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投票者模型

In the mathematical theory of probability, the voter model is an interacting particle system introduced by Richard A. Holley and Thomas M. Liggett in 1975.[1]

In the mathematical theory of probability, the voter model is an interacting particle system introduced by Richard A. Holley and Thomas M. Liggett in 1975.

voter model coexists on the graph with two clusters

thumb|right|voter model coexists on the graph with two clusters

One can imagine that there is a "voter" at each point on a connected graph, where the connections indicate that there is some form of interaction between a pair of voters (nodes). The opinions of any given voter on some issue changes at random times under the influence of opinions of his neighbours. A voter's opinion at any given time can take one of two values, labelled 0 and 1. At random times, a random individual is selected and that voter's opinion is changed according to a stochastic rule. Specifically, for one of the chosen voter's neighbors is chosen模板:Clarify according to a given set of probabilities and that individual's opinion is transferred to the chosen voter.

One can imagine that there is a "voter" at each point on a connected graph, where the connections indicate that there is some form of interaction between a pair of voters (nodes). The opinions of any given voter on some issue changes at random times under the influence of opinions of his neighbours. A voter's opinion at any given time can take one of two values, labelled 0 and 1. At random times, a random individual is selected and that voter's opinion is changed according to a stochastic rule. Specifically, for one of the chosen voter's neighbors is chosen according to a given set of probabilities and that individual's opinion is transferred to the chosen voter.

An alternative interpretation is in terms of spatial conflict. Suppose two nations control the areas (sets of nodes) labelled 0 or 1. A flip from 0 to 1 at a given location indicates an invasion of that site by the other nation.

An alternative interpretation is in terms of spatial conflict. Suppose two nations control the areas (sets of nodes) labelled 0 or 1. A flip from 0 to 1 at a given location indicates an invasion of that site by the other nation.

Note that only one flip happens each time. Problems involving the voter model will often be recast in terms of the dual system模板:Clarify of coalescing模板:Clarify Markov chains. Frequently, these problems will then be reduced to others involving independent Markov chains.

Note that only one flip happens each time. Problems involving the voter model will often be recast in terms of the dual system of coalescing Markov chains. Frequently, these problems will then be reduced to others involving independent Markov chains.

= 定义 =

A voter model is a (continuous time) Markov process $\displaystyle{ \eta_t }$ with state space $\displaystyle{ S=\{0,1\}^{Z^d} }$ and transition rates function $\displaystyle{ c(x,\eta) }$, where $\displaystyle{ Z^d }$ is a d-dimensional integer lattice, and $\displaystyle{ c( }$•,•$\displaystyle{ ) }$ is assumed to be nonnegative, uniformly bounded and continuous as a function of $\displaystyle{ \eta }$ in the product topology on $\displaystyle{ S }$. Each component $\displaystyle{ \eta \in S }$ is called a configuration. To make it clear that $\displaystyle{ \eta(x) }$ stands for the value of a site x in configuration $\displaystyle{ \eta(.) }$; while $\displaystyle{ \eta_t(x) }$ means the value of a site x in configuration $\displaystyle{ \eta(.) }$ at time $\displaystyle{ t }$.

A voter model is a (continuous time) Markov process \eta_t with state space S=\{0,1\}^{Z^d} and transition rates function c(x,\eta) , where Z^d is a d-dimensional integer lattice, and c( •,• ) is assumed to be nonnegative, uniformly bounded and continuous as a function of \eta in the product topology on S . Each component \eta \in S is called a configuration. To make it clear that \eta(x) stands for the value of a site x in configuration \eta(.) ; while \eta_t(x) means the value of a site x in configuration \eta(.) at time t.

The dynamic of the process are specified by the collection of transition rates. For voter models, the rate at which there is a flip at $\displaystyle{ \scriptstyle x }$ from 0 to 1 or vice versa is given by a function $\displaystyle{ c(x,\eta) }$ of site $\displaystyle{ x }$. It has the following properties:

The dynamic of the process are specified by the collection of transition rates. For voter models, the rate at which there is a flip at \scriptstyle x from 0 to 1 or vice versa is given by a function c(x,\eta) of site x . It has the following properties:

1. $\displaystyle{ c(x,\eta)=0 }$ for every $\displaystyle{ x \in Z^d }$ if $\displaystyle{ \eta \equiv 0 }$ or if $\displaystyle{ \eta \equiv 1 }$
2. $\displaystyle{ c(x,\eta)=c(x,\zeta) }$ for every $\displaystyle{ x \in Z^d }$ if $\displaystyle{ \eta(y)+\zeta(y)=1 }$ for all $\displaystyle{ y \in Z^d }$
3. $\displaystyle{ c(x,\eta)\leq c(x,\zeta) }$ if $\displaystyle{ \eta\leq \zeta }$ and $\displaystyle{ \eta(x)=\zeta(x)=0 }$
4. $\displaystyle{ c(x,\eta) }$ is invariant under shifts in $\displaystyle{ \scriptstyle Z^d }$
1. c(x,\eta)=0 for every x \in Z^d if \eta \equiv 0 or if \eta \equiv 1
2. c(x,\eta)=c(x,\zeta) for every x \in Z^d if \eta(y)+\zeta(y)=1 for all y \in Z^d
3. c(x,\eta)\leq c(x,\zeta) if \eta\leq \zeta and \eta(x)=\zeta(x)=0
4. c(x,\eta) is invariant under shifts in \scriptstyle Z^d
1. c (x，eta) = 0对于 z ^ d 中的每个 x，如果 eta equiv0或者如果 eta equiv1 # c (x，eta) = c (x，zeta)对于 z ^ d 中的每个 x，如果 eta (y) + zeta (y) = 1对于 z ^ d c (x，eta)中的所有 y，如果 eta lezeta 和 eta (x) = zeta (x) = 0 # c (x，eta 不变式 d 处于移位之下

