quantum mechanics part 1 – waves as particles ch 28
TRANSCRIPT
quantum mechanics
part 1 – waves as particles
ch 28
Hmm!?
Another thing to play with (P.Lenard 1900)key to Cat scans (lets diagnose the problem)
I, f, V, light intensity I vs f I vs V VStop vs f
cathode raysLight frequecy=f
photoelectric
diagnostic test : I vs f
cathode raysLight frequecy=f
diagnostic tests : I vs intensity
cathode raysLight frequecy=f
diagnostic test: I vs V
cathode raysLight frequecy=f
a new test? VStop
VStop
None of the “cathode rays”make it
I=zero
I=Big
I=mediumI=small
use it- diagnostic test : VStop vs f
cathode raysLight frequecy=f
summarize the symptoms cathode ray “current”
only for f>f0 f0 depends on metal of
cathode cathode ray “current” ~
light intensity No delay in current
appearance for “forward V” , flat
current “backward V”, a VStop VStop ~ f
Diagnosis
Note: The WRONG diagnosis when something new is
going on is often difficult to understand. It goes topsy-turvy to try to get it right.
The RIGHT diagnosis may sound crazy but is often simple and straightforward ONCE YOU ACCEPT THE NEW IDEA
the WRONG diagnosis
Light “heats” up the electrons and makes them more easily jump out of the pool (metal)
I don’t get it….
Lets go back and see…
Symptomscathode ray “current” only
for f>f0
f0 depends on metal of cathode
cathode ray “current” ~ light intensityNo delay in current
appearance
for “forward V” , flat current“backward V”, a VStop
VStop ~ f
Metal – electrons stuck – W=work function characteristic of metal Light – little balls with
energy If you don’t hit the
electrons in the metal with E>W, the don’t get out. Critical frequncy then using E=hf is
f0 =W/h.
A CRAZY IDEA - Einstein
LIGHT IS A PARTICLE with energy E=hf Problems of course Why does it behave like a wave …. Note: These things are called photons
h a constant – Plank’s constant
Lets go back and see…
Symptoms cathode ray “current” only
for f>f0
f0 depends on metal of cathode
cathode ray “current” ~ light intensity No delay in current
appearance for “forward V” , flat
current “backward V”, a VStop
VStop ~ f
Metal – electrons stuck – W=work function characteristic of metal Light – little balls with
energy If you don’t hit the
electrons in the metal with E>W, the don’t get out. Critical frequncy then using E=hf is
f0 =W/h.
Lets go back and see…
Symptoms cathode ray “current” only
for f>f0
f0 depends on metal of cathode
cathode ray “current” ~ light intensityNo delay in current
appearance for “forward V” , flat
current “backward V”, a VStop
VStop ~ f
The more light balls “photons” above f0 the more electrons
We also don’t need time to “heat up”
Lets go back and see…
Symptoms cathode ray “current” only
for f>f0
f0 depends on metal of cathode
cathode ray “current” ~ light intensityNo delay in current
appearance for “forward V” , flat
current “backward V”, a VStop
VStop ~ f
Once we are collecting the electrons, we can’t get any more by turning up the voltage, we just make them go faster
Lets go back and see…
Symptoms cathode ray “current” only
for f>f0
f0 depends on metal of cathode
cathode ray “current” ~ light intensityNo delay in current
appearance for “forward V” , flat
current“backward V”, a VStop
VStop ~ f
If we reverse the voltage though we can make them go backwards. When eV=hf we get nothing so this is VStop
Voila!
WHY DOES it changethe wavelength???
compton
Just do collisions + relativity + f=c/λ
0 (1 cos )e
h
m
It works!
The phototube – the heart of a catscan
Cat Scanning
quantization
part 2 –particles as waves
ch 28
The world – a) matter - atoms
nucleus of tightlypacked protons and neutrons. Protons-positive charge
electrons orbitingnegative charge
charges come in bitse=1.602 x 10-19C
“quantized”
The World: LightWe have a crazy new idea
LIGHT IS A PARTICLE …. These things are called photons
h a constant – Plank’s constant
h=6.625 x 10-34 J-s 1eV=1.602 x 10-19J
E=hfE=hf
How is light made?
absorbed?
idea!
Trouble: Why discreet??Emission spectrum
Absorption spectrum
2 2
1~
1 1m n
Like a Merry go round that can only go 1mph, 3mph, 5mph andNOTHING IN BETWEEN
Why do we exist at all????
Bohr 1
electron as wavein a circle
2πr=nλelectron
Crazy possibility #2 – Bohr: electrons are waveselectrons are waves
Quantization of orbitals
So only certain orbitals are OK, and NOTHING in between.We require 2πr=nλelectron (whatever λelectron is…) States are stationary
For orbitals in between, the electron goes around and interferes destructively with itself.
