Is there a way to translate quantum bits in a traditional numerological system?

In summary: The problem is that, unlike a classical bit, you cannot directly observe its value: measuring a qubit along some axis turns it into a classical bit. If the value was originally near the north pole, then the qubit is more likely to turn into a 1. If it was near the south pole, then it was more likely to turn into a 0.This is something that a system of conventional bits would need years to model for even some small number of qubits.
  • #1
Levi Porter
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Are quantum bits just a form of a ternary numeral system?

If something can be 0, 1, or both simultaneously, isn't the superposition just another equal value?

If the superposition of 0 and 1 were literally individual, combined, and possibly something different simultaneously, then it seems that a quantum bit could be an operational function of a numeral system base of 3, 4 or 5, I would guess.

Is there a way to translate quantum bits in a traditional numerological system?
 
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  • #2
Individual q-bits are sphere-valued. If you choose an axis, then 0 and 1 are the north and south pole of the sphere.
 
  • #3
Hurkyl said:
Individual q-bits are sphere-valued. If you choose an axis, then 0 and 1 are the north and south pole of the sphere.

Thanks for the response Hurkyl.

Does sphere-valued mean that information, pulses, or values occupy the entire volume of a sphere?

If the poles are just a predetermined and or adjustable locational points of reference including superposition, then it still seems that it could be correspondent to a numeral system?
 
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  • #4
The problem is that, unlike a classical bit, you cannot directly observe its value: measuring a qubit along some axis turns it into a classical bit. If the value was originally near the north pole, then the qubit is more likely to turn into a 1. If it was near the south pole, then it was more likely to turn into a 0.

(Incidentally, all orientations I have used are an arbitrary choice)

The art of quantum computing is to get your qubits to interact without measuring them while the computer is running -- you only want to measure them at the very end, at which point you are very likely to see the answer you wanted.


Of course, when writing your algorithm, you would design it to use the full set of values. But I have no idea how it would correspond to a numeral system.

Incidentally, I don't see how classical bits correspond to a numeral system either.
 
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  • #5
Hurkyl said:
Incidentally, I don't see how classical bits correspond to a numeral system either.

I could be misinterpreting...anyway, my understanding was that conventional bits are generated by electric pulses of either on or off, translated into or as a 1 or 0, and used as a code assigned to all digital information originating from a functionality of two possibilities referred to as the binary numeral system or base-2 number system.

At least this is my understanding from http://en.wikipedia.org/wiki/Binary_numeral_system
 
  • #6
Levi Porter said:
Are quantum bits just a form of a ternary numeral system?

Not really, for one because upon measurement a qubit will always have the value 0 or 1. For example, if i have 4 bits then i can store [itex]2^{4}[/itex] states, and pass any of them to you and you'd be able to recognize which state i had passed you (0100, 0011, etc). On the other hand if i have 4 qubits, then any of the [itex]3^{4}[/itex] quantum states i send you will always collapse to one of [itex]2^{4}[/itex] states, so although we can interpret a qubit as a base-3 digit, we can't use it as base-3 digit for storing and passing information.

In quantum computation we also use more information for each qubit than can be stored ina base-3 digit. For example rather than having the three states:
0, 1, 01
In quantum computation you might have:
[x] 1, [y] 0
Where [itex]x^{2}[/itex] gives the probability of the qubit being 1 and [itex]y^{2}[/itex] gives the probability of it being 0.

Therefore a qubit can have any number of states defined by [x y], where [tex]x^{2} + y^{2} = 1[/tex].

In operation, a quantum machine isn't equivalent to a base-3 machine either (which is actually equivalent to a base-2 machine, as they both can be modeled by a turing machine). When a quantum computer with n qubits performs an operation, it's performing an operation on 2^5 states (the number of states that the system of 5 qubits is in a superposition of). An operation changes the probabilities of each possible state, the one with the highest probability being the most likely to be revealed upon measurement.

This is something that a system of conventional bits would need years to model for even some small number of qubits.

Scott Aaronson has in his blog a pretty good explanation of Peter Shor's quantum factoring algorithm using simple terms:
http://scottaaronson.com/blog/?p=208

I particularly like one of the explanations given in one of the comments:
http://scottaaronson.com/blog/?p=208#comment-10026
 
  • #7
Thanks for the reply and links Job

-Job- said:
... so although we can interpret a qubit as a base-3 digit, we can't use it as base-3 digit for storing and passing information.

So is it safe to say that it is unlikely in the near future that quantum computers will replace conventional computers?

It seems like their usage will, or are primarily used for code breaking, encryption development, or for being able to more accurately measure the cosmos.

I still can't grasp in layman or non mathematical terminology how an algorithm can simultaneously look at a series of numbers hundreds or thousands of digits in length without having some sort of lineage, chronology, or individual identifiers to each digit.

Q-bit computers remind me of the Z Machine that is trying to build a sun on Earth so to speak. It can produce about 80 times the entire world's output of electricity for billionths of a second.

It seems the challenge with both is to keep them going and harness their results.
 
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1. Can quantum bits be translated into traditional numerical values?

Yes, it is possible to convert quantum bits (qubits) into traditional numerical values, such as binary or decimal numbers. This process is known as measurement and involves extracting information from a quantum system.

2. How is this translation done?

The translation of quantum bits into traditional numerical values is done through a process called quantum state tomography. This involves performing a series of measurements on the qubits and using statistical analysis to determine the state of the quantum system, which can then be represented as a traditional numerical value.

3. Is this translation accurate?

Yes, the translation of quantum bits into traditional numerical values is accurate. However, due to the nature of quantum mechanics, there may be some uncertainty in the measurements, leading to slight variations in the translated values.

4. Can quantum bits be translated into any numerical system?

Yes, quantum bits can be translated into various numerical systems, such as binary, decimal, or even hexadecimal. The choice of numerical system depends on the specific application and the desired level of precision.

5. Are there any limitations to this translation process?

There are some limitations to the translation of quantum bits into traditional numerical values. For instance, the accuracy of the translation may decrease as the number of qubits increases, and there may be some computational limitations for large-scale quantum systems. Additionally, the translation may be affected by external factors, such as noise and interference.

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