林尚廷
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Quantum
Computing
| |
Quantum
Computing
一些先備知識、觀念
Introduction
What is
Quantum Computing
說說看你們對量子運算的看法吧！
優質影片欣賞
Quantum Computer
首先量子運算要有量子電腦
QC is fast??
量子電腦比較快嗎？
如果是，那量子電腦為什麼比較快？
那我們為什麼需要量子電腦？
有什麼開發量子電腦的當務之急？
它的原理是什麼？
我們要怎麼編寫量子電腦程式？
Principles
量子位元
同時是0也是1 (????)
Principles
光偏振：電場振動方向
Principles
測量：光束通過偏振片
一個光子通過偏振片 (?)
Principle
量子位元
垂直偏振
水平偏振
斜向偏振
Applications
量子電腦可以促進以下領域的發展：
- 科學領域的新突破
- 挽救生命的藥物
- 快速診斷疾病的機器學習方法
- 可以使設備和結構更高效的材料
- 退休後的理財策略
- 快速引導如救護車等資源的演算法
- 為人類謀幸福
Game
Qubit & Concepts
Qubit
經典位元 (binary digit, bit)：一個0或1
量子位元 (quantum bit, qubit)：
0或1，或0和1的疊加態
\(|\psi \rangle =\alpha |0\rangle +\beta |1\rangle \quad \alpha ,\beta \in {\mathbb {C}}\)
其中\(|\alpha| ^ 2\)是測量它之後觀測到\(0\)的機率；\(|\beta| ^ 2\)是測量它之後觀測到\(1\)的機率，所以
\(|\alpha| ^ 2 + |\beta| ^ 2 = 1\)
Qubit - matrix
一個量子位元也可以表示成一個向量：
$$\begin{bmatrix} \alpha\\ \beta \end{bmatrix}$$
\(|\alpha| ^ 2\)是測量它之後觀測到\(0\)的機率；
\(|\beta| ^ 2\)是測量它之後觀測到\(1\)的機率
Qubit - matrix
量子位元也可以表示0或1：
$$|0\rangle = \begin{bmatrix} 1\\ 0 \end{bmatrix}, |1\rangle = \begin{bmatrix} 0\\ 1 \end{bmatrix}$$
所以它理論上可以做到任何經典運算能做的事
例如加減乘除
bloch sphere
或許你覺得矩陣很複雜：
但我覺得這其實更複雜 QQ
bloch sphere
Superposition
不同於經典位元，量子位元可以這樣
$$|+\rangle = \begin{bmatrix} \frac{1}{\sqrt 2}\\ \frac{1}{\sqrt 2} \end{bmatrix}$$
$$|-\rangle = \begin{bmatrix} \frac{1}{\sqrt 2}\\ \frac{-1}{\sqrt 2} \end{bmatrix}$$
Entanglement
兩個量子位元有時可以纏結在一起
Entanglement
為了表示兩個量子位元纏結
我們必須引進雙量子系統
如果其中一個量子是\(|\psi \rangle\)，另一個量子是 \(|\delta \rangle\) 那麼兩個量子寫在一起就是 \(|\psi\rangle\otimes|\delta\rangle\)
\begin{bmatrix} a \\ b \end{bmatrix} \otimes \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} a \begin{bmatrix} c \\ d \end{bmatrix} \\ b \begin{bmatrix} c \\ d \end{bmatrix} \end{bmatrix} = \begin{bmatrix} ac \\ ad \\ bc \\ bd \end{bmatrix}
Entanglement
比如說其中一個量子是\(|+ \rangle\)，另一個量子是 \(|0 \rangle\) 那麼兩個量子寫在一起就是 \(|+\rangle\otimes|0 \rangle = |+0\rangle\)
\begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix} \otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ 0\\ \frac{1}{\sqrt 2} \\ 0 \end{bmatrix}
Entanglement
有時候雙量子的量子態無法拆成兩個獨立的量子態
就稱為量子纏結
\begin{bmatrix} ? \\ ? \end{bmatrix} \otimes \begin{bmatrix} ? \\ ? \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ 0 \\ 0 \\ \frac{1}{\sqrt 2}\end{bmatrix}
Entanglement
兩個量子位元纏結之後無法用bloch sphere表示
只能用 qsphere 同時表示兩個位元
Simple
Quantum Gates
Flip Gate
量子運算也有NOT閘，叫做 Pauli X gate，用途就是把0變成1，反之亦然
\(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\)
Flip Gate
它的作用很像沿 X 軸轉 180 度
\(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\)
Hadamard gate
達到這種奇怪狀態的方法是 \(H\) 閘，右邊那種狀態叫做疊加態
