Deeper than primes

[doronshadmi]

What is that question supposed to mean?

By comprehensive science everything is under research, including the researched, the researcher and the research methods.

 

[doronshadmi]

Let's correct something about http://www.internationalskeptics.com/forums/showpost.php?p=7436675&postcount=16193 analogy.

This part

Furthermore, if A = "closed door" and B = "opened door", then (AB) = "indeterminate under superposition" and (A,B) = "determinate under superposition collapse".

has to be under "(" ")" because it is not related to that particular analogy, but it demonstrates the general principle of superposition and superposition collapse, no matter what (AB) superposition or (A,B) superposition collapse are involved.

---------

By using the general principle of superposition and superposition collapse, "C=A+B" expression is actually DS (A,B,C) under F (1,1,1) under 3-Uncertainty x 3-Redundancy Distinction-Tree.

Here is a typical response ( http://www.internationalskeptics.com/forums/showpost.php?p=5980338&postcount=10029 ) of a person that can't get the difference between (ABC) and (A,B,C) (which are under 3-Uncertainty x 3-Redundancy Distinction-Tree):

The single silt pattern is an expression of the Local aspect of the Non-local/Local Linkage, where the double slit pattern is an expression of the Non-local aspect of the Non-local/Local Linkage, and how Non-local/Local Linkage is expressed as:

a) Asymmetry of no superposition of ids (A,B)(known as Certainty).

b) Symmetry of superposition of certain ids (A,A)(known as Redundancy).

c) Symmetry of superposition of ids (AB)(known as Uncertainty).

Nope, just more of your word salad Doron.

a),b),c) is taken as some case of DS (A,B,C) under F (1,1,1) under 3-Uncertainty x 3-Redundancy Distinction-Tree, but The Man can't get that because his best is closed under (A,B,C) (he can't deal with the uncertainty of superposition of ids, which in this case is DS (ABC) under F (3,0,0) under 3-Uncertainty x 3-Redundancy Distinction-Tree).

As about left-hemisphere\right-hemisphere case:

Persons that say NO in "Collapsing wavefunctions?" column of http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics table, are left-only hemisphere persons that have no tolerance to Uncertainty, Indeterminate, Vagueness, etc. ...

Persons that say YES in this column have tolerance to Uncertainty, Indeterminate, Vagueness, etc. ... because they are not using only their left hemisphere.

As can be seen (according to this table) the majority of scientific community has no tolerance to Uncertainty, Indeterminate, Vagueness, etc. ...

 

[doronshadmi]

As about Mind-Body problem ( http://en.wikipedia.org/wiki/Mind-body_dichotomy ), please look at:

http://en.wikipedia.org/wiki/Henry_Stapp

http://www3.tsl.uu.se/~flechl/paper/philo/stapp.pdf

 

[doronshadmi]

Here is an important part, taken from http://www3.tsl.uu.se/~flechl/paper/philo/stapp.pdf:

It seems Stapp would have to modify his opinion about the unity of consciousness if he wants to keep the possibility of free will and conserve the laws of quantum theory at the same time.

Once again, the state of unity is not some opinion at the level of thoughts, as shown (by analogy) in http://www.internationalskeptics.com/forums/showpost.php?p=7255966&postcount=15594 .

Unity is the actual non-subjective state of mind ( as explained in http://www.internationalskeptics.com/forums/showpost.php?p=7241076&postcount=15569 ) exactly because it is the source of any possible expression, whether it is called mind or body.

 

[The Man]

C is a strict result of A strict and B strict.

In other words you are still get only strict things (in this case you are using the particular case of DS (A,B,C) under F (1,1,1), which are under 3-Uncertainty x 3-Redundancy Distinction-Tree).

Nope in the exact same words I used...

Again you really need to learn what a wave function is, instead of just trying to conflate your nonsense with quantum mechanics by simply saying “collapsed”. Oh, and a “single variable” can represent, well, “more than one variable” including some operations like addition, as in C=A+B. That’s what makes variables useful. Learn some basic algebra.

