Intro to Classical & Quantum Waves
Charles Hailey
Transcribed by Ron Wu
This is an undergraduate course. O↵ered in Fall 2013 at Columbia
University. Required Course textbooks: Fitzpatrick, Oscillations and
Waves : an introduction; Harris, Modern Physics.
Contents
1 Classical Waves
1.1 Transverse Standing Wave . . . .
1.2 Traveling Wave . . . . . . . . . .
1.3 2D Wave . . . . . . . . . . . . . .
1.4 Complex Notations . . . . . . . .
1.5 Energy Conservation . . . . . . .
1.6 Transmission & Reflection Waves
1.7 Time Evolutions of Strings . . . .
1.8 Energy Stored in the Mode . . . .
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3
3
7
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10
12
15
16
2 Wave Applications
2.1 Transmission lines . . . . . . . .
2.2 Impedance Matching . . . . . .
2.3 Loaded Transmission Line . . .
2.4 2D, 3D Waves . . . . . . . . . .
2.5 Elastic Sheets . . . . . . . . . .
2.6 Wave Guides . . . . . . . . . .
2.7 Geometric Optics: Snell’s Law .
2.8 Group Velocity Revisit . . . . .
2.9 Dispersive Waves: Plasma Wave
3 Semi-Classical Waves
3.1 Ultraviolet Catastrophe . .
3.2 Fourier Analysis . . . . . .
3.3 Bandwidth Theorem . . .
3.4 Incoherence v.s Coherence
3.5 Interference . . . . . . . .
3.6 Di↵raction . . . . . . . . .
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4 Quantum Waves
4.1 Photons . . . . . . . . . . . . . . . . . .
4.2 Bohr Model . . . . . . . . . . . . . . . .
4.3 Matter Waves . . . . . . . . . . . . . . .
4.4 Uncertainty Principle . . . . . . . . . . .
4.5 Schrodinger Equation . . . . . . . . . . .
4.6 Infinite Square Well . . . . . . . . . . . .
4.7 Bound States . . . . . . . . . . . . . . .
4.8 Scattering, Tunneling . . . . . . . . . . .
4.9 WKB, alpha Decay . . . . . . . . . . . .
4.10 Hydrogen Wave function & Spin . . . . .
4.11 Afterword: Counterfactual Measurement
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1 Classical Waves
1.1 Transverse Standing Wave
Lecture 1
(9/4/13)
A beaded string is fixed at two ends. Some vertical motion is going on on the
string. The vertical displacements are small so that small angle approximation is
valid. Each beat is a apart and the position is labeled
xi = ia
First and last beats
y(x0 , t) = y0 = 0 y(xN +1 , t) = yN +1 = 0 8t
T =tension in the string. Consider the forces on ith beat
one T is pulling from the right due to the i + 1 beat, make an angle ✓i+1
another T is pulling from the left due to the i
1 beat, make an angle ✓i
1
Hence the total vertical force on ith beat is
mÿ = T sin ✓i
T sin ✓i+1 = T (✓i+1
1
✓i 1 )
On the other hand
✓i+1 = tan ✓i+1 =
yi+1
xi+1
and
✓i
1
=
yi
yi
yi+1 yi
=
xi
a
yi
a
1
Thus
T
ma
Guess solution is a product of space & time factors
ÿi = w02 (yi+1
2yi + yi 1 ) w02 ⌘
yi (xi , t) = A sin kxi cos(wt
3
)
(1.1)
(1.2)
so that 1) for fixed time, see a wave guide in space; 2) for fixed space, see oscillation
in time.
A = amplitude
2⇡
k =
wave number
w = 2⇡⌫ angular freq
= phase
Substitute (1.2) to (1.1)
w2 sin kxi = w02 [sin k(xi + a)
2 sin kxi + sin k(xi
= 2w02 sin kxi (cos ka
a)]
1)
or
ka
(1.3)
2
a relation between w and k is known as dispersion relation. Hence (1.2) is a
solution to (1.1) i↵ (1.3) holds.
What controls k? The boundary.
Boundary can be fixed or free, (freely slide up and down). For the case that
2 ends are fixed. The sin kx factor in (1.2) automatically satisfies one boundary
condition. The other is
yN +1 = 0
w = 2w0 sin
so
k(N + 1)a = n⇡ =) kn =
n⇡
(N + 1)a
n = 1, 2, 3, ..., N + 1
As we will see the n label is called the nth mode. When n = 1, the standing wave
has 1/2 of its full in the clamp. If n = 2, one full wave in the clamp. The reason
n has to stop at N , because at this point
N +1
=
2⇡
N +1
=
2a
kN +1
N +1
4
that is each beat is o↵ phase to its neighbor, and it is the shortest wavelength one
can generate on a N beat string.
What controls ? The initial condition. If the initial wave front is A sin kx,
= 0.
In summary: in our class there is a common routine
1) Guess the functional form of di↵erential equation
2) Plug guess solution and find dispersion relation
3) find condition on k that satisfies boundary condition.
Now study uniform string, so no beats, continuum string.
Change
yi (xi , t) ! y(x, t)
a !
x
m ! ⇢ x
⇢ =length density
yi+1 ! y(x + x, t)
so (1.1) becomes
ÿ(x, t) =
T y(x + x, t)
⇢
2y(x, t) + y(x
x2
hence putting
v2 =
@y
@ 2y
2 y(x, t) + @x x +
=v
@t2
Or commonly write
1 @2y
2 @x2
x2
x, t)
T
⇢
2y(x, t) + y(x, t)
x2
@y
@x
@ 2y
1 @ 2y
=0
@x2 v 2 @t2
equation of wave, with traveling speed v.
Lecture 2
(9/9/13)
yn = An sin
n⇡x
cos(wn t
(N + 1)a
5
x+
1 @2y
2 @x2
x2
= v2
@ 2y
@x2
(1.4)
n)
is known as the nth normal mode solution. Why are they called? Because we will
show energy is stored in modes and because modes represent natural frequency of
system, which arises to resonance.
The study of the string immediately generalized to other systems of wave
equation:
EM wave in vacuum (transverse)
@ 2E
@x2
1 @ 2E
=0
c2 @t2
@ 2B
@x2
1 @ 2B
=0
c2 @t2
p
c = 1/ µ0 ✏0
Sound Wave (longitudinal)
@ 2P
@x2
1
kT
M
@ 2P
=0
@t2
P pressure, =specific heat, M =molecular mass.
Sound wave in solid (or compression wave)
@ 2P
@x2
1 @ 2P
⇣ ⌘ 2 =0
Y
@t
⇢
Y =Young’s modules
Transmission Line
@ 2V
@ 2V
LC
=0
@x2
@t2
Plug y = A sin kx cos(wt
) to wave equation, we get
w = kv
(1.5)
dispersion relation, connecting spatial change with time.
2⇡⌫ =
2⇡
v =) v = ⌫ =
T
Two assumptions in deriving (1.4): 1) small amplitudes, so the system is
linear (small angle approximation) 2) no dissipation. In reality rubber band gets
6
hot, so (1.5) is not exact.
Experiment that products the standing wave on the string is to take an electrical vibrator moving up and down, attack the string to one end (this end is
moving with vibrator but we still consider this as a fixed end, because it moves
relatively small wrt the anti nodes), and the other end is hanging over a weight on
a pulley so that the tension is known. Then adjust the vibrator frequency s.t.
L=n
2
=) ⌫ =
v
=
p
T /⇢
2l/n
to achieve standing wave.
The mathematics describing longitudinal waves are exactly the same, the displacement y becomes pressure P (or more precisely Gauge pressure, P̃ = P Pa =
ambient pressure atmospheric pressure). Node maximum y becomes maximum
pressure. Anti node becomes 0 gauge pressure. Opposite to string boundaries.
1.2 Traveling Wave
We turn a function f (x) into a traveling wave
f (x) ! f (x
vt)
so at t = 0 for a point xs on the wave with displacement f (xs ). The same displacement appears at later time t at x = xs + vt. So
f (x
vt) travel to the right
f (x + vt) travel to the left
One can check y = f (x ± vt) solves the wave equation for any f (x), but we
are only interested in certain f (x). Primarily
y = y0 cos kx k =
2⇡
Periodic function. That is because many interesting physical system is periodic.
7
So
y(x, t) = y0 cos k(x
vt) w = kv
This is the most common equation for a traveling wave.
Snapshot at t = t0 , see y v.s. x a sinusoidal wave. Fix position x = x0 , see x0
moving up and down. If slicing out the t axis, get y v.s. t a sinusoidal oscillation.
1.3 2D Wave
~ wave) & sphere wave (e.g. water surface
Two ideal models: plane wave (e.g. E
bubble).
One way to illustrate plane wave on a piece of paper is to draw some surface
of constant amplitude (or called surface of constant phase), for fixed time.
"
"
"
#
#
#
"
"
"
Similarly draw concentric circles to illustrate sphere wave and the source is
at the center of the circles. One can get plant wave from point source of sphere
wave by getting far away from the source.
1.4 Complex Notations
Wave can interfere with themselves. From the wave equation, we see that if y1 , y2
are two solutions, then y1 + y2 are too solution.
Consider superposition of two waves traveling to the left and right,
8
<2A cos kx cos wt for + constructive
y = A cos(wt kx) ± A cos(wt + kx) =
:2A sin kx sin wt for
destructive
(1.6)
showing that one can construct standing wave solution from traveling waves.
Physically it means, refereed to experimental setup in the end of 1.1, that the
vibrator sends a wave to the right and gets reflected by the weight over the pulley,
8
whether should take ± depends whether it has a ⇡ phase change. We will discuss
them later.
Schrodinger equation has only complex traveling solution, so we need to study
complex notation
z = x + iy
z⇤ = x
iy
<z = x =z = y
z ⇤ z = x2 + y 2 = |z|2 modules
Second way writing z 2 C
z = rei✓
r=
p
|z|2
y
x
tan ✓ =
z⇤ = r 1e
i✓
Since ei✓ = cos ✓ + i sin ✓, write
ỹ = Aei(wt
kx)
w = kv
ỹ denotes complex solution. For classical problem take the real part at the end of
calculation, but for QM solutions are intrinsic complex, need to figure out a rule
to connect complex solution to the real world.
Back to 1.6, we use complex notation to manipulate. First complexize
ỹ = Aei(wt
kx)
+ Aei(wt+kx) = 2Aeiwt
e
ikx
+ eikx
= 2Aeiwt cos kx
2
then take real part
y = <ỹ = 2A cos wt cos kx
The other one
ỹ = Aei(wt
kx)
Aei(wt+kx) = 2iAeiwt
9
e
ikx
2i
eikx
=
2iAeiwt sin kx
take real part
y = <ỹ = 2A sin wt sin kx
Lecture 3
(9/11/13)
We will see that complex form is more useful for dissipation. Then dispersion
relation comes simply from complex form.
1.5 Energy Conservation
As wave moves down the string, it takes energy with it. Oscillation up and down
gets KE
✓ ◆2
ˆ l
1
@y
KE =
⇢
dx (dm = ⇢dx)
@t
0 2
Tension, T , $ Potential energy. dx is the natural length of the string between x
and x + dx. ds is arc length of that segment of string from x to x + dx
UE =
=
l
ˆ
T (ds
ˆ0
T(
dx)
p
dx2 + dy 2
dx)
1 @y
T [dx(1 + ( )2 )
2 @x
✓ ◆2
ˆ
1 @y
=
T
dx
2 @x
E=
✏ energy density
ˆ
⇢
2
✓
=
ˆ
@y
@t
◆2
dx +
T
2
ˆ
✓
@y
@x
◆2
dx
dx =
ˆ
✏dx
(1.7)
" ✓ ◆
✓ ◆2 #
2
1
@y
@y
✏=
⇢
+T
2
@t
@x
Figure out rate of energy going down the string
I⌘
@y @y
T
@x} @t
| {z
Fy
By the small angle approximation Fy = T tan ✓ = T sin ✓ =force of tension in y
10
direction, so I has unit of power.
Then we get continuity equation for energy
@✏ @I
+
=0
@t @x
Indeed
⇢yt ytt + T yx yxt
⇢
T yx yxt = yt T ( ytt
T
T yxx yt
Then
d
dt
yxx ) = 0
ˆ
E
Or
1
ytt
v2
l
@I
✏dx +
dx
=0
@x
0
0
| {z } | {z }
ˆ
l
yxx ) = yt T (
I(l) I(0)
Ė = something flow in at x = 0
st flow out at x = l
Hence I is the rate of energy stored in the string.
