γ =
1
nr + ` + 1
so that pnr ,` (r) terminates at anr q nr . Finally, we get
unr ,` (r) = N pnr ,` (r)r `+1 e
−a
r
0 (nr +`+1)
Enr ,` = −
ER
(nr + ` + 1)
2
For each `, there is an infinite number of discrete energy states as nr goes from
0 to ∞.
nr = 0
− E9R
nr = 0
nr = 1
− E4R
nr = 0
−ER
`=0
s
6.4
`=1
p
`=2
d
Zeeman effect - Perturbing the Hamiltonian Ĥ
We introduce the total angular momentum as the sum of the orbital angular
momentum and the spin. We had before
p
` (` + 1) ~
m` = −`, −` + 1, ..., `
L =
p
s (s + 1) ~
ms = −s, −s + 1, ..., s
S =
Now, we add
J =
p
j (j + 1) ~
mj = −j, −j + 1, ..., +j
The question now is, how to get j from ` and s. The possible values of j range
from j = ` + s and j = |` − s| with ∆j = 1. e.g. if we have ` = 2, s = 21 , then
j = 52 or j = 23 and therefore
5
5 3 1
mj = − , − , − , ...,
2 2 2
2
or
3 1 1 3
mj = − , − , ,
2 2 2 2
We also dealt earlier with the magnetic moment, which depends on whether the
angular momentum is coming from spin or orbital angular momentum and so
we can write general formulae
µ = IA =
q
L
2m
µz = −g
e
~mj
2me
This g here is called the Landé g factor, which is given by
g = 1+
j (j + 1) − ` (` + 1) + s (s + 1)
2j (j + 1)
41
6
The Hydrogen Atom
14.06.2006
so g = 1 if s = 0 and g = 2 if ` = 0. Now we can consider an atom placed in a
magnetic field. The electron acqires energy
Emag = −µ · B
If we have arranged things, so that |B| = Bz , then
Emag = −µz B = gµB mj B
µB =
e~
2me
where µB is the Bohr magneton. We had
Ĥψnr ,`,m = Eψnr ,`,m
Ĥ = Ĥ0 + mj gµB B
| {z }
const
The effect is, that all enery levels are split into 2j + 1 components. This is called
the Zeeman effect.
6.5
Time-dependent perurbations - Radiative transitions
If Ĥ changes with time, we get transitions between “stationary states”. Imagine
the unperturbed eigenfunctions Ψ0k and eigenvalues ε0k . In this case, we have
the time-indipendent form of the Schrödinger equation
i~
∂ 0
Ψ (t) = Ĥ0 Ψ0k (t) = ε0k Ψ(t)
∂t k
We’d expect then, that the general solution of this is a superposition
Ψ0 (t) =
X
X
a0k Ψ0k (t) =
k
a0k e−
iε0
kt
~
ψk0
k
Now we perturb Ĥ = Ĥ0 + V̂ (t). We can write
i~
Assume that
³
´
∂
Ψ(t) = Ĥ0 + V̂ (t) Ψ(t)
∂t
Ψ(t) =
X
ak (t)Ψ0k (t) =
k
X
ak (t)e−
iε0
kt
~
ψk0
k
and substitute into the Schrödinger equation
¶
´
X µ dak (t) iε0
X³
iε0
iε0
kt
kt
− k ak (t) e− ~ ψk0 =
i~
ε0k + V̂ (t) ak (t)e− ~ ψk0
dt
~
k
k
which can be simplified to
X µ dak (t) ¶ iε0k t
X
iε0
kt
i~
e− ~ ψk0 =
V̂ (t)ak (t)e− ~ ψk0
dt
k
k
Now we multiply by another state ψn0 ∗ and integrate over d3 r
i~
X µ dak (t) ¶
k
and it follows
dt
e−
iε0
kt
~
Z
|
ψn0 ∗ ψk0 d3 r =
{z
δnk
}
X
k
ak (t)e−
iε0
kt
~
Z
ψn0 ∗ V̂ ψk0 d3 r
Z
X
d
i(ε0n −ε0k ) ~t
i~ an (t) =
ψn0 ∗ V̂ ψk0 d3 r
ak (t)e
dt
k
42
6
The Hydrogen Atom
19.06.2006
If we assume that the initial state were an eigenfunction ψj , we get
Z
0 t
0
d
ψn0 ∗ V̂ ψj0 d3 r ei(εn −εj ) ~
i~ an (t) =
dt
and therefore
i
an (t) = −
~
where we define
Z
Vnj :=
Zt Z
0
0
t0
ψn0 ∗ V̂ ψj0 d3 r ei(εn −εj ) ~ dt0