You are using again DS (A,B) under F (1,1), exactly because A is positive OR negative.

Certainly not, I never cited any of your DS BS, and "A" represents both positive and negative values as cited.

So even with “B” strictly a single positive value, “A” isn’t. Whether “A” is positive or negative is simply indeterminate.

You seem to be (perhaps deliberately) confusing the fact that "A" can be ascribed a "positive OR negative" value for any non-zero instance of "B" with some restriction on "A" not representing both a positive and a negative value for any non-zero instance of "B".


Your assertion wasn't that a variable could be a singular value at some particular instance, it was that it must be a singular value at any instance. That A²=B has, simultaneously, both positive and negative solutions for "A" given any instance of a non-zero value of "B" demonstrates that assertion to be patently false. In fact unless "A" represented both a positive and a negative value simultaneously for any non-zero instance of "B" you could not choose between a "positive OR negative" value of "A" at that instance of "B".

Under DS (AB) superposition A is simultaneously positive AND negative, but it is not a contradiction since (AC) is non-strict.

As you claim your BS "superposition" does not use the principle of superposition, your BS "superposition" is still without any, well, superposition whatsoever.

Not simultaneously, The Man, not simultaneously, and if it is a sum (as done in C=A+B), then the sum is some strict single value.

Fine then by all means please tell us the minimum amount of time that must pass, when B=25, between when A=-5 and when A=5?

Or perhaps at what time of day -5²=25 but 5² doesn't?


In case you have not realized "B" can represent a non-zero time (say in munities). So with both positive and negative solutions for B^-2 as "A", "A" quite literally is both positive and negative at the same time (simultaneously).

Oh, and A²=B isn't "a sum (as done in C=A+B)" that should have been obvious even to just you.

Since you have claimed "if it is a sum (as done in C=A+B), then the sum is some strict single value" please tell us the "strict single value" of "C"?

The Man, you are still closed under strictness and simply can't get out no matter what twisted maneuvers are done by you with variables.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

Only if (AB) is transformed into (A,B) by classical Fourier transform, and the result is not equivalent to the state of superposition, exactly because any measurement actually changes the superposition into set of strict values (a state vector) (the superposition is collapsed into a strict state).

Read the actual definition in article you quoted Doron not just the introduction...

( http://en.wikipedia.org/wiki/Quantum_Fourier_transform )

The quantum Fourier transform is the classical discrete Fourier transform applied to the vector of amplitudes of a quantum state.

http://en.wikipedia.org/wiki/Quantum_Fourier_transform#Definition

Your claim...

...there is no equivalence between the classical Fourier transform that acts on a vector (which is a set of strict values) and the natural non-strict state of quantum superposition.

is demonstrably false by the very article you cited but apparently never bothered to actually read or even just look at the definition.

No wonder that you can't get http://www.internationalskeptics.com/forums/showpost.php?p=7431694&postcount=16185.

Evidently you just don't understand coherence and thus de-coherence, but that is no wonder to most others here.

At least try to get out of your please.

It's your hole Doron and you just keep burring yourself by digging it deeper.

 

[doronshadmi]

Let us distinguish between (AB) and (A,B) or (ABC) and (A,B,C).

Fine then by all means please tell us the minimum amount of time that must pass, when B=25, between when A=-5 and when A=5?

Time is not involved here, you simply use (A,B) superposition collapse of the variables themselves if you wish to use them by some expression (A²=B, in this case)

Oh, and A²=B isn't "a sum (as done in C=A+B)" that should have been obvious even to just you.

Since you have claimed "if it is a sum (as done in C=A+B), then the sum is some strict single value" please tell us the "strict single value" of "C"?

