# Superposition of qubits in quantum computer

I was looking for a quantum physics section but I did not find any, so I decided to post my question here.

I've been reading articles and watching some videos on quantum computing these last few days, but there's something that I cannot understand. Qubits can have the state of 1 and 0 like binary computers, but they can also have a third state which is a quantum superposition of 1 and 0, which allows us to analyse more data at once; 2^n qubits. Now I saw in some places that there's only a third state in which a qubit could be at, and in other articles/videos that besides the 1 and 0,it can be any number between 0 and 1, which makes several states and not only a third. For example it could be 0,1 0,2. Etc

I got confused because now I do not know if there's actually a third state or several states in which the qubits could be in.

Also, if a qubit could by any number between 0 and 1, does it mean that it can be 0,1 and 0,12 at the same time, or is there a limit in the decimal cases the numbers can have ?

Cheers !

## 3 Answers

The two possible basis states are $\vert 0\rangle$ and $\vert 1\rangle$. $0$ and $1$ are in fact any two binary values, and could be understood as spin up or spin down, or as right polarized or left polarized or any other two properties of a quantum system that are mutually exclusive.

They are basis states in the sense that the measurement of a qubit will yield either $0$ OR $1$ (or spin up/down or right/left polarized), but not both because the outcomes are mutually exclusive.

There is no such thing as a $\vert 0.12\rangle$ state in the sense that qubits are built from states with only two outcomes, which are declared conventially to be $0$ and $1$.

This doesn't prevent the qubit from being in an arbitrary combination $$\vert\psi\rangle=\cos\theta \vert 0\rangle + e^{i\varphi}\sin\theta \vert 1\rangle\, , \tag{1}$$ in which case the outcome $0$ will occur with probability $\cos^2\theta$ and the outcome $1$ will occur with probability $\sin^2\theta$ (when measured in the so-called computational basis). The phase $e^{i\varphi}$ is important for the manipulation of the qubit using gates, which can change the state and affect the probabilities of outcome.

Since the parameters $\theta$ and $\varphi$ in $\vert\psi\rangle$ can take continuous values, there is not so much one third state as there is a continuous infinity of such third states of the form $\vert\psi\rangle$, one state for each one the continuous values of $\theta$ and $\varphi$. Note that any possible state of a qubit is equivalent to a state of the form (1): again the key thing is that only two mutually exclusive outcomes can occur.

• The three parameters of the equation (1) that could take any infinity of values, could these values be any number or any letter for example : 1000, 23A, 567 etc ? At the end of your answer you said that two mutually exclusive outcomes can occur, but when we actually measure the final state, or get the final state (I don't know exactly how to put it in other jwords) this final state is only one and not two right ? – Pedro Gomes Jun 8 '17 at 22:11
• When I said 1000, 23A I mean another spin state, instead of up and down. (In the middle of up and down) – Pedro Gomes Jun 8 '17 at 22:12
• I'm not sure I follow. There are 2 parameters (angles) in (1): $\theta$ and $\varphi$, ranging between $0\le \theta\le \pi$ and $0\le \varphi\le 2\pi$. You cannot get anything but spin up or down. You may choose (for instance) the direction along which you measure spin, which is the same as choosing a basis other than the computational basis. The outcome of a measurement is a number (an eigenvalue of a hermitian operator), not a state. Immediately after the measurement the state will be in an eigenstate of this operator, the eigenstate corresponding to the measured eigenvalue. – ZeroTheHero Jun 8 '17 at 22:21
• I finally got it, read you initial answer around five times. Thank you m8 – Pedro Gomes Jun 9 '17 at 17:15

A qubit will generally be in a state $$|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$$ where $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^2 + |\beta|^2 = 1$. This is a superposition of the $|1\rangle$ and the $|0\rangle$ state. The system being in a state 1 is only a special case of this, namely $\alpha = 1, \beta = 0$, similarly for the state 0. So there is, in fact, an uncountably infinite number of states the system can be in.

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

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

All the best, B.