Quantum Computing: Introduction

Qubits 

A qubit is a well-defined quantum two-level system with states $|0\rangle$ and $|1\rangle$. Unlike a classical bit that is either 0 or 1, a qubit can be in a superposition of both $|0\rangle$ and $|1\rangle$ as \begin{equation}|\text{qubit}\rangle\equiv\cos\frac{\theta}{2} |0\rangle +\sin\frac{\theta}{2} e^{i\phi}|1\rangle\,.\end{equation} Notes:
  • Quantum states are the same up to an overall phase factor. To remove the ambiguity of the overall phase factor, we use the conversion that the coefficient in front of $|0\rangle$ is always a real positive number. $\phi$  is the relative phase difference between $|0\rangle$ and $|1\rangle$.
  • Each qubit state can be one-to-one mapped to a point on the Bloch sphere. Note that the polar angle $\theta$ in the spherical coordinate system takes the value in $[0, \pi]$. To ensure the coefficient in front of $|0\rangle$ is always a real positive number, we parameterize the qubit state as $\frac{\theta}{2}$ rather than $\theta$. As a result, $|0\rangle$ and $|1\rangle$ are the north and south pole on the Bloch sphere, respectively. 

Quantum gates

A quantum gate performs reversible operation on qubits via unitary transformations:
  • 1-qubit gate: to create superpositions. A prevalent example is the Hadamard gate: \begin{equation} H\,|0\rangle\equiv\frac{|0\rangle + |1\rangle}{\sqrt{2}}\,,\quad H\,|1\rangle\equiv\frac{|0\rangle - |1\rangle}{\sqrt{2}}\,.\tag{1}\end{equation}
  • 2-qubit gate: to generate entanglements. The most important one is the controlled-NOT gate or CNOT gate: $|0\rangle|0\rangle|\rightarrow|0\rangle|0\rangle,\,|0\rangle|1\rangle\rightarrow |0\rangle|1\rangle,\,|1\rangle|0\rangle\rightarrow|1\rangle|1\rangle,\,|1\rangle|1\rangle\rightarrow|1\rangle|0\rangle$, or $|x\rangle|y\rangle\rightarrow |x\rangle|x\oplus y\rangle$ in short.
Any multiple-qubit gate can be built from a sequence of 1-qubit gates and the CNOT gate (universality).

Consider a bit string $x\in\{0, 1\}^n$: $x\equiv x_1x_2\cdots x_n$ with $x_i\in \{0, 1\}$ from $i$ from 1 to n, we denote the n-qubit state $|x_1\rangle|x_2\rangle\cdots|x_n\rangle$ by $|x\rangle_n$ for illustration. If performing Hadamard gate on each qubit in such state, we have \begin{equation} H^{\otimes n} |x\rangle_n \equiv\prod_{i=1}^n H\,|x_i\rangle =\prod_{i=1}^n \frac{|0\rangle + (-1)^{x_i} |1\rangle}{\sqrt{2}} = \sum_{y\in \{0,1\}^n}\frac{(-1)^{x\cdot y}}{2^{n/2}}|y\rangle_n\,.\tag{2}\end{equation}, where $x\cdot y\equiv x_1y_1\oplus x_2y_2\oplus\cdots\oplus x_ny_n$. Especially, \begin{equation} H^{\otimes n} |0\rangle_n = \frac{1}{2^{n/2}} \sum_{x\in\{0,1\}^n}|x\rangle_n\,,\tag{3}\end{equation} which is the equally weighted sum over all computational basis elements.

Quantum Parallelism

Suppose $f:\{0,1\}^n\rightarrow\{0,1\}^m$ is a classical function that acts on a bit string of n bits and outputs a bit string of m bits. In quantum world, we can linearly extend the classical function $f$ to a unitary transformation defined by \begin{equation} U_f\left(|x\rangle_n |y\rangle_m\right)=|x\rangle_n | y \oplus f(x)\rangle_m\,,\tag{4}\end{equation} Especially, \begin{equation}  U_f\left(|x\rangle_n |0\rangle_m\right)=|x\rangle_n | f(x)\rangle_m\,.\end{equation}
Now $U_f$ is reversible because $U_f^2=1$: \begin{equation}U_fU_f\left(|x\rangle_n |y\rangle_m\right)=U_f\left(|x\rangle_n | y \oplus f(x)\rangle_m\right)=|x\rangle_n | y \oplus f(x)\oplus f(x)\rangle_m=|x\rangle_n |y\rangle_m\,.\end{equation}

Quantum parallelism is to use the size of the Hilbert space to evaluate a classical function $f$ for many values simultaneously: \begin{equation} U_f \left(H^{\otimes n} \otimes I_m\right) \left(|0\rangle_n |0\rangle_m\right) = U_f \left(\frac{1}{2^{n/2}}\sum_{x\in \{0,1\}^n}|x\rangle_n|0\rangle_m\right) = \frac{1}{2^{n/2}}\sum_{x\in \{0,1\}^n}|x\rangle_n|f(x)\rangle_m\,.\tag{5}\end{equation}
That is, a single action of $U_f$ prepare a state that has the information of $f(x)$ for all $x$.