For y = A cos(wt kx)
I = T kA2 w sin2 (wt
time average
kx)
1
hIi = T kA2 w
2
(1.8)
Recall
⌦
2
sin (wt
↵
1
kx) =
T
ˆ
t+T
sin2 (wt0
t
⌦
kx)dt0 = cos2 (wt
↵ 1
kx) =
2
Using the natural characters of the string
1 w
1p
1
hIi = T A2 w =
⇢T w2 A2 = Zw2 A2
2 v
2
2
Define impedance
p
(1.9)
T
= ⇢v
v
Later we will do two strings transmission and reflection, Z will be a deciding factor,
Z=
⇢T =
11
since w and A in (1.9) are independent of medium. What is impedance?
Z=
T @y
T
T
Fy
= @x = @y@x =
v
vy
@t
@t
Fy
Z
Hence high Z means need more force to vibrate the string, so the string is sti↵er.
Here we consider y = A cos(wt kx) wave traveling to the right. If we do
y = A cos(wt + kx), we will get
vy =
hIi =
Now plug y = A cos(wt
1
Zw2 A2
2
kx) into(1.7)
hEi =
11 2 2
11 2 2
⇢w A l +
k A Tl
22
22
(1.10)
The two terms on the right are the same, agrees what we know about harmonics
hKEi = hU Ei.
One can define the average energy flux
h⌧ i = h✏i v =
hEi
v
l
~ wave
E.g. E
1
flux = ✏0 E 2 c
2
Later we will do mode decomposition, which has many applications including
Hawking radiation by the black holes. We will revisit the energy of the wave.
1.6 Transmission & Reflection Waves
Feynman said that if our civilization is about to end and there is only one piece
information can be stored and passed on to future ETs that should be “particles
are waves”. Because QM was developed based on the idea of wave mechanics.
Particularly we will see the transmission and reflection waves go hand to hand in
12
QM. In quantum scattering problem
w ⇠ energy
k ⇠ momentum
|A|2 ⇠ probability
Suppose two strings with di↵erent impedance (⇢1 > ⇢2 impedance mismatch)
are connected together. A wave, incident wave yi , is traveling on string 1 and
passing to string 2. What is reflected wave yr ? transmitted wave yt ?
yi = Ai ei(k1 x
wt
i)
or
yi = Ai ei(k1 x
wt)
allow Ai to be complex. Many statements proved above will not change much, like
(1.8) becomes
1
hIi = T k|A|2 w
2
Similarly use complex amplitude
yr = Ar ei(
k1 x wt)
yt = At ei(k2 x
wt)
Frequencies are the same. It is equal to the vibrator’s frequency regardless of the
sti↵ness of the string. kr = ki because w = kv, since w, v are the same for string
1.
At the interface x = 0, there are two matching conditions: y displacement
@y @y
must be continuous, in QM it means probability is continuous; energy flux T @x
@t
must be continuous, in QM it means flux of particles go in out should be the same,
or
@y
Fy = T
must be continuous
@x
if not, then at x near 0, dm ! 0 so there are 1 acceleration.
13
From the two conditions above
At x = 0,
@(yi + yr )
@x
=
x!0
Since Z ⇠ 1/v ⇠ k
Therefore
Take Z2 ! 1
yi + yr = yt =) Ai + Ar = At
@yt
@x
(Ai
=)
k1 Ai + k1 Ar =
Ar )Z1 = At Z2
Ar
Z1 Z2
At
2Z1
=
and
=
Ai
Z1 + Z2
Ai
Z1 + Z2
vy =
k 2 At
x!0+
(1.11)
Fy2
At
! 0 or
!0
Z2
Ai
no much oscillation transmitted to heavy cable.
Ar
=
Ai
Lecture 4
(9/16/13)
1
total reflection with a ⇡ phase shift.
So for this case, on string 1 we have superposition of two opposite moving
waves, assume = 0
ỹ1 = Ai ei(wt
kx)
Ai ei(wt+kx) = 2iAi eiwt sin kx
Real part
y1 = 2Ai sin wt sin kx
kn l = n⇡ due to fixed 2 ends boundary.
By (1.9), we compute ratio energy reflected
R=
1
Z w2 |Ar |2
2 1
1
Z w2 |Ai |2
2 1
14
=
✓
Z 1 Z2
Z 1 + Z2
◆2
(1.12)
ratio of energy transmitted
1
Z w2
2 2
1
Z w2
2 1
T =
|At |2
|Ai |
2
=
4Z1 Z2
(Z1 + Z2 )2
(1.13)
Check
R+T =1
1.7 Time Evolutions of Strings
Powerful technique: modal decomposition. You will see it is used in QFT, excitation of black holes, and black body radiation, etc.
General solution
y(x, t) =
1
X
An sin
n=1
⇣ n⇡v
n⇡x
cos
t
l
l
n
⌘
y(x, 0) = y0 (x) can be any given shape
8
< 2A x
x < 2l
l
y0 (x) =
: 2A (1 x) x > l
l
2
need another initial condition
vy (x, 0) = v0 (x)
Suppose v0 (x) = 0 8x string start out at rest
y0 (x) =
1
X
An sin
n=1
9 1 many constants An ,
n
to compute. Recall orthogonality
n
2
l
n⇡x
cos
l
ˆ
0
l
sin
n⇡x
m⇡x
sin
dx =
l
l
15
nm
Consider
2
l
ˆ
0
l
N ⇡x
2
y0 (x) sin
dx =
l
l
We will show
N
ˆ lX
1
0
1
X
N ⇡x
dx =
An
n sin
l
n=1
n⇡x
An sin
cos
l
n=1
nN
cos
n
= AN cos
(1.14)
= 0 for any N , so
2
=
l
AN
ˆ
= 2A
l
y0 (x) sin
0
N ⇡x
dx
l
sin N2⇡
N⇡ 2
2
So An = 0 if n is even. One can see that higher mode have smaller amplitude
due to convergent requirement. The lowest frequency, i.e. lowest n gives highest
amplitude.
Why N = 0?
@y(x, t)
= v0
@t
t=0
1
X
An
n=1
Similarly
2
0=
l
ˆ
0
l
N ⇡x
2
dx =
v0 (x) sin
l
l
since AN = 0 is not good,
N
n⇡v
n⇡x
sin
sin
l
l
ˆ lX
1
An
0 n=1
n
= v0
n⇡v
n⇡x
sin
sin
l
l
= 0 for all N .
1.8 Energy Stored in the Mode
From (1.10), we see that energy stored in each mode is
hEn i / A2n
16
n
sin
N ⇡x
AN N ⇡v
dx =
sin
l
l
(1.15)
N
N
and from (1.15), (1.15) one can define
cn
sn
2
=
l
l
n⇡x
y0 (x) sin
dx = An cos n
l
0
ˆ l
2
N ⇡x
=
v0 (x) sin
dx = An sin
n⇡v 0
l
ˆ
n
For the case considered above
A2n = s2n + c2n
The total energy
hEi =
1
X
1
n=1
2
⇢lwn2 A2n
2 Wave Applications
2.1 Transmission lines
The most common type of transmission line is coaxial cable that is made of two
conductors, the code grounded and the shell connecting to power line, or the
other way. Below is a structure diagram of transmission line, the bottom line is
grounded.
L
./
...
./
|
=
|
= C
|
|
./
|
=
|
=
|
|
...
If the code has radius a and shell has radius b, one can show the capacitance
C=
2⇡✏r ✏0
capacitance
=
b
unit length
ln a
Let L be inductance per unit length. Consider one unit of cable, assuming current
17
flowing from left to right is positive direction,
!I(x,t)
!I(x+ x,t)
./
"
V (x, t)
#
|
=
|
"
V (x + x, t)
#
Then V across inductor is
VL = L
@I
@t
so
V (x, t)
V (x + x, t) = L
@I(x, t)
@t
or
V (x + x, t) V (x, t)
=
x
denote L/ x = L be the unit inductance, hence
@V
=
@x
L
L @I
x @t
@I
@t
On the other hand current across the capacitor is
IC =
@
@
Q = CV (x + x, t)
@t
@t
similarly denote C/ x = C be the unit capacitance,
I(x, t)
so
I(x + x, t) = C
@I
=
@x
Therefore
@ 2V
@x2
C
LC
18
@V (x, t)
@t
@V
@t
@ 2V
=0
@t2
(2.1)
p
and same wave equation for I. So v = 1/ LC. One solution is
V = V0 cos(kx
By (2.1)
wt) w = kv
w
I = V0 C cos(kx
k
wt)
suggests
r
V
1
L
Z=
=
=
(2.2)
I
Cv
C
One can also get above expression from the analogy RLC oscillator with a MassSpring oscillator
mass m $ inductance
sti↵ness k $ 1/C capacitance
damping factor $ R
1
1
KE mv 2 $ magnetic energy LI 2
2
2
1 2
Q2
UE kx $ electric energy
2
2C 2
In RHS of (2.2) both quantities are per unit length, and they cancel out, so
later when we do resistors, we put
Z=R
make no reference to unit length.
One can get impedance of free space
Z0 =
r
µ0
= 377⌦
✏0
and impedance of dielectric medium by arguing that
v=p
1
c
c
=p =
µ0 ✏ 0 ✏ r
✏r
n
19
define index of refraction
n=
then
Z=
r
p
✏r
µ0
Z0
=
✏r ✏0
n
(2.3)
For completeness what can we say about reflection ratio R and transmitted
ratio T when light is passing through interface of two mediums, n1 , n2 normally
(require normality because we derived R and T from 1D wave.)
From (2.3), (1.12) and (1.13)
R=
✓
n2 n1
n2 + n1
◆2
T =
4n1 n2
(n1 + n2 )2
2.2 Impedance Matching
Given two transmission lines with impedance Z1 , Z3 . Is it possible to insert a line
l, Z2 in between Z1 , Z3 so that 100% energy transfer from Z1 to Z3 ? Yes. The idea
is to have destructive interference.
Let the interface of Z1 and Z2 be x = 0
in line Z1
in line Z2
in line Z3
i(k1 x wt)
i(k2 x wt)
moving to right A1 e
A2 e
A3 ei(k3 (x l) wt)
moving to left B1 ei( k1 x wt) B2 ei( k2 x wt)
Lecture 5
(9/18/13)
A1 , A2 , A3 , B1 and B2 can be complex. Factor out extra term eik3 l from A3 makes
calculation easier.
x=0
A1 + B1 = A2 + B2
@y
T @x
is continuous
Z1 (A1
B1 ) = Z2 (A2
B2 )
x=l
A2 eik2 l + B2 e
20
ik2 l
= A3
Z2 (A2 eik2 l
B2 e
ik2 l
) = Z3 A3
Four equations for A1 , A2 , A3 , B1 and B2 , 5 unknowns, want
T =
1
Z w2
2 3
1
Z w2
2 1
|A3 |2
|A1 |2
=1
(2.4)
One possible solution is to choose
⇡
2
() l =
2
4
k2 l =
absorb i into A3 , the last two of the four equations simplify to
8
<A
2
B 2 = A3
:Z (A + B ) = Z A
2
2
2
3 3
then
2A2 = (1 +
2B2 = (
2A1 = (1 +
4A1 =
✓
(1 +
Z3
)A3
Z2
Z3
Z2
1)A3
Z2
)A2 + (1
Z1
Z2
Z3
)(1 + ) + (1
Z1
Z2
Z2 Z 3
)(
Z1 Z 2
Z2
)B3
Z1
◆
✓
◆
Z2 Z 3
1) A3 = 2
+
A3
Z1 Z 2
By (2.4)
Z3
1
=
Z1
4
✓
Z2 Z3
+
Z1 Z2
◆2
p
p
Z22 + Z1 Z3
=) 2 Z1 Z3 =
=) Z22 2Z2 Z1 Z3 +Z1 Z3 = 0
Z2
Therefore
Z2 =
p
Z 1 Z3
(2.5)
This is known as matching impedance by a quarter wave transformer. There is
a analogy in QM, called Ramsauer-Townsend e↵ect, that particles passing through
21
some inert gas without any reflection.
Clearly (2.5) works for strings too
⇢2 =
p
⇢1 ⇢3
2.3 Loaded Transmission Line
Suppose a coaxial cable Z is connected to a source at x = 1 with frequency w
and it is loaded to a resistor R = ZL at x = 0. So the voltage pulse traveling down
the cable to the device. Let
V+ = voltage pulse traveling to the right
V
= voltage pulse reflected back to the left
I+ = current pulse traveling to the right
I
= current pulse reflected back to the left
Suppose I+ = I0+ cos(wt
kx), I = I0 cos(wt + kx), we know
V+ = ZI+
V =
ZI
At x = 0
ZL (I+ + I ) = V+ + V = Z(I+
hence
I )
I0
Z ZL
=
I0+
Z + ZL
So the ratio of power absorbed by the load to the power sent down the line is
T =1
✓
I0
I0+
◆2
=
4ZZL
(Z + ZL )2
What happen if we short the circuit, ZL = 0?
