Transcript a 1 - University of San Francisco

University of San Francisco
Modern Physics for Frommies II
The Universe of Schrödinger’s Cat
Lecture 8
3 March 2010
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Agenda
•Administrative Matters
•Applications
•Quantum Computing and Quantum Cryptography
•LASERs and Holography
•Summing Up
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Administrative Matters
Next Physics and Astronomy Colloquium
Today, Wednesday, 3 March 2010 at 4 PM
Dr. Seth Shostak, SETI Institute
Strategies in the Search for Extraterrestrial Intelligence
Refreshments at 3:30 PM
Harney Science Center Room 127
Topic:
The movie on particles and waves shown at the last class is
available on line.
Waves and Particles, Annenberg, The Mechanical Universe
http://www.learner.org/resources/series42.html
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Quantum Computing
Direct use of quantum mechanical phenomena, e.g. superposition
and/or entanglement, to perform operations on data.
Potential to be exponentially faster than classical computers, i.e
computers on your desk today.
Do not allow the computation of functions that are not
theoretically computable by classical machines, i.e. they do not
alter the Church-Turing thesis. The gain is only in efficiency.
Experiments have been carried out in which quantum
calculations were executed on a very small number of qubits
(quantum bits)
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Q-Computers differ from other computers (transistors, photonics,
DNA) in that it uses specifically QM resources like entanglement. It
is conjectured that only Q-computers are capable of achieving the
exponential advantage mentioned above.
What is it:
Classical computer: Memory of bits. 0 or 1
Q-computer: 0 or 1 or any superposition of these is a qubit.
Two qubits can be in any superposition of four states.
In general, n qubits can be in an arbitrary superposition of 2n
states simultaneously. A classical computer can only be in one of
these 2n states at any one time.
A Q-computer operates by manipulating qubits through a fixed
sequence of quantum logic gates. Sequence = quantum algorithm
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How does a classical computer add two numbers
Two bit registers each having 22 = 4 different states
Binary Decimal
00
0
01
1
10
2
11
3
Employ an arrangement of Boolean logic gates called a full
adder.
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Half Adder:
XOR
AND
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  A B
Cout  A  B
7
Full Adder:
A0
½
B0
adder
½
C0
A1
B1
C1
adder
½
1
0
adder
Add 1 + 1 which in binary is 01 + 01.
01 + 01 =10 which in decimal is 2
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Bits vs.qubits:
Classical computer with a 3-bit state register
State of the machine is a probability distribution over 23=8 3-bit strings:
000, 001, 010, 011, 100, 101, 110, 111
Deterministic computer is in exactly one of these states with probability
1
Probabilistic computer can be in any one of several different states.
Probabilities given by 8 non-negative numbers ai. Normalization
requires
8
a
i 1
i
1
Now consider a 3-qubit quantum computer. Similarly described
by an 8-D vector (a1 … a8), sometimes called a ket in Dirac
notation.
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Now the normalization requirement becomes
8
a
i 1
i
2
1
Additionally, the ai are complex numbers, i.e. they have a phase as
well as a magnitude.
Im
ai
ai  (Re  ai )   Im  ai 
2
2
2
qi
 Im  ai  
qi  Arc tan 

 Re  ai  
ai  ai eiqi  ai  cosqi  i sin qi 
Re
States being added together will undergo interference. This is a
key difference between classical and quantum computing.
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Superposition of 2 states
a1  a2
|a2>
d
Re  a1  a2
q
|a1>
Im  a1  a2
  Re  a   Re  a 
  Im  a   Im  a 
1
2
1
2
Magnitudes and phases determine the superposition
Hologram vs. Photograph
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Back to our 3-qubit string
If we measure the 3 qubits the probability of measuring a
particular string will equal the sum of the squared magnitudes of
that strings coefficients. Our quantum superposition has collapsed
to a classical state.
Semi-classically, we can think of the system as being one of the 3bit strings, we just don’t know which one.
Quantum mechanically, we need to keep track of all 2n complex
coefficients in order to see how the quantum system evolves in time.
This is a staggering task!