2πr=nλelectron
Now lets see if we can figure out the energies of these orbitals
Starting StuffRelativity
EinsteinE=hf
Waves c=fλ
MechanicsF=macircular motion
_
E&M
2 2 2 4E p c m c 2mva
r
1 22
0
1
4
q qF
r
2πr=nλelectron
But what is λelectron???
back to light as particleRelativitymass of light?? mass of photon?? =0E=pc=hf=hc/λ SO p=p=h/h/λλ or λλ=h/p=h/p
Now to electron as waveλλelectronelectron=h/p=h/pelectronelectron
2 2 2 4E p c m c
Now lets see if we can figure out the energies of these orbitals (Hydrogen)
F=ma
2πr=nλe
2 2
20
1
4
e mv
r r
2 2
2
ee e
n n h nr
p m v
hwhere
from here we get L=mvr=nh/2π the quantization of angular momentum
here we get2
2
0
1 1
8 2
emv K
r
2 2 22
0 0 0
42 2
2 2 2 22 20 0 0
0 2
4
2 2 2 20
1 1 1 1
2 4 8 4
1 1 1
8 8 324
Rydberg's 1 13.613.6
Constant 32
e
e
e
e
e e eE K U m v
r r r
m ee e
r nnm e
m e eVeV E
n
quantization of the energy levels
We get 2 2
0 24
e
nr
m e
quantized energy, radius, velocity
2
2
13.6
B
eVE
n
r a n
What about the spectral lines?Emission spectrum
Absorption spectrum
2 2
1~
1 1m n
Balmer forumla
Only photons with certain energies allowed!
bohr photons
Energy of the photonsthe Balmer formula
2 2
2 2
1 1
So we get the Balmer formula!!!
1 1
photon n m
hcE E E R
m n
hc
Rm n
Binding energy, ionization energy
2
13.6
| |
n
n
Z eVE
nE Binding energy
13.6 eV
13.6/n2 eV
The Davison Germer Experiment
If electrons are really waves, they should diffract.
The Davison Germer Experiment
Quantum MechanicsPart 3
Ch 28
Waves or Particles? We found out
electrons are particles light is waves
Later we found out light is particleselectrons are waves
What gives?Neither?Both?
Lets think about the description Particles
position (x), velocity (v), energy (E), etc Light
Electric field (E), and magnetic field (B) Strategy
Let’s First see if we can make Light look like a particle (sort of) Light has to be like a ball (a small packet)
Then see if we can make electrons like waves Make a “matter field” thing analogous to the
Electric field E(x)~Asin(kx-ωt) remember k=2π/λ and ω=2πf
Light as particles How can we make a wave pulse?
Drum beat turn light on and off; ie make it blink
beats! Can we add up some waves of different frequency to
get little light “particles”? http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=35 (50,54)
In fact, we can get any shape we want by adding enough waves
http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=33
How many frequencies to make a nice square pulse?
wave f(x)=sin ωnx ωn~n n=1, 3, 5…
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-1 -0.5 0.5 1
-1
-0.5
0.5
1
1 frequency
4 frequencies
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-1 -0.5 0.5 1
-1
-0.5
0.5
1
8 frequencies
1000 frequencies
More square
More frequencies
So a wave packet or particle is a sum over lots of waves of different frequencyBut since E=hf
what is its energy????
heisenberg uncertainty principle…. better (more square) position ~smaller Δx
worse Δp= ΔE/c= hΔf/cHEISENBERG: ΔxΔp h/2
it will turn out this affects EVERYTHING Problem with signal transmission
Heisenberg Uncertainty Principle
for instance
So is this a problem in life? SO
if we know how fast we are going, we have no idea where we are
If we know where we are, we don’t know how fast we are moving
Real life Car at the corner of Chicago and University and
1001 MPH=44.70.447 m/s. Car weighs 2000 kg Δp=2000 kg .447 m/s~1000 kg-m/s Δx h/2/Δp=6.625 x 10-34/2/1000= 3 x10-37 m Good enough to say you are at Chicago and University
and not on the freeway – you get a ticket (a big one)
ΔxΔp h/2
What about for an atom? Take E=-13.6 eV ΔE=1 eV
ΔE/E = 1/13.6 ~ .1E=p2/2m so Δp/p =½ΔE/E so Δp/p=.05 p=2mE= 2(9x10-31)(13.6)(1.602x10-19)=2x10-24
Δp=.05 p=5x10-2 2x10-24= 6x10-25
Δx h/2/Δp= 6.625 x 10-34/2/10-25 ~ 3x10-9msize of atom 10-10m
no idea where it is. So its spread out
s orbital
d orbitals
p orbital
ΔxΔp h/2
Now Electrons as waves? Maybe just like light. A nice localized electron is
a mixture of many energies A nice energy electron is in a beam. We don’t
know exactly where it is. First lets take a slow
motion view
PROBABILITIES?? Electrons similar?
