I hear a train a comin’

I hear the train a comin‘

It's rollin' 'round the bend,

…

But that train keeps a-rollin',On down to San Antone.

Johnny Cash

The Quantum

quantum from quantus: latin for how much If quantum theory had been initiated by native

English-speakers at the turn of the 21st century, I suspect that they would have used the word “pixel” pixel mechanics or the physics of the energy pixel.

Quantized means certain discrete amounts Synonymous with “pixillated”

In this section, we will first learn about why they had to abandon classical mechanics and think about pixels of energy

Masters of the Universe

One last thing to realize, the 19th century physicists considered the picture of physics as nearly complete with no new surprises… just some loose ends to tie up I sometimes think that 21st century

physicists think the same thing…

Train wrecks

We will discuss three experiments which essentially “train wrecked” classical physics. I use the words “train wreck” because 19th

century physics was an unstoppable locomotive with a well defined path going into the future

And then someone put three little coins on the rails…

Blackbody radiation

A “blackbody” absorbs 100% of the radiation that strikes it and emits NONE backKind of a blackhole of radiation without the

nasty gravitational effectsBlacker than the nastiest t-shirt of the

nastiest Goth after 3 days in the mud at Bonneroo…

What the Masters wanted

Observation: all materials when heated changed to a dull red color at 11000C and that color shifts toward blue as temperature increases By all materials, I mean EVERYTHING, metal, salt, you, me,

your cat… They wanted to find an ideal black body: perfectly

absorbs, perfectly emits.. For 40 years, they tried to calculate the blackbody

radiation pattern from 1st principles The practical problem was to develop a measurement

standard the new-fangled “light bulb” at the German bureau of standards.

How to make a blackbody

Make a box with a small hole

The radiation which enters the box is perfectly absorbed except for a small amount which exits the hole

It follows from Prévost's theory of exchanges of 1792 that the best radiation absorber - the black body, is also the best radiation emitter.

Turning on the light

So a light bulb in the box will also give off “blackbody radiation”.

Under equilibrium conditions, the dominant frequency of  this radiation depends only on the temperature. The radiation curves were shown to be very similar to Maxwell's velocity distributions of heated gas molecules in a sealed container.

Measuring the Frequency as a function of temperature

Note that it doesn’t go to zero here

Several Frequencies

Note that the wavelength of the maximum goes decreases as T increases

IT NEVER GOES TO ZERO, NOT FOR ANY TEMPERATURE

Explaining It

In 1879, Joseph Stefan found that power emitted was proportional to T4 (Stefan-Boltzman law) P=T4

Wien’s displacement law for the peak value was formulated in 1894 pT=constant

Both of these are not derivable from 1st principles but only phenomenological descriptions Similar to W=mg describes

weight on Earth but not gravitation

The “Ultraviolet Catastrophe”

Lord Rayleigh and Sir Jeans (1900) calculated the energy distribution using the same theoretical assumptions as Maxwell had done with his kinetic theory of gasses.

They applied the statistical physics method to waves by analogy with Maxwell's gas particles using the equipartition of energy, i.e. they assumed that the total energy of radiation is distributed equally among all possible vibration frequencies.

In other words, energy is a continuous distribution over all values

Rayleigh-Jeans Law

4

8),(

kT

TI

Blows up at ultraviolet wavelengths

Planck’s Idea

Planck explained this by assuming that atoms are small oscillators

He used two postulates1. Energy (either emitted or absorbed)

frequency E f or using a constant, h, E=hf

2. No arbitrary energies … discrete energies only. E= n*h*f where n= 0, 1, 2, .. No ½ values!

Walking the Planck…

1

8),(

5 kT

hc

e

hcTI

•At T=0 K, the oscillators still oscillate (Note I ≠ 0, zero point energy•At high T,

Law Jeans-Rayleigh iswhich

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),(

1...1

8

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45

55

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kThchc

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hc

PhotoElectric Effect– Einstein’s Only Nobel Prize

In the 1900s, scientists noted that if a beam of light is shown on the anode, a current is produced

Classically, we expect that if we increase the intensity of the light, we will increase the intensity of the current

However, the observed phenomenon was that the current flow was basically constant with light strength, yet varied strong with the wavelength of light such that there was a sharp cutoff and no current flow for long wavelengths no matter the intensity!

