Physics 214 Lecture 10

1
A vast time bubble has been projected into the
future to the precise moment of the end of the
universe. This is, of course, impossible.
--D. Adams, The Hitchhikers Guide to the
Galaxy
There is light at the end of the tunnel. --
proverb
The light at the end of the tunnel is just the
light of an oncoming train. --R. Lowell
2
Physics Colloquium
Special (Optional) Lecture

• Quantum Information
• One of the most modern applications of QM
• quantum computing
• quantum communication cryptography,
  teleportation
• quantum metrology
• PGK will give a special 214-level lecture on this
  topic
• Sunday, Feb. 17
• 4 pm, 151 Loomis
• Attendance is optional, but encouraged.

• Spin Based Test-beds for Quantum Information
  Processing
• Professor David G. Cory Massachusetts Institute
  of Technology
• Department of Nuclear Science and Engineering
• February 14, 2008400 pm, Room 141 Loomis Lab

3
Next week Laboratory 4
4
Overview of the rest of the course This week
the time-dependent Schr. Eq. Tunneling,
radioactive decay, Schrodingers cat, . . . Next
week quantum states in 3 dimensions, electron
spin H atom, exclusion principle, periodic
table of atoms Last week molecules and solids,
consequences of Q. M. Metals, insulators,
semiconductors, superconductors, lasers, . .
Good web site for animations http//www.falstad.co
m/qm1d/
5
Barrier Penetration and Tunneling
x
6
Overview

• Tunneling of quantum particles
• Scanning Tunneling Microscope (STM)
• Nuclear Decay
• Solar Fusion
• Tunneling of atoms in the NH3 molecule the NH3
  Maser

7
Leaky Particles Revisited

• Due to barrier penetration, the electron
  density of a metal actually extends outside the
  surface of the metal!

• Assume that the work function (i.e., the energy
  difference between the most energetic conduction
  electrons and the potential barrier at the
  surface) of a certain metal is F 5 eV. Estimate
  the distance x outside the surface of the metal
  at which the electron probability density drops
  to 1/1000 of that just inside the metal.

using
8
Application Tunneling Microscopy

• Due to the quantum effect of barrier
  penetration, the electron density of a material
  extends beyond its surface

One can exploit this to measure the electron
density on a materials surface
Na atoms on metal
http//www.quantum-physics.polytechnique.fr/en/
9
Tunneling Key points

• In quantum mechanics a particle can penetrate
  into a barrier where it would be classically
  forbidden
• We saw the effect in lecture 8, on the wave
  function Y(x) for a finite square well
• We showed that in the barrier where E lt U, Y(x)
  has an exponential form D1 e-Kx D2 eKx
• We showed that either D1 or D2 0, but we did
  not solve the equations too hard!
• You did it using the computer in Lab 3
• The most important result was that the
  probability of finding the particle in the
  barrier region decreases as e-2Kx in the barrier
• This is a property of waves e.g., total
  internal reflection of light- wave decays
  exponentially
• Today we consider a related problem a particle
  approaching a barrier and tunneling through the
  barrier
• The results are very similar, and again part of
  the problem is too hard to solve completely in
  this course
• The probability of the particle tunneling through
  the barrier is proportional to e-2KL where L is
  the width of the barrier

10
Tunneling Through a Barrier (1)

• Find Transmission Coefficient, T,the
  probability an incident particle tunnels through
  the barrier. The particle can berepresented by
  a packet witha narrow range of k, coming in only
  from the left (positive k).For a single k, how
  much transmitted density is there, compared to
  the incident density?

We solve for y in each of the three regions shown.
reflected
in
transmitted
11
Tunneling Through a Barrier (2)

• A1 doesnt matter- it just sets the average
  intensity of the incoming wave, and what were
  after is just the fraction that goes through.
  Multiplying A1 by a constant would just multiply
  everything by the same constant, and not affect
  ratios.
• So that leaves A2, B1, B2, and C1 as the FOUR
  UNKNOWNS.
• Next we would need to apply the continuity
  conditions for both y and dy/dx at the
  boundaries x 0 and x L to determine the A, B,
  and C coefficients ? FOUR EQUATIONS.

• In general the tunneling coefficient T can be
  quite complicated (due to the contribution of
  amplitudes reflected off the far side of the
  barrier).

• However, in many situations, the barrier width L
  is much larger than the decay length 1/K of the
  penetrating wave KL gtgt 1. In this case we can
  set the B1 term to zero, and the reflected
  amplitude A21, making the approximate solution
  relatively simple.