I0 = I0+
22
We will see V lagging I by 900 , commonly seen if you work with oscilloscopes.
Ṽ = V0 ei(wt
I˜ = I0 ei(wt
kx)
kx)
V0 ei(wt+kx) =
2iV0 eiwt sin kx
+ I0 ei(wt+kx) = 2I0 eiwt cos kx
<Ṽ = V (x, t) = 2V0 sin kx sin wt
=I˜ = I(x, t) = 2I0 cos kx cos wt
What happen if we put ZL = Z?
Then
I =0
no reflection. All energy goes down the line. From power source perspective, it is
generating energy loaded to a 1 long cable.
The idea above is the starting point of microchannel plates (MCPs), which can
capture sequential images of charged particle events at very high speeds. MCP is
a slab with a regular array of tiny slots. Slots are parallel to each other. A particle
or photon that enters one of the channels is guaranteed to reach the other end,
because the load
ZL = Z
Cascade of electrons that propagates through the channel amplifies the original
signal by several orders of magnitude. MCPs were widely used in magnetic plasma
(nuclear weapon) and laser plasma (e.g. take a snapshot of the plasma every 1
picosecond see how it cools down)
2.4 2D, 3D Waves
Plane Wave
Lecture 6
(9/23/13)
For example electric filed propagating in x̂ direction and vibrating ŷ direction
Ey = y0 cos(wt
kx)
How to describe a plane wave traveling along not just x̂ but n̂ with wave
23
length ?
Clearly n̂ is orthogonal to the wave front. Let ~r be some arbitrary position
point vector. Let d~ be the shortest distance from the origin to the wave front that
~r ends. Let
~k = 2⇡ n̂
then
~k · ~r = kd ⌘
phase di↵erence between the sheet where ~r ends and the phase at the origin,
indeed because if d = , kd = 2⇡ no phase di↵erence. Then assuming phase at
the origin is wt, thus
~ = ~0 cos(wt ~k · ~r)
(2.6)
For those mathematical less picky, can see (2.6) is immediately true. Because
plane wave is independent of coordinate, choose x̂ along k̂ and ŷ along ~0 then
kx) has to become (2.6).
0 cos(wt
We can work backward to see what kind of 3D wave equation that will give
solution (2.6). That is
@x2 + @y2 + @z2
(~r, t)
1 2
@ (~r, t) = 0
v2 t
or 2D wave equation
@x2 + @y2
(~r, t)
1 2
@ (~r, t) = 0
v2 t
where ~r is a 2D vector.
Cylindrical Wave
Write Laplace in cylindrical
@2
@2
1@
1 @2
+ 2 +
+
@z 2
@⇢
⇢ @⇢ ⇢2 @✓2
24
1 @2
=0
v 2 @t2
(2.7)
for most application
is independent of z and ✓, so we get
@2
1@
+
2
@⇢
⇢ @⇢
1 @2
=0
v 2 @t2
(2.8)
one approximation solution is
0
(⇢, t) = p cos(wt
⇢
Ignoring order
@2
@⇢2
1@
⇢ @⇢
!
!
1
,
⇢↵
k⇢)
↵ > 2, because the order of pde (2.8) is 2
k 2 cos(wt k⇢) 3 cos(wt k⇢)
+
p
⇢
4
⇢5/2
cos(wt k⇢) k sin(wt k⇢)
+
2⇢5/2
⇢3/2
k⇢)
Therefore if we ignore cos(wt
,
4⇢5/2
p
1/ ⇢ in the amplitude is that
p0
⇢
ˆ
⇢
cos(wt
✓
0
p
⇢
◆2
k sin(wt k⇢)
⇢3/2
k⇢) satisfies (2.8). The reason of
d✓
the energy carried by the wave crest should be conversed.
Spherical Wave
✓
◆
✓
◆
1 @
1
@
@
1
@2
2@
r
+
sin
✓
+
r2 @r
@r
r2 sin ✓ @✓
@✓
r2 sin2 ✓ @ 2
If assume
is spherical symmetric,
@2
2@
+
2
@r
r @r
1 @2
=0
v 2 @t2
= (r, t), we only need to solve
1 @2
=0
v 2 @t2
(2.9)
compare this to (2.8), there is a 2 in front of @ /@r. That is because (2.8) is
circular symmetric, then in general spherical coordinates in N dimensions Laplace
25
is
@2
N 1 @
1 @
=
+
+ ... = N 1
2
@r
r @r
r
@r
✓
r
N 1
@
@r
◆
+ ...
One particular solution to (2.9) is
=
0
r
cos(wt
kr)
Indeed
2@
r @r
@2
@r2
2k sin(wt kr) 2 cos(wt kr)
r2
r3
2
k cos(wt kr) 2 cos(wt kr)
!
+
r
r3
!
2k sin(wt
r2
kr)
The 1/r factor in the amplitude makes total energy flux on a sphere conserved
4⇡r
2
✓
0
r
◆2
2.5 Elastic Sheets
The technique of solving higher dimensional wave leads to QM.
Consider a rectangular 2D sheet a long b wide, clump all sides. We solve
(2.7), use separation of variables
= X(x)Y (y)T (t)
plugging into (2.7), and dividing
out
1 d2
1 d2
1 1 d2
X+
Y
=
T
X dx
Y dy 2
v 2 T dt2
RHS is a function of t, while LHS is a function of x, y. Since x, y, t are
independent variables, RHS and LHS must be constants, denoted k 2 (the reason
26
to put negative constant is to have periodic solutions). Put
1 d2
T =
T dt2
k2v2 =
w2
Or
d2
T = w2 T
2
dt
which is a typical eigenvalue equation. It has eigen solution
T ⇠ e±iwt
Similarly
8
1 d2
>
>
X = k12
>
X
< dx
1 d2
Y = k22
Y
dy 2
>
>
>
:k 2 + k 2 = k 2
1
=)
2
X ⇠ e±ik1 x
Y ⇠ e±ik2 y
(2.10)
Boundary conditions
(0, y, t) = (a, y, t) = 0
(x, 0, t) = (x, b, t) = 0
Thus
nm (x, y, t)
k1 =
Lecture 7
(9/25/13)
n⇡
a
=
sin k1 x sin k2 y cos(wt
)
r⇣ ⌘
m⇡
n⇡ 2 ⇣ m⇡ ⌘2
k2 =
wnm =
+
v
b
a
b
0
(n, m) mode number. The energy of the sheet is just like before (1.10) / w2 , so
the lowest energy state is w11 , w12, w21 , or ... depending on how big a, b are. If
a = b, then the lowest energy state is w11 , and for other energy levels there may
be many degenerated states.
We have seen that (2.10) gives continuous available k’s and the boundary
conditions impose discrete allowed k’s.
Another example: circular sheet
Suppose a sheet of radius a is clumped along the edge. Clearly we have
27
circular symmetry so use polar coordinate, we solve (2.8). Separation of variable
(⇢, t) = ˜(⇢) cos wt
Plugging into (2.8)
d2 ˜ 1 d ˜
+
=
d⇢2
⇢ d⇢
Change variable z =
w2 ˜
v2
w
⇢
v
d2 ˜ 1 d ˜ ˜
+
+ =0
dz 2
z dz
This is Bessel’s equation of 0th order. If we have ✓ dependence, we will get higher
order Bessel equations. The solutions are Bessel function
˜(z) = J0 (z) =
ˆ
⇡
cos(z sin ✓)d✓
0
Thus
w
(⇢, t) = J0 ( ⇢) cos(wt
v
The boundary condition implies w is discrete,
)
w
a = zj
v
where zj ’s are the jth zeros of Bessel function. Therefore
j (⇢, t)
= J0 (
zj
⇢) cos(wt
a
)
2.6 Wave Guides
Consider a 1 long strip of width b is clumped at two sides y = 0 and y = b.
Suppose a wave is propagating in the sheet with
˜i ⇠ ei(wt
28
~ki ·~
r)
or
i (x, y, t)
= < ˜i ⇠ cos(wt
k1 x
k2 y)
(2.11)
~ki = (k1 , k2 )
once it reaches the other side, it gets reflected
r (x, y, t)
⇠ cos(wt
k1 x + k2 y)
(2.12)
~kr = (k1 , k2 )
so the total wave propagating in the sheet
˜(~r, t) = A1 ei(wt
~ki ·~
r)
+ A2 ei(wt
~kr ·~
r)
One of the boundary conditions says
˜(0, t) = 0 =) A1 =
A2
So
˜(x, y, t) = Aeiwt e
ik1 x
e
ik2 y
e
ik1 x ik2 y
e
=
A2iei(wt
k1 x)
sin k2 y
By the other boundary condition
k2 =
n⇡
b
so
n⇡
y
b
We get standing wave in y direction, traveling wave in x direction.
For given w, k12 + k22 = w2 /v 2
n (x, y, t)
= < ˜ = 2An sin(wt
r⇣ ⌘
w 2
k1 =
v
k1 x) sin
⇣ n⇡ ⌘2
b
(2.13)
(2.14)
Compare (2.11) and (2.13). They have two di↵erent wave length. The incident
29
wave has wave length = 2⇡
and it is impinging to the side at angle ✓; the wave
k
guide is traveling along x has wave length 1 = 2⇡
. Clearly
k1
k1 = k sin ✓
so
1
=
sin ✓
1 is called the guide wavelength. We know the velocity of prorogation v = w/k
is determined by the vibrating material. And v is derived from either (2.11) or
(2.12).
But there are two more concepts of velocity: phase velocity and group velocity.
Phase velocity in this case is the actual velocity of the moving crest of the
total wave guide (2.13), so
vp =
w
w 1
v
=
=
>v
k1
k sin ✓
sin ✓
and group velocity is the actual velocity that carries energy. Since the energy is
carried by either (2.11) or (2.12) the e↵ective velocity of the moving energy is only
the x component of speed of (2.11), i.e.
vg = v sin ✓ < c
From this specific case, we get some general equations that are always true
w
ki
dw
=
dki
(vp )i =
(vg )i
Indeed if we are interested in x direction
vgx =
Lecture 8
dw
d
d
=
vk =
vk1 sin ✓ = v sin ✓
dk1
dk1
dk1
We will revisit group and phase velocities later.
Another interesting feature of (2.14) is that there is a cuto↵ frequency,
(10/7/13)
30
w
n⇡
v
b
If w < ⇡v
, k1 is imaginary, the wave will die out. We will see that in the case EM
b
wave guide passing in metal sheet.
2.7 Geometric Optics: Snell’s Law
We will show Snell’s law on the base that lights are EM waves.
(Griffiths Problem 9.16) Suppose the coordinate are chosen so that x goes up
z goes right and y out of paper. Let z < 0 be filled with dielectric medium n1 and
~ k ŷ, B
~ lie in the plane of xz. So
z > 0 filled with n2 . Send in plane light wave E
~ki lie in the plane of xz. Suppose ~ki makes angle ✓i with the normal axis, z axis.
~ wave
Consider B
8
< ei(wt ~ki ·~r) + e(wt ~kr ·~r) z < 0
i
r
=
: ei(wt ~kt ·~r)
z>0
t
~ wave to analyze because
where i , r and t are complex vectors. We choose B
~ is continuous across the boundary, while as continuity for E
~ wave depends on
B
whether charges built up at the interface. Then at z = 0
ie
i(wt ~ki ·~
r)
+
re
i(wt ~kr ·~
r)
=
te
i(wt ~kt ·~
r)
for all t,x, and y. This can be true only if
8
<k = k = k
ix
rx
tx
:k = k = k
iy
ry
ty
Since kiy = 0, reflection and refraction happen in the same xz plane.
Because k = wv = wc n, |~ki | = |~kr |, then
kix = krx =) sin ✓i = sin ✓r
31
(2.15)
and |~ki |/n1 = |~kt |/n2
kix = ktx =) n1 sin ✓i = n2 sin ✓t
we derived Snell’s law.