$$\frac{1}{\sqrt 2} \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2}\end{bmatrix}$$
Hadamard gate
它就是沿 \(X=Z\) 軸轉 180 度
$$\frac{1}{\sqrt 2} \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2}\end{bmatrix}$$
CNOT Gate
cnot gate 是雙量子位元邏輯閘
它的作用是：
如果一個特定 bit 是1，則會反轉\((X)\)另一個 bit
所以如果這個特定 bit 是疊加態，另一個 bit 會與之纏結
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}
Rroblems
我知道很難想像 QAQ 所以來做例題吧
Description
我們假設所有 qubit 的初始狀態都是 \(|0\rangle\)
嘗試對這些 qubit 使用\(X, H, CNOT\)使它們變成指定狀態
Rroblem 1
|1\rangle
Solution 1
X
Rroblem 2
|+\rangle
Solution 2
H
Rroblem 3
|-\rangle
Solution 3
X
H
Rroblem 4
\frac{|00\rangle + |11\rangle}{\sqrt 2}
Solution 4
X(0)
CNOT(0,1)
Entanglement
最簡單的纏結可以經由疊加態產生
首先將其中一個 bit 變成疊加態
\begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix} \otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ 0\\ \frac{1}{\sqrt 2} \\ 0 \end{bmatrix}
然後再把兩個 bit 丟進 cnot gate
cnot gate 的作用是：
如果一個特定 bit 是1，則會反轉另一個 bit
所以如果這個特定 bit 是疊加態，另一個 bit 只能跟它纏結在一起了
$$\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2} \\ 0\\ \frac{1}{\sqrt 2} \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ 0 \\ 0 \\ \frac{1}{\sqrt 2}\end{bmatrix}$$
Entanglement
Rroblem 5
\frac{|00\rangle - |11\rangle}{\sqrt 2}
Rroblem 6
\frac{|01\rangle + |11\rangle}{\sqrt 2}
Rroblem 7
\frac{|00\rangle - |01\rangle - |10\rangle + |11\rangle}{2}
Break time
Quantum
Programming
量子電腦動起來
Build a
Quantum Circuit
Python is powerful... and fast;
plays well with others;
runs everywhere;
is friendly & easy to learn;
is Open.
Python
Google Colab
線上Python整合開發環境
提供免費GPU
多樣的套件
簡單, 快速且
免費
Qiskit
!pip install qiskitfrom qiskit import *
from qiskit.visualization import *
from qiskit.tools import job_monitor
# 2 量子位元 (qubits) / 2 經典位元 (classical bits)
circuit = QuantumCircuit(2, 2)
# 將qubit通過hadamard閘
circuit.h(0)
# CNOT, Controlled NOT.
# 如果第一個qubit是1，則將第二個qubit翻轉 (flip)
circuit.cx(0, 1)
# 測量qubit的值到對應的經典位元
circuit.measure([0, 1], [0, 1])A Simple Circuit
circuit.draw() # ASCII ART!!!!!!A Simple Circuit
circuit.draw(output='mpl') # use matplotlib
A Simple Circuit
Execute the Circuit
IBM Q Experience
IBM Q Experience
登入並取得 Token
IBM Q Experience
登入並取得 Token
# please paste your token here
IBMQ.save_account("TOKEN")# load your account after save it
IBMQ.load_account()Select backend
provider = IBMQ.get_provider("ibm-q")
backend = provider.get_backend("ibmq_burlington")
It's time to run code
job = execute(circuit, backend, shots=512)
job_monitor(job) # 監控有多少工作正在排隊稍候以下訊息出現 ...
Job Status: job has successfully run
counts = job.result().get_counts()
plot_histogram(counts)Is the result in expected ?