12
Tunneling Through a Barrier (3)
The case where KLgtgt1, Tltlt1.

• This is nearly the same result as in the leaky
  particle example! Except for G
• G slightly modifies the
• transmission probability
• G arises from the fact
• that the amplitude at
• x 0 is not a maximum

13
Tunneling Through a Barrier (4)
where

• The main effect for big barriers is the decaying
  exponential

• T depends on the energy below the barrier (U0-E)
  and on the barrier width L.

• The plot illustrates how the transmission
  coefficient T changes as a function of barrier
  width L, for two different values of the particle
  energy.

Some references omit G. We will state when you
can ignore G.
14
Example Barrier Tunneling in an STM

• Lets consider a simple problem
• An electron with a total energy of E6 eV
  approaches a potential barrier with a height of
  Uo 12 eV. If the width of the barrier is
  L0.18 nm, what is the probability that the
  electron will tunnel through the barrier?

15
Example Barrier Tunneling in an STM

• Lets consider a simple problem
• An electron with a total energy of E6 eV
  approaches a potential barrier with a height of
  Uo 12 eV. If the width of the barrier is
  L0.18 nm, what is the probability that the
  electron will tunnel through the barrier?

Question What will T be if we double the width
of the gap?
16
Lecture 10, Act 1
Consider a particle tunneling through a barrier.
1. Which of the following will increase the
likelihood of tunneling? a. decrease the
height of the barrier b. decrease the width of
the barrier c. decrease the mass of the
particle 2. What is the energy of the particles
that have successfully escaped? a. lt initial
energy b. initial energy c. gt initial
energy
17
Lecture 10, Act 1 - Solution
Consider a particle tunneling through a barrier.
1. Which of the following will increase the
likelihood of tunneling? a. decrease the
height of the barrier b. decrease the width of
the barrier c. decrease the mass of the
particle 2. What is the energy of the particles
that have successfully escaped? a. lt initial
energy b. initial energy c. gt initial
energy
Reducing m or U0 will reduce K
Although the amplitude of the wave is smaller on
the far side of the barrier, no energy is lost
in the tunneling process.
18
Example Al wire contacts

• Everyday problem
• Youre putting the electrical wiring in your
  new house, and youre considering using Aluminum
  wiring, which is cheap and a good conductor.
  However, you also know that aluminum tends to
  form an oxide surface layer (Al2O3) which can be
  as much as several nanometers thick.

This layer could cause a problem in making
electrical contacts with outlets, for example,
since it presents a barrier of roughly 10 eV to
the flow of electrons in and out of the Al.
Your requirement is that your transmission
coefficient across any contact must be T gt 10-10,
or else the resistance will be too high for the
high currents youre using, causing a fire risk.
Should you use aluminum wiring or not? (You can
neglect G here.)
Compute L
Oxide is thicker than this, so go with Cu wiring!
(Al wiring in houses is illegal for this reason)
19
Example Electrons in Nanoscale devices
Current measurement by real-time counting of
single electrons JONAS BYLANDER, TIM DUTY
PER DELSING
40 fA
80 fA
120 fA
Electrons that successfully tunnel through the 50
junctions are detected using a fast single
electron transistor (SET).
20
Tunneling and Radioactivity
In large atoms (e.g., Uranium), the nucleus can
be unstable to the emission of an alpha particle
(2p 2n), which is a 4He nucleus. This form of
radioactivity is a tunneling process, involving
transmission of a 4He nucleus from a low-energy
valley through a barrier to a lower energy
outside.

• y leaks out from C through B to outside, A the
  particle tunnels out
• The leakage is slow, so y just outside the
  barrier stays negligible.
• The shape of y remaining in B-C shows almost no
  change just its amplitude goes down.
• The rate at which probability flows out is just
  proportional to the probability thats left (by
  linearity) exponential decay in time.
• Measurement If you look at any one radioactive
  nucleus, you see it decay at some particular
  random time. An ensemble of many nuclei is needed
  to see the exponential decay. After one
  half-life, half of the nuclei will have
  decayed after two, half of the remainder
  decay,...

21
a-Radiation Illustrations of the enormous range
of decay rates in different nuclei

• Consider a very simple model of a-radiation
• Assume the alpha particle (m 6.64 x 10-27 kg)
  is trapped in a nucleus which presents a square
  barrier of width L and height Uo. To find the
  decay rate we consider
• (1) the attempt rate at which the alpha
  particle of energy E inside the nucleus hits the
  barrier (2) the tunneling probability for an
  alpha particle with energy E each time the
  particle hits the barrier. For this order of
  magnitude calculation you may neglect G. Here
  we useBecause of the exponential this factor
  can vary enormously!