Einstein takes the next step…

Based on Planck’s successful theory, he argues that energy changes between oscillators as E=hf

First, Einstein assumed that some work (or energy) must be utilized to remove the electron from the surface,

Next, once the electron is free, it may have some kinetic energy.

So ½ mv2+ =E , the total mechanic energy of the system The Planck oscillators are giving off energy in amounts of hf. 1/2mv2+=hf

Maximum Kinetic Energy

½ mv2 = hf – In words, when light of frequency, f, is shown

on an object which has an electron binding energy of , if hf is larger than , the electron is freed with a kinetic energy of hf- .

To keep these electrons from moving away, you will have to have a potential energy of eV0

where V0 is a potential difference between the anode and cathode

Einstein’s words

If the derived formula is correct then, then V0, when represented in cartesian coordinates as a function of the frequency of the incident light, must be a straight line whose slope is independent of the nature of the emitting light.

Millikan Corroborates This in 1916

And YOU corroborated it in Physics 302

Another Observation

When scientists passed a large electric field through gases with low vapor pressure, they excited the gasses to visible light

Transmitting this light through a diffraction grating, they produced the following spectra

In More Detail

What is the problem?

Note that each of the spectra has a single line at various wavelengths

There is not a continuum of wavelengths but discrete lines!

Again, this suggests that energy is taking only particular values or is moving in small packets.

Einstein coined the word “photon” for energy packet.

The Bohr Model

To explain this, Bohr suggested that the orbits of electrons would happen only at certain radii

His reasoning was the following

1. Assume that a Coulomb attraction what holds the electron to a central positive nucleus Zeq

eq

wherer

qqkFC

2

1

221

r

qqkUC

21

The Bohr Model, cont’d

2. The electron moves in circular orbits

3. Instead of the infinity of orbits which would be possible in classical mechanics, it is only possible for an electron to move in an orbit for which its orbital angular momentum L is and integral multiple of

4. Energy is emitted (absorbed) when an electron moves from a high (low) orbit to a low (high) orbit

Energy Quantization

2

22

22

422

RadiusBohr

2Energy

mkZe

nr

n

eZmkE

n

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Derive This!

Hint

For circular orbits, ma=mv2/r13.6 eV = mk2e4/22

Need some stuff from Relativity

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De Broglie and Matter Waves

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De Broglie Waves

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then

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Then each orbital circumference is a quantized wavelength

n=4 atom

Davisson, Germer, etc

So matter has wavelengths (the earth’s is on the order of 10-64 m)

Davisson and Germer confirmed De Broglie’s hypothesis

Neutron scattering (see Dr. Barzilov) is based on these ideas

The Wave Function

The Wavefunction

It is a mathematical function which specifies a wave disturbance at point in space and time

Consider the following:

An electromagnetic wave is described by Є(r,t)= Є = Є0

cos((k·r-t)) or Є0 sin((k·r-t))

The energy, E, at small volume, dv, is defined byE= Є2 dv or E Є2 dv

Now E is made up of batch of N photons, and we can find N by dividing E by hfN = E/hf Є2 dv

Intensity

The intensity, I, of the photons is the number of photons in a region of space (essentially N) I Є2 dv

Planck says that there are NO ½ photons so if dv is very small, N better be integer

How can we eliminate this possibility?

A way out…

Consider if we measure 0 photons and then we measure 1 photon. On average, we have ½ photon, right?

So a way out of our dilemma is to realize that Є2 dv is counting the average number of photons in a given volume! Є2 dv = Average number, N, of photons in

volume dv

Waves is waves….

From de Broglie, we know that on a fundamental level, matter behaves like light i.e. it has a wavelength and is a wave but instead of Є2 dv, we will use 2

dv Probable # of electron etc in a volume 2 dv

However, there is 1 tiny problem…

Euler Angles

Sines and cosines are made of complex numbers!

i

ee

ee

ii

ii

2sin

2cos

How to get rid of complex number

Consider the number z=(a+bi) It has a “complex conjugate” of z*=(a-bi)

For a complex number, z, it’s complex conjugate is defined as z* And Re(z*)=Re(z) and Im(z*)= - Im(z)

If I multiply z by z* then (a-bi)* (a+bi) is equal to a2 + b2 .

