I do a bit of hobby programming and I often search the internet for little oddities that are fun to ponder over. I have read a few passages that try to explain quantum computing to the layman like myself. I have read of the Qubit, the more 'power' version of the bit, and its bad habit of being in superposition. This, to me, sounds as if it sits halfway between 1 and 0.

So, I reason that one can create a qu-binary number with these; something resembling a ternary number, made from 0's, 1's and 1/2's (or Q's). I have read that a quantum computer has more 'power' when it comes to computation because one qu-value is a possibility between at most n^2 regular values in n bits. I have constructed a little problem with this value when you try to store a specific set of regular values in a qu-value.

Imagine a value is a superposition between 2 and 3. In qu-binary, I would write "10 or 11 -> 1Q", as the last bit is "both". OK, so this works. But what about real values 2, 3 and 4 in superposition? in my ternary notation "QQQ" is potentially any of the possibilities 0 through 7, and so actually represents a whole lot more values than I want!?

My question is, how does it really work? Am I thinking about it all wrong? Because this is how the whole subject of Quantum computing looks like from the outside. Or is this an example of quantum computing's non-determinism? I assume all bits are completely isolated from one another and have no qu-knowledge of any other. Maybe something obscure like quantum gates sharing information between bits could explain the problem; or if the bits represent continuous probabilities. I don't know. Could someone explain it for me?

  • $\begingroup$ A qubit is in no way analogous (let alone equivalent) to a classical bit with many values. $\endgroup$ Commented Aug 3, 2014 at 9:06
  • $\begingroup$ Then what is it good for? $\endgroup$
    – bimmo
    Commented Aug 3, 2014 at 10:17
  • 1
    $\begingroup$ A qubit carries much more information of a completely different type than any finite number of bits, yet when one measures it, it is just one bit. Quantum computers, when constructed and made reasonably large, are able to quickly solve problems that no bit or trit-based computer could solve in thousands of years. $\endgroup$ Commented Aug 3, 2014 at 17:03
  • $\begingroup$ So this is what you get when you try to hack the universe. It is still not clear to me, of course, what it all means. If only the 'information' could be simplified into a data type that a programmer could easily understand and utilized. $\endgroup$
    – bimmo
    Commented Aug 4, 2014 at 7:45
  • $\begingroup$ Dear @bimmo, a programmer who only understands or wants to understand the classical information, whether digital or analog, will probably never understand (certainly not easily) quantum computers which work qualitatively differently than classical computers. To understand quantum computers, one has to penetrate into quantum mechanics, at least at a basic level. $\endgroup$ Commented Aug 4, 2014 at 7:55

2 Answers 2


You need to be a bit (pardon the pun) more strict about the size of the (Hilbert) space you're playing with. A qu_bit can be in a superposition of two (pure) states, but not more. For this reason, "real values 2, 3 and 4 in superposition" doesn't make sense. To draw an analogy to the binary system you mentioned, it's as if you're trying to stuff large numbers into a bit.

This restriction appears more clearly in the visual representation of a pure state, the Bloch sphere.

Secondly, you need to be careful when drawing analogies between bits and qubits. For example, your statement "all bits are completely isolated from one another and have no qu-knowledge of any other" is wrong in the general case when there may be entanglement between qubits.

I started learning quantum computing with some CS knowledge, and this was a very helpful reference. I think it'll get you started on the right path.

  • $\begingroup$ So an entire register (value) cannot reliably contain (more than two) values in its superposition, or is 'superposition' not even an applicable concept here? If not, how is a qu-register any different to a regular one? I know that qubits are not necessarily used on their own, so how would a "16 qubit computer" store numbers in its registers? What type of algorithms actually utilize quantum effects? Can I supply the computer with quantum-fuzzy input? so many questions. $\endgroup$
    – bimmo
    Commented Aug 3, 2014 at 7:00
  • $\begingroup$ all of these questions are natural, and like I said, that book would answer many of them. A qubit would physically be something like an electron spin, or a photon polarization state, or some other interesting system. $\endgroup$
    – eqb
    Commented Aug 3, 2014 at 8:30
  • $\begingroup$ To clarify, I was trying to stuff more than one of the 000, 001, 010, 011 ... 111 values into a single 3 bit value. Something tells me that it isn't that simple. $\endgroup$
    – bimmo
    Commented Aug 4, 2014 at 7:47
  • $\begingroup$ Hey btw it's customary to choose one of the answers as an answer to your question. $\endgroup$
    – eqb
    Commented Aug 5, 2014 at 3:32
  • $\begingroup$ I can't tell if either is the "right" one. The votes aren't helping much. But seeing as your are so kind... $\endgroup$
    – bimmo
    Commented Aug 8, 2014 at 6:23

A qubit is a (complex) vector, with $2$ complex coordinates, you may write it :

$$\vec q = \alpha \vec 0 + \beta \vec 1 \tag{1}$$ Here $\alpha$ and $\beta$ are complex quantities and $\vec 0, \vec 1$ is a basis, so you have $\vec 0.\vec 0= \vec 1.\vec 1=1$, and $\vec 1.\vec 0 = \vec 0.\vec 1=0$. The qbit is generally normed to $1$, so you have $|\alpha|^2+|\beta|^2=1$. A standard notation, instead of writing $\vec q$ is to note it $|q\rangle$ (ket). An other notation is to write the complex conjugate vector $(\vec q)^*$ as $\langle q|$ (bra). A last useful notion is the "hermitian product" : $\langle q'|q\rangle = (\vec q')^*.\vec q$. The orthonormality of the basis could be written now : $\langle 0|0\rangle = \langle 1|1\rangle = 1$, $\langle 0|1\rangle = \langle 1|0\rangle = 0$

So you see that this has nothing to do with classical bits.

The most interesting consequence is that you may study $2$ entangled qbits, where the vectorial nature of the qbits have important consequences on quantum correlations, which cannot be reproduced by classical correlations.

  • $\begingroup$ Squiggly lines! So a qubit can be generalized into a single 2D complex (4D real) dimensional 'direction'? Please excuse my naivety. Year 10 math has nothing on this stuff. I dare not ask how two or more interact! Final thoughts: bra+key="(" $\endgroup$
    – bimmo
    Commented Aug 4, 2014 at 7:40
  • $\begingroup$ A qbit is a 2-dimensional complex vector, normed to $1$. That's all the story. $\endgroup$
    – Trimok
    Commented Aug 4, 2014 at 7:51
  • $\begingroup$ Nothing to do with þͤ olde goode bit, you said? Quantum information is different from classical information, but its units of measurement are the same. Qubit behaves differently, but it contains the same (quantitative) amount of information as bit. $\endgroup$ Commented Aug 21, 2014 at 15:09
  • $\begingroup$ @IncnisMrsi : I meant that the vectorial structure of quantum states that we see in qbits has nothing to do with classical bits. And you can make some operations with qbits which are impossible with classical bits. The defniition of entropy and information is also different (with qbits, you will use Von Neumann entropy, and the entropy of a composite system can be lower than the entropy of any of its parts). See also the Bennett's laws of Quantum Information $\endgroup$
    – Trimok
    Commented Aug 22, 2014 at 8:44
  • $\begingroup$ I know about this entropy stuff (it is easy to attain with entanglement), but this doesn’t prove your point. We can encode N (and no more) classical states in a quantum system of N states and then take them back with 100% probability. And we can extract log N (and no more) classical information from a quantum system of N states. $\endgroup$ Commented Aug 22, 2014 at 8:52

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