Again The Man, you still do not grasp that (AB) or (ABC) are the superposition of the variables themselves, where (A,B) or (A,B,C) are the superposition collapse of these variables, which enables expressions like A²=B or C=A+B.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

Again, stop simply force your inability to distinguish between (AB) (which is an example of variables under superposition) and (A,B) (which is an example of variables under superposition collapse) onto others.

The rest of your post is derived from your inability.

More details can be found in http://www.internationalskeptics.com/forums/showpost.php?p=7443804&postcount=16202.

I never cited any of your DS BS

You can't, exactly because your BS covers your mind, as can be seen all along this thread.

Your assertion wasn't that a variable could be a singular value at some particular instance, it was that it must be a singular value at any instance.

A²=B or C=A+B are expressions of superposition collapse of the variables themselves.

 

[doronshadmi]

Look as this:

the quantum Fourier transform acts on a quantum state, whereas the classical Fourier transform acts on a vector, so the quantum Fourier transform can not give a generic exponential speedup for any task which requires the classical Fourier transform.

( http://en.wikipedia.org/wiki/Quantum_Fourier_transform )
In other words, since superposition is involved in non-strict quantum states that are collapsed into some strict state by some measurement, there is no equivalence between the classical Fourier transform that acts on a vector (which is a set of strict values) and the natural non-strict state of quantum superposition (in this case the members of the vector are under superposition (their ids are uncertain)).

Here is the Hebrew version of this part

מאידך, בעוד הפעולה ה"קלאסית" פועלת על וקטור של ערכים, הפעולה הקוונטית הינה על מצב קוונטי הנמצא בסופרפוזיציה, כאשר המקדמים מתארים את ערכי הווקטור הקלאסי השקול. מכיוון שלא ניתן לבצע מדידה ישירה של ערכי המקדמים בסופרפוזיציה קוונטית, השקילות אינה מלאה, ולא ניתן לקבל שיפור מעריכי עבור ביצוע ההתמרה באופן כללי​

and its translation to English is:

"On the other hand,
(,מאידך)
while the "classic" operation operates on vector's values,
(,בעוד הפעולה ה"קלאסית" פועלת על וקטור של ערכים)
the quantum operation is on a quantum state that is found in superposition,
(,הפעולה הקוונטית הינה על מצב קוונטי הנמצא בסופרפוזיציה)
when the coefficients describe the values of the equivalent classical vector.
(.כאשר המקדמים מתארים את ערכי הווקטור הקלאסי השקול)
Since direct measurement of the coefficients' values of quantum superposition can't be done,
(,מכיוון שלא ניתן לבצע מדידה ישירה של ערכי המקדמים בסופרפוזיציה קוונטית)
the equivalence is incomplete,
(,השקילות אינה מלאה)
and there is no exponential improvement for the transformation's implementation in general.
(.ולא ניתן לקבל שיפור מעריכי עבור ביצוע ההתמרה באופן כללי)"

The quantum Fourier transform is the classical discrete Fourier transform applied to the vector of amplitudes of a quantum state.

( http://en.wikipedia.org/wiki/Quantum_Fourier_transform )
In other words, by transforming (AB) superposition into (A,B) superposition collapse, the classical discrete Fourier transform applied to the vector of (strict) amplitudes of a quantum state (which is not under superposition anymore).

 

[doronshadmi]

Please look at http://techie-buzz.com/science/quantum-teleportation-of-photons-achieved.html .

In this article bits are expressed by N, where qubits are expressed by 2^N.

But the base value can be k = 2 to (any base value > 2), where qubits are expressed by k^N.

For more details, please look at http://www.sciencemag.org/content/332/6027/330.full.pdf?ijkey=oM68vTRllELO6&keytype=ref&siteid=sci

 

[epix]

Both hemispheres are used -- that's why they are included in the hardware. But there is that "third hemisphere," which can solve problems without the "learn and repeat" process. The best test for the third hemisphere is based on something that relates to 3 -- like the cube. So here is the question: What is in that box?

