Transcript Slide 1

Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers: Ron Folman, Daniel Rohrlich
Teaching Assistants: Ben Yellin, Shai Bartal
Week 12. Quantum mechanics – propagation on a crystal lattice •
Fermi level/momentum/energy • Kronig-Penney model and band
gaps • GHZ and Bell’s theorem • quantum computing and
cryptography.
Sources: Feynman Lectures III, Chap. 13;
Ashcroft and Mermin, Solid State Physics, Chap. 8;
Tipler and Llewellyn, Sect. 10.6;
selected papers to be distributed.
Propagation on a crystal lattice
Some solids – including crystals and most metals – are periodic
lattices. A periodic lattice obeys the Bragg condition for
diffraction of waves from solids:
d
θ
θ
2d sin θ = nλ
Propagation on a crystal lattice
Some solids – including crystals and most metals – are periodic
lattices. A periodic lattice obeys the Bragg condition for
diffraction of waves from solids:
d
θ
θ
θ
2d sin θ = nλ
Electron diffraction
X-rays on zirconium oxide
electrons
Electrons on gold
Neutron diffraction
Diffraction of X-rays on
a single NaCl crystal
Diffraction of neutrons on
a single NaCl crystal
Propagation on a crystal lattice
To understand some generic features of electron conduction on a
crystal lattice, let’s model a 1D crystal, i.e. a lattice with a
periodic potential.
V(x)
Ion core
d
x
The exact shape of the periodic potential will not matter.
Propagation on a crystal lattice
To understand some generic features of electron conduction on a
crystal lattice, let’s model a 1D crystal, i.e. a lattice with a
periodic potential.
V(x)
x
d
The exact shape of the periodic potential will not matter.
Propagation on a crystal lattice
We know how to solve Schrödinger’s equation for each barrier,
but how do we connect up the solutions for all the barriers?
V(x)
e

 reiKx
iKx
x
teiKx 
d
K
2mE

Propagation on a crystal lattice
We know how to solve Schrödinger’s equation for each barrier,
but how do we connect up the solutions for all the barriers?
V(x)
iKx
 te
e
iKx
reiKx 
d
K
2mE

x
Propagation on a crystal lattice
Bloch’s theorem (1D): The eigenstates of
 2 d 2

Eψ( x)  
 V ( x) ψ( x)
2
 2m dx

,
where V(x) is periodic with period d, can be written
ψnk ( x)  eikx unk ( x)
,
where unk(x) is periodic with period d, i.e. unk(x+d) = unk(x).
That is , every eigenstate has the property
ψ( x  d )  eikd ψ( x)
.
Proof of Bloch’s theorem in three steps:
ˆ
ip d / 
e
1. The operator
shifts any function f(x) by d.
Proof: e
ipˆ d / 
f ( x)  e


ipˆ d / 
1
2

~
f (k )e ikx dk
~
ˆ
f (k )e ipd /  e ikx dk

~
1
ikd ikx
f
(
k
)
e
e
dk

f
(
x

d
)
2 
1
2
.
ipˆ d /  ˆ
ipˆ d / 
[
e
,
H
]

[
e
, V ( x)]  0.
2. Therefore
3. Therefore the eigenstates of Hˆ can be written as eigenstates of
ˆ
e ipd / , which are ψnk ( x)  eikx unk ( x).
Fermi level/momentum/energy
We will use Bloch’s theorem to find the relation between the
momentum k and the energy E of an electron in a crystal lattice.
But we ask first, How many momentum states are there in a
crystal of length D?
Assume that electron wave functions vanish outside the crystal.
To be orthogonal, the eigenfunctions of
pˆ must be
1, ei 2x / D , ei 4x / D , ei6x / D , ei8x / D , ,
in 1D, using “periodic boundary conditions” (x + D ↔ x).
If electrons were bosons, we could put them all in the lowestmomentum state.
Fermi level/momentum/energy
We will use Bloch’s theorem to find the relation between the
momentum k and the energy E of an electron in a crystal lattice.
But we ask first, How many momentum states are there in a
crystal of length D?
Assume that electron wave functions vanish outside the crystal.
To be orthogonal, the eigenfunctions of
pˆ must be
1, ei 2x / D , ei 4x / D , ei6x / D , ei8x / D , ,
in 1D, using “periodic boundary conditions” (x + D ↔ x).
But electrons are fermions, so we can put just one in each
momentum state.
Fermi level/momentum/energy
In the ground state, all the momentum states are filled up to the
state with the highest momentum magnitude kF, which is called
the Fermi momentum. The corresponding highest energy EF,
which equals E F
Fermi energy.

