C. Zhong

### Distinguishing states

Definition: A set of states $\{\left|\psi_i\right>\}$ defined on a Hilbert space $H_s$ is called distinguishable if there exists a measurement system $\{M_i\}$, such that $||M_i\left|\psi_i\right>||=\delta_{ij}$ for any $i$ and $j$.

### Covariance

Covariance is defined, for two random variables, as the formula
$Con(X,Y)=E[(X-E(X))(Y-E(Y))]=E(XY)-E(X)E(Y)$

### Zero knowledge confirmation

In graph theory, there is a concept called isomorphism for different graphs. Say we have two graphs A and B, they are called isomorphic to each other if one can find a one to one map from one graph to the other (mapping point to point, line to line). Determining the isomorphism of two graphs turns out to be a NP hard problem.

Suppose you have a quantum computer, and are given two graphs A and B. With the QC, You are able to find the map easily which proves A and B are isomorphic. The question is: are you able to convince other people that A is isomorphic to B without telling them the map?

### The center of mass energy in GR

In collision experiment, a meaningful quantity is the so called center of mass energy, which is the energy measured in the center of mass frame. It is given by (for two particle with the same mass)
$E=\sqrt{2m}\sqrt{1-g_{ab}u^a_1 u^b_2}$, which can easily shown below.
$P^a=P^a_1+P^a_2$ ,
$P^aP_a=-m^2=-E^2+p^2$ ,
$P^aP_a=(P^a_1+P^a_2)(P_{1a}+P_{2a})=-m^2-m^2+2m^2g_{ab}u_1^au_2^b$,
In the center of mass frame $p=0$, which gives

$E=\sqrt{2m}\sqrt{1-g_{ab}u^a_1 u^b_2}$.

### Fisher information

Fisher information 衡量一个随机变量x包含一个未知量$\theta$信息的多少。设$f(x;\theta)$为关于x的概率密度函数(依赖于$\theta$)。$f(x;\theta)$也叫做关于$\theta$的likelihood function。Score被定义为$\frac{\partial}{\partial\theta}\ln f(x;\theta)$。可以证明score的平均值等于零。而score的方差被定义为Fisher information.
$F(\theta)=\left=\left<(\frac{\partial\ln f(x;\theta)}{\partial\theta})^2\right>$

Cramer Rao bound:

An unbiased estimator $\hat{\theta}(x)$, which satisfies
$\int(\hat{\theta}(x)-\theta)f(x;\theta)dx=0$, can faithfully tell the real value of $\theta$. The above relation must be true for all $\theta$, thus
$\frac{\partial}{\partial \theta}\int(\hat{\theta}(x)-\theta)f(x;\theta)dx=0$ must also be true. So we get
$\rightarrow \int(\hat{\theta}(x)-\theta)\frac{\partial f(x;\theta)}{\partial\theta}dx-\int f(x;\theta)dx=0\\ \rightarrow \int(\hat{\theta}(x)-\theta)\frac{\partial f(x;\theta)}{\partial\theta}dx=1 \\ \rightarrow \int(\hat{\theta}(x)-\theta)f(x;\theta)\frac{\partial \ln f(x;\theta)}{\partial\theta}dx=1 \\ \rightarrow 1=(\int(\hat{\theta}(x)-\theta)f(x;\theta)\frac{\partial \ln f(x;\theta)}{\partial\theta}dx)^2 \\ \rightarrow 1\le \int[(\hat{\theta}(x)-\theta)]^2f(x;\theta)dx\cdot\int f(x;\theta)[\frac{\partial \ln f(x;\theta)}{\partial\theta}]^2dx$
The inequality follows from the Schwarz-Cauchy inequality. We see the relation
$\int[(\hat{\theta}(x)-\theta)]^2f(x;\theta)dx\ge\frac{1}{\int f(x;\theta)[\frac{\partial \ln f(x;\theta)}{\partial\theta}]^2dx}\\ \rightarrow \Delta\theta^2\ge\frac{1}{F(\theta)}$, which is the so called Cramer Rao bound. It gives the lower bound of the uncertainty for the estimator.

A similar concept in the quantum case can be derived using quantum fisher information. Give later.

### The drag force

Consider an spherical object moving in gas with a velocity given by $\vec{u}= u\hat{n}$. If a gas particle with mass $m$ hits the object and regularly scattered into the angle $\theta$, there is a momentum transfer along the unit vector direction, which is simply given by
$\Delta P=\hat{n}\cdot (\vec{P}_i-\vec{P}_f)=P(1-\cos\theta)$
with $\vec{P}_i=-P\hat{n},|\vec{P}_f|=P$. If that process happens in a short time $\Delta t$, the object will experience a force from the gas particle
$F=\frac{P(1-\cos\theta)}{\Delta t}$.
We assume the total scattering cross section for the object $\sigma=\pi r^2$, there are $N$ particles in the volume $\sigma v\Delta t$, with ($v=P/m$), and all these particle will emerge with some probability in the solid angle after scattering. Then we have
$F=\frac{v\int NP(1-\cos\theta)\frac{d\sigma}{d\Omega}d\Omega}{\sigma v\Delta t }=\int \rho v^2 (1-\cos\theta)\frac{d\sigma}{d\Omega}d\Omega$, with the mass density $\rho=\frac{Nm}{v\Delta t \sigma}$.

The next step is to take the average of the gas particle distribution. One assumes the Maxwell distribution, and keep in mind the velocity of the moving object. Simple integral yields the drag force.

### Open problems

1,) It is still a major open question in quantum information to decide whether Entanglement Formation is a fully additive quantity, i.e.
$E_F(\rho^{AB}\otimes\sigma^{AB})=E_F(\rho^{AB})+E_F(\sigma^{AB})$, which is equivalent to the strong superadditivity of entanglement formation.