The third hemisphere . . . . It looks that the aliens took up the challenge.
http://www.cropcircleconnector.com/2011/furzeknoll/furzeknoll2011a.html

This is not a difficult puzzle: Just label the three hemispheres. If you use critical and rational thinking, then you label them with A, B, A, coz the size of them calls for it -- they are two identical objects with one different in the middle. Hence A, B, A. But critical and rational thinking has been plagued by arbitrary crap, and so you need to chose another approach. For example, go with the motif, take a sphere and cut it in half by the plane. The result is SPH / ERE.

But what happens if you cut the sphere exactly SP / HERE? This cut on 2 and 4 letters is supposed to be done coz

a) 2 + 4 = 6
b) 2 x 4 = 8

where 6 is the day and 8 is the month.

Furze Knoll, Bishop Cannings, Wiltshire. Reported 6th August.

http://www.cropcircleconnector.com/2011/furzeknoll/furzeknoll2011a.html

And so if someone asks you to cut a sphere here, the natural question is to ask where exactly. If the answer doesn't come, the initial request has been deemed sufficient for you to know where.



I think the aliens stopped by a wrong planet.

(I wonder how the aliens solve my puzzle that asks what's inside that box.)

 

[doronshadmi]

I think the aliens stopped by a wrong planet.

The aliens left because they are allergic to wheat.

 

[doronshadmi]

More about 2^N Quantum Computing and qubits' superposition:

http://www.dailygalaxy.com/my_weblo...new-record-register-with-14-quantum-bits.html

( see also http://www.internationalskeptics.com/forums/showpost.php?p=7447672&postcount=16208 )

 

[doronshadmi]

The einselected states lack coherence, and therefore do not exhibit the quantum behaviours of entanglement and superposition.

Since only quasi-local, essentially classical states survive the decoherence process, einselection can in many ways explain the emergence of a (seemingly) classical reality in a fundamentally quantum universe (at least to local observers).

( http://en.wikipedia.org/wiki/Einselection )
In other words, real superposition is not lack coherence, where by this coherence even the used variables are under superposition (their ids are uncertain).

Uncertainty x Redundancy Distinction-Trees can be useful for Quantum Darwinism ( http://en.wikipedia.org/wiki/Quantum_Darwinism ).

 

[The Man]

Correction

In case you have not realized "B" can represent a non-zero time (say in munities). So with both positive and negative solutions for B^-2 as "A", "A" quite literally is both positive and negative at the same time (simultaneously).

Highlighting added

B^-2 should have been B^1/2, B^-2 would have been the inverse of the square of “B” while B^1/2 would be the square root of “B”, I apologize for any confusion.

 

[The Man]

Let us distinguish between (AB) and (A,B) or (ABC) and (A,B,C).

OK,

Time is not involved here, you simply use (A,B) superposition collapse of the variables themselves if you wish to use them by some expression (A²=B, in this case)

If you are going to claim simultaneity or the lack of it, then time is quite specifically involved as simultaneous means ‘at the same time’. You said…

Not simultaneously, The Man, not simultaneously, and if it is a sum (as done in C=A+B), then the sum is some strict single value.

So again…

Fine then by all means please tell us the minimum amount of time that must pass, when B=25, between when A=-5 and when A=5?

Or perhaps at what time of day -5²=25 but 5² doesn't?


In case you have not realized "B" can represent a non-zero time (say in munities). So with both positive and negative solutions for B^-2 as "A", "A" quite literally is both positive and negative at the same time (simultaneously).

Again The Man, you still do not grasp that (AB) or (ABC) are the superposition of the variables themselves, where (A,B) or (A,B,C) are the superposition collapse of these variables, which enables expressions like A²=B or C=A+B.

Again Doron you simply refuse to grasp that your “superposition”, by your own assertions, is not a superposition and that you simply using the word “collapse” will not mystically imbue your non-superposition “superposition” with some quantum like properties whatever you may think they are.

Again, stop simply force your inability to distinguish between (AB) (which is an example of variables under superposition) and (A,B) (which is an example of variables under superposition collapse) onto others.

