deBroglie Waves

• But what is the wavelength of an electron?!

• For photons, it was known that photons have momentum E= pc= hc/

p=h/ =h/p

• deBroglie proposed that this

is also true for massive particles (particles w/mass)!

=h/p = “deBroglie wavelength”

(He also proposed E = hf ). We’ll talk about this later.)

(momentum)

p

(wavelength)

A few more lengths • What would this make the wavelength of matter?

– =h/p = “deBroglie wavelength”

– Remember KE = ½ mv2 or gmc2

– h=6.6x10-34 J s

Thing Energy (eV)

Energy (J) Speed (m/s)

Momentum (kg m/s)

Wavelength (m)

Photoelectron 1 1.6x10-19 6x105 5.4x10-25 1.2x10-9

Rutherford a (4 amu)

2x106 3.2x10-13 2x107 1.3x10-19 5x10-15

Billiard ball (0.2 kg)

6x1019 10 10 2 3x10-34

Thermal C60 molecule 1.2x10-24 kg

0.16

2.4x10-20

200

2.4x10-22

2.8x10-12

Wave uncertainty

• In general waves are a bit “fuzzy”

• The exact location of a wave packet is somewhat uncertain. It has a distribution which has both an average value 𝑢0 and an

uncertainty Δ𝑢 = 𝑢 − 𝑢02 : which is the standard

deviation and means the average over that quantity.

• This is a general statement about all waves

• If a signal is made up of

Ψ 𝑢 = 𝐴𝑛𝑒𝑖𝑣𝑛𝑢

𝑛

• Then there is an uncertainty relationship:

Δ𝑢Δ𝑣 ≥1

2

But you already know this, maybe, from radio signals.

Heisenberg Uncertainty Principle

• In math: Δx·Δpx ℏ

2

• In words: Position and momentum cannot both be

determined completely precisely. The more precisely one is determined, the less precisely the other is determined.

• Should really be called “Heisenberg Indeterminacy Principle.”

• This is weird if you think about particles, not very weird if you think about waves.

Heisenberg Uncertainty Principle

Δx small Δp – only one wavelength

Δx large Δp – wave packet made of lots of waves

Δx medium Δp – wave packet made of several waves

Δx·Δpx ℏ

2

What about time dependence?

• They can be made up of different frequencies in time or wave-vectors (k or frequency in space) at the same time so:

Δ𝜔Δ𝑡 ≥1

2

Remember for light : E = hf =hω

2𝜋= ℏ𝜔

∆𝐸∆𝑡 ≥ℏ

2

This means the uncertainty in time and energy are related. So a state with well defined energy has a long “lifetime”

Review ideas from matter waves: Electron and other matter particles have wave properties.

See electron interference

If not looking, then electrons are waves … like wave of fluffy cloud.

As soon as we look for an electron, they are like hard balls.

Each electron goes through both slits … even though it has mass.

(SEEMS TOTALLY WEIRD! Because different than our experience. Size scale of things we perceive)

Electrons & other particles described by wave functions () Not deterministic but probabilistic

Physical meaning is in ||2 = *

||2 tells us about the probability of finding electron in

various places. ||2 is always real, ||2

is what we measure

If all you know is fish, how do you describe a moose?

LASERs!

What’s so special about LASER light?

A) It doesn't diffract when it goes through two slits.

B) All the photons in the laser beam oscillate in-

phase with each other.

C) The photons in the laser beam travel a little bit

faster because they all go the same direction.

D) Laser light is pure quantum light, and therefore,

cannot be described with classical EM theory.

E) Laser light is purely classical light, and therefore,

it is incompatible with the photon picture.

How light interacts with atoms

Laser: Stimulated emission to clone photon many times (~1020/s)

Light Amplification by Stimulated Emission of Radiation

e 1

2

absorption of

light: Atom absorbs

one photon

in

e

1

2

stimulated

emission:

clone the photon

-- A. Einstein

1916

in out

e

1

2

spontaneous

emission of light:

Excited atom emits

one photon.

out

Surprising fact: Chance of stimulated

emission of excited atom EXACTLY the

same as chance of absorption by lower

state atom. Critical when making a laser.

e

Atom in excited state

e

Atom in ground state Photon

Legend:

Identical phase

Identical energy

Identical momentum

Stimulated emission

e

Atom in excited state

e

Atom in ground state Photon

Legend:

Random phase

Random direction

Similar energy (as absorbed photon)

Spontaneous emission

To increase the number of photons when going through the atoms,

more atoms need to be in the upper energy level than in the lower.