0
εnj := ε0n − ε0j
ψn0 ∗ V̂ ψj0 d3 r
If the perturbation is constant and applied for some time t, we can write
an (t)
=
=
i
− Vnj
~
−
Define
Zt
e
iεnj
i
= − Vnj
~
´
Vnj ³ iεnj t
~ − 1
e
εnj
where as t → ∞
π
2~ δ(εnj ),
dt
0
0
Pjn := |an (t)|
behaves as
t0
~
2
|Vnj |2
=
4 sin2
ε2nj
³
εnj t
2~
tε2nj
sin2
µ
µ
¶¯t
~ iεnj t0 ¯¯
~
e
¯
iεnj
0
εnj t
2~
¶
´
that is
→ lim
t→∞
2π
Pjn (t)
2
=
|Vnj | δ(εnj )
t
~
If V̂ is time varying, we cannot take outside of the integral
is interacting with the dipole of the atom, e.g.
V̂
R
dt0 , e.g. the E-wave
= (er̂)E0 cos ωt
where er̂ is the dipole of the atom. So it is exactly as before, but now
δ → δ (εnj ± ~w)
and the transition probability is
lim
t→∞
Pjn (t)
2π
2
2
=
|Vnj | |A(ωnj )|
t
~
ε
2
where |A(ω)| is the strength of perturbation at ω = ~nj . We only get transi2
tions, if ~ω = |εnj | and |Vnj | 6= 0. The interaction with the dipole of the atom
is proportional to r̂ which is the operator of r and not the unit vector. Look at
Z
ψn r̂ψj d3 r
Note that the operator r̂ changes sign when r → −r, therefore Vnj = 0 unless
ψn and ψj have opposite parity.
2p
2s
1s
43
6
The Hydrogen Atom
6.6
19.06.2006
Some (small) complicating effects
So far, we have considered Ĥ
• non-relativistic
• only with Coulomb forces
• assuming, that the mass of the proton was ∞.
6.6.1
The reduced mass effect
The mass of the proton mp is not ∞ and therefore the proton and the electron
orbit arround a common center of mass. The energy is
E =
p2p
p2e
e2
+
−
2me
2mp
4πε0 r
In the center of mass frame, we have pe = pp = p and
p2
e2
−
2µ 4πε0 r
E =
µ =
me mp
me + m p
where µ is the reduced mass. The net effect of this is to change the length and
energy scales
me
a0
µ
a00 =
µ
ER
me
0
ER
=
En0 =
µ
En
me
For the electron-proton atom (i.e. Hydrogen 1 H), we have
µ
me
1836
1837
=
or for Deuterium 2 H (e− and p+ n0 ) we get
µ
me
6.6.2
3671
3672
=
Spin-orbit coupling
This is a modification of Ĥ due to the magnetic field produced by the orbital
motion interacting with µ from the spin. The electron sees an “orbiting” proton
with period
2πr
me r 2
τ =
= 2π
v
Le
This means, that effectively we have a current
I =
e
τ
⇒
B =
µ0 I
2r
=
µ0 eLe
2r 2πme r2
and we have
µe = −2
and therefore
Emag = −µ · B =
=
e
S
2me
e2
L·S
4πε0 m2e c2 r3
This isn’t completely correct. Correct would be
Emag =
where L ∼ S ∼ ~ and
Emag ∼
e2
L·S
8πε0 m2e c2 r3
e2 ~2
4πε0 m2e c2 a30
44
e
Le
4πε0 me c2 r3
= α 4 me c 2
6
The Hydrogen Atom
6.6.3
19.06.2006
Relativist effects
We know the relativistc energy
ε =
p
m2e c4
+
p 2 c2
p2
1
= me c +
−
2me
8
2
µ
p
me c
¶4
me c2 + ...
where p2 = E2me ∼ α2 m2e c2 and therefore p ∼ αme c. This means that the
relativistic correction is of order
ε =
α4
me c 2
8
45
7
Quantum mechanics of identical particles
7
19.06.2006
Quantum mechanics of identical particles
In deterministic Classical Physics we keep track of even identical particles. In
Quantum Physics it is probabilistic, so it’s impossible to keep track of individual
particles.
