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