Again stop simply lying about your “superposition” being a superposition, which you claim yourself you do not use.

The rest of your post is derived from your inability.

More details can be found in http://www.internationalskeptics.com/forums/showpost.php?p=7443804&postcount=16202.

All of your posts derive from your deliberate ignorance of mathematics and your apparently compulsive need to lie about what you claim your own notions do not use.

You can't, exactly because your BS covers your mind, as can be seen all along this thread.

Then don’t lie about what you now claim you know I was not and could not have been using. Continuing this compulsive lying of yours about what you assert, what some article asserts and what others have asserted will not help you and fools evidently only you.

A²=B or C=A+B are expressions of superposition collapse of the variables themselves.

No “A²=B” is an expression where the square of variable “A” is set equal to variable “B” and “C=A+B” is an expression where variable “C” is set equal to the sum of variables “A” and “B”. Again please learn what a superposition is, why it is in reference to linear systems and what the collapse of a state vector means. Simply lying and conflating your nonsense with what you would like it to mean has gotten you nowhere for 20 some odd years and will continue to do just that. Waste your entire life if you want Doron, but there are people here and elsewhere that can and have been trying to help you. The choice however remains entirely yours.

 

[The Man]

Look as this:

( http://en.wikipedia.org/wiki/Quantum_Fourier_transform )
In other words, since superposition is involved in non-strict quantum states that are collapsed into some strict state by some measurement, there is no equivalence between the classical Fourier transform that acts on a vector (which is a set of strict values) and the natural non-strict state of quantum superposition (in this case the members of the vector are under superposition (their ids are uncertain)).

Here is the Hebrew version of this part

and its translation to English is:

"On the other hand,
(,מאידך)
while the "classic" operation operates on vector's values,
(,בעוד הפעולה ה"קלאסית" פועלת על וקטור של ערכים)
the quantum operation is on a quantum state that is found in superposition,
(,הפעולה הקוונטית הינה על מצב קוונטי הנמצא בסופרפוזיציה)
when the coefficients describe the values of the equivalent classical vector.
(.כאשר המקדמים מתארים את ערכי הווקטור הקלאסי השקול)
Since direct measurement of the coefficients' values of quantum superposition can't be done,
(,מכיוון שלא ניתן לבצע מדידה ישירה של ערכי המקדמים בסופרפוזיציה קוונטית)
the equivalence is incomplete,
(,השקילות אינה מלאה)
and there is no exponential improvement for the transformation's implementation in general.
(.ולא ניתן לקבל שיפור מעריכי עבור ביצוע ההתמרה באופן כללי)"

Again read the whole article particularly the definition instead of just the introduction.

( http://en.wikipedia.org/wiki/Quantum_Fourier_transform )
In other words, by transforming (AB) superposition into (A,B) superposition collapse, the classical discrete Fourier transform applied to the vector of (strict) amplitudes of a quantum state (which is not under superposition anymore).

The article defines a Quantum Fourier transform in none of your “other words”. Stop lying about what the articles you cite, but evidently never actually read, say. Here is a clue to help you in the future Doron, if you have to use your own “other words” then there is a good chance those “other words” are not what the article says.

Please show where that article specifically claims a “superposition collapse” or “a quantum state (which is not under superposition anymore)”, otherwise stop just lying about what the article does assert.


http://en.wikipedia.org/wiki/Qubit

Again please at least learn something…



http://en.wikipedia.org/wiki/Qubit#Qubit_states

Qubit states
A pure qubit state is a linear superposition of the basis states. This means that the qubit can be represented as a linear combination of and :

Oh look “A pure qubit state is a linear superposition of the basis states” that being the superposition you specifically claim you do not use.



http://en.wikipedia.org/wiki/Qubit#Operations_on_pure_qubit_states

Operations on pure qubit states
There are various kinds of physical operations that can be performed on pure qubit states.[citation needed]
• A quantum logic gate can operate on a qubit: mathematically speaking, the qubit undergoes a unitary transformation. Unitary transformations correspond to rotations of the Bloch sphere.
• Standard basis measurement is an operation in which information is gained about the state of the qubit. The result of the measurement will be either , with probability | α | 2, or , with probability | β | 2. Measurement of the state of the qubit alters the values of α and β. For instance, if the result of the measurement is , α is changed to 1 (up to phase) and β is changed to 0. Note that a measurement of a qubit state entangled with another quantum system transforms a pure state into a mixed state.