7.1
Wave functions for identical particles
Consider two particles p, q, then we have a “2-particle wave-function”
Ψ (r p , r q , t)
The probability of finding p at a and q at b is
2
|Ψ (a, b, t)| d3 ad3 b
If we have two identical particles, we get
|Ψ (a, b, t)|
2
= |Ψ (b, a, t)|
2
This enables us to say that
Ψ (a, b, t) = eiδ Ψ (b, a, t)
where e−iδ can be a “phase factor”. Then
Ψ (a, b, t) = e2iδ Ψ (a, b, t)
⇒
eiδ = ±1
Therefore, the wave function of two identical particles must be symmetric or
anti-symmetric under exchange of p and q. Later, the symmetric ones, we will
call Bosons and the anti-symmetric ones Fermions.
Tale the Hamiltonian of the harmonic oscillater, assuming that p and q do not
depend on each other
1
~2 ∂ 2
1
~2 ∂ 2
−
+ mω 2 x2p + mω 2 x2q
2m ∂x2p
2m ∂x2q
2
2
¡
¢
The eigenfuncions are ψn with En = n + 21 ~ω
Ĥ(xp , xq ) = −
1) If p and q are in exactly the same state (nq = np ), we get
E = 2En = (2n + 1) ~ω
and the wave-function
t
Ψ(xp , xq , t) = ψn (xp )ψn (xq )e−i(2En ) ~
Verify that it is a solution of
ĤΨ (xp , xq , t) = EΨ (xp , xq , t)
In this case, Ψ is symmetric under the p to q exchange operator. We can
not find an anty-symmetric solution, if p and q are in the same state.
2) If p and q are in different states and also distinguishable (denoted by D
superscript), we will have
E = En + En0 = (n + n0 + 1)~ω
We can construct two solutions
t
ΨD
1 (xp , xq , t)
=
ψn (xp )ψn0 (xq )e−i(En +En0 ) ~
ΨD
2 (xp , xq , t)
=
ψn (xq )ψn0 (xp )e−i(En +En0 ) ~
t
and any linear combination will be a solution. In particular, we can write
D
ΨD (xp , xq , t) = c1 ΨD
1 (xp , xq , t) + c2 Ψ2 (xp , xq , t)
2
where |c1 | is the probability that p is in state n and q in state n0 and
2
|c2 | is the probability that q is in state n and p in state n0 .
46
7
Quantum mechanics of identical particles
21.06.2006
3) If we have n and n0 but p and q are now indistinguishable, then
|c1 |
2
= |c2 |
2
=
1
2
Now we can construct symmetric and anti-symmetric wave functions
ΨS (xp , xq , t)
=
ΨA (xp , xq , t)
=
t
1
√ (ψn (xp )ψn0 (xq ) + ψn (xq )ψn0 (xp )) e−i(En +En0 ) ~
2
t
1
√ (ψn (xp )ψn0 (xq ) − ψn (xq )ψn0 (xp )) e−i(En +En0 ) ~
2
Its interesting to look at ΨD , ΨS , ΨA for xp = xq = x0 . We find
t
ΨS (x0 , x0 , t)
= ψn (x0 )ψn0 (x0 )e−i(En +En0 ) ~
√
t
=
2 ψn (x0 )ψn0 (x0 )e−i(En +En0 ) ~
ΨA (x0 , x0 , t)
= 0
ΨD (x0 , x0 , t)
This means, that relative to the case of distinguishable particles, we can say
that the symmetric ΨS is twice as likely to have both particles in the same
place at the same time and the anti-symmetric ΨA never has the two particles
in the same place at the same time. For examples, see the exercice sheet.