Rough estimate with E 5 to 10 MeV the alpha
particle makes about 1021 attempts per second
(nuclear diameter/velocity)
22
a-Radiation Example 1

• Polonium-212 atom (Z 84), which presents a
  square barrier of width L 9.4 x 10-15 m (9.4
  fermi) and height Uo 26 MeV. What is the
  tunneling probability for an alpha particle with
  energy 9 MeV?

23
Lecture 10, Act 2
We just looked at Polonium, with an effective
barrier width of 10 fermi, and saw a tunneling
probability of 10-15. Now consider Uranium,
which has a similar barrier height, but an
effective width of about 20 fermi. Estimate
the tunneling probability in Uranium a.
10-30 b. 10-14 c. 10-7
24
Lecture 10, Act 2
We just looked at Polonium, with an effective
barrier width of 10 fermi, and saw a tunneling
probability of 10-15. Now consider Uranium,
which has a similar barrier height, but an
effective width of about 20 fermi. Estimate
the tunneling probability in Uranium a.
10-30 b. 10-14 c. 10-7
Think of it this way there is a 10-15 chance to
get through the first half of the barrier, and a
10-15 chance to then get through the second
half. Alternatively, when we double L in this
is equivalent to squaring the transmission T.
Now lets do it for real
25
a-Radiation Example 2

• Uranium-235 atom (Z 92), which presents a
  square barrier of width L 21 x 10-15 m (21
  fermi) and height Uo 28 MeV. What is the
  tunneling probability for an alpha particle with
  energy 4.5 MeV?

Using 1021 attempts at the barrier per second,
the probability of escape is about 5 x 10-18 per
second. Therefore the decay time for an
a-particle is about (5 x 10-18 s-1)-1 2 x
1017 seconds 1010 years!
26
FYI The sun!
The solar nuclear fusion process starts when two
protons fuse together (they eventually become a
helium nucleus, which fuses with another one,
releasing two energetic protons).
The temperature of the suns core is 1.3 107
K. This corresponds to an average kinetic energy
kBT 2 x 10-16 J (kB Boltzmans constant)
How much energy would they need classically to
touch (approach to a distance the size of the
nucleus, 1 fm)? They are both charged, and
therefore repel each other U(r) (1/4pe0)xe2/r
(9x109)x(1.6x10-19 C)2/10-15 m 2 x 10-13
J Thus, classically, the protons in the sun
very rarely (never, in practice) have enough
thermal energy to overcome their coulomb
repulsion. (see Phys 213) How do they fuse
then? By tunneling through the last part of the
coulomb barrier!
27
Tunneling through a barrier Revisited
Consider a situation in which a particle (e.g.,
an electron or an atom) can be in either of two
wells separated by a potential barrier. Is the
particle on the left or right?
Both! If the barrier is finite, the wave
function extends into both wells
What is the state with the next highest energy?
Why is this state higher in energy?
28
Tunneling through a barrier Revisited
Consider a situation in which a particle (e.g.,
an electron or an atom) can be in either of two
wells separated by a potential barrier.
If the barrier is high then Y is small inside the
barrier The probability for finding the particle
is very similar for the two states energy
difference E2 E1 is small
E2
E1
What is the time dependence if the Probability to
find the particle is highest in the left well at
time t0? As discussed in Lect. 9, the particle
oscillates between the wells with half an
oscillation period, T h/(E2-E1)
Two-well simulation Qmdbw
29
Another Tunneling Example The NH3 Molecule

• The following example will bring together several
  things youve learned up to this point. Consider
  the ammonia (NH3) molecule

