Download Exam 3 review

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

(wavelength)
• 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
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
0.16
2.4x10-20
200
2.4x10-22
2.8x10-12
Thermal C60
molecule
1.2x10-24 kg
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 − 𝑢0 2 : 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 =
𝑛 𝐴𝑛 𝑒
𝑖𝑣𝑛 𝑢
• Then there is an uncertainty relationship:
1
Δ𝑢Δ𝑣 ≥
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
medium Δp – wave packet made of several waves
Δx
large Δp – wave packet made of lots of 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:
1
Δ𝜔Δ𝑡 ≥
2
hω
2𝜋
Remember for light : E = hf =
= ℏ𝜔
ℏ
∆𝐸∆𝑡 ≥
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)
If all you know is fish, how do you describe a moose?
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
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 inphase 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
e
2
out
1
spontaneous
emission of light:
Excited atom emits
one photon.
2
in
e
1
absorption of
light: Atom absorbs
one photon
e
in
2
out
1
stimulated
emission:
clone the photon
-- A. Einstein
1916
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.
Laser: Stimulated emission to clone photon many times (~1020/s)
Light Amplification by Stimulated Emission of Radiation
Stimulated emission
Identical phase
Identical energy
Identical momentum
Legend:
Photon
Atom in ground state
Atom in excited state
e
e
Spontaneous emission
Random phase
Random direction
Similar energy (as absorbed photon)
Legend:
Photon
Atom in ground state
Atom in excited state
e
e
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
A) Use photons with hf < ΔE
B) Use photons with hf = ΔE
ΔE
e
excited
C) Use photons with hf > ΔE
D) Use very strong lamp with hf ≈ ΔE.
E) Will never get population inversion in this system.
not excited
No population inversion in 2 level atom!
e
2
out
1
spontaneous
emission of light:
Excited atom emits
one photon.
2
in
e
1
absorption of
light: Atom absorbs
one photon
e
in
2
out
1
stimulated
emission:
clone the photon
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
Mirror
Half-silvered 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 (2r = n), gives:
2r = nh/p
• Or, rearranging:
pr = nh/2
L = nħ
• deBroglie EXPLAINS quantization of angular
momentum, and therefore EXPLAINS quantization of
energy!
Three strings:
Case I: no fixed ends
(infinite length)
y
x
y
Case II: one fixed end
(infinite length)
x
y
Case III: two fixed end
(finite length)
x
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
Three strings:
Case I: no fixed ends
y
x
y
Case II: one fixed end
x
y
Case III, two fixed end:
x
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
PE
Only certain energies allowed
Quantized energies
Any energy allowed
Boundary Conditions
 standing waves,
fixed wavelength
No Boundary Conditions
 traveling waves,
any wavelength allowed
The Schrodinger Equation
   ( x, t )
 ( x, t )

 U ( x , t )  ( x , t )  i
2
2m x
t
2
2
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.
The Schrödinger equation in 1D
   ( x, t )
 ( x, t )

 U ( x , t )  ( x , t )  i
2
2m x
t
2
Mass of
particle
2
Potential
space and time
coordinates
Complex i,
with i2 = -1
The Schrödinger equation in 1D
   ( x, t )
 ( x, t )

 U ( x , t )  ( x , t )  i
2
2m x
t
2
2
KE   ( x, t )  PE   ( x, t )  E   ( x, t )
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.
   ( x, t )
 ( x, t )

 U ( x , t )  ( x , t )  i
2
2m x
t
2
2
Most physical situations (e.g. H atom) no time dependence in V!
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)
   ( x)

 U ( x) ( x)  E ( x)
2
2m x
2
2
“Time independent Schrödinger equation”
Solving the Schrodinger equation for electron wave in 1D
   ( x)

 U ( x) ( x)  E ( x)
2
2m x
2
2
time independent
eq.
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
|(x)|2dx =1
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!
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.
Example: simplest case, free space U(x) = const.
   ( x)

 U ( x) ( x)  E ( x)
2
2m x
2
2
   ( x)