7.2
Exchange symmetry with spin
p
s (s + 1)~ with zRecall that we had the spin angular momentum S =
component Sz = ms ~. Lets say, we have a spin wave function χ(p) with
χ(p) =
s
X
cms χs,ms (p)
ms =−s
We can combine this with the (normal) spatial wave function ψ, to get
Φn,`,m` ,s,ms = ψn,`,m` χs,ms
Now look at two indistinguishable particles in exactly the same state. We will
again find a wave function
Φ(p, q) = Φn,`,m` ,s,ms (p)Φn,`,m` ,s,ms (q)
and again, ΨA for the same states does not exist.
But what about particles in different states? Can we represent the wave function
as product of two single particle states?
ΨS (p, q)
´
1 ³
= √ Φn,`,m` ,s,ms (p)Φn0 ,`0 ,m0` ,s0 ,m0s (q) + Φn,`,m` ,s,ms (q)Φn0 ,`0 ,m0` ,s0 ,m0s (p)
2
ΦA (p, q)
´
1 ³
= √ Φn,`,m` ,s,ms (p)Φn0 ,`0 ,m0` ,s0 ,m0s (q) − Φn,`,m` ,s,ms (q)Φn0 ,`0 ,m0` ,s0 ,m0s (p)
2
which is exactly the same approach as we had before. But we could also have
the product of two two-particle states as
Φ2,S (p, q)
2,A
Φ (p, q)
Φ2,A (p, q)
Φ2,S (p, q)
=
ψ S (p, q)χS (p, q)
= ψ A (p, q)χS (p, q)
= ψ S (p, q)χA (p, q)
= ψ A (p, q)χA (p, q)
47
7
Quantum mechanics of identical particles
05.07.2006
Example 7.1: Consider two particles with s1 , s2 = 21 and remember that if we
have the combination of two spin-numbers s1 , s2 then S = |s1 − s2 |, ..., s1 + s2 .
Therefore, we get S = 1 = s1 + s2 with mS = −1, 0, 1 or S = 0 = |s1 − s2 | with
mS = 0. Lets call χ 12 ,+ 12 = χ+ and χ 21 ,− 12 = χ− , then
χS1,1 (p, q)
=
χS1,0 (p, q)
=
χS1,−1 (p, q)
=
χA
0,0 (p, q)
=
χ+ (p)χ+ (q)
1
√ (χ+ (p)χ− (q) + χ+ (q)χ− (p))
2
χ− (p)χ− (q)
1
√ (χ+ (p)χ− (q) − χ+ (q)χ− (p))
2
¤
As we already mentioned, the nature gives us then two types of particles, namely
Bosons (photon, 4 He, ...) and Fermions (e− , p+ ). Lets simply state the rules:
The exchange symmetry depends on the spin of the particle. In particular,
particles with integer total spin must have a symmetric wave function under
particle exchange. Particles with half integer spin must have anti-symmetric
two-particle wave-functions under the action of changing the two particles. The
Bosons have integer spin and therefore a symmetric Ψ. The Fermions have half
integer spin and therefore an anti-symmetric Ψ.
Consequences for Fermions:
1) Two Fermions (e.g. two electrons) can never have absolutely identical
quantum states, because we can only make symmetric ΨS out of identical
wave functions.
2) When the two-particle spin state is anti-symmetric χA , then Ψ must be
symmetric ΨS . This means, that the “electrons like to be together”. We
call it Covalent Bond.
3) When the spin is symmetric χS , then Ψ must be anti-symmetric, which
means that the electrons are never together. We call this rigidity of matter.
Consequences for Bosons: We can always make symmetric wave-functions from
two or more identical states. This is used by Lasers, the Bose-Einstein Condensate and superfluidity.
48
8
8
More Complicated Atoms
05.07.2006
More Complicated Atoms
For more complicated atoms, we’ll have to modify what we’ve got for the Hatom. Multiple e atoms will have a Hamiltonian H that includes the term due
to the electron-electron interaction, e.g. for He we have
H = −
where
V (r p , r q ) = −
¢
~2 ¡ 2
∇p + ∇2q + V (r p , r q )
2m
2e2
2e2
e2
−
+
4πε0 rp
4πε0 rq
4πε0 |r p − r q |
with two’s in the first two terms, because the nucleus has order Z = 2. What
we do, to deal with those more complicated atoms is to modify the central
potential. Close to the nucleus, the potential is something like
V (r) = −
Ze2
4πε0 r
Very far away, we could take
−e2
4πε0 r
V (r) =
We combine these two potentials by
V (r) = z(r)
−e2
4πε0 r
z(r) =
¡
r
(z − 1)e− a + 1
¢
a =
1
a0
2
Therefore, we solve for a single electron to get the eigenfunctions. Once we’ve
got the eigenfunctions, we can “populate” them with the electrons. Then there
are two changes:
• At low n, the energy eigenvalues are much lower (by a factor of z 2 , where
one z is because of the potential V (r) and the other z because of a0 ∝ z1 ).
