wanted dead and alive abbreviated

8/12/2019 Wanted Dead and Alive Abbreviated

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Wanted: Schodingers Cat: Dead and Alive

Howard A. Blair

Syracuse [email protected]

January 15, 2013

Howard A. Blair (Syracuse University) QuantumComputing 01/15/13 1 / 14
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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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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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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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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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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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