 E ( x)
2
2m x
2
Smart choice:
constant U(x)
 U(x)  0!
2
Solution:
 ( x)  A cos kx, or:  ( x)  B sin kx
2 2
with:
k
E
2m
No boundary conditions
 not quantized!
So we found the solution for the time-independent
Schrödinger equation for the special case with U(x) = 0:
 ( x)  A cos kx ,
2
2m
k  E,
2
p  k
k, and therefore E, can take on any value.
So almost have solution, but remember still have to
include time dependence:

( x, t )   ( x) (t )
 (t )  e
 iEt / 
A bit of algebra, and identity ei = cos + isin gives:
(x,t)  Acos(kx  t)  Ai sin(kx  t)
(Solution of time-independent Schrödinger eqn. with U(x)=0)
Which free electron has more kinetic energy?
a. 1., b. 2., c. same
big k = big KE
1.
2.
x
small k = small KE
x
If U=0, then E = Kinetic energy.
So first term in Schröd. Eq. is always just kinetic energy!
2
2



(x)

E

(x)
2
2m x
More Negative Curvature  KE. Bending  tighter = more KE
• Review for exam 3 part 2.
Let’s do an estimate in real life!
0.5
0.0
-0.5
-1.0
-1.5
0
Small “non-rigid box” accidentally
made by jamming an STM tip into a
surface.
2
4
6
8
10 12 14
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 ed. 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
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.
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!
U(x)
U(x)
0
0
L
Good Approximation:
“Infinite square well” or “rigid box”
Electrons never get out of wire
(x<0 or x>L) =0.
x (OK when Energy << work function)
NOTE:
Book uses “rigid box” for “infinite square well”
Last class: “infinite square well”
U(x)
U(x) =
Energy
(x)
n=3
For 0<x<L:
   ( x)

 E ( x)
2
2m x
2
n=2
n=1
0
∞, x ≤ 0
0, 0<x<L
∞, x ≥ L
0
Solution:
2
With B.C.: (0)=(L) =0
L
x
( x, t )   ( x ) (t ) 
2
nx iEnt / 
sin(
)e
L
L
p 2 n 2 22
kn = n / L, En 

, (n=1,2,3,4 …)
2
2m
2mL
( x, t ) 

2
nx iEnt / 
sin(
)e
L
L
Quantized: kn = n/L
n=2
Quantized: En  n
What you expect classically:
Electron can have any
energy
Electron is localized
If it has KE, it bounces
back and forth between the
walls
2 2


2
2mL2
 n 2 E1
What you get quantum
mechanically:
Electron can only have specific
energies. (quantized)
Electron is delocalized
… spread out between 0 and L
Electron does not “bounce”
from one wall to the other.
4.7 eV
mathematically
U(x) = 4.7 eV for x<0 and x>L
U(x) = 0 eV for 0<x<L
0 eV
0
L x
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.
The Finite Square Well
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!
Energy
U(x)
0
~Work function
L
Thin wire along x-axes
x
Outside well
(E<V):
(Region I)
On HW
Outside well (E<V):
(Region III)
d  III ( x)
d  II ( x)
2
2

a
 III ( x)
 k  II ( x)
2
2
dx
dx
 II ( x)  C sin(kx) D cos(kx)
 III ( x)  Be ax
2
2
Energy
4.7 eV
Inside well (E>V):
(Region II)
Eelectron
U=0 eV
0
L
x
ψ(x): smooth & continuous!
Regions I and III:

d 2
 ( x)  0 
 2  0 (curves upward)
dx
or:
or:
x
2
d
 ( x)  0 
 2  0 (curves downward)
dx
Boundary conditions
Energy
4.7 eV
Eelectron
U=0 eV
0
L
1) ψ(x) has to be continuous:  II ( L)   III ( L)
2) ψ(x) has to be smooth:  ' II ( L)   ' III ( L)
3) ψ(|x|
)
0 (required for normalization)
(similar conditions for x=0)
x
d 2 ( x) 2m
 2 (U  E ) ( x)
2
dx

“Classically allowed”
(ψ(x): sinusoid)
Energy
4.7 eV
Eelectron
U=0 eV
0
L
“Classically forbidden” regions
(ψ(x): exponentially decaying)
Electron is delocalized … spread out.
Some small part of wave is where total
energy is less than potential energy!
x
 (L)
 ( L) *1 / e
1/a
Eelectron
0
L
d 2 ( x) 2m
2

