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I am Computer Science student and learning about quantum computing. But, I have a problem in understanding Bit and Qubit relationship. A bit with 2 bits = 4 states 00,01,10,11--- 1 state at a time. How is Qubit different? Will it also contain 4 states with 2 bits? How will the states get processed paralalley?

Please, ignore my 'novice'ness in this topic but this is hell interesting.

I came across "How are qubits better than classical bit?" but the frequency of this discussion is higher than my frequency.


So, I have gathered something, please rectify me if I am wrong.

The only difference is in their states. So, 2 bits will be 00,01,10,11. For 2 Qubits it will also be same i.e 00,01,10,11.

Will the diagram look like this:

enter image description here

Now, the probability that which state will it enter depends on the superposition of 0 and 1.

This explains: "When the compass needle points north, that is like a qubit being in the state $∣0 \rangle$, and when the compass needle points east, that is like a qubit being in the state $∣1 \rangle$. But a compass needle can also point northeast. The direction northeast is neither north nor east, but it is a superposition of equal parts north and east: if you add a north-pointing vector and an east-pointing vector of equal magnitude, you will get a vector that points northeast. Similarly, the qubit state $\frac{1}{\sqrt{2}}(∣0\rangle+∣1\rangle)$ is neither $∣0 \rangle$ nor $∣1 \rangle$, but it is a superposition of equal parts $∣0 \rangle$ and $∣1 \rangle$." - David

enter image description here

Where am I wrong?

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  • $\begingroup$ Two qubits can also represent four different (basis) states, but the difference is that it can represent them all simultaneously. Whereas classical bits must be one or another. $\endgroup$ – lemon Feb 9 '15 at 6:49
  • $\begingroup$ Well, thanks for the reply. Now if you can explain "represent them all simaltaneously" a bit. $\endgroup$ – Matt Cube Feb 9 '15 at 9:09
  • $\begingroup$ A kind of example will be of great help! $\endgroup$ – Matt Cube Feb 9 '15 at 9:25
  • $\begingroup$ Qubits are often represented as the spin states of particles or molecules. Are you familiar with quantum superposition? $\endgroup$ – lemon Feb 9 '15 at 10:04
  • $\begingroup$ Yes, an object that is in two states at once! Schrödinger cat experiment. But how qubit states related to it. thanks! $\endgroup$ – Matt Cube Feb 9 '15 at 11:13
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After several discussions on that topic, I came to think that it is more appropriate to separate those different types of "bits" into four instead of two classes:

  • A deterministic classical bit is an ordinary bit can be either in state $\left|0\right>$ or in state $\left|1\right>$. No other states allowed.

  • A random classical bit can be in a "unknown" state with probabilities $p_i$: $$p_0\left|0\right>+p_1\left|1\right>,\quad \text{where}\quad p_0+p_1=1$$

  • A pure qubit, can be in a quantum superposition of the two states: $$\alpha\left|0\right>+\beta\left|1\right> ,\quad \text{where}\quad \left|\alpha\right|^2+\left|\beta\right|^2=1$$ (I suspect that it is important to stress here that $\alpha$ and $\beta$ above are complex numbers.)

  • Finally, a mixed qbit can be in a state that combines quantum and classical uncertainties. Such states are described by a density matrix formalism: $$\rho_{00}\left|0\right>\left<0\right|+\rho_{11}\left|1\right>\left<1\right|+\rho_{01}\left|0\right>\left<1\right|+\rho_{01}^*\left|1\right>\left<0\right|$$ with $\rho_{ij}$ being positive-definite, hermitian matrix of trace 1.

You can spot the difference between those definitions just by looking at the number of dimensions of corresponding spaces:

  • For a deterministic classical bit it is 0-dimensional. Just two points.

  • For a random classical bit we have a 1-dimensional interval in $\{p_0, p_1\}$ space

  • For a pure qubit the space is a manifold defined by the implicit equation $\left|\alpha\right|^2+\left|\beta\right|^2=1$. This manifold is 2-dimensional and is well-known as a Bloch sphere.

  • Finally, the space of a mixed qbit states is represented by a 3-dimensional set. That actually corresponds to an interior of a Bloch sphere (a "Bloch ball" if you want).