http://www.colorado.edu/physics/2000/schroedinger/two-slit2.html
http://www.colorado.edu/physics/2000/schroedinger/two-slit3.html
??
applet run 2 slits
So this is what“really happens”
Now for a quantum leap Intensity for Light ~ (Electric Field)2
Intensity is like probability density DEFINE A MATTER FIELD ψ(x)
P(x)=| ψ(x)|2
This thing (a wave function) will tell us what matter is
the square of it will give a probability density
20
02avg
EPowerI S
Area c
What dowavefunctionslooklike?
ψ can be negative
P=|ψ|2 is always positive
Double slit for electrons: the probability
Double slit for electrons: the wave function
Normalization Now if we have 1 particle (say), it had better be
somewhere. So P(x) has to normalize to 1
2( ) | ( ) | 1P x dx x dx
So what is an electron? A matter wave
packet Its mostly at
position x0
It has a median velocity v
However there is an smearing of both position and velocity no neither of these are exact
x0
The Schoedinger Equation There are physics laws which are encoded
in equations e.g. F=ma; Einitial=Efinal etc Is there a rule for ψ ????? Schoedinger ~ 1924
Why do we believe itBecause it works!
2
2 2
( ) 2[ ( )] ( )
d x mE U x x
dx
What is the program? Justifying it
This isn’t the real reason we believe itreal reason is because of experiment
Making models of real life and solving Periodic Table of the elementsMolecular binding (chemistry) tunneling (decay) transistorsZero point energy of the vacuum!…
Stuff you know! Mechanics
Waves
( , ) sin( )
2
D x t A kx t
k vk
0
0
0 0
cos( )
cos( )
E kx t
B kx t
E cB
E x
B z
^
^
E=hf E=hf p=h/p=h/λλ
Electricity and Magnetism
Modern Physics ideas
221
2 2
p mv
pK mv
mK E U
This is the 1-D eqn. We should really use a 3-D version (in this class we will stick to 1-D)
what we need to do is to find appropriate U(x) which describe real situations and then solve for ψ(x)
2
2 2
( ) 2[ ( )] ( )
d x mE U x x
dx
Boundary ConditionsWe have to require some things
about ψ(x) since P(x)= ψ2
ψ(x) must be continuousψ(x)0 as x and x- ψ(x)=0 in areas it should not beψ(x) must be normalized (i.e. its total
probability = 1 or 100%)
What kinds of problems do we want to solve?
Atom (to get the Bohr model) U(X)=-e2/r (Black line)
in 3D Too hard –make it 1-D too hard (approx with the
red line) Still to hard – approx as a
1D box with infinitely high walls. Maybe not such a good
approx, but perhaps we will see some basic things like energy only having some specific values i.e. energy quantization
U
x, r
Particle in a box – a toy model 1-D draw potential
U(x)=0 0<x<L U(x)= x=0; x=L
ψ(x)=0 if x0 or x L
Solutions to sin, cos choose sin since sin(0)=0 try Asin(kx)
U(x)
x
L
2
2 2
( ) 2[ ( )] ( )
d x mE U x x
dx
infinitely strong
2
2
( )( )
d xC x
dx
Walls infinitely strong
Energyquantized!
So what does it mean?
There is NO E=0 state!a zero-point energyConsistent with heisenbergmomentum>0
So have we learned anything? we modeled an coulomb
potential as an infinite well we used 1-D instead of 3-D
BUT we got quantized energies quantized radii The shapes of the wave
functions don’t tell us much about real life
Maybe if we go to a 3-D model with a real 1/r coulomb potential we might get something that makes sense
Do it
Get the right energies!E=-13.6ev/n2
Get orbital shapesChemistry!
each n, has many orbital shapesdepending on the angular momentum
it Gets pretty complicated
The correspondence principle If you put a ball in a box and let it rattle around,
the probability of finding it anywhere in the box is pretty much the same
How does this square with QM?? Answer: For large quantum number n the
probabilities approach the classical value
0.2 0.4 0.6 0.8 1
0.5
1
1.5
2
0.2 0.4 0.6 0.8 1
0.5
1
1.5
2P(x) n=10 P(x) n=100 Now its pretty flat ~classical
P(x) n=1 P(x) n=2 P(x) n=3
A finite potential well U=0 0<x<L U=U0 x<0 or x>L Particle with energy E
E can be >U0 Free particle
energy enough to puncture walls
<U0 Particle “trapped in well” X<0 and X>L is the
“classically forbidden region”
Walls are not infinitely strong
particle energy E
throw ball out of hole example
What is this Outside the box stuff? If we put a particle in a
box (say a match box) it doesn’t get out.