k F 2

2 me
in the simplest cases, is called the
Fermi level/momentum/energy
In 3D, the momentum states are vectors with components kx, ky,
kz taking values
1, ei 2x / D , ei 4x / D , ei6x / D , ei8x / D , .
We can think of triplets (kx,ky,kz), e.g. (4π/D,–16π/D,6π/D ), as
points on a “reciprocal lattice”. When states are occupied up to
the Fermi momentum, the triplets lie inside a ball of radius kF.
The surface of this ball is called the Fermi surface.
kz
kF
[From here.]
ky
kx
Compare Maxwell-Boltzmann and Fermi-Dirac statistics to see
the effect of the Pauli exclusion principle:
TF = 500 K
PMB ( E )  e  E / k BT
PFD ( E ) 
1
e E / k BT  1
Compare Maxwell-Boltzmann and Fermi-Dirac statistics to see
the effect of the Pauli exclusion principle:
PFD
PMB
Average occupancy vs.
average kinetic energy
PMB
PFD
PMB ( E )  e  E / k BT
PFD ( E ) 
1
e E / k BT  1
Exercise: What is the minimum energy of a population of N
fermions?
Exercise: What is the minimum energy of a population of N
fermions?
Solution: N 
(2 spinstates)(4 /3) (k F ) 3
(2 /D)
3

( k F D) 3
3 2
 2 (k F ) 2
 2 (3 2 N ) 2/3
(3 2 N )1/3

.
so k F 
and E F 
2me
D
2me D 2
Exercise: What is the minimum energy of a population of N
fermions?
Solution: N 
(2 spinstates)(4 /3) (k F ) 3
(2 /D)
3