Need a “Population inversion”

(This is the hard part of making laser, b/c atoms jump down so quickly.)

Nupper > Nlower, more cloned than eaten.

Nupper < Nlower, more eaten than cloned.

How to get population inversion?

e

e excited not excited

ΔE A) Use photons with hf < ΔE

B) Use photons with hf = ΔE

C) Use photons with hf > ΔE

D) Use very strong lamp with hf ≈ ΔE.

E) Will never get population inversion in this system.

No population inversion in 2 level atom! e

1

2

spontaneous

emission of light:

Excited atom emits

one photon.

out

e 1

2

absorption of

light: Atom absorbs

one photon

in

e

1

2

stimulated

emission:

clone the photon

in out

Equal probability

Population inversion means: More atoms are in the excited

state than in the ground state.

As soon as we have the same number of atoms in the excited

state as in the ground state, the probability of creating an

excited atom is same (or smaller, when considering spontaneous

emission) as the probability of having stimulated emission!

Can never reach population inversion in 2-level atom!

Summary

Half-silvered mirror Mirror

out

LASERs need:

• Population inversion Gain

• Optical feedback (optical resonator) Coherent light

deBroglie Waves

• Substituting the deBroglie wavelength (=h/p) into the condition for standing waves (2r = n), gives:

2r = nh/p

• Or, rearranging:

pr = nh/2

L = nħ

• deBroglie EXPLAINS quantization of angular momentum, and therefore EXPLAINS quantization of energy!

x

y

x

y

x

y

Case I: no fixed ends

(infinite length)

Case II: one fixed end

(infinite length)

Case III: two fixed end

(finite length)

Three strings:

For which of these cases do you expect to have only certain

wavelengths allowed… that is for which cases will the allowed

wavelengths be quantized?

A. I only B. II only C. III only D. more than one

x

y

x

y

x

y

Case I: no fixed ends

Case II: one fixed end

Case III, two fixed end:

Three strings:

For which of these cases, do you expect to have only certain

frequencies or wavelengths allowed… that is for which cases will

the allowed frequencies be quantized.

a. I only b. II only c. III only d. more than one

Last class: quantization came when we applied 2nd boundary

condition, bound on both sides (answer c).

Electron bound

in atom (by potential energy) Free electron

Only certain energies allowed

Quantized energies Any energy allowed

PE

Boundary Conditions

standing waves,

fixed wavelength

No Boundary Conditions

traveling waves,

any wavelength allowed

t

txitxtxU

x

tx

m

),(),(),(

),(

2 2

22

The Schrodinger Equation

Once at the end of a colloquium Felix Bloch heard Debye saying something like: “Schrödinger, you are not working right now on very important problems…why don’t you tell us some time about that thesis of deBroglie, which seems to have attracted some attention?” So, in one of the next colloquia, Schrödinger gave a beautifully clear account of how deBroglie associated a wave with a particle, and how he could obtain the quantization rules…by demanding that an integer number of waves should be fitted along a stationary orbit. When he had finished, Debye casually remarked that he thought this way of talking was rather childish…To deal properly with waves, one had to have a wave equation.

t

txitxtxU

x

tx

m

),(),(),(

),(

2 2

22

The Schrödinger equation in 1D

Mass of particle

Potential Complex i,

with i2 = -1 space and time coordinates

t

txitxtxU

x

tx

m

),(),(),(

),(

2 2

22

The Schrödinger equation in 1D

This is just the conservation of energy… And let’s see if it will work for non-plane wave solutions for different potentials. The answer is it does and it does extremely well.

),(),(),( txEtxPEtxKE

Simplification #1 when U(x) only (or U(x,y,z) in 3D):

(x,t) separates into position part dependent part (x) and

time dependent part (t) =exp(-iEt/ħ). (x,t)= (x)(t)

Plug in, get equation for (x)

“Time independent Schrödinger equation”

Most physical situations (e.g. H atom) no time dependence in V!

)()()()(

2 2

22

xExxUx

x

m

t

txitxtxU

x

tx

m

),(),(),(

),(

2 2

22

Solving the Schrodinger equation for electron wave in 1D

1. Figure out what U(x) is, for situation given.

2. Guess or look up functional form of solution.

3. Plug in to check if ’s, and all x’s drop out, leaving equation

involving only bunch of constants; showing that trial solution

is correct functional form.

4. Figure out what boundary conditions must be to make sense

physically.

