diff --git a/src/master/COMP90084.md b/src/master/COMP90084.md
index 60fc2387..805f475c 100644
--- a/src/master/COMP90084.md
+++ b/src/master/COMP90084.md
@@ -161,7 +161,7 @@ tag:
   - 基于量子力学的计算机学说
   - 量子化学，量子深度学习
   - 后量子密码学
-- Physical Qbits:
+- Physical Qubits:
   - Particles with polarization(photon) 光子
   - Trapped ions 离子阱
   - Cold Atoms 冷原子
@@ -194,7 +194,19 @@ tag:
   - being simultaneously in all states
   - detect the state with probability $x_i = \frac{|c_i|^2}{\sum_{i=0}^{n} |c_i|^2}$
   - when observed, it will collapse to one of the basic states
-- 设$\Omega$是一个可观测量，$|\psi\rangle$是一个量子态。如果测量的结果是特征值$\lambda$，则测量后的量子态将始终是对应于$\lambda$的特征向量。
+
+- 時間反演對稱
+  - $UU^\dagger = I$
+  - $U^\dagger U = I$
+  - $|\psi\rangle = U|\phi\rangle$
+  - $\langle\psi| = \langle\phi|U^\dagger$
+  - $\langle\phi|\psi\rangle = \langle\phi|U^\dagger U|\psi\rangle$
+- 模不变
+  - 量子态的模的表示: $\langle\psi|\psi\rangle$
+  - $\langle\psi|\psi\rangle = \langle\phi|U^\dagger U|\phi\rangle = \langle\phi|\phi\rangle$
+  - $|\psi\rangle$ 和 $|\phi\rangle$ 的模相等
+
+- 设$\Omega$是一个可观测量，$|\psi\rangle$是一个量子态。如果测量的结果是特征值$\lambda$，则测量后的量子态将始终是对应于$\lambda$的特征向量
 
 - superposition 叠加: $|\psi\rangle \longmapsto \alpha|0\rangle + \beta|1\rangle$
   - $\alpha$ and $\beta$ are complex amplitudes 复振幅
@@ -207,10 +219,15 @@ tag:
   - determine how likely a start state will change to an end state (after measurement)
 - entanglement 纠缠
   - connect two qubits using a gate
-  - Bell state: $|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$
-  - Bell state: $|\Phi^-\rangle = \frac{|00\rangle - |11\rangle}{\sqrt{2}}$
-  - Bell state: $|\Psi^+\rangle = \frac{|01\rangle + |10\rangle}{\sqrt{2}}$
-  - Bell state: $|\Psi^-\rangle = \frac{|01\rangle - |10\rangle}{\sqrt{2}}$
+- Bell state
+  - 量子比特间的强纠缠关系
+  - $|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$
+    - 两个量子比特要么都在状态 $|0\rangle$，要么都在状态 $|1\rangle$，且这两种情况的概率相等
+    - 最大纠缠态，两个量子比特的状态之间没有相位差，量子态无法分解成两个量子态的直积
+  - $|\Phi^-\rangle = \frac{|00\rangle - |11\rangle}{\sqrt{2}}$
+    - 两个量子比特的状态之间存在一个相位差
+  - $|\Psi^+\rangle = \frac{|01\rangle + |10\rangle}{\sqrt{2}}$
+  - $|\Psi^-\rangle = \frac{|01\rangle - |10\rangle}{\sqrt{2}}$
 
 ## 3. Quantum Architecture, Classical, Reversible and Quantum Gates
 
@@ -222,15 +239,40 @@ tag:
   - AND: 1 if both are 1
     - $\begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$
   - OR: 1 if either is 1
+    - $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \end{bmatrix}$
   - NAND: 0 if both are 1
+    - NOT $\star$ AND
+    - $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{bmatrix}$
   - XOR: 1 if either is 1, but not both
-
-- bloch sphere 用于表示量子比特状态的几何图形
+    - NOT $\star$ OR
+    - $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix}$
+- all Quantum gates are reversible
+- Toffoli gate: 3 qubits, 2 control and 1 target
+  - if both control qubits are 1, then the target qubit is flipped
+  - $\begin{array}{} \ 000 \ 001 \ 010 \ 011 \ 100 \ 101 \ 110 \ 111 \\ \begin{array}{c} 000 \\ 001 \\ 010 \\ 011 \\ 100 \\ 101 \\ 110 \\ 111 \end{array} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \end{array}$
+  - $$
+- Fredkin gate: 3 qubits, 1 control and 2 target
+  - if the control qubit is 1, then the target qubits are swapped
+  - $\begin{array}{} \ 000 \ 001 \ 010 \ 011 \ 100 \ 101 \ 110 \ 111 \\ \begin{array}{c} 000 \\ 001 \\ 010 \\ 011 \\ 100 \\ 101 \\ 110 \\ 111 \end{array} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \end{array}$
+
+- bloch sphere 布洛赫球: 用于表示量子比特状态的几何图形
   - X, Y axis: $0 \leq \theta \leq 2\pi$
   - Z axis: $0 \leq \phi \leq \frac{\pi}{2}$
+  - Pauli X: 沿 X 轴旋转 180 度
+  - Pauli Y: 沿 Y 轴旋转 180 度
+  - Pauli Z: 沿 Z 轴旋转 180 度
+
+- phase shift gate:
+  - $R(\theta) = \begin{bmatrix} 1 & 0 \\ 0 & e^{\theta} \end{bmatrix}$
+- Deutch gate:
+  - if both control qubits are 1, then apply a phase shift on target
+
 - Jyā, koti-jyā and utkrama-jyā 印度的三角函数
 - no-cloning theorem: cannot copy an arbitrary unknown quantum state
   - 证明：假设可以复制，那么可以通过两个相同的量子比特，得到一个新的固定的量子比特。但这是不可能的
+- transportion
+  - take $|\psi\rangle$ in first system and nothing in second system, then teleport $|x\rangle$ to second system
+  - $T(|\psi\rangle \bigotimes |0\rangle) = |\psi\rangle \bigotimes |0\rangle$
 