Essentially I have eliminated any imaginary component and z*z is always 0

Probability density

Probable number of electrons |2 dv where |2

= Probable number of electrons per volume then is |2 dv / dv or just |2

|2 is called a probability density

V

dv

dv

dv

2

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Normalizing a wave function

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N

Back to the “salad days” of Physics 270

f2 frequency,angular ffrequency T

1f frequency, T period

2k number, wave wavelength

Wave speed is defined as speed at which the wave crest moves v=/T =f = /k

Also recall: E=hf = and p=h/= k

Waves

A sine wave travels in space repeating itself forever

A real wave travels in space repeating itself for a definite period i.e. an ocean wave

Certainly matter waves behave more like ocean waves than sine waves

Matter Waves

Consider an electron in a cathode ray tubeDefinitely the

electron is localized in space

Call it a wave pulse or wave packet

Fourier Analysis

Any function can be written as a sum of sines and cosines The black wave

packet is the sum of all the waves below

Note that waves below represent a large number of frequencies

Fourier Analysis

nk

xkAxf

xn

Axf

n

nnn

nn

2

cos)(

or

2

cos)(

0

0

Ancos(knx)

When n=1, that frequency called the fundamental

Making a square wave using continuous waves

Fourier Transform

dkekAxf

i

xkAxf

ikx

nnn

)(2

1)(

is dk) (i.e.k in changes finiteover

expression general a then sincose Since

cos)(

i

0

•A(k) is a set of amplitudes: the amplitude of the wave with wave number k is A(k)•Thus we have created a “spectra” of frequencies which make up the wave that we see i.e. f(x)

What about reversing the process?

What if I wanted to create a spectrum of frequencies from my wave which is a function of x?

Note the differences, this has a negative near the i and is integrated over dx!

dxexfkA ikx)(2

1)(

A simple example

My function,f(x) has only 1 frequency, 0

Obviously, the Fourier transform must be a discrete spectrum

FT(f(x))

Note that my wave has continuous values

Now I will transform my single frequency into a wave

FT-1(g())

Some Notation

In this class, we will use FT to represent

Note that this is transforming from position domain to wave number domain

This is its inverse Fourier Transform, FT-1

))(()(2

1)( 1 kAFTdkekAxf ikx

))(()(2

1)(

xfFTdxexfkA ikx

Time Domain to Frequency Domain

We can also transform between these domains (or representations)

This is its inverse Fourier Transform, FT-1

A FT is often used in reducing noise in audio recordings A recording is FT’ed. The amplitudes of undesirable

frequencies are zeroed. Then a FT-1 is performed on the result

))(()(2

1)( 1

AFTdeAtf xi

))(()(2

1)(

tfFTdtetfA ti

Consider the following behavior

A small wave packet (small x) has a corresponding big k and vice-versa

An Aside

Do you remember that a cosine function of a traveling wave is written:

cos(kx-t)Did you notice that FT(x)=k and

FT(t)=?Have you considered what might happen

if you took the Poisson bracket of the cosine function for the variables k and x and and t, respectively?

Now, the big finish

Consider the following picture

Focus only on the terms of ±

At the origin, is in phase with the wave

At 1 wavelength from origin, these waves are 2 apart

At N wavelengths, the distance between peaks is 2N

Continuing this line of reasoning

At N wavelengths, the distance between peaks is 2N

At /2, they add destructively It is possible that at N cycles, they add

destructively at x Nso that2N/2xNor N x/x/ /2x(/2)2

Nearly there

2x

Thus24

xor 4

x

and

2

d

dk thus

d

dk

magnitude its toequal is change of rate average the

constant is d

dk of change of rate sinstanteou theSince

2

d

dk and

2

22

2

2

k

k

kk

k

If I use RMS uncertainties

x*k ½ And correspondingly,t* ½ If p=h/ = k then p = k x* p /2And E= implies t* /2x* p /2 and t* /2 are called the

Heisenberg Uncertainty Principle

We’re Not in Classical land any more

Consider a particle in a volume with a width (x) of a

Then p *a /2 or p /2a If the average kinetic energy, <KE> is defined

as <p2/2m> and is constant then <p2/2m> p2/2m

So <KE> 2/8ma2

This implies that there is kinetic energy even at 0K !