We will continue to derive some relations between the amplitudes which will
~ field too
be useful later. We need to look at E
8
<E = Ẽ ei(wt ~ki ·~r) ŷ
i
0i
:B = 1 Ẽ ei(wt ~ki ·~r) ( cos ✓ x̂ + sin ✓ ẑ)
i
i
i
v1 0i
8
<E = Ẽ ei(wt ~kr ·~r) ŷ
r
0r
:B = 1 Ẽ ei(wt ~kr ·~r) (cos ✓ x̂ + sin ✓ ẑ)
r
i
i
v1 0r
8
<E = Ẽ ei(wt ~kt ·~r) ŷ
t
0t
:B = 1 Ẽ ei(wt ~kt ·~r) ( cos ✓ x̂ + sin ✓ ẑ)
t
v2
0t
t
t
Maxwell implies two boundary conditions
k
k
k
Ez!0 = Ez!0+
k
Bz!0 = Bz!0+
then
Ẽ0i + Ẽ0r = Ẽ0t
1
1
1
Ẽ0i ( cos ✓i ) + Ẽ0r cos ✓i = Ẽ0t ( cos ✓t )
v1
v1
v2
Therefore we find
Ẽ0r =
1 ↵
Ẽ0i
1+↵
cos ✓t n2
↵=
=
cos ✓i n1
Ẽ0t =
2
Ẽ0i
1+↵
p
(n2 /n1 )2 sin2 ✓i
cos ✓i
Since ↵ is always positive, Ẽ0t is always in phase with Ẽ0i .
If n2 > n1 ,
(n2 /n1 )2 > 1 = sin2 ✓i + cos2 ✓i =) ↵ > 1
32
(2.16)
Hence incident light from n1 reflected at n2 with n1 < n2 will gain an extra ⇡
phase. If n2 < n1 , no extra phase. This result is the generalization of the 1D case
we studied before (1.11).
2.8 Group Velocity Revisit
Combine two 1D waves of same amplitude and similar frequencies
1
Let
w̄ =
= a cos(w1 t
w1 + w2
2
w=
k1 x)
w1
2
w2
= a cos(w2 t
k2 x)
k 1 + k2
2
k=
k̄x) cos( wt
kx)
k̄ =
2
k1
k2
2
then
1
+
2
= 2a cos(w̄t
(2.17)
If w1 w̄ w2 = 2w̄w ⌧ 1, we get phenomena of beats. The graph y v.s. t for fixed x
has envelope given by cos( wt
kx) and oscillation inside given by cos(w̄t k̄x).
The beat frequency we hear is twice of the envelope frequency w.
Tbeat =
1
⌫2 |
2|⌫1
We find the motion of the envelope is
vg =
=
w
k
w1
k1
(2.18)
w2
=v
k2
because w = kv, linear dispersion relation. So is the phase velocity
vp =
w̄
=v
k̄
This means that as the wave moves both the oscillation pattern and the envelope
move at the same speed, thus the sharp of the wave stays the same. This will not
be the case if w = w(k) 6= kv, non-linear, vg = v = vp is not likely. Later we will
33
see another example that as wave moves it attenuates (2.23).
We can generalize (2.18) to
vg =
dw
dk
Suppose n = n(w)
Lecture 9
k=
(10/9/13)
then
n(w)w
c
dk
n w dn
= +
dw
c
c dw
Since vp = c/n,
vg =
c
dw
vp
n
=
=
w
dn
dn
dk
1 + n dw
1 + wn dw
(2.19)
In vacuum n = 1, dn/dw = 0, so for EM wave vg = vp .
Continue (2.19), for normal dispersion
dn
> 0 =) vg < vp
dw
For anomalous dispersion
dn
< 0 =) vg > vp
dw
Light through prism has spectrum: red, orange, yellow, green, blue. If prism
is made of anomalous glass, the spectrum is reversed.
If n = n( ), from (2.19),
dn
dn d
=
dw
d dw
2⇡c
wn
2⇡c
=
w2 n
=
d
=
dw
w
so (2.19) becomes
vg =
vp
1
dn
nd
⇡ vp (1 +
34
dn
)
nd
2.9 Dispersive Waves: Plasma Wave
We now study the radiation of plasma due to excitation by the external field.
~
Think electrons are bounded to the nuclei like a string k and under external E
field, e are making oscillations. Let m be the mass of e , x be the position
coordinate, then
P = qx
is the dipole momentum. The intensity of radiation
I / (P̈ )2
First we want to show that the relative permittivity ✏r of the material that
made of molecule with density n = #moles/volume, and dipole momentum of a
molecule P is
nP
✏r = 1 +
(2.20)
E✏0
One way to derive the equation above is to fill such material into a parallel
planes capacitor with separation d and area of two planes A. Then the electric
field inside is
Ef
f
Enet = Ef ree Ebound =
=
✏r ✏0
✏r
where
f
is charge density that disposed on the planes. Then
Ebound = (✏r
1)Ef =
b
✏0
(2.21)
The bound charges are polarized by the free charges, and their e↵ects are canceled
in the internal of the capacitor, only those populated near the planes contribute to
Ebound . They makes a thin layer of width db , which is half of the distance between
+ and poles of the polarized molecule, because the another half of the charges
are immersed in the internal and hence canceled. Thus
b
=
qb
qn2Adb
=
= nq2db = nP
A
A
(2.22)
q2db = P , dipole momentum. Combining (2.21) and (2.22) gives (2.20).
Now back to the situation we discussed in the beginning of this section. Ex-
35
ternal generating field E = E0 eiwt acts on plasma, equation of motion for one
e
m
mẍ =
ẋ kx + qE
T
we add a damping force term m
ẋ, because without it resonance goes to 1.
T
Suppose we look for wave solution
X̃ = X0 eiwt
p
put w0 = k/m, natural frequency, in many cases it is the quantum transition
frequency. Then
✓
◆
iw
q
2
w X0 + X0 + w0 X0 eiwt = E0 eiwt
T
m
2
qE0 /m
w2 + w02 +
X0 =
iw
T
X0 is the average displacement, plugging into (2.20), with P = qX0
✏r = 1 +
nq 2
m✏0
w2 + w02 +
define plasma frequency, wp
nq 2
m✏0
wp2 =
then
✏r = 1 +
and
ñ =
p
wp2 (w02
(w2
| 0
w2 )
w 2 )2 +
iw
T
w2
T2
i
{z
1 wp2 (w02 w2 )
✏r = 1 +
2
2 (w02 w2 )2 + w
T2
|
{z
}
nR
36
wp2 w
T
(w02
w 2 )2 +
1
i
2 (w2
| 0
w2
T2
wp2 w
T
}
2
w 2 )2 + w
T2
{z
}
nI
Now we see that the external generating field E = E0 eiwt becomes wave
E = E0 ei(wt
= E0 ei(wt
w
nx)
c
w
n x)
c R
e
w
n x
c I
(2.23)
The graph of =ñ show that it has resonance near w = w0 , hence the wave die out
w2
fastest when w = w0 . One can graph <(ñ 1), at w = 0, <(ñ 1) = 2p then
q
it goes up reaches its peak at w = w02 wT0 , then falls down <(ñ 1) = 0 at
q
w = w0 , then continues down reaches bottom at w = w02 + wT0 , then up but stays
q
q
negative and asymptotically approaches 0. Hence w 2 ( w02 wT0 , w02 + wT0 )
dnR
<0
dw
anomalous dispersion.
3 Semi-Classical Waves
3.1 Ultraviolet Catastrophe
1900 there were 2 clouds, ultraviolet catastrophe from black body radiation and
undetectable aether from Michelson-Morley experiment.
(10/14/13)
Consider a conductor square box with side l. Since E = 0 on the faces of the
~ field inside
box, the E
Lecture 10
En1 ,n2 ,n3 (x, y, z, y) = E0n1 ,n2 ,n3 sin
n1 ⇡x
n2 ⇡y
n3 ⇡z
sin
sin
sin wt
l
l
l
r⇣
n 1 ⇡ ⌘ 2 ⇣ n 2 ⇡ ⌘ 2 ⇣ n 3 ⇡ ⌘2
w = kc = c
+
+
l
l
l
r⇣
w
cn1 ⌘2 ⇣ cn2 ⌘2 ⇣ cn3 ⌘2
⌫=
=
+
+
2⇡
2l
2l
2l
Typical value l ⇠ 1m, c = 3 ⇥ 108 m/s, n ⇠ 107 , so
⌫ ⇠ 107
37
(3.1)
adjacent frequency are closed spaced
c
⌫
= cl = 10
⌫
n
l
7
This allows to calculate things in continuum limit. In the modal space, x̂ axis cn2l1 ,
ŷ axis cn2l2 , and ẑ axis cn2l3 . A small cube encloses one discrete dot in the modal
space, the volume of the cube
⇣ c ⌘3
c3
=
2l
8V
and the volume in the modal space from frequency ⌫ to ⌫ + d⌫ is
1
4⇡⌫ 2 d⌫
8
So the number of states from ⌫ to ⌫ + d⌫ is
1
4⇡⌫ 2 d⌫
8
c3
8V
=V
4⇡⌫ 2 d⌫
= dN
c3
(3.2)
If we divide dN/V = dn, we get number of states per volume from ⌫ to ⌫ + d⌫.
Equipartition theorem from statistical mechanics says each degree of freedom
of the system at equilibrium temperature T gets kT /2 of the total energy of the
system. k is Boltzmann constant.
E.g. SHO
p2
1
+ kx2
2m 2
has two degree of freedom
kT
KE = U E =
2
E.g. monatomic atom, non interacting gas, no potential
1
3
E = N m(vx2 + vy2 + vz2 ) = kT N
2
2
In our case the box is at temperature T . The atoms and molecules of the conductor vibrating, and radiation are emitted, The populated radiation are standing
38
waves. Each mode has energy
Energyn1 ,n2 ,n3 / E02n1 ,n2 ,n3 + B02n1 ,n2 ,n3
Treat each mode (n1 , n2 , n3 ) as SHO, so each has
kT
2
2⇥2⇥
one 2 is from E and B fields, and the other 2 is from polarization, because the
amplitude in (3.1) has two polarization directions.
So the total energy per volume from frequency 0 to 1 is
U=
ˆ
1
ˆ
u(⌫)d⌫ =
0
1
(2kT )
0
4⇡⌫ 2 d⌫
=1
c3
(3.3)
This is UV catastrophe. It says even before the microwave oven is on, it contains
1 radiation.
Planck had the experimental curve u(⌫) v.s. ⌫. For small ⌫,
u(⌫) ⇠ ⌫ 2
agrees (3.3). For large ⌫
u(⌫) ⇠ ⌫ 3 e
h⌫/kT
suggested by Wien. So it is called Wien tail.
h = 6.67 ⇥ 10
27
erg·s
is Planck constant. Planck said “I am going to fit the 2 ends”
u(⌫) ⇠
⌫3
eh⌫/kT
1
(3.4)
and his explanation was that equipartition assumes E ⇠ kT , since T changes
continuously, E changes continuously. Planck said No. Only
nh⌫
n = 1, 2, 3, ...
39
Lecture 11
goes into each mode.
Now by (3.4)
U=
(10/16/13)
ˆ
1
0
u(⌫)d⌫ < 1
Planck said that counting in (3.2) showed that higher ⌫, higher E, and most
states are available, but states with high v.s low energy are not equally likely being
occupied. Planck said the probability of occupying states with energy E is
P (E) = N e
E/kT
(3.5)
Planck distribution. N is normalization
ˆ 1
1
P (E)dE = N kT = 1 =) N =
kT
0
The reason Planck made such guess (3.5) because
hEi =
1
ˆ
EP (E)dE = kT
(3.6)
0
agrees classical result for SHO. The important feature about (3.5) is that the lowest
energy has the highest probability being occupied.
For the blackbody radiation, in addition, Planck said energy was quantized
Em = mh⌫
for each mth mode. So Planck distribution is
P (Em ) = N e
mh⌫
kT
Find normalization
1=
X
m
P (Em ) =
1
X
Ne
mh⌫
kT
=N
m=0
That is
P (Em ) = (1
e
40
h⌫
kT
)e
mh⌫
kT
1
1
e
h⌫
kT
One can show that similar thing happens to (3.6) when T is large, so we
return to classical result.
hEi =
@
@↵
kT
=
lim
@
kT
@↵
h⌫e
h⌫
kT
↵!1
=
1
=
e
h⌫
h⌫
!
↵=1
↵
X
✓
1
mh⌫
↵
kT
e
1
e
h⌫
kT
(1
◆
↵
e
(1
h⌫
kT
e
)
h⌫
kT
)
h⌫
kT
e kT
1
h⌫
1+
h⌫
kT
(3.7)
1
= kT
We can now correctly count number of states with proper weight factor
(Planck distribution) and the correct energy levels (not kT , but Em ), back to
(3.3)
u(⌫)d⌫ =
=
8⇡⌫ 2 d⌫ X
Em P (Em )
c3
m
8⇡⌫ 2 d⌫ h⌫
8⇡h
⌫3
=
h⌫
c3 e kT
c3 eh⌫/kT
1
1
d⌫
This is still semi-classical argument, stating from string and imposing quantization.