( k F D) 3
3 2
 2 (k F ) 2
 2 (3 2 N ) 2/3
(3 2 N )1/3

.
so k F 
and E F 
2me
D
2me D 2
(2me ) 3/2 D 3
dN

The “density of states” dN/dEF is
dEF
2 2  3
EF .
Exercise: Find the 2D versions of EF and dN/dEF.
Exercise: Given that electron densities in metals are of order
1022/cm3, find the speed of the fastest electrons.
Kronig-Penney model and band gaps
So far, we have not discussed E(p), the relation between the
energy E and the momentum p = ħk of electrons in crystals. But
E(p) has dramatic consequences.
E
Allowed
band
E
E
Forbidden band
Allowed
band
k
k
Free electron
Electron in crystal
Kronig-Penney model and band gaps
So far, we have not discussed E(p), the relation between the
energy E and the momentum p = ħk of electrons in crystals. But
E(p) is remarkable, and has dramatic consequences.
E
Allowed
band
E
Forbidden band
Allowed
band
k
Free electron
EF
conductor
Electron in crystal
k
Kronig-Penney model and band gaps
So far, we have not discussed E(p), the relation between the
energy E and the momentum p = ħk of electrons in crystals. But
E(p) is remarkable, and has dramatic consequences.
E
Allowed
band
Forbidden band
E
EF
Allowed
band
k
Free electron
insulator
Electron in crystal
k
Kronig-Penney model and band gaps
So far, we have not discussed E(p), the relation between the
energy E and the momentum p = ħk of electrons in crystals. But
E(p) is remarkable, and has dramatic consequences.
E
Allowed
band
Forbidden band
E
EF
Allowed
band
k
Free electron
Where does the “band gap” come from?
insulator
Electron in crystal
k
Kronig-Penney model and band gaps
The “Kronig-Penney” model is a 1D lattice of square potential
wells with lattice spacing is d.
V(x)
x
-d
d
2d
3d
Kronig-Penney model and band gaps
The “Kronig-Penney” model is a 1D lattice of square potential
wells with lattice spacing is d. The energy E of an electron as a
function of its wave number k is shown below for an electron in
this lattice. Energy gaps appear when k = nπ/d.
0
π/d 2π/d 3π/d
k
The gap arises from standing waves that resonate through the
lattice at k = nπ/d. For n = 1, for example, the waves are
ψ  ( x)  (eix / d  e ix / d ) / 2  cos(x / d ) ,
ψ  ( x)  (eix / d  e ix / d ) / 2i  sin(x / d ) .
The figure shows the spatial probability distribution of both waves.
|ψ–|2
|ψ+|2
traveling
wave
x
d
Prob. 8.1 in Solid State Physics by Ashcroft and Mermin is a
mathematical analysis of band gaps in 1D. Assume a potential V(x)
such that V(–x) = V(x) and V(±d/2) = 0 and V(x+d) = V(x). There
are two scattering solutions, ψl(x) and ψr(x), with the following
behaviors at x = ±d/2:
ψl ( x)  eiKx  reiKx ,
x  d / 2 ,
 teiKx ,
x  d / 2;
V(x)
eiKx 
 reiKx
teiKx 
d
2mE
K

x
Prob. 8.1 in Solid State Physics by Ashcroft and Mermin is a
mathematical analysis of band gaps in 1D. Assume a potential V(x)
such that V(–x) = V(x) and V(±d/2) = 0 and V(x+d) = V(x). There
are two scattering solutions, ψl(x) and ψr(x), with the following
behaviors at x = ±d/2:
ψl ( x)  eiKx  reiKx ,
 teiKx ,
and
ψ r ( x)  teiKx ,
 e iKx  reiKx ,
x  d / 2 ,
x  d / 2;
x  d / 2,
x  d / 2.
We now define a general solution ψ(x) = Aψl(x) + Bψr(x) and
impose two conditions:
ψ(x  d )  eikd ψ(x)
[from Bloch's theorem] ,
and
ψ'( x  d )  eikd ψ'( x) [by differentiating].
Imposing these conditions on ψ(x) we can derive the following:
t 2  r 2 iKd 1 iKd
cos kd 
e
 e
,
2t
2t
and since in general rt* is pure imaginary and |r|2+|t|2 = 1, we can
also derive
cos(Kd   )
cos kd 
,
|t |
where t = |t|eiδ . Now if |t| < 1, there are no solutions for k when
Kd+δ ≈ nπ. This is the origin of the band gaps.
Exercise: For weak scattering (|t| ≈ 1, δ ≈ 0), derive the width of
the band gap.
Solution: The gap occurs for cos Kd > |t|. Since |r|2+|t|2 = 1 and
|t| ≈ 1, we can approximate |t| ≈ 1–|r|2 /2. The relevant values of K
are those for which Kd ≈ nπ. For those values, we can approximate
cos Kd as 1 – |Kd–nπ |2/2. Thus, the limits of the gap occur when
|Kd±nπ| = |r|, i.e. when K = (|r|±nπ)/d. Since the energy is
we have
2
E 
2 me
2K 2
E
,
2me
 | r |  n

d

2
|
r
|

n

2 2 n | r |
 

 
 
