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8/12/2019 Wanted Dead and Alive Abbreviated
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Wanted: Schodingers Cat: Dead and Alive
Howard A. Blair
Syracuse Universityblair@ecs.syr.edu
January 15, 2013
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Leibniz gave us two things important for computer
science
The Principle ofInertia: Everything remains in the state in which it
is, unlesscausedto change state.
The Principle of theIdentity of Indiscernibles: Two things are
the actually the same thing if, and only if, they have exactly the
sameessentialproperties. (HUGE)
Imagine all possible properties that anything could have are
arranged in a sequence:
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And then came Mach
Ernst Machs positivist/empiricist view: Stick to observables. If
with the help of the atomic hypothesis one could actually establish
a connection between several observable properties which
without it would remain isolated, then I should say that this
hypothesis was an economical one. (i.e: The atom concept is,maybe, a useful hack.
Machs positivist/empiricist view, which remains at the heart of
contemporary physics, leads right back into metaphysics. What
we observe hasprofound metaphysical implications: Forsomething to exist it must have properties observable in
principle. (Theres the rub!)
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Quantum States:
Put a 0 bit into a Hadamard gate, and observe the output, you get 0
half the time, 1 the other half. Feed the unobserved output through
another Hadamard gate you get back the 0. Same thing happens with
a 1. How can this be?Howard A. Blair (Syracuse University) QuantumComputing 01/15/13 5 / 14
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Superposition:
The circular polarization property of a photon is a qubit: its stateis aquantum state: a0 +b1, called awave function.
a, bare twisted square roots of probabilities:
a0 +b1= ei
p0 +ei
(1
p) 1
The squares of the absolute values ofaandbare probabilities.
The absolute value operation removes twist. e.g.
ei
p
=
p
Two of the quantum states of a qubit are regularly seen:
1 0+ 0 1=0 and 0 0+ 1 1=1
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Leibniz and Machs principles imply:
That which cannot be observed,in principle, does not exist.
Question: Can we observe superposition states?
Answer: It depends.
Question: On what?
Answer: On what questions of nature we ask; i.e. on what what
we choose to measure.
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The rules
Observe a qubit thats in state a0 +b1.
The result of your observation will be either 0 or 1.
The probability of getting 0 is |a|2
and of getting 1 is |b|2
.(Wave functioncollapse.) After the observation the photon is
either
in state0or in state1
depending on whether0or1was observed.
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2 photons: Entanglement
a00 +b01 +c10 +d11Suppose we have two photons, the first in state a0 +b1and thesecond in statec0 +d1.
The combined system is in theseparablestate
(a0 +b1) (c0 +d1) =ac00 +ad01 +bc10 +bd11
What about states like
1
2 00 + 1
2 11The quantum states of the separate qubits arent there. Therefore
those qubits do not separately exist.
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Quantum Teleportation
The Setup 1: On an apparatus in her lab in New York, Alice has a
qubit in some quantum statea0 +b1. She does not know thequantum state of her qubit. She wants to send the qubit to Bob
who is in Paris. Specifically, she wants to cause the quantum state
of the qubit storage apparatus in Bobs lab in Paris to become
a0 +b1.The Setup 2: Alice and Bob have prepared a pair of entangled
qubits in state1
200 +
12
11
on a pair of single qubit storage devices, one of which they each
took to their labs.
By what we have said so far, there are no separate qubits stored
on their devices from the entangled pair.
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Quantum Teleportation
Initially, there are three qubit storage devices in play: (1) holdsAlices qubit that she wants to teleport; (2) Alices qubit from the
entangled pair; (3) Bobs qubit from the entangled pair. The state
of the three devices is partially entangled like this:
a2
000 + a2
011 + b2
100 + b2
111
Alice sends her qubit that she wants to teleport and her qubit from
the entangled pair through an entanglement device (a
controlled-notgate.) The state of the three devices becomes:
a2
000 + a
2011 +
b2
110 + b
2101
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Quantum Teleportation
Alice then sends her qubit (the one to be teleported) through aHadamard gate:
0 ( 12
0 + 1
21) 1 ( 1
20 1
21)
Substituting on the first qubit, the state of the three partially
entangled devices is now:
a
2(
1
20 +
1
21)00 +
a
2(
1
20 +
1
21)11
+ b
2(
12
0 12
1)10 + b
2(
12
0 12
1)01
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Quantum Teleportation
which simplifies to
a
2000 +
a
2100 +
a
2011 +
a
2111
+b
2010 b
2110 +
b
2001 b
2101
(and now for more quantum wierdness):
= 12
00(a0+b1)+ 12
01(a1+b0)+ 12
10(a0b1)+ 12
11(a1b0)
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Quantum Teleportation
Alice measures the qubits in her lab. Lets suppose she gets the10result.
The wave function on all three qubits then collapses to
10(a0
b1)
The probability that Bob would see0is |a|2, and that he would see1is |b|2. If Alice tells Bob the results of her measurement, Bobcan appropriately rotate his qubit into the original state of Alices
qubit.
Notice that the probability distribution on the potential results for
Bob to obtain is dependent onbothAlices original qubitand the
results of Alices measurements.
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