Modern interpretation of (3.7) is that h⌫ is photon energy and h⌫1 is the
e kT 1
number of photons. This is first time for us to think waves as particles. Later we
will treat particles as waves. It was easily accepted by 19 century hero’s that EM
waves as particles, but many of them (including Einstein) didn’t fully embrace
that e.g. e are waves.
41
3.2 Fourier Analysis
The idea of particles are waves builds on forming wave packets. Suppose we add
N waves each has frequency w di↵erence from each other
R=
N
X1
a cos[(w + n w)t]
n=0
and
N w=
w = ”Bandwidth”
di↵erence between the upper and lower frequencies in a continuous set of frequencies, is so narrow that
w
⌧1
(3.8)
w
One can show that
R̃ = a
X
e
iwt in wt
e
= ae
N wt
i 2
sin N 2wt
eiN wt
iwt e
=
ae
wt
1 ei wt
ei 2 sin wt
iwt 1
2
so
R⇡a
sin
wt
2
wt
2
cos w̄ = aN
sin
wt
2
wt
2
cos w̄t
1
w̄ = w + (N 1) w
2
which is about the middle of the upper and lower frequencies. Because of (3.8),
sin 2wt
is the envelope similar to (2.17) and if we put R into a traveling wave
wt
2
R=
N
X1
a cos[(w + n w)(t
n=0
w(t
2
sin
x
)] = aN
w(t
v
x
)
v
x
)
v
cos w̄(t
2
x
)
v
(3.9)
we have a moving envelope.
Above example motivates Fourier.
Recall we can write any function f (x) with periodicity L
1
a0 X
2⇡nx
2⇡nx
f (x) =
+
an cos
+ bn sin
2
L
L
n=1
42
(3.10)
Similar we can write any function f (t) with periodicity T
f (t) =
1
a0 X
2⇡nt
2⇡nt
+
an cos
+ bn sin
2
T
T
n=1
Let
wn =
so
2⇡n
= nw
T
1
a0 X
f (t) =
+
an cos wn t + bn sin wn t
2
n=1
As showed before, the energy stored in the nth mode is / a2n + b2n .
We can use complex notation
1
X
f (t) =
iwn t
dn e
(3.11)
n= 1
with the identifications
an = d n + d
bn = i(dn
or
8
<d =
n
:d =
n
and
n
d
an ibn
2
a|n| +ib|n|
2
n
n)
0
n<0
En / a2n + b2n = 4|dn |2
and
1
f˜(w) = dn =
T
T /2
ˆ
f (t)eiwn t dt
T /2
hence the Fourier transform of f (t), because
ˆ
T /2
e
iwn t iwm t
e
=
T /2
43
8
<0
:T
n 6= m
n=m
(3.12)
Similarly we can write (3.10)
f (x) =
=
1
a0 X
+
an cos kn x + bn sin kn x
2
n=1
1
X
dn eikn x
(3.13)
n= 1
where
1
f˜(k) = dn =
L
ˆ
L/2
ikn x
f (x)e
dx
L/2
hence the Fourier transform of f (x).
What if kn or wn in (3.13) or (3.11) change continuously, so the summaLecture 12
tion has to become integral. For this to converge, there are some mathematical
(10/21/13)
requirements for f but we won’t care about them.
Take (3.11) for example,
wn =
2⇡n
= w0 n
T
now label w0 n ! w a continuous variable.
dw = wn+1
We guess
wn = w0 =
1
1 X
f (t) = lim
T dn e
|{z}
T !1 2⇡
n= 1 ˜
| {z
} f (w)
´
8
<f (t) =
:f˜(w) =
1
2⇡
´1
´1
1
f˜(w)e
44
iwt
iwt
f (t)eiwt dt
1
2⇡
T
2⇡
T
|{z}
dw
dw
(3.14)
Let’s verify
¨ 1
1
0
f˜(w) =
e iwt f˜(w0 )eiw t dw0 dt
2⇡
1
ˆ 1
ˆ T
0
0 1
=
dw
lim
dtei(w w)t f˜(w0 )
2⇡ T !1 T
1
|
{z
}
2 sin[(w0 w)T ]
w0 w
Change of variable y = (w0
w)T
✓
◆
ˆ 1
2
sin y
y
dy
lim f˜( + w)
T !1
2⇡ 1
y
T
✓
◆
ˆ 1
2
sin y ˜
=
dy
f (w)
2⇡ 1
y
|
{z
}
f˜(w) =
⇡
= f˜(w)
Indeed (3.14) is right. One can simplify the derivation above by associating
1 sin[(w0
T !1 ⇡
w0
lim
w)T ]
= (w0
w
w)
One application of Fourier in electronics is the amplifier.
A(w)Vin (t) = Vout (t)
Since A(w) is multiplied to the frequency components due to the mechanism of
the amplifier, we first Fourier Vin in frequency space, then multiply A(w), then
inverse Fourier back to Vout (t). Suppose Vin is a square wave, we know from Fourier
analysis, the Fourier of square is made of a few sine and cosines and a cheap sound
system that only picks up the first few leading terms doesn’t resemble or duplicate
Vin very well, if Vout happens to have a lot of sharp edges.
45
3.3 Bandwidth Theorem
If one studies atom spontaneous or stimulated transitions radiations, the theory
predicts that transition occurs at a specific frequency w0 , hence I(w), the detected
intensity of radiations due to that transition should a spark or (w0 ), but actually
one gets a Gaussian like curve around w0 with some width. Why?
That is because physically no radiation goes on forever. If radiation E(t) =
E0 cos w0 t, t 2 ( 1, 1), then Fourier of E(t) would indeed be (w0 ). But in
reality each atom only radiates in a finite interval ⌧ , called characteristic time.
Typical ⌧ ⇠ 10 9 s, ⌫0 ⇠ 1017 s 1 .
E(w) =
ˆ
⌧ /2
E0 cos w0 teiwt dt
⌧ /2
⌧ /2
eiw0 t + e iw0 t iwt
E0
e dt
2
⌧ /2
⇢
sin[(w0 + w) ⌧2 ] sin[(w0 w) ⌧2 ]
= E0 ⌧
+
(w0 + w) ⌧2
(w0 w) ⌧2
=
ˆ
(3.15)
which is made of two broad curves around ±w0 . Normally one is only interested
in positive frequency. But the two curves are identical, meaning that if we are
interested in energy flux, energy per unit volume per time pass unit area
1
F = ✏0 E 2
2
we only have to calculate it using the positive w0
F (w) / (E0 ⌧ )
or
F (w)
=
F (w0 )
2
⇢
⇢
sin[(w0 w) ⌧2 ]
(w0 w) ⌧2
sin[(w0 w) ⌧2 ]
(w0 w) ⌧2
2
2
because recall limx!0 sin x/x = 1.
Observe (3.15), the first zero next to the peak w0 is at
(w0
w)
46
⌧
=⇡
2
so the width of the curve is
w=
2⇡
⌧
this is called Bandwidth theorem.
Application: we know Laser has very narrow frequency. So it has a metastable
state, whose transition time is very long. In our words, it oscillates for long time.
Another application: student cheapo oscilloscope. Bandwidth ⌫ = 100MHz,
then
1
⌧=
= 10 8 s = 10ns
⌫
so it cannot detect oscillations that last less than 10ns. Good oscilloscope, Bandwidth ⌫ = 1GHz, ⌧ = 1nsec.
Similarly one can get Bandwidth theorem in spatial space. From
w=
kv
multiplied by ⌧
⌧| {zw} =
k |{z}
v⌧
2⇡
Get uncertainty principle in QM
x
x k = 2⇡
Since
k=
2⇡
we get
2
2
x=
called coherence length, which is the length of the coherent radiation spreads. We
will use it in optics.
We have seen that we can freely do Fourier in time space or spatial space. Both
Lecture 13
give Bandwidth theorem and the connection between the two spaces is dispersion
(10/23/13)
relation.
47
3.4 Incoherence v.s Coherence
Consider a focusing lens that takes N light rays (think this as dividing a light
beam into N parallel rays) into a point. Suppose the light rays are from light
bulbs, i.e. incoherent sources
Etotal =
N
X
E0 ei(wt
kx+
r)
= E0 ei(wt
kx)
r=1
r
=
E02 [
X
r
+ i sin
2
r + cos
r
r
is roughly equally distributed.
r )(cos
s
(sin2
ei
r
E0 is the same by equipartition theorem.
I / |Etotal |2
XX
= E02
(cos
X
r) +
r=s
X
i sin
r
(cos
r
sin
r6=s
|
= N E02
r)
s
+ sin
{z
r
cos
s )]
(3.16)
}
0
The zero is because first fix r, then sum s, there are equal number positive and
negative s except r so
X
cos
X
sin
r
sin
s
=
r
cos
s
=
X
(cos
r
r
sin
r
X
sin
s
cos
s
s
r6=s
so
cos
r
r6=s
similarly
X
sin
s
+ sin
r
cos
s)
=
X
sin 2
r
=0
r
r6=s
The result (3.16) i.e.
I = N I0
shows no interference, if the sources is incoherence, because I0 is 1/N of intensity
of the total light before entering the lens.
48
If the source is coherence, above calculation will give
If ocused / N 2 E02
or
If ocused = N 2 I0
(3.17)
(3.17) doesn’t imply that if our initial choice of N was 1, then If ocused would
be 1. N is limited by the fact that if N is too big, then our construction would
allow even parallel rays to interfere each other, so the initial light intensity would
be intensified by our method of calculation. Below example shows the properness
of choosing N .
Consider coherence source sending parallel rays to a screen and focusing lens
focuses rays into a point. We put a glass index of refraction n, thinness l, onto
the path of half of the rays in front of the lens. So we intentionally create optical
path di↵erence (note: like example before the lenses themselves are designed so
that rays get focused without creating any optical path di↵erence.)
= k1 l
k2 l =
2⇡
(n
1)l
0
k1 =
wn
c
k2 = k0 = wave number in vacuum
or
1
=
0
n
=
0
= wavelength in vacuum
Thus
Etotal = E0 ei(wt
= E0 ei(wt
I / 1 + ei
1+e
i
k1 l)
k1 l)
+ ei(wt
1 + ei
= 2(1 + cos
k2 l)
eik0 x
eik0 x
) = 4 cos2
2
so
If ocused = 4I0 cos2
49
2
(3.18)
cf (3.17); while if the source is incoherent. Putting a glass changes nothing
If ocused = 2I0
where I0 is half of the intensity of the total light.
3.5 Interference
One can do the whole business of optics using Huygen’s geometric constructions.
It will match qualitatively to results obtained from Maxwells equations. We will
show the equivalence of the two when we study single slit di↵raction.
2-Slit Interference
In an introductory course, we learned that parallel rays that are parallel to the
axis of lens will be focused to the focal point. In fact lens has property that focus
(10/28/13)
parallel rays that are not parallel to the axis of lens. They will focused to a point
on the focal plane. One can prove this fact geometrically by drawing two lines
among the parallel rays: one goes to the center of the lens; the other goes to the
focal point before entering the lens.
Consider two slits separated by d, plane wave light enter the slits. A focusing lens placed behind the slits. Observe interfere on the screen. Ignoring any
background signals produced by lights going through the slit without interference.
Mathematically it is the same as (3.18), commonly write
Lecture 14
(✓)
2
I(✓) = I0 cos2
I0 here is the maximum intensity on the screen, of course happens at ✓ = 0.
(✓) = kd sin ✓ =
2⇡d sin ✓
where ✓ is the angle of elevation from the point of observation to the midpoint of
the 2-slit. One can find the maxima at
(✓)
= m⇡
2
50
In fact
I(✓) = I0 cos
2
✓
⇡d sin ✓
◆
is not completely correct. Because experimentally one will see that the central
peak has the maximum intensity. The correct expression takes into account the
width of the slit a
I(✓) = I0 cos2
✓
⇡d sin ✓
◆
sin ⇡a sin ✓
⇡a sin ✓
!2
(3.19)
the extra term is due to single slit di↵raction, which we will discuss later.
N-Slit Interference
Same setup, N slits. Similar math
(✓) = kd sin ✓
where ✓ is the angle of elevation from the point of observation to the midpoint of
the N slit. N d should be small so that from the lens prospective, the N rays are
still near axis parallel rays.
E = E0 + E0 ei + Ee2i + ...
N
X1
1 eiN
= E0
ein = E0
1 ei
n=0
I = I0
sin N2
N sin
2
!2
Nx
because limx!0 sin
= N , where I0 maximum intensity on the screen. Of course
sin x
just like (3.19) we may want to multiply the single slit e↵ect. But if we deal with
small angle i.e. ⇡a sin ✓ ⌧ 1 then the single slit factor ⇡ 1.
