On Tue, Aug 21, 2018 at 10:05 PM Bruce Kellett <bhkell...@optusnet.com.au>
wrote:

> From: Jason Resch <jasonre...@gmail.com>
>
>
> On Tue, Aug 21, 2018 at 7:43 PM Brent Meeker <meeke...@verizon.net> wrote:
>
>>
>>
>> On 8/21/2018 3:37 PM, Jason Resch wrote:
>>
>>
>>
>> On Tue, Aug 21, 2018 at 5:00 PM Brent Meeker <meeke...@verizon.net>
>> wrote:
>>
>>>
>>>
>>> On 8/21/2018 2:40 PM, agrayson2...@gmail.com wrote:
>>>
>>>
>>>> If I start a 200 qubit quantum computer at time = 0, and 100
>>>> microseconds later it has produced a result that required going through
>>>> 2^200 = 1.6 x 10^60 = states (more states than is possible for 200 things
>>>> to go through in 100 microseconds even if they changed their state every
>>>> Plank time (5.39121 x 10^-44 seconds), then physically speaking it *
>>>> *must** have been simultaneous.  I don't see any other way to explain
>>>> this result.  How can 200 things explore 10^60 states in 10^-4 seconds,
>>>> when a Plank time is 5.39 x 10^-44 seconds?
>>>>
>>>
>>> It's no more impressive numerically than an electron wave function
>>> picking out one of 10^30 silver halide molecules on a photographic plate to
>>> interact with (which is also non-local, aka simultaneous).
>>>
>>>
>> Well consider the 1000 qubit quantum computer. This is a 1 followed by
>> 301 zeros.
>>
>>
>> What is "this".  It's the number possible phase relations between the
>> 1000 qubits.  If we send a 1000 electrons toward our photographic plate
>> through a 1000 holes the Schrodinger wave function approaching the
>> photographic plate then also has 1e301 different phase relations.  The
>> difference is only that we don't control them so as to cancel out "wrong
>> answers".
>>
>>
>
> The reason I think the quantum computer example is important to consider
> is because when we control them to produce a useful result, it becomes that
> much harder to deny the reality and significance of the intermediate
> states. For instance, we can verify the result of a Shor calculation for
> the factorization of a large prime.
>
>
> Someone else is interested in factorizing primes?
>

My background is more in algorithms and cryptography, than physics or
quantum mechanics.


>
>   We can't so easily verify the statistics of the 1e301 phase relations
> are what they should be.
>
>
>> This is not only over a googol^2 times the number of silver halide
>> molecules in your plate, but more than a googol times the 10^80 atoms in
>> the observable universe.
>>
>> What is it, in your mind, that is able to track and consistently compute
>> over these 10^301 states, in this system composed of only 1000 atoms?
>>
>>
> Are you aware of anything other than many-worlds view that can account for
> this?
>
>
> Yes.
>

We're listening.


>
> Also note that you can only read off 200bits of information (c.f. Holevo's
>>> theorem).
>>>
>>>
>> True, but that is irrelevant to the number of intermediate states
>> necessary for the computation that is performed to arrive at the final and
>> correct answer.
>>
>>
>> But you have to put in 2^200 complex numbers to initiate your qubits.  So
>> you're putting in a lot more information than you're getting out.
>>
>
> You just initialize each of the 200 qubits to be in a superposition.
>
>
>> Those "intermediate states" are just interference patterns in the
>> computer, not some inter-dimensional information flow.
>>
>
> What is interference, but information flow between different parts of the
> wave function: other "branches" of the superposition making their presence
> known to us by causing different outcomes to manifest in our own branch.
>
>
> The superposition exists in our branch.
>

That is one way of using the term branch.  Another is that a superposition
is when multiple branches can interfere because everything else about them
is the same.


>
> Also, many quantum algorithms only give you an answer that is probably
>> correct.  So you have to run it multiple times to have confidence in the
>> result.
>>
>
> I would say it depends on the algorithm and the precision of the
> measurement and construction of the computer.  If your algorithm computes
> the square of a randomly initialized set of qubits, then the only answer
> you should get (assuming perfect construction of the quantum computer)
> after measurement will be a perfect square.
>
>
>>
>> Quantum computers will certainly impact cryptography where there's heavy
>> reliance on factoring primes and discrete logarithms.  They should be able
>> to solve protein folding and similar problems that are out of reach of
>> classical computers.  But they're not a magic bullet.  Most problems will
>> still be solved faster by conventional von Neumann computers or by
>> specialized neural nets.  One reason is that even though a quantum
>> algorithm is faster in the limit of large problem size, it may still be
>> slower for the problem size of interest.  It's the same problem that shows
>> up in classical algorithms; for example the Coppersmith-Winograd algorithm
>> for matrix multiplication takes O(n^2.375) compared to the Strassen
>> O(n^2.807) but it is never used because it is only faster for matrices too
>> large to be processed in existing computers.
>>
>
> So where do you stand concerning the reality of the immense number of
> intermediate states the qubits are in before measured?
>
>
> Brent can answer for himself. But from my point of view the idea that the
> quantum computer works by doing a large number of classical computations in
> parallel in different "universes" is overly naĩve.
>

This is exactly what Shor's algorithm does, for the first and second parts
of the algorithm:
https://en.wikipedia.org/wiki/Shor%27s_algorithm#Finding_the_period


> Shor's algortithm centres on a fast Fourier transform implemented by
> interference; it certainly doesn't simply calculate all classical
> possibilities directly in parallel.
>

It does. But if it ended there, only one branch would get the right answer
at the time of measurement.   To make the probability high that all
branches learn something useful from the parallel computation, the third
step of Shor's algorithm uses the Fourier transform.

Jason

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