5. Figure out values of constants to meet

boundary conditions and normalization

6. Multiply by time dependence (t) =exp(-iEt/ħ) to have full

solution if needed. Ψ(x,t) STILL HAS TIME DEPENDENCE!

|(x)|2dx =1

-∞

∞

time independent

eq.

)()()(

)(

2 2

22

xExxUx

x

m

For a U(x) where does an electron want

to be?

Electron wants (will fall to) to be at position where

a. U(x) is largest

b. U(x) is lowest

c. KE > U(x)

d. KE < U(x)

e. where elec. wants to be does not depend on U(x)

Electron wants to be at position where

a. U(x) is largest

b. U(x) is lowest

c. Kin. Energy > U(x)

d. Kin. E. < U(x)

e. where elec. wants to be does not depend on U(x)

U(x)

x

electrons always want to go to position

of lowest potential energy, just like

ball going downhill.

c. and d. not right, because actual value of U(x) is

arbitrary, so can choose bigger or smaller than KE.

)()(

2 2

22

xEx

x

m

)()()()(

2 2

22

xExxUx

x

m

Example: simplest case, free space U(x) = const.

Smart choice:

constant U(x)

U(x) 0!

kxAx cos)(

Em

k

2

22

Solution:

with:

kxBx sin)( , or:

No boundary conditions

not quantized!

kxAx cos)(

2

2mk 2 E kp

So almost have solution, but remember still have to

include time dependence:

)()(),( txtx /)( iEtet

A bit of algebra, and identity ei = cos + isin gives:

k, and therefore E, can take on any value.

(x,t) Acos(kxt) Aisin(kxt)

So we found the solution for the time-independent

Schrödinger equation for the special case with U(x) = 0:

, ,

(Solution of time-independent Schrödinger eqn. with U(x)=0)

2

2m

2

x2(x) E(x)

If U=0, then E = Kinetic energy.

So first term in Schröd. Eq. is always just kinetic energy!

Which free electron has more kinetic energy?

a. 1., b. 2., c. same

big k = big KE

small k = small KE

More Negative Curvature KE. Bending tighter = more KE

1.

2.

x

x

• Review for exam 3 part 2.

Let’s do an estimate in real life!

-1.5

-1.0

-0.5

0.0

0.5

14 12 10 8 6 4 2 0

Small “non-rigid box” accidentally made by jamming an STM tip into a surface.

Profile of the box (along the blue line) all heights are in nm.

Using the uncertainty relationship estimate the energy of a bound state (if any) a) 10 eV b) 1 eV c) 0.1 eV d) <0.01 eV

0.7

0.6

0.5

0.4

-0.2 -0.1 0.0 0.1

Moore’s Law

The end of Moore's law?

As feature density goes up, device and line sizes must get smaller and smaller. Semiconductor chips are made with

optical lithographic techniques.

Current linewidths are 28-32 nm. These are essentially

at the diffraction limit for optical techniques using visible

light. Problem making features much smaller.

As device sizes get smaller and smaller, then intrinsic

quantum effects will get more and more important. This

may be good or bad.

Quantum dots, nanotubes, and all of “nanotechnology”.

Nanotechnology: how small does a wire have to be

before movement of electrons starts to depend on size

and shape due to quantum effects?

How to start?

Need to look at?

a. size of wire compared to size of atom

b. size of wire compared to size of electron wave function

c. Spacing between wires compared to wavelength of e-

d. Energy level spacing compared to thermal energy, kBT.

e. something else (what?) or more than one of the above.

Nanotechnology: how small does a wire have to be

before movement of electrons starts to depend on size

and shape due to quantum effects?

How to start?

Need to look at Energy level spacing compared to

thermal energy, kBT. kB=Boltzmann’s constant

Typically focus on energies in QM.

Electrons, atoms, etc. hopping around with random energy kBT.

kBT >> than spacing, spacing irrelevant. Quantum does not

play a big role. Quantum effects = notice the discrete energy

levels.

Quantum effects

not critical

Quantum effects

critical

+

PE

+ + + + + + + +

1 atom many atoms

but lot of e’s

move around

to lowest PE

repel other electrons = potential energy near that spot higher.

as more electrons fill in, potential energy for later ones gets

flatter and flatter. For top ones, is VERY flat.

+

PE for electrons with most PE. “On top”

As more electrons fill in, potential energy for later

ones gets flatter and flatter.

For top ones: U(x) is VERY flat.

+ + + + + + + + + + + + + + + +

PE

Good Approximation:

“Infinite square well” or “rigid box”

Electrons never get out of wire

(x<0 or x>L) =0.