51
Total Reflection
This is destructive interference and it is simpler to be found by Snell’s law. Consider a piece of fiber optic, which is made of 2 kinds of glass. Clad wrap the core.
no air
ncore > nclad
Send light into the cable at incident angle ✓,
nair sin ✓ = ncore sin ✓1
nair = 1
then the light will hit the clad from inside at incident angle ✓2 = ⇡/2
to have no light refracted into clad glass,
ncore sin ✓2 = nclad sin
✓1 , we want
⇡
2
this gives the smallest ✓2 . Hence the largest ✓ is ✓critical
sin ✓critical = ncore cos ✓2 = ncore
s
1
✓
nclad
ncore
◆2
q
= n2core
n2clad
Thin Film
Before we do thin film, we need to understand why it only works for thin films not
thick films.
Michelson interferometer consists two highly polished mirrors, one fixed, one
is movable. Plane wave light beam (may be produced by a point source putting at
the focal point of a lens) hits a half-silvered mirror, splits into two beams. Both
then hit the mirrors and get reflected and recombined to give interference, forming
fringes. One may put addition glass in one of paths to cancel any discrepancy in
the optical path di↵erence due to one beam may travel longer in the half-silvered
mirror.
The experimentalists may move the movable mirror, and find the fringes
pattern moves too. It gives a precise way to measure distance. Interestingly
fringes disappear if the movable mirror, is moved beyond certain distance. This
is because interference patter is produced by coherent lights. If the optic path
52
di↵erence is beyond the coherence length
2
cannot produce interference. Therefore we require the thinness of the film to be
within the coherence length.
Consider a thin film of thinness t, refraction index n. The medium above and
Lecture 15
below the film is air. So we have 2 interfaces to consider. Send plane wave light to
(10/30/13)
the film from the air with incident angle ✓. Consider two light rays that are coincide
after they get reflected by the film. One is reflected right at the first interface; the
other had penetrated the film and gets reflected at the bottom interface. (note:
it is common mistake to draw one incident ray and two parallel reflected rays.
It should be two parallel incident rays and one reflected ray, although math is
the same.) Recall (2.16) reflected light from n1 < n2 medium gain extra phase.
Refraction doesn’t gain phase. So the optical path di↵erence of the two rays
sin ✓ = n sin ✓t
=
=
2⇡
2⇡
✓
2t
n
cos ✓t
2tn cos ✓t
It is the same as (3.18), so
2t tan ✓t sin ✓
⇡=
◆
4⇡tn p
1
I(✓) = I0 cos2
⇡
(sin ✓/n)2
2
maxima at 2 = m⇡.
Many other variations: inclined plane, Newton’s rings, ...
53
(3.20)
⇡
3.6 Di↵raction
Single Slit Di↵raction
Slit width a, focusing lens behind the slit. In the introductory course, we learn to
divide the bean into 2. The top half balances the bottom half. If it happens to be
that
a
sin ✓ = m
m = 1, 3, 5, ...
(3.21)
2
2
We get destructive interference, hence minima. This formula also reveals that if
a
we will not distinguish minima and maxima, hence no di↵raction, only see a slit
image on the screen.
We now want to do a better job. First applying Huygen’s principle to a
general situation, consider a plane wave moving in z direction. Point A to the
one of the points on the constant wave phase plane, I , by Huygens it products a
spherical wave
~
e ik·~r
d A=g I
dS
r
g is the fraction of I that spherical wave element shares, which will give the
right normalization. dS is the area element that A occupies on the phase plane.
Consider a point A0 on another constant phase plane in distance l ahead of I , so
A0
Another way to calculate
waves from I , so
A0
A0
=
Ie
ikl
is to think
=g
I
¨
1
1
e
(3.22)
A0
is generated by the spherical
i~k·~
r
r
dS
(3.23)
We can reasonably argue that only these spherical waves that are near behind
A will contribute to A0 , so we say that x2 + y 2 is small, so approximately
0
r=
p
1
l2 + x2 + y 2 ⇡ l[1 + (x2 + y 2 )]
2l
54
and
1
1
⇡
r
l
thus
A0
ikl ¨ 1
e
=g
I
l
dxdye
1
{z
|
equating to (3.22)
ik
(x2 +y 2 )
2l
}
2l
⇡ ik
ik
i
=
2⇡
We now do single slit di↵raction. Let (⇠, ⌘) be slit coordinate, a2 < ⇠ < a2 ,
1 < ⌘ < 1. Let (x, y) be screen coordinate. And (0, 0) in both coordinates
systems are aligned in z direction. To use (3.23), we need
g=
~r =
p
(x
⇠)2 + (y
r
2x⇠
R2
⌘)2 + l2 n̂
where n̂ is pointing from (⇠, ⌘) in the slit to (x, y) on the screen. Let R2 = x2 +y 2 +l2
r=R
2
1
2y⌘ ⇠ 2 + ⌘ 2
+
R2
R2
2
If we ignore ⇠ R+⌘
2 , i.e. treat ⇠ ⌧ x, ⌘ ⌧ y , we get Fraunhofer di↵raction. That is
far field approximation. While Fresnel di↵raction, near-field approximation, will
keep this term. We do far field
x⇠
R
r⇡R
y⌘
R
So by (3.23), we get
(x, y) =
i
I
R
e
ikR
ˆ
a/2
d⇠e
a/2
|
{z
sin kax
2R
kx
2R
/
sin kax
2R
kx
2R
55
ik x⇠
R
ˆ
1
y⌘
d⌘eik R
1
{z
}
}|
⇠ (y)
I = I0
Lecture 16
(11/6/13)
sin kax
2R
kx
2R
!2
sin ⇡a sin ✓
= I0
⇡ sin ✓
!2
with sin ✓ = pl2x+x2 , this justifies (3.19).
One can draw above I, for three cases: ⇡ a, ⌧ a, and
a. One will
see that only ⇡ a gives nice di↵raction pattern. ⌧ a gives a very narrow peak,
and
a gives a very broad peak. That is because if
a, the first zero
⇡a sin ✓
=
⇡
2
can never be met.
Circular Hole Di↵raction
Hole of radius a, do the similar integral (3.23) in polar, get
I = I0
✓
J1 (k✓a)
k✓a
◆2
where Jn , Bessel’s equation
1
Jn (u) =
2⇡in
ˆ
2⇡
eiu cos ein d
0
The zeros of J1 at u = 3.83, 7.02, ....
The first minimum
3.83
✓1 =
= 0.61
2⇡ a
a
or in term of the opening angle
✓ = 2✓1 = 1.22
a
(3.24)
This gives Rayleigh Criteria, saying that two point sources are resolved when the
principal di↵raction maximum of one image coincides or separated from the first
minimum of the other. So better resolution, bigger a, that means larger lens.
Another application of (3.24), at night, infrared camera is used. So
is
longer, so the resolution gets worse.
56
An old sage says spy satellite sees license plate. Can it see? No, A satellite
is D altitude from the earth and it can resolve the object of size on earth
S=D ✓=
400km ⇥ 1.22 ⇥ 4000Ȧ
= 8cm
2m
Babinet’s Principle
Not only hole gives di↵raction, but also an opaque disk gives di↵raction. And if
we fill the hole with an opaque disk, we get nothing.
hole
+
disk
= 0 =)
disk
=
hole
Idisk = Ihole
4 Quantum Waves
4.1 Photons
Photoelectric E↵ects
1905 good year for Einstein. He figured out special relativity, photoelectric e↵ects
and Brownian motion implied existence of atoms.
Two observations in front of Einstein
Observation 1: There was a minimum ⌫ to eject e from a metal. E.g. N a,
500Ȧ ejects, 600Ȧ noting.
Observation 2: There is no time leg from shinning lights on metal to ejection
of e .
Both contradict wave theory
energy = intensity ⇥ time
independent on ⌫, and depending on t.
Einstein postulated photons:
1) light composed photons
2) interacts with 1 e
57
3) transfers energy h⌫
The maximum energy of the ejected e
Emax e = h⌫
this is maximum because the ejected e may not be right on the surface. may
have to penetrate in some depth and the ejected e is scattered before coming out
of the metal. , work function, energy used to break binding.
The experimental setup is to put metal on one of the electrodes and space
between another electrode. Shin light on the metal. Put adjustable voltage V to
the electrodes. Use Vmax to stop current, from there to determine Emax
eVmax = Emax
The graph V v.s. ⌫.
Vmax =
h
⌫
e
e
The measurement is kind of tricky. Initially data gave V (⌫) ⇠ V 1/2 . It was
Millikan, being the first doctoral student graduated from our department, who
obtained clean surface metals, and got a straight line. Millikan later got his Nobel
prize for oil drop experiment.
X rays Production
This is the reverse of photoelectric e↵ect. Use a secondary voltage to heat up the
wire. When wire gets hot, e detaches. Then use another high voltage V source
to accelerate those e . The place e takes o↵ is called cathode, and then e hit
the metal target, called anode, at very high energy. Not only current is produced,
but also photons come out with high frequency from the metal.
Energy of the e before hit the metal
Ee = eV
58
the maximum energy transfer to photons
eV = Ee = Emax = h⌫ =
hc
min
or
hc
(4.1)
eV
This is in fact what one will get if he measures the intensity of X rays v.s. of
the X rays. See that no X rays with less than min . This contradicts classical
wave theory that predicts all can be produced.
min
=
Compton Scattering
We have seen two phenomena, photoelectric e↵ect and X rays, that give definite
support that lights are particles. Some historical remarks. X rays were discovered
(11/11/13)
before photoelectric e↵ects, but X rays were not understood until photoelectric
e↵ects were understood 1905. 1924 Compton explained Compton e↵ect. By that
time people had well accepted lights are particles, but Compton got Nobel prize
anyway.
Suppose e at rest, hits e then and e scatter with angles ✓, with
respect to the incident axis.
Lecture 17
energy conservation
h⌫ + mc2 = h⌫ 0 + Ee0
momentum conservation
p = p0 cos ✓ + p0e cos
0 = p0 sin ✓
By E 2 = p2 c2 + m2 c4
photon p =
p0e sin
h⌫
c
Thus
(h⌫/c
0
0
h⌫ 0 cos ✓/c)2 + (h⌫ 0 sin ✓/c)2 = (pe )2 = E e2 /c2
= (h⌫ + mc2
59
m2 c2
h⌫ 0 )2 /c2
m2 c2
therefore
⌫0 =
⌫
1+
h⌫
(1
mc2
cos ✓)
This is an example of elastic scattering. This happens when photon energy
h⌫ ⌧ the rest energy of e = mc2 = 511keV.
Experiment to verify above is to send photon of known energy h⌫ to a metal
target, measure scatter photons ✓ and ⌫ 0 . How to prepare h⌫? Use X ray setup
(4.1). How to measure frequency of scattering light and precisely the scattering
angle? Use crystal of Bloch spectrum, which we will discuss later.
We now fully establish particle nature of light. We need to reconcile particle
wave duality. Because wave interferes but particles don’t. This requires quantum
mechanics.
4.2 Bohr Model
Rutherford 1910-1912 used ↵ particle, He nucleus, to hit metal sheet. ↵ particles
were obtained from ↵ decay, which we will discuss later in detail.
U ! Th + ↵
Rutherford found that while most ↵ passing through without any deflection,
there were some ↵ got almost reflected back, so he believed that those reflected
back ↵ must hit the nuclei of the metal. This contracts the early model that nuclei
and e s were uniformly distributed inside of the atom. If it were the case, ↵ would
jiggle through like Brownian and not reflected back.
Thus Rutherford proposed that nuclei is highly concentrated. If the size
of atom is 1Ȧ, then size of nuclei is 10 5 Ȧ. This model immediately created a
problem. If e orbits around nuclei, dipole momentum
I / P̈
2
6= 0
hence it radiates energy, hence atoms are unstable.
Motivated by planetary model, Bohr said
1) All classical law of physics are valid except those pertaining to radiation
60
2) radiation is only emitted when e jump from one planetary orbit to another
3) angular momentum is quantized
L = n~ = n
h
2⇡
Bohr’s rule solved the mystery put forward by a high school teacher, Balmer:
the Rydberg formula for Hydrogen
1
=R
if
1
n2i
1
n2f
!
(4.2)
The radiated photon wavelength if is related to the orbital labels, where R is
Rydberg constant.
Because
mv 2
ke2
1
= 2
k=
r
r
4⇡✏0
L = mvr = n~
That is
8
<mv 2 = ke2
r
: m 2 v 2 = n 2 ~2
=) rn =
r2
where
a0 =
n 2 ~2
= n 2 a0
e2 mk
~2
= Bohr radius
e2 mk
one can replace m by reduced mass µ to make more precise. a0 = 0.5Ȧ.
