11
$\begingroup$

Our current computers use bits, so they use the binary numeral system. But I heard that the future quantum computers will use qubits instead of simple bits.

Since in the word "qubit" there is the word "bi" I first thought that this meant that quantum computers would use binary (base 2).

But then I heard that qubits had three possible states: 0, 1, or a superposition of 0 and 1. So I then thought that this must mean that they will use ternary (base 3).

But then I saw that one qubit can hold as much information as two bits. So I thought that this maybe mean that they will use quaternary (base 4).

So which numeral system will the future quantum computers use: binary, ternary or quaternary?

$\endgroup$
  • 1
    $\begingroup$ A qbit is neither of the three. Note furthermore that "bit" is sometimes used as a synonym for "one symbol" (i.e. systems with base other than two might use the term, too). $\endgroup$ – Raphael Oct 7 '14 at 12:12
  • $\begingroup$ roughly speaking the inputs/outputs are binary but the intermediate calculations are in qubit superpositions wrt R.s comment "none of the above" $\endgroup$ – vzn Oct 7 '14 at 15:40
13
$\begingroup$

The other answers are nice, but none address the question: what numeric base(s) might quantum computers use? I will answer in two parts: first, the question is a little subtle, and second, you may use any numeric base, and then you work with qutrits or in general with qudits, which lead to qualitatively new intuitions! Or at any rate, I will try to make the case that they do.

A quantum bit isn't just a $0$ or a $1$, it's a bit more complex than that. For example, a quantum bit may be in the state $\sqrt{\frac{1}{4}}|0\rangle+\sqrt{\frac{3}{4}}|1\rangle$. When measured, you will measure the outcome $0$ with probability $\frac{1}{4}$ and the outcome $1$ with probability $\frac{3}{4}$. The 'superposition' you talked about is $\sqrt{\frac{1}{2}}|0\rangle+\sqrt{\frac{1}{2}}|1\rangle$, but in general any pair of complex numbers $a$ and $b$ will do, as long as $a^2+b^2=1$. If you have three qubits, then you can entangle them, and the state will be

$$a_0|000\rangle + a_1|001\rangle+a_2|010\rangle+a_3|011\rangle+a_4|100\rangle +a_5|101\rangle+a_6|110\rangle +a_7|111\rangle$$

But when you measure this three-qubit system, your measurement outcome is one out of these 8 states, that is, three bits. This is this really weird dichotomy where on the one hand quantum systems seem to have this exponential state space, but on the other hand we only seem to be able to 'get at' a logarithmic part of the state space. In 'Quantum Computing Since Democritus', Scott Aaronson probes this question by matching off several complexity classes to try and understand how much of this exponential state space we can exploit for computation.

Having said that, there is an obvious complaint to the answer above: all the notation is in binary. Qubits are in a superposition of two base states, and entangling them doesn't change that much, because three qubits are in a superposition of $2^3$ base states. It's a legitimate complaint, because one usually thinks of $\texttt{unsigned int}$ as a number, and only remembers that it is implemented as a 32-bit string as an afterthought.

Enter the qutrit. It is a vector in $\mathbb{C}^3$, in other words, it consists of three base states rather than two. You operate on this vector with a $3\times 3$ matrix, and all the usual things done in quantum computing don't change much, because any operation expressed in terms of qutrits can be expressed in terms of qudits, so it's really just syntactic sugar. But some problems are much easier to write down and/or think about when expressed as qudits instead of entangled qubits. For example, a variation of the Deutsch-Josza problem might ask, given an oracle for a function $f\colon \{0,\ldots, kn-1\}\rightarrow \{0,\ldots, k-1\}$, is this function constant or balanced, given that one is promised to be the case? This function naturally takes one $k$-qudit register as input. To solve it, you must apply a Fourier transform to this $k$-qudit, like so: (if this goes over your head, don't worry, it's just for illustration)

$$|a\rangle \mapsto \sum_{u=0}^{k-1}e^{i2\pi\frac{au}{k}}|u\rangle$$

If you want to express this in binary, you end up with a gate that does this on numbers $0\ldots k-1$ and acts trivially (does nothing) on all numbers $\geq k$, which is slightly less contrived than doing it this way. Similarly, consider a Bernstein-Vazirani variation where the oracle computes an in-product in some radix $r$. If $r=2$, then we know how to do it. But if $r=5$, the problem is much easier to solve by hand using several $5$-qudit registers. Some problems are easier if you have several different qudit registers, e.g. one $5$-qudit register and one $2$-qudit register.