No-cloning theorem

There is no unitary transformation $U$, such that for arbitrary state $|\psi\rangle$ such that $U|\psi\rangle |0\rangle = |\psi\rangle |\psi\rangle$. That is, we cannot copy a quantum state without changing it.
The proof is simple. Suppose there exists such operation, then $U|\psi\rangle |0\rangle = |\psi\rangle |\psi\rangle$ and $U|\phi\rangle |0\rangle = |\phi\rangle |\phi\rangle$. By the linearity, we have $U\,\left(a|\psi\rangle + b|\phi\rangle\right)|0\rangle=a U |\psi\rangle|0\rangle + b U |\phi\rangle|0\rangle=a|\psi\rangle |\psi\rangle+b|\phi\rangle |\phi\rangle$. But on the other hand, by the definition, we have $U\,\left(a|\psi\rangle + b|\phi\rangle\right)|0\rangle= \left(a|\psi\rangle + b|\phi\rangle\right)\left(a|\psi\rangle + b|\phi\rangle\right)$. The two results are clearly not equal for general $a, b, \psi, \phi$.

Quantum algorithms in early ages

Deutsch Algorithm

Deutsch algorithm is the first quantum computing algorithm that is faster than the best classical algorithm, although not exponentially faster. It is proposed to solve a toy problem: Given a function $f:\{0,1\}\rightarrow\{0,1\}$ that maps a bit to a bit, whether $f(0)=f(1)$ or not? A classical algorithm has to evaluate the function $f$ twice. But the quantum algorithm call the function $f$ only once: \begin{eqnarray}|0\rangle|1\rangle&\xrightarrow{H\otimes H}& \frac{1}{2}\left(\sum_{x\in \{0,1\}}|x\rangle\right)\left(|0\rangle-|1\rangle\right) \\ &\xrightarrow{U_f} & \frac{1}{2} \sum_{x\in \{0,1\}}|x\rangle \left(|0\oplus f(x)\rangle-|1\oplus f(x)\rangle\right) = \frac{1}{2}\left(\sum_{x\in \{0,1\}}(-1)^{f(x)}|x\rangle\right)\left(|0\rangle-|1\rangle\right)\\ &\xrightarrow{H\otimes I}&\frac{1}{2\sqrt{2}} \left(\sum_{x,y\in \{0,1\}}(-1)^{f(x)+x\cdot y}|y\rangle\right)\left(|0\rangle-|1\rangle\right)\end{eqnarray} The first qubit is finally in the state $\left((-1)^{f(0)}+(-1)^{f(1)}\right)|0\rangle+\left((-1)^{f(0)}-(-1)^{f(1)}\right)|1\rangle$. The measurement of the first qubit is $|0\rangle$ if $f(0)=f(1)$ and $|1\rangle$ otherwise. 

Bernstein-Vazirani Algorithm

As an extension to the Deutsch algorithm, Bernstein-Vazirani algorithm is to find out the secrete bit string $s\in \{0, 1\}^n$ in the boolean function $f_s(x)\equiv s\cdot x$. The classical solution is to call the function $f_s$ n times. For each $i$ from 1 to n, one feed in a bit string $x$ with only $x_i=1$ and others zero, and the output of $f_s$ is then $s_i$. The quantum solution only requires ONE call to the function $f_s$: \begin{eqnarray}|0\rangle_n |1\rangle&\xrightarrow{H^{\otimes (n+1)}}&\frac{1}{2^{(n+1)/2}} \left(\sum_{x\in\{0,1\}^n}|x\rangle_n\right)\left(|0\rangle-|1\rangle\right) \\ &\xrightarrow{U_{f_s}}& \frac{1}{2^{(n+1)/2}} \sum_{x\in\{0,1\}^n}|x\rangle_n\left(|f_s(x)\rangle-|1\oplus f_s(x)\rangle\right) = \frac{1}{2^{(n+1)/2}} \left(\sum_{x\in\{0,1\}^n}(-1)^{s\cdot x}|x\rangle_n\right)\left(|0\rangle-|1\rangle\right) \\ &\xrightarrow{H^{\otimes n}\,\otimes I}&|s\rangle_n\frac{|0\rangle-|1\rangle}{\sqrt{2}}\,.\end{eqnarray}