Therefore
1
En = mv 2
2
ke2
=
r
ke2
=
2rn
ke2
=
2a0 n2
where
1 k 2 me4
=
2 n 2 ~2
ke2
1
↵=
= fine structure constant =
~c
137
This agrees exactly to quantum mechanics calculation.
Since
h⌫ = Enf Eni
61
1 mc2 2
↵
2 n2
we derived (4.2)
1
if
1 ke2
=
2⇡~c 2a
| {z 0}
↵
4⇡a0
✓
1
ni
1
nf
◆
This derivation works for any hydrogen like atom with replacement
e2 ! Ze2
But fails miserably for others, e.g. He
4.3 Matter Waves
de Broglie was the first person to evoke symmetry argument: if waves ! particles,
then particles ! waves.
To complete his program, he needed
1) a formula for converting particles to wave: Broglie wavelength
2) a propagation equation of the matter waves: Schrodinger equation.
Broglie interpreted Bohr third rule
L = n~
as wave wraps around a circle
n = 2⇡r
so
=
2⇡r
2⇡~
h
h
=
=
=
n
n~/r
L/r
p
Broglie generalized this to that all parties have wavelength
=
or
p=
h
=
h
p
~2⇡
62
= ~k
Then for free particle
p2
~2 k 2
=
2m
2m
and the associated photon with that energy
E=
E = h⌫ = ~w
(4.3)
(4.4)
thus
~k 2
2m
dispersion relation, we can get group velocity
w(k) =
vg =
dw
~k
p
=
=
dk
m
m
agrees particle definition.
Davisson–Germer Experiment verified Broglie. Shin e beam to crystal obtained some image as shinning X ray. Crystal has latter layer, so similar to thin
film (3.20), but no refraction index hence ✓t = ✓ and no extra ⇡ phase, so we get
Bragg’s law
n = 2t cos ✓
(4.5)
where t is the latter spacing. But unlike thin film, the direction of latter plane
may not be obvious (see Eisberg, Quantum Physics of Atoms, Molecules, Solids,
Nuclei, and Particles. figure 3.33). From the angle of deflection, i.e. the angle
between incident beam and reflected beam , then ✓ = /2.
Davisson–Germer used energy e 54eV, t = 0.91Ȧ spacing and detect =
50o . By (4.5)
= 2 · 0.91Ȧ · cos 25o = 1.65Ȧ
Using de Broglie
=
h
h
=p
= 1.65Ȧ
p
2mE
In perfect agreement.
63
4.4 Uncertainty Principle
Lecture 18
(11/13/13)
It says one can measure conjugate variable simultaneously with an uncertainty
given by
~
2
~
2
~
2
x px
y py
z pz
Let’s see what they mean.
Example 1:
Suppose we shin e beam through a slit with width a, then we know that at
the moment it passes the slit its vertical position has uncertainty less than a. Since
e is wave, it will form an di↵raction pattern on the screen. Hence at the moment
it passes the slit it had some momentum in vertical direction too, which can be
calculated. Since e will likely fall within the first minimum of the di↵raction
pattern, cf (3.21)
py
h/p
= ✓⇠ =
=)
p
a
xy
xy py ⇠ h
Example 2: Heisenberg Microscope
Suppose we use ordinary microscope to see e . There is a “microscope” L
distance above an e . It sees the e by photons, with a resolution
✓ = 1.22
D
⇠
D
where D is the size of the lens, and is the wavelength of the photons. So the
uncertainty in the horizontal direction of the e is
xe = L ✓ ⇠ L
D
The reason it sees the e because many photons beam are sent near and parallel
to the lens axis to illuminate the object, in this case is e . So some photons are
64
scattered by the e and enters the lens. For those photons to enter the lens their
horizontal momentum must not greater than
px ⇠
ED
h⌫ D
hD
=
=
c L
c L
L
By conservation of momentum, the e must recoil and gain some horizontal
momentum in opposite direction
pxe ⇠ h
D
L
Thus
hD
=h
D L
Example 3: Simple harmonic oscillation
Suppose a mass is hang and oscillating by a spring k, assume that the average
x is 0 and average p is 0, so
x e p xe ⇠ L
E=
By uncertainty principle
( p)2 1
+ k( x)2
2m
2
p ⇠ ~/ x
E=
~2
1
+
k( x)2
2
2m( x)
2
One can find that the minimum E is at
x=
✓
~2
mk
◆ 14
and
r
k
m
This is exactly equal to what we will get from quantum mechanics.
Example 4:
Hydrogen
( p)2 ke2
~2
ke2
E=
=
2m
x
2m( x)2
x
Emin = ~w = ~
65
minimum at
x=
~2
= a0
kme2
and
Emin =
ke2
2a0
Both in perfect agreement to quantum mechanics.
Example 5:
By uncertainty principle, e cannot get in the nuclei, then x ! 0, p ! 1.
This explanation is more fundamental than Bohr rules, but still doesn’t answer
why orbiting causes no radiation.
One can get E v.s. t version of the uncertainty principle, by E =
E t=
p p
mv p t
t=
=
m
m
p x
Example 6: Lamb Shift
Due to vacuum fluctuations, e e+ are created within
tainty principle, which causes screening
k
Ze2
e
r
p2
2m
h
t allowed by uncer-
hr
mc
4.5 Schrodinger Equation
Quantum Postulates
1) To each classical observable assign a quantum operator which will operator
the wave function, e.g. x̂, p̂, Ê
Â
2) The result of operating on with operator  must be a physically possible
result if is an eigenstate. The results are eigenvalues.
3) The definition relative to classical physics not involving derivative are unchanged
~ @
x ! x̂
p ! p̂ =
i @x
66
p̂2
+ V (x̂)
2m
Ê =
L̂z = x̂p̂
Example: Free Particle
Lecture 19
Ê =
(11/18/13)
p̂x̂
p̂2
2m
find eigenfunction try
Êei(kx
wt)
=
~2 @ 2 i(kx
e
2m@x2
wt)
=
~2 k 2 i(kx
e
2m
wt)
Hence
~2 k 2
2m
We have other quantity for E = ~w, cf (4.3), (4.4), so put Hamiltonian
E=
Ĥ = i~
so for
= ei(kx
@
@t
wt)
(Ĥ
Ê) = 0
In general if potential presents
i~
~2 @ 2
+ V (x)
2m @x2
@
=
@t
Or 3D
@
~2 2
=
r + V (~x)
@t
2m
We solve 1D Schrodinger equation by separation of variables. This works only
if V has no t dependence.
(x, t) = (x)f (t)
i~
get
i~ 0
f =
f
~2
2m
00
+V
Since left depends on t, right depends on x, both should be constant = E
f =e
67
iEt/~
Or
(x, t) = (x)e
where (x) solves
~2
2m
00
iEt/~
(x) + V (x) = E
Time independent Schrodinger equation. The solution is normally chosen to be
real. With one exception in hydrogen wave function, complex is more useful.
Because the eigenfunctions of the rotation operator are complex.
Quantum Interpretation Postulates
1) The probability of a particle being at position x to x + dx is
| (x)|2 dx
The implies
that
has to be proper normalized,
ˆ
or 3D
ˆ
| (x)|2 dx = 1
| (x, y, z)|2 dxdydz = 1
And also means that
| |2 ⇠
1
[length]
In the spirit of (3.12), we put
2) Given an eigenfunction ua with eigenvalue a of an observable  and a
general state vector , then the probability of measure and obtain a is
P (a) =
call overlapping integral.
E.g. If a = x̂, then ua (x) = (x
ˆ
2
u⇤a (x) (x)dx
a), so
P (a) = | (a)|2
68
agreeing 1). Another example up (x) = eipx/~ momentum eigenfunction, i.e.
~ @ ipx/~
e
= peipx/~
i @x
Measure
2
whose | (x)| has deviation
˜(p) =
ˆ
u⇤p (x)
1
(x) = pp
2⇡
e
x2 /4
2
x
x
x
(x)dx =
ˆ
1
e
ipx/~
1
2
1
pp
2⇡
e
x2 /4
x
2
dx ⇠ e
p2
x /~
whose ˜(p) = Probability of measuring (x) and obtain momentum p, has
deviation p = ~/2 x . Hence we get the minimum of uncertainty
x p
=
~
2
3) The expectation value of observable on a general state
ˆ
D E ˆ
 = aP (a) =
is
(x)⇤ Â (x)dx
E.g.
hx̂i =
ˆ
(x)⇤ x (x)dx
hp̂i =
ˆ
(x)⇤
4.6 Infinite Square Well
V =
8
<0
L
2
~ @
(x)dx
i @x
<x<
:1 otherwise
Physical example like p n in nuclear like in a box.
69
L
2
(4.6)
First consider parity x !
x, since V ( x) = V (x) symmetric potential
~2 @ 2 (x)
+ V (x) (x) = E (x)
2m @x2
change x !
(4.7)
x
~2 @ 2 ( x)
+ V ( x) ( x) = E ( x)
2m @( x)2
then
~2 @ 2 ( x)
+ V (x) ( x) = E ( x)
2m @x2
hence ( x) is a solution of (4.7) too. Since (4.7) is linear, we can also construct
solutions of (4.7) have even or odd parities.
(x) + ( x) even sol
(x)
Now solve (4.6) for
( x) odd sol
L/2 < x < L/2
~2 @ 2 (x)
= E (x)
2m @x2
or
00
Hence
n
and
=
8
<
q
2
=
cos n⇡x
even
L
L
q
=
2
:
sin n⇡x
odd =
L
L
(11/25/13)
n = 1, 3, 5, ..
n = 2, 4, 6
~2 k 2
~2 ⇡ 2 2
=
n
2m
2mL2
This example shows some very important general rules
1) Ground state is always even, if the potential is symmetric.
En =
Lecture 20
k2
70
(4.8)
2) Eigenstates are stable, because full wave function
= e
iEt/~
| (x, t)|2 = | (x)|2
independent of time. Hence dipole momentum
hP i = e
ˆ
n,m |
x|
2
dx
is independent of time. This shows why e orbits without radiation.
But if we put e in the box in state
n,m (x, t)
ne
iEn t/~
+ b2 |
m|
=a
+b
me
+ 2ab|
n
iEm t/~
then
|
n,m |
2
= a2 |
n|
m | cos wt
where w = En ~Em .
Compute dipole momentum
hP i = e
ˆ
x|
n,m |
2
dx ⇠ cos wt + terms independent of t
Hence
I⇠
d2
hP i =
6 0
dt2
such e will radiate.
If eigenstates are truly stationary, then how come spontaneous radiative decay happen in nature? That is because 0 point vibration of vacuum, the true
Hamiltonian is not truly time independent.
3) Selection rule: prohibit electric dipole transition unless change parity. Refer to (4.8) clearly
ˆ
x
even
even
=0
71
ˆ
x
odd
odd
=0
This doesn’t mean no transition can happen between these states. Higher poles
can happen, but they are weak.
4.7 Bound States
We study two more bound states: finite square well and simple harmonics oscillator.
Finite Square Well
e energy
✏>
Need to solve
Let
V0 look for bound solutions
8
< V
0
V =
:0
8
<
:
~2
2m
00
~2
2m
00
=
V0 > 0
|x| > a
V0 =
|x| < a
✏
|x| > a
✏
2m
(V0
~2
Look for even or odd solutions
k2 =
✏)
↵2 =
even
b1 cos kx
c1 e ↵x
c1 e↵x
|x| < a
x>a
x<a
At x = ±a, is continuous,
For the even solution
|x| < a
0
2m
✏
~2
odd
b2 sin kx
c2 e ↵x
c2 e↵x
is continuous.
b1 cos ka = c1 e
b1 k sin ka =
72
↵a
c1 ↵e
↵a
So
↵
tan ka = =
k
where
=
2ma2
V0 .
~2
p
(ka)2
ka
Put y = ka
tan y =
p
y2
y
One can draw the LHS and the RHS. And see that if
(n
1)⇡ 
< n⇡
there are n di↵erent solutions of y, so n di↵erent energies
✏n =
Lecture 21
(12/2/13)
~2 k 2
2m
V0 =
~2 yn2
2ma2
V0
In fact if the well is shallower or narrower, i.e. smaller , less number of y,
but no matter how small is, there is at least one bound state.
Question: the solution has a piece in |x| > a region. That is possible to find
e outside of well, hence e will have negative kinetic energy. What does negative
kinetic energy mean? Outside of well = e±↵x , so
x⇠
1
↵
By uncertainty
p ⇠ ~↵
so the uncertainty in energy when the e is outside
( p)2
~2 ↵ 2
E=
⇠
=✏
2m
2m
hence the uncertainty in energy as large as the energy itself, then negative kinetic
energy has no definite confirmation.