QM tells us that there is a small probability that the particle is OUTSIDE THE BOX! (the classically forbidden region)
Penetration depth=1/κ related to probablility for
particle to “tunnel” outside box
particle energy E
classically forbidden region
Walls are not infinitely strong
( ) ( ) /
( )
' '
xR
x L x L
x Ge
G e G e
02
2 ( )m U E
How big is the effect?
find how far a cube of ice with mass 10g sits outside the glass
02
( )
2 ( )
xR x Ge
m U E
h0=10 cm h=9cm
U0=mgh0
E=mghU0 -E=mg(h0-h)
=.01 (9.8 )(.1-.09)=.00098J
U0E
The well
3134 2
2(.01)(.00098)4.4 10
(1 10 )
penetration depth = 1/κ=2x10-32 myour ice stays in your glass (we will calculate a probability later)
the effect is big on atomic scales
What do the wave function look like? electron in a 1eV potential well
Energies are shifted relative to inifinite well wave function penetrates into classically “forbidden region”
So to summarize QM has told us that particles penetrate
into classically forbidden regionsTunnelingdecay
Crazy – your car has a small probability of being OUTSIDE your garage
Covalent bonds Model of H atom
particle in a box E=-13.6 ev (n=1 state) top of the box be at zero width of the box is ~ 2aB~0.1 nm
2 H atoms for a bond Separation =.12 nmmatch solutions at
boundary
Ge-κx
outside
solution
MatchsolutionsAsin(kx)
inside
solution
2
2
2 ( )
24.2
2 (0 )
m E Uk
where U eV
m E
Now solve for the new energies
E=-17.5 eVE=-9 eV
n=1 n=2
prob density prob density
Total energy – the covalent bond If the total energy is negative, then the
particles are bound. pp repulsive energy =
n=1 E=12 eV-17.5eV = -5.5eV covalent bond!!
n=2 E=12 eV – 9 eV=+3eV doesn’t bind
2
0
112 .12
4
eeV r nm
r
TunnelingClassical Physics
Quantum Physics
toy model of decay
so AR=ALe-w/
w=width of barrierremember =1/κ
Ptunnel=|AR|2/|AL|2=e-2w/
Nuclear Decay and Spontaneous Fission Problem
similar nuclei have radically different lifetimes
U238 4.5B years U234 244 K years different by 20,000
Answer: Decay is from tunneling, exceedingly sensitive to width of barrier P~e-1=0.4 P~e-11 = 1.7x10-5
different by 20,000
nucl
ear
pote
ntia
l
repulsive coulomb barrier
alpha particle tunnelsthrough barrier
w
Scanning Tunneling Microscope
Ptunnel=e-2w/
tunneling probability is very sensitive to w
STM images (100 nm 100 nm) collected from Au nanoclusters on a TiO2(110) substrate. They were grown by depositing 1 ML of Au and annealing to 600°C.
STM of Gold Nanoclusters in Ultra-High Vacuum
STM of Iron foil in a Uranyl Nitrate Solution
The surface of an iron foil was monitored with in situ STM in a solution containing uranyl nitrate. The 500x500 nm images show the rough surface, characteristic of a native iron oxide, becoming smoother as the reaction proceeds.
DNA
Wolfgang SchonertGSI
Which of these can be located more precisely?
Thought questions For the n=2 state the particle is most likely to be found
A. in the middle
B. about ¼ of the way from either end
C. at one of the ends
D. the same everywhere
The maximum kinetic energy of photoelectrons depends on
a. the frequency of the light.
b. the intensity of the light.
c. the number of photons that reach the surface per second.
d. the number of quanta.
e. the speed of light.
A photon collides with an electron. After the collision the wavelength of the scattered photon is
a. greater than or equal to the initial wavelength.
b. equal to the initial wavelength.c. less than or equal to the initial
wavelength.d. greater than the initial wavelength.e. less or greater depending on the
scattering angle.
The energy of a photon isA. Proportional to its frequency
B. Proportional to its wavelength
C. independent of its wavelength
D. independent of frequency
The wavelength of an electron isA. Inversely proportional to its momentum
B. independent of momentum
C. proportional to its momentum
D. the question is meaningless