d
 
 
me d 2
2
as the width of the band gap.
Electrons perform miracles!
In superconductivity (which
occurs at temperatures as high
as 160 K), currents encounter
no resistance.
Electrical resistance suddenly
vanishes as a superconductor
is cooled below its critical
temperature Tc.
normal metal
superconductor
Electrons perform miracles!
In superconductivity (which
occurs at temperatures as high
as 160 K), currents encounter
no resistance.
Superconductors expel all
magnetic flux (Meissner
effect)
Electrons perform miracles!
In superconductivity (which
occurs at temperatures as high
as 160 K), currents encounter
no resistance.
Superconductors expel all
magnetic flux (Meissner
effect)
The explanation for superconductivity put forward by J. Bardeen,
L. Cooper and J.R. Schrieffer [BCS] in the Physical Review 106
(1957) 162 is based on “bosonic” pairs of electrons having k (with
spin up) and –k (with spin down) that are far apart but interact via
the lattice.
GHZ and Bell’s theorem
Is the uncertainty principle a fundamental limit on what we can
measure? Or can we evade it? Einstein and Bohr debated this
question for years, and never agreed.
GHZ and Bell’s theorem
Is the uncertainty principle a fundamental limit on what we can
measure? Or can we evade it? Einstein and Bohr debated this
question for years, and never agreed.
Today we are certain that uncertainty will not go away.
Quantum uncertainty is even the basis for new technologies such
as quantum cryptology.
It may be that the universe is not only stranger than we imagine,
but also stranger than we can imagine.
GHZ and Bell’s theorem
In 1935, after failing for years to defeat the uncertainty principle,
Einstein argued that quantum mechanics is incomplete.
The famous “EPR” paper
GHZ and Bell’s theorem
In 1935, after failing for years to defeat the uncertainty principle,
Einstein argued that quantum mechanics is incomplete.
Note that [x, ˆp] ≠ 0, but [x2–x1, pˆ 2+pˆ 1] = [x2, pˆ 2] – [x1, pˆ1] = 0.
That means we can measure the distance between two particles
and their total momentum, to arbitrary precision.
So we can measure either x2 or p2 without affecting Particle 2 in
any way, via a measurement on Particle 1 (and vice versa).
That means both x2 and p2 are simultaneously real.
Quantum mechanics has no place for simultaneous x2 and p2 , so
quantum mechanics is incomplete.
GHZ and Bell’s theorem
Some reactions to the EPR argument:
Bohr (1935) The uncertainty principle still applies.
Pauli (1954) “One should no more rack one's brain about the
problem of whether something one cannot know anything about
exists all the same, than about the ancient question of how many
angels are able to sit on the point of a needle. But it seems to me
that Einstein's questions are ultimately always of this kind.”
Bell (1964) Quantum mechanics contradicts EPR!
GHZ and Bell’s theorem
We will consider a theorem similar to Bell’s, proved in 1988 by
D. Greenberger, M. Horne, and A. Zeilinger (GHZ).
It describes three spin-½ particles prepared in one laboratory and
sent to three different laboratories, operated by Alice, Bob, and
Claire. For simplicity, we assume the particles are not identical.
GHZ and Bell’s theorem
Setting for Greenberger-Horne-Zeilinger (GHZ) paradox:
Alice
A
Bob
B
time
O
space
Claire
C
GHZ and Bell’s theorem
We will consider a theorem similar to Bell’s, proved in 1988 by
D. Greenberger, M. Horne, and A. Zeilinger (GHZ).
It describes three spin-½ particles prepared in one laboratory and
sent to three different laboratories, operated by Alice, Bob, and
Claire. For simplicity, we assume the particles are not identical.
The spin state of the three particles is
GHZ 