Simulator
backend = Aer.get_backend("qasm_simulator")job = execute(circuit, backend, shots=512)
# job_monitor(job)
counts = job.result().get_counts()
plot_histogram(counts)Statevector
backend = Aer.get_backend("statevector_simulator")job = execute(circuit, backend)
# job_monitor(job)
statevec = job.result().get_statevector()
plot_bloch_multivector(statevec)發生什麼事？
Statevector
circuit = QuantumCircuit(2, 2)
circuit.h(0)
circuit.cx(0, 1)
# circuit.measure([0, 1], [0, 1])
job = execute(circuit, backend)
# job_monitor(job)
statevec = job.result().get_statevector()
plot_bloch_multivector(statevec)發生什麼事？
Qsphere
circuit = QuantumCircuit(2, 2)
circuit.h(0)
circuit.cx(0, 1)
# circuit.measure([0, 1], [0, 1])
job = execute(circuit, backend)
# job_monitor(job)
statevec = job.result().get_statevector()
plot_state_qsphere(statevec)Unitary
backend = Aer.get_backend("unitary_simulator")job = execute(circuit, backend)
# job_monitor(job)
unitary = job.result().get_unitary()
print(unitary)發生什麼事？
Simulators
# quantum assembly simulator
plot_histogram(execute(circuit, backend=Aer.get_backend("qasm_simulator"), shots=1024).result().get_counts())# statevector simulator
plot_bloch_multivector(execute(circuit, Aer.get_backend("statevector_simulator")).result().get_statevector())# unitary simulator
print(execute(c, Aer.get_backend("unitary_simulator")).result().get_unitary())# qsphere simulator
plot_state_qsphere(execute(circuit, Aer.get_backend("statevector_simulator")).result().get_statevector())Quantum Gates
Quantum Gates
Rotation Gates
circuit = q.QuantumCircuit(1)
# 翻轉 (flip) 閘
circuit.x(0)
circuit.y(0)
circuit.z(0)Rotation Gates
circuit = q.QuantumCircuit(1)
# hadamard 閘，通常拿來製造疊加態
circuit.h(0)
Rotation Gates
circuit = q.QuantumCircuit(1)
# 旋轉閘：axis(angle, bit)
circuit.rx(Math.pi, 0)
circuit.ry(Math.pi/2, 0)
circuit.rz(1.8, 0)Controlled Gates
circuit = q.QuantumCircuit(2)
# axis(control, target)
# 如果控制位元是1，就翻轉目標位元
circuit.cx(0, 1)
circuit.cy(0, 1)
circuit.cz(0, 1)如果控制位元是疊加態呢 ?????
Controlled Gates
circuit = q.QuantumCircuit(2)
# axis(angle, control, target)
# 如果控制位元是1，就旋轉目標位元 (target)
circuit.crx(Math.pi, 0, 1)
circuit.cry(Math.pi * 2, 0, 1)
circuit.crz(0, 0, 1)Controlled Gates
circuit = q.QuantumCircuit(3)
# toffoli閘, 量子電腦的NAND閘
# axis(control, control, target)
# 如果兩個控制位元都是1，就翻轉目標位元
circuit.ccx(0, 1, 2)
Measurements
circuit = q.QuantumCircuit(2, 2)
# 做某些事
circuit = do_something(circuit)
# 然後量它
# [list of qubits], [list of classical bits]
circuit.measure([0, 1], [0, 1])測量是量子運算唯一的不可逆操作
Measurements
這是做啥用的？
Measurements
跑跑看就知道了！
qr = QuantumRegister(1, 'qr')
cr = ClassicalRegister(1, 'cr')
c = q.QuantumCircuit(qr, cr)
# do something
c.measure([0], [0])
c.x(0).c_if(cr, 1)
# do something
c.measure([0], [0])
c.draw()Build your own circuit NOW !!!
Real Dice
I, at any rate, am convinced that He does not throw dice. -- Albert Einstein
Pseduorandomness
大家都知道電腦的隨機數都不是真隨機
// C++ 偽隨機數生成器
static long long seed = 1;
long long rand(void) {
return seed = (seed * 998244353) % 1000000009;
}Quantum RNG
但是量子電腦是真亂數
# 單位元真亂數產生器
c = q.QuantumCircuit(1, 1)
c.h(0)
c.measure([0], [0])H |0\rangle = |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt 2}
Quantum RNG
更多位元
# 雙位元真亂數產生器 (0~3)
c = q.QuantumCircuit(2, 2)
c.h([0, 1])
c.measure([0, 1], [0, 1])|+\rangle \otimes |+\rangle = \frac{|00\rangle + |01\rangle + |10\rangle + |11\rangle}{\sqrt 4}
Real Dice
DICE = \frac{|000\rangle + |001\rangle + |010\rangle + |011\rangle + |100\rangle + |101\rangle}{\sqrt 6}
How to do this ???
A Simple Case
DICE_3 = \frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt 3}
A Simple Case
DICE_3 = \frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt 3}
import math
c = QuantumCircuit(2, 2)
c.ry(2 * math.asin(1 / math.sqrt(3)), 0)
c.x(0)
c.ch(0, 1)
c.measure([0, 1], [0, 1])
c.draw()Quantum
Algorithms
一堆有趣的東西
Deutsch Jozsa Algorithm
Constant? Balanced?
