# Basic Quantum Computing with Least Physics Possible

## Classical Probability Bit

We will start by talking about classical probability bit. Here is the one bit representation $p_0$, $p_1$ are the probability of being $0$, $1$ respectively.

$\left( \begin{array}{c} p_0 \\ p_1 \end{array} \right) \\s.t.\ p_0+p_1=1,\ p_0,p_1\in\mathbb{R},\ 0\leq p_0,p_1\leq1$

And here is the n-bits representation

$\left( \begin{array}{c} p_{0\cdots00} \\ p_{0\cdots01} \\ p_{0\cdots10} \\ \vdots \\ p_{1\cdots11} \end{array} \right)\in\mathbb{R}^{2^n}\\ s.t.\ \sum p_x=1,\ p_x\in[0,1]\in\mathbb{R},\ x\in\{0,1\}^n$

So we can do some transformation to the bit:

$\left( \begin{array}{c} p_0 \\ p_1 \end{array} \right) \xrightarrow{I} \left( \begin{array}{c} p_0 \\ p_1 \end{array} \right)$ $\left( \begin{array}{c} p_0 \\ p_1 \end{array} \right) \xrightarrow{NOT} \left( \begin{array}{c} p_1 \\ p_0 \end{array} \right)$

Or if we have two bits

$\left( \begin{array}{c} p_0 \\ p_1 \end{array} \right), \left( \begin{array}{c} p_0 \\ p_1 \end{array} \right)$

and want to do an XOR with two dependent bits

$\left( \begin{array}{c} p_{00} \\ p_{01} \\ p_{10} \\ p_{11} \end{array} \right) \xrightarrow{XOR} \left( \begin{array}{c} p_0 \\ 0 \\ 0 \\ p_1 \end{array} \right)$

But it is not possible (physically) to do something like this, which kind of does XOR with two independent bits:

$\left( \begin{array}{c} p_{00} \\ p_{01} \\ p_{10} \\ p_{11} \end{array} \right) \xrightarrow{XOR'} \left( \begin{array}{c} p_0\centerdot p_0 \\ p_0\centerdot p_1 \\ p_1\centerdot p_0 \\ p_1\centerdot p_1 \end{array} \right)$

In other words, physically, in our universe, we can only do $l_1$ norm linear tranformation.

## Quantum Bit (qubit)

For qubit, we are able to do $l_2$ norm. A single qubit is difine as follows:

$\left( \begin{array}{c} \alpha \\ \beta \end{array} \right)\in\mathbb{C}^2\\ s.t.\ \lvert\alpha\rvert^2+\lvert\beta\rvert^2=1,\ \lvert\alpha\rvert:=\sqrt{\alpha\centerdot\overline{\alpha}}\\ Prob[m=0]=\lvert\alpha\rvert^2=\left( \begin{array}{c} 1 \\ 0 \end{array} \right)\\ Prob[m=1]=\lvert\beta\rvert^2=\left( \begin{array}{c} 0 \\ 1 \end{array} \right)$

Certainly we can do transformation to one qubit. In this case, the transformation we have is $T\in\mathbb{C}^{2\times2}$. Before talking about transformations, we first introduce k-qubits as follows

$\left( \begin{array}{c} \alpha_{0\cdots00} \\ \alpha_{0\cdots01} \\ \alpha_{0\cdots10} \\ \vdots \\ \alpha_{1\cdots11} \end{array} \right)\in\mathbb{C}^{2^k}\\ s.t.\ \sum_{x\in\{0,1\}^k} \lvert\alpha_x\rvert^2=1$