(OK when Energy << work function) 0

U(x

)

x 0 L

U(x)

Approximation: the infinite square well

Exact Potential Energy curve (U)

“Finite square well” or “non-rigid box”

small chance electrons get out of wire

(x<0 or x>L)~0, but not exactly 0!

NOTE: Book uses “rigid box” for “infinite square well”

)()(

2 2

22

xEx

x

m

With B.C.: (0)=(L) =0

Last class: “infinite square well”

0

Energ

y

x 0 L

U(x) U(x) =

∞, x ≤ 0 0, 0<x<L ∞, x ≥ L

Solution:

2

2222

22 mL

n

m

pEn

kn = n / L, , (n=1,2,3,4 …)

n=1

n=2

n=3

(x)

/)sin(

2)()(),(

tiEneL

xn

Ltxtx

For 0<x<L:

/)sin(

2),(

tiEneL

xn

Ltx

Quantized: kn = n/L

Quantized: 1

2

2

222

2En

mLnEn

n=2

What you expect classically:

Electron can have any

energy

Electron is localized

If it has KE, it bounces

back and forth between the

walls

Electron is delocalized … spread out between 0 and L

What you get quantum mechanically:

Electron can only have specific energies. (quantized)

Electron does not “bounce”

from one wall to the other.

mathematically

U(x) = 4.7 eV for x<0 and x>L

U(x) = 0 eV for 0<x<L 0 eV 0 L

4.7 eV

What does that say about boundary condition on (x) ?

A. (x) is about the same everywhere

B. (x) 0 everywhere, except for 0<x<L

C. (x) = 0 everywhere, except for 0<x<L

D. (x) 0 for 0<x<L, and (x)=0 everywhere else.

E. Can’t tell anything yet. Need to find (x) first.

x

Need to solve for exact potential energy curve:

Finite U(x): small chance electrons get out of

wire, so: (x<0 or x>L)~0, but not exactly 0!

Energ

y

0 L

U(x)

The Finite Square Well

~Work function

Thin wire along x-axes

x

U=0 eV 0 L

4.7 eV

Energ

y

x

Eelectron

x

III Bex a )(

Inside well (E>V):

(Region II)

)()( 2

2

2

xkdx

xdII

II

)()( 2

2

2

xdx

xdIII

III a

Outside well (E<V):

(Region III)

)cos()sin()( kxDkxCxII

Outside well

(E<V):

(Region I)

On HW

downward) (curves 00)(

upward) (curves 00)(

2

2

2

2

dx

dx

dx

dx

Regions I and III:

x or: or:

ψ(x): smooth & continuous!

U=0 eV 0 L

4.7 eV

Energ

y

x

Eelectron

)( )( LL IIIII 1) ψ(x) has to be continuous:

2) ψ(x) has to be smooth:

3) ψ(|x| ) 0 (required for normalization)

Boundary conditions

)(' )(' LL IIIII

(similar conditions for x=0)

)()(2)(

22

2

xEUm

dx

xd

Electron is delocalized … spread out.

Some small part of wave is where total

energy is less than potential energy!

U=0 eV 0 L

4.7 eV E

nerg

y

x

Eelectron

“Classically forbidden” regions (ψ(x): exponentially decaying)

“Classically allowed”

(ψ(x): sinusoid)

0 L

Eelectron

wire

How far does wave extend

into this “classically

forbidden” region?

)()()(2)( 2

22

2

xxEUm

dx

xda

Measure of penetration depth: 1/a

when decreases by factor of 1/e

For U-E = 4 eV, 1/a ~ 0.1 nm (very small ~ an atom!!!)

)()( LxBex a Big a -> quick decay:

Small a -> slow decay:

)(L

eL /1*)(

1/a

)(2

2EU

m

a

Consider the ground state wavefunction for an electron

in an infinite square well of length L and a finite square

well of length L (same L for both cases). What can you

say about the ground state energies?

0

Energ

y

x 0 L

U(x)

a)They are the same

b) Finite well has lower energy.

c) Infinite well has lower energy.

2

22

12mL

E

Remember for the infinite square well:

0

Energ

y

x 0 L

V(x)

a)They are the same

b) Finite well has lower energy.

c) Infinite well has lower energy.

Consider the ground state wavefunction for an electron

in an infinite square well of length L and a finite square

well of length L (same L for both cases). What can you

say about the ground state energies?