• The degeneracy between ` states is broken
More complicated atoms
H-Atom
Note: 3d > 4s
Now assign electrons to these n, `, m states, 2 at a time (↑↓), to get
• 1s: n = 1, ` = 0, m = 0
⇒ 2 electrons
• 2s: n = 2, ` = 0, m = 0
⇒ 2 electrons
• 2p: n = 2, ` = 1, m = −1, 0, 1
⇒ 6 electrons
• 3s: n = 3, ` = 0, m = 0
⇒ 2 electrons
• 3p: n = 3, ` = 1, m = −1, 0, 1
⇒ 6 electrons
• 3d: n = 3, ` = 2, m = −2, −1, 0, 1, 2
⇒ 10 electrons
49
Stichwortverzeichnis
A
Q
absorption coefficiant . . . . . . . . . . . . 5
angular frequency . . . . . . . . . . . . . . . 11
quasi classical state . . . . . . . . . . . . . 28
R
B
Rayleigh Jeans . . . . . . . . . . . . . . . . . . . 8
reduced mass . . . . . . . . . . . . . . . . . . . 44
rigidity of matter . . . . . . . . . . . . . . . 48
Rydberg energy . . . . . . . . . . . . . . . . . 40
black body . . . . . . . . . . . . . . . . . . . . . . . 6
Bohr radius . . . . . . . . . . . . . . . . . . . . . 40
Born. . . . . . . . . . . . . . . . . . . . . . . . . . . .14
Boson . . . . . . . . . . . . . . . . . . . . . . . . . . 46
S
C
Schrödinger equation . . . . . . . . . . . 17
Schrödinger Equation . . . . . . . . . . . 12
standard deviation . . . . . . . . . . . . . . 13
Stefan-Boltzmann constant . . . . . . . 6
collapse of wave function . . . . . . . . 17
commutator . . . . . . . . . . . . . . . . . . . . 31
Compton Scattering. . . . . . . . . . . . . .3
Covalent Bond . . . . . . . . . . . . . . . . . . 48
U
D
ultraviolet catastrophe . . . . . . . . . . . 8
degeneracy. . . . . . . . . . . . . . . . . . . . . .19
W
E
wavelength . . . . . . . . . . . . . . . . . . . . . 11
wavenumber . . . . . . . . . . . . . . . . . . . . 11
effective force . . . . . . . . . . . . . . . . . . . 38
effective potential . . . . . . . . . . . . . . 38 f
energy density . . . . . . . . . . . . . . . . . . . 6
expectation value . . . . . . . . . . . . . . . 13
Z
Zeeman effect . . . . . . . . . . . . . . . . . . . 42
F
Fermions . . . . . . . . . . . . . . . . . . . . . . . 46
fine structure constant . . . . . . . . . . 40
frequency . . . . . . . . . . . . . . . . . . . . . . . 11
G
Gamow energy . . . . . . . . . . . . . . . . . . 25
H
Hamiltonian operator . . . . . . . . . . . 16
Heisenbergs microscope . . . . . . . . . . 9
Hermitian . . . . . . . . . . . . . . . . . . . . . . 30
L
Landé g factor . . . . . . . . . . . . . . . . . . 41
P
parity. . . . . . . . . . . . . . . . . . . . . . . . . . .39
period . . . . . . . . . . . . . . . . . . . . . . . . . . 11
phase velocity . . . . . . . . . . . . . . . . . . 11
photon . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
Plank’s constant . . . . . . . . . . . . . . . . . 9
Plank’s formula . . . . . . . . . . . . . . . . . . 9
Poisson Distribution . . . . . . . . . . . . 28
50