and our transformation becomes $T\in\mathbb{C}^{2^k\times2^k}$

## Some Notations

$v= \left( \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_d \end{array} \right)\in\mathbb{C}^{d}\\ v^{\dagger}=(\bar{\alpha_1},\cdots,\bar{\alpha_d})\\ <v,w>=v^{\dagger}\centerdot w=\overline{<w,v>}=<v|w>\\ v\perp w\Leftrightarrow v^{\dagger}w=0\\ \Vert v\Vert=\sqrt{\sum_{i=1}^{d}\alpha_i \bar{\alpha_i}}=v^{\dagger}v\\ T= \left( \begin{array}{c} T_{11} & \cdots & T_{1d} \\ \vdots & \ddots & \vdots \\ T_{d1} & \cdots & T_{dd} \end{array} \right)\\ \bar{T}= \left( \begin{array}{c} \overline{T_{11}} & \cdots & \overline{T_{d1}} \\ \vdots & \ddots & \vdots \\ \overline{T_{1d}} & \cdots & \overline{T_{dd}} \end{array} \right)\\ \left( \begin{array}{c} \alpha_0 \\ \alpha_1 \end{array} \right) \otimes \left( \begin{array}{c} \alpha_0 \\ \alpha_1 \\ \end{array} \right)= \left( \begin{array}{c} \alpha_0\beta_0 \\ \alpha_0\beta_1 \\ \alpha_1\beta_0 \\ \alpha_1\beta_1 \\ \end{array} \right)\\ \mathbb{C}^k\otimes\mathbb{C}^l=\mathbb{C}^{kl}\\ if\ |\tilde{\Phi}>=U|\Phi>,|\tilde{\Psi}>=U|\Psi>,then\ <\Phi|\Psi>=<\tilde{\Phi}|\tilde{\Psi}>and\ <\tilde{\Phi}|=<\Phi|U^{\dagger}\\ |0>:= \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)\\ |1>:= \left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)\\ |+>:= \left( \begin{array}{c} 1/\sqrt{2} \\ 1/\sqrt{2} \\ \end{array} \right)=1/\sqrt{2}(|0>+|1>)\\ |->:= \left( \begin{array}{c} 1/\sqrt{2} \\ -1/\sqrt{2} \\ \end{array} \right)=1/\sqrt{2}(|0>-|1>)\\ H:=1/\sqrt{2} \left( \begin{array}{c} 1 & 1 \\ 1 & -1 \\ \end{array} \right)\\ H|0>=|+>\\ H|1>=|->\\ H|+>=|0>\\ H|->=|1>\\ |X,Y>:=|X>\otimes|Y>\\ if\ f:\{0,1\}^k\rightarrow\{0,1\},then\ |X,Z>\xrightarrow{U_f}|X,Z+f(X)>$

## Quantum Transformations

$T$ is a quantum transformation iff

• $T$ is linear: $T\in\mathbb{C}^{d\times d}$
• $T$ is norm preserving $\Leftrightarrow\forall v\in\mathbb{C}^d,\Vert v\Vert^2=\Vert Tv\Vert^2\Rightarrow T\centerdot T^{\dagger}=I$

## Basic Quantum Gates (Between 2/3 qubits)

$SWAP:= \left( \begin{array}{c} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array} \right)\\ i.e.\ (X,Y)\xrightarrow{SWAP}(Y,X)\\ CNOT:= \left( \begin{array}{c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array} \right)\\ i.e.\ (X,0/1)\xrightarrow{CNOT} \left\{ \begin{array}{rcl} (X,0/1) & \mbox{if} & X=0 \\ (X,1/0) & \mbox{if} & X=1 \\ \end{array}\right.\\ CCNOT:= \left( \begin{array}{c} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ \end{array} \right)\\ i.e.\ (X,Y,Z)\xrightarrow{CCNOT}(X,Y,Z\oplus X\wedge Y)\\ (X,Y,0)\xrightarrow{CCNOT}(X,Y,X\wedge Y)\\$

Note: all transformations are inversible. Specifically, for the above 3 trans., the inverse of them are themselves. Namely, apply it twice and we will be back to the original position.

To implement some classical function $f$, we need the input bits as well as some extra bits for input to have extra space to work with.

## One Small Quantum Algorithm Example

Here we will talk about a small example of quantum algorithms, just to get a feel of how quantum gates works and why it is sometimes more efficient than classical algorithms.

Deutsch Problem

Given $$f\:\{0,1\}\rightarrow\{0,1\}$$, we ask if $$f(0)=f(1)$$. Classically we have to evaluate twice: both $$f(0)$$ and $$f(1)$$. Quantumly we can solve it with one transformation and evaluate it.

$input:\vert0,1>$

1. $\vert0,1>\xrightarrow{H}\vert+,->$
2. $\vert+,->\xrightarrow{U_f}???$

So

$|b,->\xrightarrow{U_f}\cdots\rightarrow|(-1)^{f(b)}b,->\\ And\\ |+,->\xrightarrow{U_f} 1/\sqrt{2}((-1)^{f(0)}\centerdot|0>+(-1)^{f(1)}\centerdot|1>)\otimes|->\rightarrow\\ \left\{ \begin{array}{rcl} \pm|+> & \mbox{if} & f(0)=f(1) \\ \pm|-> & \mbox{if} & f(0)\neq f(1) \\ \end{array}\right.$

So if we measure the result, we will see $”0”\ iff\ f(0)=f(1)$, and $”1”\ iff\ f(0)\neq f(1)$

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