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  • $\begingroup$ Thanks Kostya, just last request, please explain it with real life example..Please! $\endgroup$ – Matt Cube Feb 9 '15 at 15:31
  • $\begingroup$ @MattCube, can I ask you... Do you have any familiarity at all with complex numbers? $\qquad\qquad\qquad$ How much is $(5+i3)\cdot 3e^{i\frac\pi4}$ ? $\endgroup$ – Kostya Feb 9 '15 at 17:24
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I'll leave out the math, partly because I'm not sure I remember it precisely enough. And then there's the formatting problem.

A bit (per Shannon's information theory) is the smallest amount of information. So in digital circuitry, it is a voltage (or lack thereof) on a TTL (or similar) gate. Where it came from (ie, random sequence or one bit of an ASCII character) is irrelevant for our purposes here. At any given time, a bit has one value only.

A qubit is implemented rather differently. It might be one of several spin states of a some elementary particle or whatever. But whatever it is, due to Heisenberg's Uncertainty Principle, it can be in any of several states (at present this is almost always 2). It can therefore "carry" more information than a conventional bit. We, of course, can't observe what it's value (or the many possible) is as the observation will cause the superposition of states to collapse. Assuming we can sufficiently cleverly read information into a system of qubits, and cause them to carry about some computation, we will have a kind of quantum computer. A company in W Canada is selling them (although there is some debate about whether its machine is really a quantum computer). Some clever people at Bell Labs (RIP) figured out how to do this, in principle, some decades ago -- or at least that it was possible, again in principle.

Since several computations are going on simultaneously, this will be rather faster than any conventional computer, using regular bits. The possible speed up has a great many people drooling and awaiting desktop units.

Computers are seriously limited, for some problems (eg, many of those cryptography), by their speed. Quantum computers, if made practical, will greatly extend the range of problems we can solve with computers.

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Hey you are asking really great questions. You are absolutely correct in your ponderings - there is more to the qubit than meets the eye. To answer your question the qubit is a unit quaternion. Let me explain as best I can.

The qubit is a 2-dimensional complex number. The complex numbers $\mathbb{C}$ are isomorphic to vectors in $\mathbb{R}^2$, whereas the qubit $|\Psi\rangle\in\mathbb{C}^2$ is isomorphic to vectors in $\mathbb{R}^4$. The quaternions are isomorphic to vectors in $\mathbb{R}^4$.

\begin{equation} |\Psi\rangle = \binom{\alpha}{\beta}\qquad |\alpha|^2+|\beta|^2=1 \end{equation}

The most common representation of the quaternions is the special unitary group of 2x2 matrices SU(2), which are the square matrices of the form:

\begin{equation} \hat{\Psi}= \begin{pmatrix} \alpha & -\beta^* \\ \beta & \alpha^* \end{pmatrix} \end{equation} with unit determinant - which effectively means $|\alpha|^2+|\beta|^2=1$.

To see the connection between the qubit and the quaternions, consider that there are two orthonormal representations of the qubit $|\Psi^\pm\rangle$ where

\begin{equation} |\Psi^+\rangle = \binom{\alpha}{\beta}\qquad |\Psi^-\rangle = \binom{-\beta*}{\alpha^*} \end{equation}

These satisfy the orthonormal relations $\langle\Psi^\pm|\Psi^\pm\rangle=1$ and $\langle\Psi^\pm|\Psi^\mp\rangle=0$.

Now the connection between the qubit and the quaternion is found by recognising that the qubit forms the columns of the SU(2) matrix.

\begin{equation} \hat{\Psi}= \begin{pmatrix} |\Psi^+\rangle & |\Psi^-\rangle \end{pmatrix} \end{equation}

As for superpositions - that is a flawed concept derived from treating the qubit (aka the quaternion) as a 2 level system and coming up with mad notions like attributing a unit vector to a spin state such as

\begin{equation} |\uparrow\;\rangle = \binom{1}{0}\qquad |\downarrow\;\rangle = \binom{0}{1} \end{equation}

Now between ourselves nothing could be further from the truth - but don't go causing a fuss, you'll wake the babies! They are off in dreamland trying to build a quantum computer. Its best to leave sleeping dogs lie as they say ;-)

If you are interested in getting deeper into the quaternion I can recommend

https://arxiv.org/abs/1601.02569

as a good starting point. (I wrote the article.)

B.

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