2
1
A

1
0
where     and     .
0
1
B

C
 
A

B

C
,
GHZ and Bell’s theorem
Alice can measure  x(A) or  (yA) on her system;
Bob can measure  x(B) or  (yB) on his system;
Claire can measure  x(C) or  (yC) on her system;
0 1
0  i
1 0 
,  y  
,  z  
.
where  x  
1 0
i 0 
 0  1
0 1  ˆ
0  i  ˆ
1 0  
ˆ
 , S y  
 , S z  
 ).
(Recall S x  
1 0 2
i 0  2
 0  1 2
The result of each  x or  y measurement can be 1 or –1.
Note  x    ,  x    ,  y   i  ,  y   i  .
GHZ and Bell’s theorem
To prove:
(A) (B) (C)
1.  x  x  x | GHZ   | GHZ
( A ) ( B) ( C )
|



2.
GHZ
x  y  y | GHZ
  (yA ) x( B) (yC) | GHZ
  (yA ) (yB) x(C) | GHZ
That is, Alice, Bob and Claire discover two rules:
(A) (B) (C)
1.  x  x  x  1
(A) (B) (C)
(A) (B) (C)
(A) (B) (C)
2.  x  y  y  1   y  x  y   y  y  x
GHZ and Bell’s theorem
But now



1   x(A) (yB) (yC)  (yA) x(B) (yC)  (yA) (yB) x(C)
  x(A) x( B) x(C)
 1 ,

 
(A) 2
because  x

(A) 2
y
 1, etc.
CONTRADICTION!

GHZ and Bell’s theorem
But now



1   x(A) (yB) (yC)  (yA) x(B) (yC)  (yA) (yB) x(C)

  x(A) x( B) x(C)
 1 ,

 
(A) 2
because  x

(A) 2
y
 1, etc.
CONTRADICTION!
This contradiction arose because we tacitly assumed, with EPR,
that observables have values whether we measure them or not.
Quantum computing and cryptography.
Introducing…the qubit (quantum bit).
Like an ordinary “classical” bit, a qubit has two states:  ,  .
But unlike an ordinary bit, a qubit can be in any superposition of
these two states!
Quantum “CNOT” gate:
Quantum computing and cryptography.
Introducing…the qubit (quantum bit).
Like an ordinary “classical” bit, a qubit has two states:  ,  .
But unlike an ordinary bit, a qubit can be in any superposition of
these two states!
In principle, a quantum computer, given a superposition of
inputs, computes the final result for each input simultaneously.
Therefore there is a great speed-up of the computation. (The
problem is that the final results, as well, are superposed.)
In 1994, P. Shor devised an algorithm for factoring large
numbers into primes, using a quantum computer. It is
exponentially faster than any classical computer.
Quantum computing and cryptography.
Shor’s algorithm puts in jeopardy all codes based on the RSA
encoding method, which is based on the difficulty of factoring
large numbers into primes.
But have no fear!
!‫תורת הקוונטים הקדימה תרופה למכה‬
There are quantum codes, and they are unbreakable!
Example of quantum cryptography: “the one-time pad”.
100011111111110110
–10010101011010100
“10001010100100010”
Bob
“10001010100100010”
+10010101011010100
100011111111110110
Alice
How Alice, Bob and Claire can share a one-time pad:
1. They share many GHZ triplets (indexed), on which they
measure σz .
2. From time to time they measure σx and σy to check that no one
is trying to interfere with their key-sharing.
Alice
A
time
10010101011010100
space
Bob
B
10010101011010100
O
Claire
C
10010101011010100
The Wall Street Journal
(European edition),
5 March, 2004, p. A7
“…new technologies such as quantum
computers...could perform calculations quickly
enough to crack today’s best encryption systems.
For that reason, military agencies in both the U.S.
and Britain are funding quantum weirdness studies;
they probably wouldn’t mind knowing how to put
their troops in two places at once, either.”
A Quantum Century
“In the 19th century, life was transformed by the
conscious application of classical
mechanics….. In this [20th] century, a similar
transformation has been wrought by
electromagnetism…. It is easy to predict that
in the 21st century it will be quantum
mechanics that influences all our lives.”
- Sir Michael Berry