Simon's Algorithm

Simon's algorithm is the first quantum computing algorithm that is exponentially faster than the best classical algorithm. It is proposed to solve another toy problem: Given a two-to-one function $f_s:\{0, 1\}^n\rightarrow \{0, 1\}^n$, find out the secrete bit string $s\in \{0, 1\}^n$ such that $f_s(x)=f_s(x\oplus s)$. A special case is that $s=0$, in which $f_s$ becomes a one-to-one map. 
Classically, we have to check up to $2^{n-1}+1$ inputs. If we are lucky, we can find out $s$ by the first two tries. But if $s=0$ or we are really unlucky, we have to check all the $2^{n/2}+1$ inputs. So a classical algorithm takes time $\Omega(2^n)$. A quantum solution solves this problem in $O(n)$ times: \begin{eqnarray}|0\rangle_n|0\rangle_n&\xrightarrow{H^{\otimes n} \,\otimes I^{\otimes n}}&\frac{1}{2^{n/2}} \left(\sum_{x\in\{0,1\}^n}|x\rangle_n\right)|0\rangle_n \\&\xrightarrow{U_{f_s}}& \frac{1}{2^{n/2}} \sum_{x\in\{0,1\}^n}|x\rangle_n |f_s(x)\rangle_n \\ &\xrightarrow{\text{Measure to } |f_s(x)\rangle_n}&\frac{1}{\sqrt{2}}\left(|x\rangle_n+|x\oplus s\rangle_n\right)\\ &\xrightarrow{H^{\otimes n}} &\frac{1}{2^{(n+1)/2}}\sum_{y\in \{0,1\}^n} \left((-1)^{x\cdot y} +(-1)^{(x\oplus s)\cdot y}\right)|y\rangle_n\,.\end{eqnarray} Finally, measuring the first register to the state $|y\rangle_n$ only if $y\cdot s=0$. Repeating $n$ times gives $n$ such linear equations, from which we can solve $s$.
Location: 美国加利福尼亚帕罗奥图

Comments

Post a Comment

Popular posts from this blog

529 Plan

This post is for knowledge sharing only. It is not intended to be investment or tax advice. 529 Plans are tax-advantaged accounts specific for education saving: Contribution are made with after-tax dollars. Growth within the account is tax-free. Withdraw for qualified education expenses is also tax free. Roth IRA First 529 Plans are very similar to Roth IRA . But Roth IRA withdraw under  the 5-year rule  is tax free for any purpose. Therefore, unless for the state tax deduction in some states, it is always better to max out Roth IRA first (through backdoor and mega backdoor  for high income earners) before funding a 529 plan. Note: there is NO California state tax deduction for any 529 plan. 529 or Not? Here are some considerations in reality before funding a 529 plan: What if no qualified education expenses In the worst case that there will be no qualified education expenses, you can still take money out for non-qualified expenses, but you have to pay: federal income tax + 10% federa
Read more

How to offset W2 tax

This post is for knowledge sharing only. It is not intended to be tax or investment advice. 1. Contribute to pre-tax 401K and HSA As well known, pre-tax 401K and health saving account (HSA) are the only a few tax shelters for W2 workers.  2. Tax-loss harvesting If your total capital losses exceed your capital gains in a tax year, you can deduct up to \$3,000 (married filing jointly) from your ordinary income. If your losses are more than $3,000, you can carry over the remaining amount to future years, continuing to offset your income until all the losses are used up. A popular method to generate these deductible losses is  tax-loss harvesting . This involves selling a security at a loss to offset gains or reduce ordinary income, and optionally purchasing a similar, but not "substantially identical", security to avoid missing out on the market. However, be aware of the  wash-sale rule , which disallows claiming a loss if you buy the same or a substantially identical security
Read more

Retirement Accounts

Image
This post is for knowledge sharing only. It is not intended to be investment or tax advice. 0. Account Overview  Table 1 Common retirement  accounts in U.S. I summarizes the common retirement accounts in Table 1:  The second row in Table 1 lists the company-sponsored accounts that are offered through employers. The third row in Table 1 lists the individual retirement accounts (IRA) that is unrelated to employments. The columns in Table 1 are about different tax advantages as described below. WARNING: because of the tax advantages, there are also limitations and specific rules imposed by IRS for early withdrawal before the retirement age. For accounts listed in the second column,  Contributions are made with before-tax dollars . The money taken out will be taxed as regular income in the year of withdrawal. For accounts listed in the third column,  Contributions are made with after-tax dollars . The money taken out is  tax free . For after-tax 401K in the fourth column, Contributions are
Read more