2

22

12mL

E

Remember for the infinite square well:

0 L

Eelectron

Review: ‘Penetration depth’

)(2

2EU

m

a

E<V: Classically forbidden region

)()( LxBex a , with

)(L

eL /)(

1/a

Penetration depth: Distance 1/a

over which the wave function

decays to 1/e of its initial value at

the potential boundary (here (L)):

For U-E = 4 eV, 1/a ~ 0.5 nm (very small ~ an atom!!!)

x

0 L

xBex a )(

)(2

2EU

m

a

What changes would increase how far the wave penetrates into the classically forbidden region?

with:

A. Decrease potential depth (= work-function of metal)

B. Increase potential depth

C. Decrease wire length L

D. Increase wire length L

E. More than one of the above

Wide vs. narrow finite potential well

2

22

12 effmL

E

L

U-E

U

)(2

2EU

m

a

Smaller U-E smaller a larger penetration depth

STM (picture with reversed voltage, works exactly the same)

end of tip always

atomically sharp

Use tunneling to measure very(!) small changes in distance.

Nobel prize winning idea: Invention of “scanning tunneling

microscope (STM)”. Measure atoms on conductive surfaces.

Application of quantum tunneling: Scanning

Tunneling Microscope 'See' single atoms!

Measure current

between tip and

sample

Current vs. Position.

• Tuneling

• Controversy

• STM now

• Superconductors

STM (picture with reversed voltage, works exactly the same)

end of tip always

atomically sharp

Radioactive decay (Quantum tunneling – George Gamow)

Nucleus is unstable ejects alpha particle (2 netrons, 2 protons)

Typically found for large atoms with lots of protons and neutrons.

Polonium-210

84 protons,

126 neutrons

Neutron (no charge) Proton (positive charge)

Nucleus has lots of protons and lots of

neutrons.

Two forces acting in nucleus:

- Coulomb force .. Protons really close

together, so very big repulsion between

protons due to coulomb force.

- Nuclear force (attraction between nuclear

particles is very strong if very close

together) … called the STRONG Force.

Potential energy curve for the α particle

Nucleus

(Z protons,

& bunch of neutrons)

New nucleus

(Z-2 protons,

bunch of neutrons)

+ Alpha particle

(2 protons,

2 neutrons)

Look at this system… as the

distance between the alpha particle

and the nucleus changes.

V=0 for r ∞

As we bring the a particle closer to the core,

what happens to potential energy?

KE

U(r)

r

Coulomb repulsion:

r

eeZk

r

qkqrU

)2)()(2()( 21

Nucleus:

(Z-2) protons

Strong attractive force

(Nuclear forces)

U(r)

x

What’s the kinetic energy of this particle inside the

nucleus?

A

B

C

D

E: Something else

Ene

rgy

What would the kinetic energy of that particle be after it

tunneled out from the nucleus?

x

A

B

C

D

E: Something else

U(r) E

nerg

y

So we found that the particle has less kinetic energy

outside than inside the nucleus. Did it loose energy?

x

KEoutside

KEinside A) Yes.

B) No.

C) Impossible to tell. Need to

solve Schröd. equ. first.

U(r) E

nerg

y

Wave function picture:

~1-10MeV of KE

outside

~100MeV

of KE inside

the nucleus

Exponential decay in the barrier

Wave function of the particle

inside the potential well: Large

KE small Wavelength

Wave function of the free particle:

‘small’ KE Large wavelength

U(r) E

nerg

y

Observe a particles from different isotopes (same # protons,

different # neutrons), exit with different amounts of energy.

30 MeV

4MeV KE 9MeV KE

The 9 MeV electron more probable…

1. Less distance to tunnel 2. U-E is smaller (smaller a) Wave function doesn’t decay as much before reaches other side … more probable!

Isotopes that emit higher energy alpha

particles, have shorter lifetimes!!!

)(2

2EU

m

a

x

U(r)

Energ

y

Solving Schrodinger

equation for this

potential energy is hard!

V(x) V(x)

Square barrier is much easier…

and get almost the same answer!

The “classically forbidden regions” are where … a. a particle’s total energy is less than its kinetic energy b. a particle’s total energy is greater than its kinetic energy c. a particle’s total energy is less than its potential energy d. a particle’s total energy is greater than its potential energy

Answer is c.

E=hc/…

A. …is true for both photons and electrons.

B. …is true for photons but not electrons.

C. …is true for electrons but not photons.

D. …is not true for photons or electrons.

E = hf is always true but f = c/ only applies to light, so E = hf ≠ hc/ for electrons.

c = speed of light!