(
U

E
)

(
x
)

a
 ( x)
2
2
dx

wire
How far does wave extend
into this “classically
forbidden” region?
 ( x)  Be
a ( x  L )
2m
a
(U  E )
2

Big a -> quick decay:
Small a -> slow decay:
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!!!)
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?
U(x)
Energy
a)They are the same
b) Finite well has lower energy.
c) Infinite well has lower energy.
0
0
L
x
Remember for the infinite square well: E1 
 22
2mL2
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?
V(x)
Energy
a)They are the same
b) Finite well has lower energy.
c) Infinite well has lower energy.
0
0
L
x
Remember for the infinite square well: E1 
 22
2mL2
Review: ‘Penetration depth’
E<V: Classically forbidden region
Eelectron
0
L
 ( x)  Be
a ( x  L )
Penetration depth: Distance 1/a
over which the wave function
decays to 1/e of its initial value at
the potential boundary (here (L)):
2m
(U  E )
, with a 
2


 (L)
 ( L) / e
1/a
x
For U-E = 4 eV, 1/a ~ 0.5 nm (very small ~ an atom!!!)
 ( x)  Be
ax
2m
(U  E )
with: a 
2

0
L
What changes would increase how far the wave penetrates
into the classically forbidden region?
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
U
U-E
E1 
 
2 2
L
2mL2eff
2m
a
(U  E )
2

Smaller U-E  smaller a  larger penetration depth
STM (picture with reversed voltage, works exactly the same)
end of tip always
atomically sharp
Application of quantum tunneling: Scanning
Tunneling Microscope  'See' single atoms!
Use tunneling to measure very(!) small changes in distance.
Nobel prize winning idea: Invention of “scanning tunneling
microscope (STM)”. Measure atoms on conductive surfaces.
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
Proton (positive charge)
Neutron (no 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
+
KE
New nucleus
Nucleus
Alpha particle
(Z-2 protons,
(Z protons,
(2 protons,
& bunch of neutrons) bunch of neutrons) 2 neutrons)
Strong attractive force
(Nuclear forces)
U(r)
Look at this system… as the
distance between the alpha particle
and the nucleus changes.
As we bring the a particle closer to the core,
what happens to potential energy?
r
Coulomb repulsion:
Nucleus:
(Z-2) protons
kq1q2 k ( Z  2)(e)(2e)
U (r ) 

r
r
V=0 for r  ∞
Energy
What’s the kinetic energy of this particle inside the
nucleus?
U(r)
D
C
B
x
A
E: Something else
Energy
What would the kinetic energy of that particle be after it
tunneled out from the nucleus?
U(r)
D
C
B
x
A
E: Something else
Energy
So we found that the particle has less kinetic energy
outside than inside the nucleus. Did it loose energy?
U(r)
KEoutside
x
KEinside
A) Yes.
B) No.
C) Impossible to tell. Need to
solve Schröd. equ. first.
Energy
Wave function picture:
U(r)
~100MeV
of KE inside
the nucleus
Exponential decay in the barrier
~1-10MeV of KE
outside
Wave function of the free particle:
‘small’ KE  Large wavelength
Wave function of the particle
inside the potential well: Large
KE  small Wavelength
Observe a particles from different isotopes (same # protons,
different # neutrons), exit with different amounts of energy.
2m
a
(U  E )
2

U(r)
Energy
30 MeV
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!
9MeV KE
4MeV KE
x
The 9 MeV electron more probable…
Isotopes that emit higher energy alpha
particles, have shorter lifetimes!!!
Solving Schrodinger
equation for this
potential energy is hard!
V(x)
Square barrier is much easier…
and get almost the same answer!
V(x)
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.
B.
C.
D.
…is true for both photons and electrons.
…is true for photons but not electrons.
…is true for electrons but not photons.
…is not true for photons or electrons.
c = speed of light!
E = hf is always true but f = c/ only applies to
light, so E = hf ≠ hc/ for electrons.