“Unitary transformations” on qubit states? That sounds familiar, oh yes a Quantum Fourier transform is a unitary transformation on qubit states.
http://en.wikipedia.org/wiki/Quantum_Fourier_transform#Unitarity

Unitarity
Most of the properties of the quantum Fourier transform follow from the fact that it is a unitary transformation.

As well as...

Quantum Fourier transform
In quantum computing, the quantum Fourier transform is a linear transformation on quantum bits, and is the quantum analogue of the discrete Fourier transform.

a linear transformation.

Funny how math works and your lying, conflation and deliberate ignorance doesn’t (except evidently on just you). Please stop being deliberately ignorant, lying and trying to conflate your nonsense with what works and instead just learn at least something Doron.

 

[doronshadmi]

Funny how math works and your lying,

Not funny how your step-by-step left-hemisphere-only reasoning prevents from you to grasp a simple notion line "superposition of the variables themselves", which actually shows how the mathematical science itself is nothing but a particular case of more comprehensive framework, which is parallel (under superposition of non-strict ids), serial (under superposition collapse of strict ids) and any possible intermediate state between them.

And again, time is not involved here.

No wonder that you can't grasp something like

because step-by-step left-hemisphere-only reasoning (which is all that is for you) is nothing but some particular case of it.

 

[doronshadmi]

Again read the whole article particularly the definition instead of just the introduction.

The definitions of the article are based only on the classical Fourier transform exactly because direct measurement of the coefficients' values of quantum superposition can't be done, as written here:

"On the other hand,
(,מאידך)
while the "classic" operation operates on vector's values,
(,בעוד הפעולה ה"קלאסית" פועלת על וקטור של ערכים)
the quantum operation is on a quantum state that is found in superposition,
(,הפעולה הקוונטית הינה על מצב קוונטי הנמצא בסופרפוזיציה)
when the coefficients describe the values of the equivalent classical vector.
(.כאשר המקדמים מתארים את ערכי הווקטור הקלאסי השקול)
Since direct measurement of the coefficients' values of quantum superposition can't be done,
(,מכיוון שלא ניתן לבצע מדידה ישירה של ערכי המקדמים בסופרפוזיציה קוונטית)
the equivalence is incomplete,
(,השקילות אינה מלאה)
and there is no exponential improvement for the transformation's implementation in general.
(.ולא ניתן לקבל שיפור מעריכי עבור ביצוע ההתמרה באופן כללי)"

and supported here:

the quantum Fourier transform acts on a quantum state, whereas the classical Fourier transform acts on a vector, so the quantum Fourier transform can not give a generic exponential speedup for any task which requires the classical Fourier transform.

( http://en.wikipedia.org/wiki/Quantum_Fourier_transform )

Persons like you, The Man, that can't grasp http://www.internationalskeptics.com/forums/showpost.php?p=7450437&postcount=16217, have no choice but to define frameworks that are limited only to strict ids.

Simply lying and conflating your nonsense with what you would like it to mean has gotten you nowhere for 20 some odd years and will continue to do just that. Waste your entire life if you want Doron, but there are people here and elsewhere that can and have been trying to help you. The choice however remains entirely yours.

Your reasoning uses a spotlight in order to research the natural life of night creatures.