Constant function：
\(f(x) = 0\)
\(f(x, y, z) = 1\)
\(f(x, y, z, ...) = \) 常數
Balanced function：
\(f(x, y, z, ...) = x+y\)
\(f(x, y, z) = cnot(x, z)\)
\(f(x, y, z, ...) = H(y)\)
你要試多少次才能確定它是哪種函數？
The Black Box
量子電路必須是可逆的!
The Black Box
# bit 0 is the input
# bit 1 is the output
def black_box(c):
c.cx(0, 1)
return cConstant ?
Balanced ?
The Black Box
# bit 0, 1 is the input
# bit 2 is the output
def black_box(c):
c.cx(0, 2)
c.cx(1, 2)
return cConstant ?
Balanced ?
The Black Box
# bit 0 is the input
# bit 1 is the output
def black_box(c):
c.z(0)
c.x(1)
return cConstant ?
Balanced ?
Deutsch Jozsa Algorithm
c = q.QuantumCircuit(2, 1)
# 將輸出位元通過NOT閘
c.x(1)
# 將所有位元通過H閘
c.h([0, 1])
# 通過 black box
c = black_box(c)
# 再一次將所有位元通過H閘
c.h([0, 1])
# 測量所有輸入位元
c.measure([0], [0])Quantum Teleportation
Teleportation
Circuit
傳送 qubit0 的量子態到 qubit 2
量子電路必須是可逆的!
Code
c = q.QuantumCircuit(3, 3)
# c.barrier()
c.h(1)
c.cx(1, 2)
c.cx(0, 1)
c.h(0)
# c.barrier()
c.measure([0, 1], [0, 1])
# c.barrier()
c.cx(1, 2)
c.cz(0, 2)
# c.barrier()WHY ?
And WHY ?
為什麼不用簡單的交換閘 (swap gate) ?
c = q.QuantumCircuit(3)
c.swap(0, 2)Bernstein-Vazirani Algorithm
Secret Box & AND
箱子裡有一個秘密的 \(n\)位元數字 \(s\)
每次詢問你可以：
問箱子一個數字 \(x\)，然後它會回答你 \(s\ \&\ x\)
最壞情況下最少需要猜多少次呢？
Secret Box & AND
# secret_number = "101001"
def secret_box_AND(c):
c.cx([5, 3, 0], 6)
return c箱子可能長成這樣
# secret_number = "????"
def secret_box_AND(c):
c.cx([3, 2, 0], 4)
return c# secret_number = "???"
def secret_box_AND(c):
c.cx(0, 3)
return c# secret_number = "?????"
def secret_box_AND(c):
c.cx([1, 2], 5)
return cSecret Box & AND
secreat_number = "101001"
def secreat_box_AND(c):
for i, istrue in enumerate(reversed(secreat_number)):
if istrue == '1':
c.cx(i, 6);
return c更加一般化的形式
Solution Circuit
Code
# 要猜的數字為6位元整數
c = q.QuantumCircuit(6 + 1, 6)
c.h([0, 1, 2, 3, 4, 5])
c.x(6)
c.h(6)
c.barrier()
c = secreat_box_AND(c)
c.barrier()
c.h([0, 1, 2, 3, 4, 5])
c.measure([0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5])More Generalized
# 要猜的數字是 [size] 位元整數
c = q.QuantumCircuit(size + 1, size)
c.h(range(size))
c.x(size)
c.h(size)
c.barrier()
c = secreat_box_AND(c)
c.barrier()
c.h(range(size))
c.measure(range(size), range(size))References
References
Quantum Computing for Computer Scientists：
https://www.youtube.com/watch?v=F_Riqjdh2oM
Quantum Computer Programming w/ Qiskit：
https://www.youtube.com/watch?v=aPCZcv-5qfA
https://www.youtube.com/watch?v=lB_5pC1MkGg
https://www.youtube.com/watch?v=_BHvE_pwF6E
Introduction to quantum programming
https://www.youtube.com/playlist?list=PLIxlJjN2V90w3KBWpELOE7jNQMICxoRwc
“Nature isn't classical, dammit, and if you want to make a simulation of nature, you'd better make it quantum mechanical, and by golly it's a wonderful problem, because it doesn't look so easy.”
----Richard Feynman
“大自然不是古典的，如果你想要模擬大自然，你最好把它變成是量子力學的，而這是一個精彩的好問題，因為它看起來並不是那麼容易。”
——理查費曼
Quantum Computing
By 林尚廷
Quantum Computing
量子運算 :: 應該是寒訓講義的東西