300 qubits define a state described by 2300 complex numbers.
2300  1090 > number of atoms in the observable universe.
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Operating a Quantum vs. a Classical Computer:
In both cases the system must be initialized, e.g. into the all zeros
string |000> corresponding to the vector (1, 0, 0 ,0 ,0, 0, 0, 0).
Classical computer logic operations are in general not reversible
e.g. consider an AND gate
If the output bit is 0, we can’t tell if the
input bit pattern was (1,00, (0,1) or (0,0)
In quantum computing the manipulations are said to be unitary, i.e.
rotations since they preserve that the sum of the squares add up to 1.
Since rotations can br undone by counter rotations, quantum
computations should be reversible.
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Finally, when the computational sequence is complete the result must
be read out.
Classically, we sample the probability distribution on the 3-bit output
register to obtain one definite 3-bit string.
Quantum mechanically, we measure the 3-qubit state, which
collapses the quantum state to a classical distribution followed by
sampling from that distribution.
This destroys the quantum state. Schrödinger’s cat cannot be
restored to its mixed (unopened box) superposition.
Many algorithms only give the correct answer with a certain
probability. By repeatedly initializing, running and measuring the
probability of getting the correct answer can be increased
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Quantum Decoherence:
Avoidance usually requires isolating the system from its environment.
Interactions with the environment cause the system to decohere,
entropy increases and information is lost.
Typical decoherence times, also called dephasing times, range
from nanoseconds at room temperature to seconds at L4He
temperatures.
There are a number of proposed schemes to overcome the
decoherence problem. Research is ongoing.
There are a very large number of quantum computing candidates.
This large number is an indication that the topic is still in its
infancy. At the same time, the large number is an indication of a
vast amount of flexibility.
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Quantum Cryptography
Problem: Find a mechanism for “Alice” to send a secure message to
“Bob” while “Eve” might or might not be eavesdropping.
The only encoding scheme that has been proven to provide perfect,
unbreakable security is the Vernam cipher or one time pad scheme.
Alice sends e-mail to Bob
Characters are sent in 7-bit clusters, e.g. a = 1100001, 4 = 0110100
etc.
Bob’s computer decodes Alice’s bit stream into letters and numbers.
Eve’s computer can do the same.
Alice produces a random string of bits, called the key, exactly as
long as the message.
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If the nth bit of the key is 0 then she sends the nth bit of her message as
is. If the nth bit of the key is 1 then she sends the nth bit of her message
reversed (0  1).
e.g. character “a”
1100001
0101101
1001100
standard character “a”
key
code for “a”
After encoding, Alice sends Bob the message and the key and he
decodes the message.. The key must be sent so that Eve cannot
intercept it along with the message.
A new key must be sent with each message. There are easy ways to
break the code if the same key is used even twice. Hence the name
“one time pad”.
Problem is not the distribution of messages but the distribution of
keys.
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Can quantum mechanics help?
Suppose Alice and Bob set up “EPR distant measurements” from
the last lecture.
+
-
+
-
source
Alice and Bob record the exits taken as each pair reaches their
analyzers.
If Alice records ++-+-, then Bob will record exactly the opposite
i.e. --+-+
Alice converts + → 1, - → 0
Bob converts + → 0, - → 1
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Alice and Bob now have the same
key
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Not so fast! Snoopy Eve can still intercept the key by inserting a
vertical analyzer between the source and Bob. Recall the repeated
measurement experiments. Eve and Bob both get the key
Exp.1: Measurement of mz and mz
E
x
p
-.
+
-
all
none
ignore
mz = + mB at first analysis and at second
Instead, Alice and Bob set up the “random distant measurement”
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Tilting S-G analyzers in 120º orientations A, B and C
R
G
R
G
A
C
B
source
(+) exit = red light, (-) exit = green light
Left detector flashes green  right atom has mz = + mB
Right detector flashes red with a probability of ½.
See next slide
Conclusion: If orientations are the same, flashes are different always.
If orientations are ignored, flashes are different with
probability ½.
This is, in fact, what is observed.
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When an atom passes through an analyzer, Alice (or Bob) records
both the analyzer orientation and the exit taken.