In summary, yes, you are free to consider other numeric bases, and in the right setting that will make your life easier, for the same reason that thinking about numbers in terms other than their binary expansion helps you with normal computers. I felt compelled to answer because while most answers explained that a qubit has something to do with two base states when measuring but infinite in principle, no answer mentioned that the OPs suggestion of using other bases is legitimate and in fact really happens (for example, in Quantum walks on graphs, Aharonov et al. use a subroutine that takes a qubit and an $n$-qudit as input)

$\endgroup$
3
$\begingroup$

Quantum computers use binary. But really, this is a simplification, and there is no simple answer of how quantum algorithms work that don't get into the mathematics of quantum physics and quantum computation. The best way for you to understand this subject area is to start by studying quantum computation. There are many excellent textbooks and tutorials out there.

Whoever told you that qubits have 3 possible states, was wrong. That's not quite how quantum mechanics works. In some sense there are infinitely many possible states... but read about quantum computation to learn the real story.

$\endgroup$
2
$\begingroup$

Bits are bits, i.e. entities that can have only one of two states, usually noted $0$ and $1$.

Quantum computing uses qbits (I suppose it stands for quantum bits). Qbits allows "superposed" bits, i.e. entities that can sort of hold several bits in the same place, theoretically (according to current state of knowledge) an unbounded number of bits.

You can easily infer from this that the number of states of a qbit is actually any power of two. But those states cannot be used exactly the way you use states in usual automata theory (where you can encode states as bit strings when you have more than two). You must see them more as representing several separate but coexisting bits on the same support, that are simultaneously computed on in a parallel computation. So this idea of seeing them as states, or as representing a digit in an alphabet of size $2^n$ is actually misleading. Understanding it as parallel computations executed simultaneously (superposed) on a single piece of hardware (single core CPU, if you will) is probably closer to what happens (as much as I understand it).

So it does stay in a binary system, albeit one that has different physical properties.

But I strongly suggest you follow the advice of D.W. and look at books and tutorials.

$\endgroup$
1
$\begingroup$

The state space of a qubit is described by a column vector $(a\ \ b)^T$ of $\mathbb C^2$ with norm equal to 1. Thus, there are uncountably many different states even a single qubit can have.

However, the above is not very useful for fault-taulerant quantum computing which is what you would need if you actually want to program anything on an existing quantum computer. Under that model, you would not be able to prepare arbitrary qubits (in the above sense), however any qubit state may be approximated with arbitrary precision. Thus, you would still have infinitely many states even for a single qubit, but they will be countably many (compared to the other case).

In both cases, you may consider the basis states to be $|0 \rangle$ and $|1\rangle$ which are just special states in $\mathbb C^2.$ Then, any other single-qubit state may be described as a superposition (simply, a linear combination in the algebraic sense) of these two states (which can be prepared by applying operations which are allowed in your model of computation).

$\endgroup$
-4
$\begingroup$

Quantum particles can be in four states. They can spin up, down and be right or left-handed. If you are measuring particles that are entangled, when you measure them they will be in some combination of those four states. If we could some how predict or use a eraser of some kind, it would seem like a good idea to use quarternary rather than binary. As it stands right now, binary is being used but in the future something different will likely take the place of binary. Quantum computers are like classical computers were in the 50s, they are HUGE, expensive and not practical. In fact they are hardly useful at this time. We still struggle with decoherence. The hope it to identify a topological quantum particle that can maintain coherence (is robust) and if that day comes, look out! The revolution with take off like a rocket. To be honest no one can tell you with certainty what Q-computers will be like in the future became when singularity occurs (about 30 years from now) all bets are off. No one can tell you what will happen past that point. Computers could take off in directions we have not even dreamed about.

$\endgroup$
  • 3
    $\begingroup$ "Quantum particles can be in four states." [citation needed] $\endgroup$ – David Richerby Jun 1 '16 at 20:43

Not the answer you're looking for? Browse other questions tagged or ask your own question.