Question: Why are bound states discrete? Outside of well, satisfies
00
= ↵2
73
and has to be normalizable, we will see only special ↵, i.e. special ✏, will make
this happen. Suppose at the boundary of the well, (a) > 0 then 00 (a) > 0 so
0
(a) has to be negative otherwise will keep growing. So 0 (a) < 0, even then
0
will become less negative because 00 (a) > 0. If ↵ is too big, or 0 reaches 0
too fast, i.e. 0 becomes positive before x ! 1, then ! 1, not normalizable.
If ↵ is too small, or 0 reaches 0 too slow, i.e.
becomes negative before 0
reaches 0, then 00 becomes negative, then 0 will keep negative, then ! 1,
not normalizable.
Simple Harmonics Oscillation
~2 d 2
1
+ mw2 x2 = E
2
2m dx
2
First choose dimensionless variables
r
mw
2E
u= x=
x
↵=
~
~w
(4.9) becomes
00
(u) = (u2
↵) (u)
The solutions are
1
2
= cn e 2 u Hn (u)
✓
◆ 12
cn = p n
⇡2 n!
n (u)
Hn Hermite polynomials
H0 (u) = 1
H1 (u) = 2u
H2 (u) = 4u2
2
H3 (u) = 8u3
12u
with
↵n = 2n + 1
74
(4.9)
or
1
En = (n + )~w
2
We see that ground state energy is
system.
n = 0, 1, 2, ...
1
~w,
2
so there’s always energy in the
4.8 Scattering, Tunneling
Step function Scattering
(Griffiths problem 2.34)
V =
8
<0
:V
x<0
x>0
0
V0 > 0
Particle coming from the left, with energy E > V0 , we get
8
<
:
00
+ k12 = 0 x < 0
00
+ k22 = 0 x > 0
2m
2m
E
k22 = 2 (E V0 )
2
~
~
This is exactly equivalent to section 1.6, impedance mismatch.
k12 =
=
8
<Aeik1 x + Be
ik1 x
x<0
:Ceik2 x
x>0
Compare to section 1.6 we don’t have to put eiwt because in QM, eiwt $ eiEt/~ .
And the same boundary conditions , 0 continuous, therefore
B
k1 k2
C
2k1
=
and
=
A
k1 + k 2
A
k 1 + k2
and reflected ratio, transmitted ratio are the same
R=
✓
k 1 k2
k 1 + k2
◆2
T =
75
4k1 k2
(k1 + k2 )
R+T =1
However the interpretations of these ratio are di↵erent. In section 1.6, R, T are
energy ratio. Here they are ratio of finding particles. Suppose initial N particles
coming from the far left. Let
F± = flux of particles to right or left
then
8
<F = N |A|2 v
+
1
:F = N |B|2 v
F+ = N |C|2 v2 x > 0
x<0
1
v1 =
p
p
2E/m, v2 = 2(E
V0 )/m, thus
B
R=
A
2
T =
r
E
V0 C
E
A
2
The identification of the particle nature and wave nature is of course by de Broglie
mv = p = ~k
Step FUnction Tunneling
Lecture 22
(12/4/13)
We continue Griffiths problem 2.34, what happens if E < V0 ? Tunneling, which is
not only impossible for classical particles, but impossible for classical wave.
Particle coming from the left, with energy E < V0 , we get
8
<
:
00
+ k12 = 0 x < 0
00
k22 = 0 x > 0
2m
2m
E
k22 = 2 (V0 E)
2
~
~
So (simply recognize that solutions can be quickly obtain by replacing k2 in the
k12 =
76
previous case by ik2 )
=
using ,
0
continuous, find
8
<Aeik1 x + Be
:Ce
ik1 x
k2 x
x<0
x>0
B
k1 ik2
C
2k1
=
and
=
A
k1 + ik2
A
k1 + ik2
and
R=
Be ik1 x
Aeik1 x
2
=
x > 0 | |2 ⇠ e
k12 + k22
=1
k12 + k22
2k2 x
T =
k2 Ce k2 x
k1 Aeik1 x
2
e
2k2 x
=
4k1 k2
e
(k1 + k2 )
2k2 x
, so the uncertainty of finding particles outside well
x⇠
1
2k2
Hence the uncertainty in T is as large as T itself, so as far as uncertainty is concern
R+T ⇡R=1
Finite Square Well Scattering
8
< V
0
V =
:0
|x| < a
|x| > a
V0 > 0
Particle coming in from the left with energy E > 0. We will see this is exactly
equivalent to the technique used in section 2.2, impedance matching.
We solve
8
< 00 + k 2 = 0 |x| > a
:
k2 =
00
+ q2 = 0
2mE
~2
q2 =
77
|x| < a
2m
(E + V0 )
~2
(4.10)
then
8
>
>
Aeikx + Be
>
<
= Ceiqx + De
>
>
>
:F eikx
ikx
x<a
iqx
|x| < a
x>a
After some exhausting exercise (Griffiths equations 2.167, 2.168)
ie 2ika (q 2 k 2 ) sin(2qa)
A
2qk cos(2qa) i(k 2 + q 2 ) sin(2qa)
e 2ika (2qk)
F =
A
2qk cos(2qa) i(k 2 + q 2 ) sin(2qa)
B =
If E
V0 , then
q2
then
B
(q 2
=
F
k 2 ⌧ 2qk
k 2 ) sin(2qa)
⌧1
2qk
complete transmission. There is another way to get complete transmission: RamsauerTownsend e↵ect. That is if 2qa = n⇡, B = 0. And by (4.10)
n 2 ~2 ⇡ 2
8ma2
this is exactly the allowed energies in a 1 square well, except it is now shifted by
V0 and n starts from some integer n0 such that En0 > 0. If n = 1 is allowed, then
choose
⇡
a=
=
2q
4
En =
V0 +
This is quarter wave transmission we discussed in section 2.2.
Finite Square Barrier Tunneling
8
<V
0
V =
:0
|x| < a
|x| > a
78
V0 > 0
Particle coming in from the left with energy V0 > E > 0.
8
>
>
Aeikx + Be
>
<
= Ceqx + De
>
>
>
:F eikx
ikx
qx
x<a
|x| < a
x>a
2mE
2m
2
q
=
(V0 E)
(4.11)
~2
~2
So (simply recognize that solutions can be quickly obtain by replacing q in
the previous case by iq),
k2 =
F =
2ika
e
2qk cosh(2qa)
(2qk)
A
i(k 2 q 2 ) sinh(2qa)
(4.12)
4.9 WKB, alpha Decay
From (4.12), if qa
1
1
cosh 2qa, sinh 2qa ! e2qa
2
then
✓
◆2
F
T =
A
2
ln T =
2q |{z}
2a + small terms
!
4qk
e
2
q + k2
2qa
then
x
Suppose instead of having one finite barrier, we have V (x), e.g. a Gaussian.
Particle coming from the left with energy < max V (x) tries to tunnel through.
What is the probability of tunneling, P ? We will approximate the tunnel region,
i.e. where V (x) > E, by many finite square barrier of width x, so the probability
of tunneling is the product of the probability, Ti , of tunneling through each finite
square barrier, i.e.
P =
ln P =
Y
X
Ti
ln Ti =
2
X
79
qi ( x)i =
2
ˆ
q(x)dx
Thus
P =e
2
´ q 2m
~2
(V (x) E)dx
=e
G
the integral is taken over classically forbidden region.
↵ decay
Lecture 23
The 1D model we develop can be used to estimate the lifetimes T of ↵ decay
-Last Lec-
A=↵+D
(12/9/13)
A parent, D daughter, ↵ =4 He ion. The potential is given by Griffiths figure 8.5.
1
r1 = 1.2fmA 3
A =number of n and ps. 1fm = 10
15
m, r2 is given by
E↵ =
kZZ↵ e2
r2
Let Z be number of protons after ↵ has passed r1 .
Assume within r1 , ↵ is moving at speed v, then it takes
2r1
sec
v
to for one ↵ particle to begin tunneling again if it didn’t succeed last time. The
probability of escape is e G . So in 1sec the number of escaped ↵ is at the order
(because there may be multiple ↵ bouncing in r < r1 )
⇠
1
2r1
v
e
G
which is the rate of tunneling, so lifetime
T ⇠
v G
e
2r1
80
We estimate G.
G ⇡
⇠
/
=
ˆ
1 r2 p
dr 2m(V (r)
~ r1
r2 |p|
Z p
E↵
E↵
Z
p
E↵
E↵ )
Experimentalists prefer T 1 half time. The relation between lifetimes and half
2
time is
T 1 = T ln 2
2
✓
◆
1
ln T ln 2
1
1
1
log10
=
=
ln
ln ln 2
T1
ln 10
ln 10
T
2
Hence log10
1
T1
2
is linear wrt ln T1 , so we can write
log10
1
= C1
T1
2
Theoretical estimates for C1 = 27
The experiment data plot
C2 Z
p
E
p
28, C2 = 1.73 MeV.
log10
1
1
v.s. p
T1
E
2
find C1 = 28.9, C2 = 1.61.
Nuclear Fusion
Fusion is the reverse of ↵ decay. It happens inside of sun
2
H + 2 H !3 He + n
2
3
H + 2 H !3 H + p
H + 2 H ! 4 He + n
81
Why aren’t there high Z fusion?
Ca + ↵ !?
Because for high Z, the Coulomb potential
Z↵ Ze2 k
r
is very high, for normal energy ↵ the probability of tunneling is too low. Even at
the center of the star,
kT = 1MeV =) T ⇠ 1010 K
still too low for high Z fusion.
4.10 Hydrogen Wave function & Spin
Solve Schrodinger in spherical coordinate
~2 2
r +V
2m
=E
becomes

✓
◆
✓
◆
~2 1 @
1
@
@
1
@2
2@
r
+ 2
sin ✓
+ 2 2
+V
2m r2 @r
@r
r sin ✓ @✓
@✓
r sin ✓ @ 2
solution
nlm (r, ✓,
) = Rnl (r)Ylm (✓, )
Ylm (✓, ) ⇠ Plm (✓)eim
Ylm spherical harmonics, Plm associated Legendre polynomials.
En =
kme4
2~2 n2
82
k=
1
4⇡✏0
=E
n = 1, 2, 3, ...
l = 0, 1, ..., n
m =
1
l, l + 1, ..., l
Groundstate n = 1, l = m = 0
R10 ⇠ e
r/2a0
1
Y00 = p
4⇡
r
3
Y10 =
cos ✓
4⇡
r
3
Y1 1 =
sin ✓e i
8⇡
r
3
Y11 =
sin ✓ei
8⇡
Probability of finding particles at volume element (r, ✓, ) is
| |2 r2 drd⌦
Probability of finding particles at radius r is
|Rnl (r)|2 r2 dr
Probability of finding particles at radial direction ✓,
|Ylm |2 d⌦
Ylm spherical harmonics are eigenstate of L2 , Lz
Lz = xpy
ypx
L2 = L2x + L2y + L2z
83
is
L̂2 Ylm = ~2 l(l + 1)Ylm
L̂z Ylm = ~mYlm
Spin
Spin is completely quantum mechanical phenomena that has no classical analogy.
Pauli said no two e s can have same quantum numbers. Since the groundstate
of hydrogen 100 is occupied by 2 e , there must be a missing quantum number.
Pauli said spin is an internal degree of freedom
s,ms
called spinor. It has the same properties as Ylm
Ŝ 2
s,ms
= ~2 s(s + 1)
Ŝz
s,ms
= ~ms
s,ms
s,ms
s is fixed for the particle, and
ms =
s, ..., s
so the groundstates
nlmms
=
100 12
or
100
1
2
2 groups of particles and supersymmetry particles
fermion
Boson
half integer spin
integer spin
Superpartner
Sfermion
integer spin
Sboson
half integer spin
4.11 Afterword: Counterfactual Measurement
Thing to think about
84
A Mach-Zehnder interferometer is used in such a way that when reaches
detector 1 always constructive interference and when reaches detector 2 always
deconstructive interference, hence no detected by detector 2.
E0 !
bean spliter
half silvered mirror
!
!
#
#
mirror
mirror
#
#
!
!
bean spliter
half silvered mirror
!
detector 1
#
#
detector 2
Now block one of the two paths
E0 !
bean spliter
half silvered mirror
!
!
#
#
mirror
mirror
#
#
!
stop
bean spliter
half silvered mirror
!
detector 1
#
#
detector 2
The interference is destroyed, so both detectors detect . The spooky thing
is that the information of the present of the stopper is conveyed via the photons
that hit detector 2. Paradoxically these photons don’t even pass the path where
the stopper are. Hence we can know things about an object without interacting
with it.
85