 - CNOT
   - control qubit, target qubit
@@ -244,6 +286,8 @@ tag:
   - $|\psi'\rangle = CNOT \cdot |\psi\rangle = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} \alpha \ \beta \ \gamma \ \delta \end{bmatrix} = \begin{bmatrix} \alpha \ \beta \ \delta \ \gamma \end{bmatrix}$
   - $|\psi'\rangle = \alpha|00\rangle + \beta|01\rangle + \delta|10\rangle + \gamma|11\rangle$
 - Hadamard
+  - 创建和破坏量子比特的叠加态
+  - $H = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$
 - Pauli X,Y,Z
   - $X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
     - X is also called NOT gate
@@ -265,15 +309,57 @@ tag:
     - eigenvectors:
       - up $|\uparrow\rangle = \psi_{z+} = |0\rangle$
       - down $|\downarrow\rangle = \psi_{z-} = |1\rangle$
+- S: 添加一个 $\frac{\pi}{2}$ 的相位
+  - $S = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}$
+- T: 添加一个 $\frac{\pi}{4}$ 的相位
+  - $T = \begin{bmatrix} 1 & 0 \\ 0 & e^{i\frac{\pi}{4}} \end{bmatrix}$
 
 ## 4. Simple Quantum Algorithms
 
+- n Hadamand gates: $H^{\bigotimes n}$
+  - $|\psi\rangle = \frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1}|i\rangle$
+- reversible function execution:
+  - control: $|x\rangle$ --> $U_f$ --> $|x\rangle$
+  - target: $|0\rangle$ --> $U_f$ --> $|f(x)\rangle$
 
 - quantum parallelism: exponential numbers were inputted simultaneously, result was given in next clock, by quantum function
 - given a quantum representation, only one result can be given, not all solutions
 - the point is: what algorithm can take advantage of quantum mechanism?
 
 ## 5. QFT and Quantum Phase Estimation
+
+- quantum mechanics can solve combinatorial optimization problems
+  - finding optimal objects that satisfy a certain condition 
+- Knapsack problem
+  - given a set of items, each with a weight and a value
+  - determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible
+  - n items
+  - $x_i$ is 1 if the item is included, 0 otherwise
+  - $v_i$ is the value of the item
+  - $W$ is the maximum weight
+  - maximize $\sum_{i=1}^{n} v_i x_i$
+  - subject to $\sum_{i=1}^{n} w_i x_i \leq W$
+- max-cut
+  - edges have weights
+  - find a cut that maximizes the sum of the weights of the edges that are cut
+- minimum vetec cover
+  - find the minimum number of vertices that cover all edges
+- simulated annealing 模拟退火
+  - a probabilistic technique for approximating the global optimum of a given function
+- local search (Monte Carlo)
+  - find solution from neighbors
+- Metropolis algorithm
+  - reproduce annealing process
+  - $E_i$ is the energy of state $i$
+  - if $E_i - E_j > 0$, change current state to state $j$ 
+  - if $E_i - E_j < 0$, change current state to state $j$ with probability $e^{\frac{E_i - E_j}{K_b T}}$
+- quadratic unconstrained binary optimization (QUBO)
+  - $min \ or \ max \sum_{i=1}^{n} \sum_{j=1}^{n} q_{ij} x_i x_j$
+  - subject to $x_i \in \{0, 1\}$
+  - $Y = X^T Q X$
+  - Q is symmetric or upper triangular
+  - NP problem
+
 ## 6. Quantum Key Distribution .
 ## 7. Quantum Principal Component Analysis, QAOA
 ## 8. Variational Quantum Circuits for Machine Learning