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

Today…

…everything you ever wanted to know

about the Schrödinger/Shroedinger

equation – but never dared to ask.

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

Let’s put Ψ 𝑥, 𝑡 = 𝐴𝑒𝑖 𝑘𝑥−𝜔𝑡 and see what we get.

2

22 ),(

2 x

tx

m

First term

First: What is the derivative of Ψ 𝑥, 𝑡 =

𝐴𝑒𝑖 𝑘𝑥−𝜔𝑡 w.r.t. x?

a) 𝜕Ψ 𝑥,𝑡

𝜕𝑥= Ψ 𝑥, 𝑡

b) 𝜕Ψ 𝑥,𝑡

𝜕𝑥= −𝑘𝑥Ψ 𝑥, 𝑡

c) 𝜕Ψ 𝑥,𝑡

𝜕𝑥= 𝑖𝑘𝑥Ψ 𝑥, 𝑡

d) 𝜕Ψ 𝑥,𝑡

𝜕𝑥= 𝑖𝑘Ψ 𝑥, 𝑡

e) 𝜕Ψ 𝑥,𝑡

𝜕𝑥= 𝑖𝑥Ψ 𝑥, 𝑡

Back to putting Ψ 𝑥, 𝑡 = 𝐴𝑒𝑖 𝑘𝑥−𝜔𝑡

),(2

)(),()(

2

),(

2

22

2

2

22

txm

ktxik

mx

tx

m

First term

Now remember 𝑘 =2𝜋

𝜆 and 𝜆 =

ℎ

𝑝. So 𝑘 =

2𝜋

ℎ/𝑝= 𝑝/ℏ .

Putting that into the first term:

),(2

),(2

)/(),(

2

22

2

22

txm

ptx

m

p

x

tx

m

),(),(

2 2

22

txKEx

tx

m

That looks like the

Kinetic energy!

t

txitxtxU

x

tx

m

),(),(),(

),(

2 2

22

The Schrödinger equation in 1D

t

txitxtxUtxKE

),(),(),(),(

Second term is just PE written as “potential” so

Same thing, for plane waves with debroglie, as:

t

txitxPEtxKE

),(),(),(

t

txi

),(

Right hand side?

Remember ω = 2𝜋𝑓, ℏ =ℎ

2𝜋 and E=hf:

Let’s put Ψ 𝑥, 𝑡 = 𝐴𝑒𝑖 𝑘𝑥−𝜔𝑡 and see what we get:

),(),()(),(

txtxiit

txi

),(),(),(

txEtxhft

txi

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

Example: What is the most general case for U(r,t) for an

electron interacting with a proton? + -

A. -ke2/r, where r is the distance between and origin.

B. -ke2/r where r is distance between + and - .

C. (-ke2/r)·(sint), with =ang. freq. of electron around proton.

D. Impossible to tell unless know how electron is moving.

E. Can’t figure out what time dependence should be.

ans. b. although potential energy will be different as

electron moves to different distance, at any given distance

will be same for all time. So U(r,t) = U(r) = -ke2/r.

H atom.

t

txitxtxU

x

tx

m

),(),(),(

),(

2 2

22

Here: k = 1/(4πε0) : The coulomb force constant

+ -

-

The 3D Schrodinger equation for hydrogen atom

t

tzyxitzyxzyxke

zm

tzyx

ym

tzyx

xm

tzyx

),,,(),,,(])/([

2

),,,(

2

),,,(

2

),,,(

2/12222

2

22

2

22

2

22

Schrodinger solved it for the hydrogen atom and got

energy levels that matched the experiments.

(We’ll get back to this.)

Simplification #1 when V(x) only (or V(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)

Seems like a good exercise for HW ;)

“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

(Will use this eqn. in all Schrödinger Eq’n problems in this class!)

Example: How to use the Schrödinger equation

Say you wanted to calculate electron waves in a thin

wire (1-D). What’s the first step you’d take?

a. figure out how many electrons will be interacting

b. figure out what general solutions will be by

plugging in trial solutions and see if it works.

c. figure out what the forces will be on the electron in

that physical situation.

d. figure out what the boundary conditions must be

on the electron wave.

e. figure out what potential energy is at different x

and t for the physical situation.

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

This approach is incredibly

successful! Explains Hydrogen atom:

Energy levels and more (bonding, selection

rules, “fine structure”, correct angular

momenta, wavefunction shapes)

Other atoms:

The structure of the periodic table

spectra of other atoms

Bonding of atoms

Electronic structure of solids

semiconductors, metals, insulators

superconductors and more

The list would take longer to enumerate

than we have time for.

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)

Using equation, probability of electron being in dx at x = L is

_______ than probability of being in dx at x = 0.

a. always bigger, b. always same, c. always smaller

d. oscillates up and down in time between bigger and smaller

e. without being given k, can’t figure out

if k positive

x = 0 x = L

ans. b. Prob ~ * = A2cos2(kx-t) +A2sin2(kx-t)= A2,

so constant and equal, for all x and t.

Electron in free space or long wire with no voltage

For one value of k, equal chance to find electron anywhere.

‘Real’ electron: wave packet, many k’s.

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

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

Curvature KE. Bending tighter = more KE

1.

2.

x

x

When is small ‘small’?

(When does quantum mechanics matter?)

Does it matter

here? How to

find out?

Moore’s Law

Why does Moore’s Law work?

Why can we get about twice the computing power

every other year for the same amount of money?

Intel and AMD (and many others) can pack

about twice the number of transistors on the

same chip-area with every new chip-generation.

A reduction of the structure size by leads to

twice the number of transistors on the same area

for about the same cost. Cool!

(Intel: 2003: 90nm, 2005: 65nm, 2007: 45nm,

2009: 32nm, 2011: 22nm)

2

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

Nanotechnology: how small (short) does a wire have to be

before movement of electrons starts to depend on size

and shape due to quantum effects?

Look at energy level spacing compared to thermal energy,

kBT= ~25 meV at room temp. kB= Boltzmann’s constant

Calculate energy levels for electron in wire of length L.

We know the spacing is big for 1 atom. What L for ~25 meV?

0 L

?

E

Figure out U(x), then figure out

how to solve, what solutions

mean physically.

)()()()(

2 2

22

xExxUx

x

m

Use time independ. Schrod. eq.

Want this to be

~25 meV. What L

do we need?

Short copper wire, length L.

What is U(x)?

0 L

Remember photoelectric effect.

Took energy to kick electron out. So wants to be inside wire.

inside is lower PE.

Everywhere inside the same?

)()()()(

2 2

22

xExxUx

x

m

+

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

PE for electrons with most PE. “On top”

How could you find out how deep the pit is for the

top electrons in copper wire?

This is just the energy needed to remove them from

the metal. That is the work function!!

work function of

copper = 4.7 eV

E

Reasoning to simplify how to solve.

Electron energy not much more than ~kT=0.025 eV.

Where is electron likely to be?

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

B. 0.025 eV<< 4.7eV. So very small chance (e-4.7/.025) an

electron could have enough energy get out.

A. zero chance

B. very small chance (<50% chance)

C. 50:50 chance

D. very likely (>50% chance)

What is chance it will be outside of well?

x