Bob and Alice run the experiment for a long time and then send each
other the list of their analyzer orientations. (Each list looks
something like BBACABBC…). No need to encode. This tells
nothing to Eve.
Bob and Alice compare lists. Most cases, analyzers were set to
different orientations, but in about 1/3 orientations were the same.
Discard exit information, ±, for cases with different orientations
and use the remaining cases to construct a key as before.
Suppose snoopy Eve tries to intercept the key, as he did before, by
placing a vertical analyzer between the source and Bob.
Now, when atoms arrive at the tilting analyzers (Alice & Bob) they
are no longer entangled. mz = +mB for one and –mB for the other.
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If the analyzers are vertical (A) this makes no difference. The atoms
still emerge from opposite exits.
If both detectors are in orientation B, there is some probability (3/8)
that the atoms will emerge from the same exit.
Alice and Bob have a previous arrangement not to use the entire
key they generated above. Instead, Bob mails the first half of his
key back to Alice. If it matches Alice’s first half then it is safe for
Alice to send her message using the second half of the key.
If the two first halves don’t match then Alice knows the key is
compromised and does not send her message to Bob.
Measurement has caused a change in the information.
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Polarization
E
If we somehow break this symmetry
E. M. radiation is linearly polarized
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Polaroid: Material with long complex molecules oriented parallel
to one another
E  E0 cos q
I  I 0 cos 2 q
(The Law of Malus)
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Add another vertical polarizer
Nothing changes. This is like the Stern-Gerlach measure mz twice
experiment.
What happens if we rotate the intermediate polarizer through some
angle, say 45°?
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Horizontal polarization has been “ regenerated” and some
transmission occurs.
This is like the tilted Stern-Gerlach experiment. Eve’s snooping, in
the form of the intermediate polarizer, affects Bob’s reception of
Alice’s key.
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LASERs
Light Amplification by Stimulated Emission of Radiation
Produces a narrow intense beam of monochromatic coherent light
monochromatic  narrow frequency width
coherent  phase is constant across beam cross section
absorption
stimulated emission
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For lasing a population inversion must occur so that emissiom
dominates over absorption.
The excited state must be a metastable state (long lifetime) so
that stimulated emission dominates over spontaneous,
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Ruby Laser
Al2O3 with small % of Al replaced with Cr
Cr does the lasing
Flash lamp   550 nm
He-Ne Laser
15% He  85% Ne
E '3 is metastable
Gas discharge
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Holography
Ordinary photography simply records the intensity as a function
of position. This yields a 2-D image.
A hologram produces 3-D images via interference, without
lenses.
Illumination of the film with a laser beam produces a 3-D
image.
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Reflected light from every point on the the object reaches every
point on the film.
Interference between the two beams allows the film to record
both the intensity and the relative phase of the light at each point.
Recall the discussion of quantum computing where information
storage involved both intensity and phase.
Note that information is stored in the pattern as a whole.
Destruction of part of the hologram does not result in discrete
destruction of part of the image but generates blurriness or loss of
detail.
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Summing Up
By no means has this been an exhaustive treatment of a very
complicated subject. What did we get done?
State of classical physics ca. 1900. and clouds on the horizon
Blackbody radiation and Planck’s quantum of action.
Einstein, the photoelectric effect and the photon
Wave particle duality and the double slit
Early atomic models
The Bohr atom
Wave mechanics
Probability vs. determinism. Heisenberg’s uncertainty principle
Particles and potentials
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The hydrogen atom in wave mechanics
Atoms with multiple electrons
Molecules
Superposition and entanglement
Applications
Glaring ommisions
Matrix approach
Solid state electronics
Nuclear physics
etc.
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Some Comments on Quantum Mechanics
God does not play dice with the universe.
- Albert Einstein
Nobody feels really comfortable with it.
- Murray Gell-Mann
I can safely say that nobody understands quantum mechanics. Any
one who claims to understand quantum mechanics is lying.
- Richard Feynman
If that electron was up your a__, you’d know where it was.
- William Woodruff.
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Here’s to you, Erwin!
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