H
N
Plane of hydrogen atoms.
The N atom in the ammonia molecule (NH3) can have
two equal configurations above or below the
plane of the H atoms, as shown above. If we view
the molecule along the H-plane, the two positions
for the N atom are shown below
NH3 Model
The nitrogen atom can tunnel between these two
equivalent positions.
30
Application Tunneling the NH3 Maser
Question 1 What are the two lowest eigenstates
of this double-well potential?
First consider two separate wells
The double-well eigenstates are (almost)
superpositions of these two single-well
wavefunctions.
Which of these has higher E? Why?
31
Application Tunneling the NH3 Maser
Question 2 Given the energy difference between
the ground and first excited states, E2 - E1
1.8x10-4 eV, estimate how long it takes for the N
atom to tunnel from one side of the NH3
molecule to the other?
As discussed in Lect. 9, this takes a half an
oscillation period, T h/(E2-E1)
Two-well simulation Qmdbw
32
Application Tunneling the NH3 Maser
Question 3 Emission of radiation between
these two lowest energy configurations of
ammonia, E1 0.004637 eV and E2 0.004655 eV
(DE 1.8x10-4 eV). What wavelength of
radiation corresponds to this transition?
Solution
l hc/(E2-E1) 1240 eVnm/1.8x10-4 eV 6888
mm 6.88 mm
33
Application Tunneling the NH3 Maser
This effect due to tunneling of the atoms in NH3
was used to create the Ammonia Maser, by C.
Townes in 1954. This was the first laser
called a maser because it emits microwave
radiation. A maser or laser works by
Stimulated Emission which causes large numbers
of atoms to radiate together in phase, leading to
very intense directional radiation.
Intense stimulated emissionof microwaves with
wavelength6888 mm 6.88 mm
Stimulated Emission is a quantum effect! More
about this later
34
Lecture 10, Act 3
You are trying to make a laser that emits violet
light (400 nm), based on the transition an
electron makes between the ground and
first-excited state of a double quantum well as
shown. Your first sample emitted at 390 nm.
What could you modify to shift the wavelength to
400 nm? a. decrease the height of the
barrier b. increase the height of the
barrier c. decrease the width of the barrier
35
Lecture 10, Act 3 - Solution
You are trying to make a laser that emits violet
light (400 nm), based on the transition an
electron makes between the ground and
first-excited state of a double quantum well as
shown. Your first sample emitted at 390 nm.
What could you modify to shift the wavelength to
400 nm? a. decrease the height of the
barrier b. increase the height of the
barrier c. decrease the width of the barrier
The frequency of the electron oscillating between
the left and right well was too high ? the
probability to tunnel was too high! You can
reduce this by increasing the barrier height.
The wavelength of the emitted photon was too low
? the frequency of the photon was too high ? the
energy splitting between the ground and
first-excited state was too large. Raising the
barrier makes the difference in energy E2-E1
smaller. Why?
36
Lecture 10, Act 3, Solution - More
As we raise the height of the central barrier,
the coupling between the two wells decreases. In
the limit of an infinite barrier, it looks like
two independent wells ? same wavefunction
curvature for both the symmetric (ground state)
and anti-symmetric (1st excited state)
wavefunctions ? same kinetic energy, i.e.,
degenerate solutions.
37
FYI If QM is lossless, how do we get
dissipation, heating, collapse of the
wavefunction, etc.???

• We discussed an example of tunneling (or lack
  thereof) leading to heating of an aluminum
  contact. But no energy is lost in the
  tunneling process (or in any quantum process!),
  so how is it that we do see heating, and
  friction?
• Answer In our simple model we were neglecting
  other parts of the system. For example, we
  ignored all impurities in the material that the
  electron could scatter off. We also neglected
  phonons, the quantum mechanical excitations
  of lattice vibrations (i.e., quantized sound).
• While none of these interactions actually loses
  energy -- energy is ALWAYS conserved -- they can
  take energy away from the particle we are
  considering (this is dissipation), distributing
  it to the unobserved degrees of freedom.
• In fact, even if the particle becomes coupled to
  these other degrees of freedom without losing
  energy, they can still cause decoherence, and a
  collapse of the wavefunction.

38
Example Problem
Suppose an electron of KE 0.1 eV approaches a
barrier. a) What is the wavelength of this
electron? b) For what barrier height will it
have a 50 chance of penetrating 1 nm into the
forbidden region? What about 1 mm?Here you can
assume that the G term is not important for our
purposes.
Solution
a) In the allowed region E h2/2ml2
1.505eVnm2/l2 0.1 eV (for an electron) l
3.9 nm b) In the forbidden region K
?(2m(U-E)) / ? e-2KL 1/2 U E (? ln2/2L)2 /
2m 0.1 eV 7.4?10-22 J 0.105 eV or
0.1 eV 7.4?10-28 J 0.100000005 eV
Use the De Broglie relation, p h/l, and the
kinetic energy E p2/2m. U is the unknown
barrier height and E 0.1 eV. The 2 in e-2KL
results from probability being Y2. Note that
all that really matters is U-E. You can also use
h2/2m 1.505eVnm2 for an electron. L 1
nm. L 1 mm. Penetration by a significant
distance isnt possible unless the energy is
nearly allowed.