Quantum Mechanics 102
Tunneling and its Applications
Interference of Waves and the
Double Slit Experiment

Waves spreading out from two points, such as
waves passing through two slits, will interfere
l
d
Wave crest
Wave trough
Spot of
constructive
interference
Spot of
destructive
interference
Interpretation



The probability of finding a particle in a particular
region within a particular time interval is found by
integrating the square of the wave function:
 P (x,t) =  |Y(x,t)|2 dx =  |c(x)|2 dx
|c(x)|2 dx is called the “probability density; the
area under a curve of probability density yields the
probability the particle is in that region
When a measurement is made, we say the wave
function “collapses” to a point, and a particle is
detected at some particular location
Particle in a box
c(x) = B sin (npx/a)
n=3
c(x)
n=2
|c(x)|2
certain wavelengths l = 2a/n are allowed
 Only certain momenta p = h/l = hn/2a are allowed
 Only certain energies E = p2/2m = h2n2/8ma2 are
allowed - energy is QUANTIZED
 Allowed energies depend on well width
 Only
What about the real world?




Solution has non-trivial form, but only certain
states (integer n) are solutions
Each state has one allowed energy, so energy is
again quantized
Energy depends on well width a
Can pick energies for electron by adjusting a
|c(x)|2
n=2
n=1
x
Putting Several Wells Together
How does the number of energy bands
compare with the number of energy levels
in a single well?
 As atom spacing decreases, what happens to
energy bands?
 What happens when impurities are added?

Quantum wells

An electron is trapped since no empty energy
states exist on either side of the well
Escaping quantum wells


Classically, an electron could gain thermal energy and
escape
For a deep well, this is not very probable
Escaping quantum wells


Thanks to quantum mechanics, an electron has a non-zero
probability of appearing outside of the well
This happens more often than thermal escape
What if free electron
encounters barrier?
Do Today’s Activity
What Have You Seen?
What happens when electron energy is less
than barrier height?
 What happens when electron energy is
greater than barrier height?
 What affects tunneling probability?

T  e–2kL
k = [8p2m(Epot – E)]½/h
A classical diode



According to classical physics, to get to the holes on the
other side of the junction, the conduction electrons must
first gain enough energy to get to the conduction band on
the p-side
This does not happen often once the energy
barrier gets large
Applying a bias increases
the current by decreasing
the barrier
A tunnel diode



According to quantum physics, electrons could tunnel
through to holes on the other side of the junction with
comparable energy to the electron
This happens fairly often
Applying a bias moves the
electrons out of the p-side
so more can tunnel in
Negative resistance



As the bias is increased, however, the energy of the empty
states in the p-side decreases
A tunneling electron would then end up in the band gap no allowed energy
So as the potential difference is increased, the current
actually decreases = negative R
No more negative resistance


As bias continues to increase, it becomes easier for
conduction electrons on the n-side to surmount the energy
barrier with thermal energy
So resistance becomes positive again
The tunneling transistor
• Only electrons with energies equal to the energy
state in the well will get through
The tunneling transistor
• As the potential difference increases, the energy levels on the
positive side are lowered toward the electron’s energy
• Once the energy state in the well equals the electron’s energy,
the electron can go through, and the current increases.
The tunneling transistor
• The current through the transistor increases as each successive
energy level reaches the electron’s energy, then decreases as the
energy level sinks below the electron’s energy
Randomness


Consider photons going through beam splitters
NO way to predict whether photon will be
reflected or transmitted!
(Color of line is
NOT related to
actual color of
laser; all beams
have same
wavelength!)
Randomness Revisited


If particle/probabilistic theory correct, half the
intensity always arrives in top detector, half in
bottom
BUT, can move mirror so no light in bottom!
(Color of line is
NOT related to
actual color of
laser; all beams
have same
wavelength!)
Interference effects



Laser light taking different paths interferes,
causing zero intensity at bottom detector
EVEN IF INTENSITY SO LOW THAT ONE
PHOTON TRAVELS THROUGH AT A TIME
What happens if I detect path with bomb?
No
interference,
even if bomb
does not
detonate!
Interpretation




Wave theory does not explain why bomb detonates half the
time
Particle probability theory does not explain why changing
position of mirrors affects detection
Neither explains why presence of bomb destroys
interference
Quantum theory explains both!



Amplitudes, not probabilities add - interference
Measurement yields probability, not amplitude - bomb
detonates half the time
Once path determined, wavefunction reflects only that
possibility - presence of bomb destroys interference
Quantum Theory meets Bomb



Four possible paths: RR and TT hit upper detector,
TR and RT hit lower detector (R=reflected,
T=transmitted)
Classically, 4 equally-likely paths, so prob of each
is 1/4, so prob at each detector is 1/4 + 1/4 = 1/2
Quantum mechanically, square of amplitudes must
each be 1/4 (prob for particular path), but
amplitudes can be imaginary or complex!

e.g.,
1
1
1 i
1 i
Y  TR 
RT 
RR 
TT
2
2
2 2
2 2
Adding amplitudes
1
1
1 i
1 i
Y  TR 
RT 
RR 
TT
2
2
2 2
2 2
1 1
Y  
0
2 2
2
2

Lower detector:

1 i 1 i
2  2i
Upper detector: Y 


1
2 2 2 2
2 2
2
2
2
What wave function would
give 50% at each detector?
Y  a TR  b RT  c RR  d TT

Must have |a| = |b| = |c| = |d| = 1/4

Need |a + b|2 = |c+d|2 = 1/2
Y
1
2 2
TR 
1
2 2
RT 
i
2 2
RR 
i
2 2
TT