Go ahead dig yourself under your spotlight

 

[The Man]

Not funny how your step-by-step left-hemisphere-only reasoning prevents from you to grasp a simple notion line "superposition of the variables themselves", which actually shows how the mathematical science itself is nothing but a particular case of more comprehensive framework, which is parallel (under superposition of non-strict ids), serial (under superposition collapse of strict ids) and any possible intermediate state between them.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

And again, time is not involved here.

So are you claiming that your “simultaneously” does not involve simultaneity? Given your “superposition” that you claim does not use superposition that would hardly be an unusual assertion from you.

No wonder that you can't grasp something like
[qimg]http://farm7.static.flickr.com/6146/6017791855_661f47be5b_b.jpg[/qimg]
because step-by-step left-hemisphere-only reasoning (which is all that is for you) is nothing but some particular case of it.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

 

[The Man]

The definitions of the article are based only on the classical Fourier transform exactly because direct measurement of the coefficients' values of quantum superposition can't be done, as written here:

No Doron the actual definition states clearly (as quoted before)…

The quantum Fourier transform is the classical discrete Fourier transform applied to the vector of amplitudes of a quantum state.

Again read the definition not just the introduction.

and supported here:

Here is the English version of the section of the introduction you keep quoting.

However, the quantum Fourier transform acts on a quantum state, whereas the classical Fourier transform acts on a vector, so the quantum Fourier transform can not give a generic exponential speedup for any task which requires the classical Fourier transform.

No support for you Doron.

Here is the full pharagragh from the introduction that you just keep quoting a single line from.

The quantum Fourier transform can be performed efficiently on a quantum computer, with a particular decomposition into a product of simpler unitary matrices. Using a simple decomposition, the discrete Fourier transform can be implemented as a quantum circuit consisting of only O(n²) Hadamard gates and controlled phase shift gates, where n is the number of qubits.[1] This can be compared with the classical discrete Fourier transform, which takes O(n2ⁿ) gates (where n is the number of bits), which is exponentially more than O(n²). However, the quantum Fourier transform acts on a quantum state, whereas the classical Fourier transform acts on a vector, so the quantum Fourier transform can not give a generic exponential speedup for any task which requires the classical Fourier transform.

Clearly Referring to the “O(n2ⁿ) gates (where n is the number of bits), which is exponentially more than O(n²)” as the reason for the inapplicability of the quantum Fourier transform to “give a generic exponential speedup for any task which requires the classical Fourier transform”.

That the Hebrew version for some reason refers to a direct measurement…

As in this alternate translation of the one line from the introduction you keep quoting.

On the other hand, in another the action H"klasit " works on vector of values, the action Hkwntit is on situation of my situated quantum Bsoprpozitsia, when the coefficients describe principled Hwktor the classic the weighed. Inasmuch as that it's impossible to execute a direct measurement of principled the coefficients Bsoprpozitsia Kwntit, the equivalence is not full, and it's impossible to receive an improvement of my assessor for execution Hhtmra in general

Is irrelevant as it specifically refers to such “direct measurement” as being “impossible”. Your own translation Doron specifically refutes your claims of “superposition collapse” or “a quantum state (which is not under superposition anymore)” as it asserts such direct measurement “can't be done”.

( http://en.wikipedia.org/wiki/Quantum_Fourier_transform )

Persons like you, The Man, that can't get grasp http://www.internationalskeptics.com/forums/showpost.php?p=7450437&postcount=16217, have no choice but to define frameworks that are limited only to strict ids.

Again stop simply trying to posit aspects of your own failed reasoning onto others.

 

[epix]

More about 2^N Quantum Computing and qubits' superposition:

http://www.dailygalaxy.com/my_weblo...new-record-register-with-14-quantum-bits.html

( see also http://www.internationalskeptics.com/forums/showpost.php?p=7447672&postcount=16208 )

I didn't help. Just can't write down the matrix for N = 4.

http://demonstrations.wolfram.com/QuantumFourierTransformCircuit/

Can you? You seem to have extended knowledge of the topic.