Property (1) says that $\displaystyle{ \eta\equiv 0 }$ and $\displaystyle{ \eta\equiv 1 }$ are fixed points for the evolution. (2) indicates that the evolution is unchanged by interchanging the roles of 0's and 1's. In property (3), $\displaystyle{ \eta\leq \zeta }$ means $\displaystyle{ \forall x,\eta(x)\leq\zeta(x) }$, and $\displaystyle{ \eta \leq \zeta }$ implies $\displaystyle{ c(x,\eta)\leq c(x,\zeta) }$ if $\displaystyle{ \eta(x)=\zeta(x)=0 }$, and implies $\displaystyle{ c(x,\eta)\geq c(x,\zeta) }$ if $\displaystyle{ \eta(x)=\zeta(x)=1 }$.

Property (1) says that \eta\equiv 0 and \eta\equiv 1 are fixed points for the evolution. (2) indicates that the evolution is unchanged by interchanging the roles of 0's and 1's. In property (3), \eta\leq \zeta means \forall x,\eta(x)\leq\zeta(x) , and \eta \leq \zeta implies c(x,\eta)\leq c(x,\zeta) if \eta(x)=\zeta(x)=0 , and implies c(x,\eta)\geq c(x,\zeta) if \eta(x)=\zeta(x)=1 .

Clustering and coexistence

The interest in is the limiting behavior of the models. Since the flip rates of a site depends its neighbours, it is obvious that when all sites take the same value, the whole system stops changing forever. Therefore, a voter model has two trivial extremal stationary distributions, the point-masses $\displaystyle{ \scriptstyle \delta_0 }$ and $\displaystyle{ \scriptstyle \delta_1 }$ on $\displaystyle{ \scriptstyle \eta \equiv 0 }$ or $\displaystyle{ \scriptstyle \eta\equiv 1 }$ respectively, which represent consensus. The main question to be discussed is whether or not there are others, which would then represent coexistence of different opinions in equilibrium. It is said coexistence occurs if there is a stationary distribution that concentrates on configurations with infinitely many 0's and 1's. On the other hand, if for all $\displaystyle{ \scriptstyle x,y\in Z^d }$ and all initial configurations, then

$\displaystyle{ \lim_{t\rightarrow \infty}P[\eta_t(x)\neq\eta_t(y)]=0 }$

It is said that clustering occurs.

The interest in is the limiting behavior of the models. Since the flip rates of a site depends its neighbours, it is obvious that when all sites take the same value, the whole system stops changing forever. Therefore, a voter model has two trivial extremal stationary distributions, the point-masses \scriptstyle \delta_0 and \scriptstyle \delta_1 on \scriptstyle \eta \equiv 0 or \scriptstyle \eta\equiv 1 respectively, which represent consensus. The main question to be discussed is whether or not there are others, which would then represent coexistence of different opinions in equilibrium. It is said coexistence occurs if there is a stationary distribution that concentrates on configurations with infinitely many 0's and 1's. On the other hand, if for all \scriptstyle x,y\in Z^d and all initial configurations, then

\lim_{t\rightarrow \infty}P[\eta_t(x)\neq\eta_t(y)]=0

It is said that clustering occurs.

= = = 群集和共存 = = = = 关注的是模型的限制行为。由于一个网站的翻转率取决于它的邻居，很明显，当所有网站的价值相同时，整个系统将永远停止变化。因此，选民模型有两个平凡的极值平稳分布，点质量 scriptstyle delta _ 0和 scriptstyle delta _ 1分别在 scriptstyle eta equv 0和 scriptstyle eta equv 1上，代表共识。要讨论的主要问题是，是否还有其他的问题，这些问题将表现出不同意见在平衡中的共存。我们说，如果有一个静态分布，它集中在具有无穷多个0和1的构型上，那么共存就发生了。另一方面，如果对于所有的 scriptstyle x，y 在 z ^ d 和所有的初始配置，那么: lim { t right tarrow infty } p [ eta _ t (x) neq eta _ t (y)] = 0据说发生了聚类。

It is important to distinguish clustering with the concept of cluster. Clusters are defined to be the connected components of $\displaystyle{ \scriptstyle \{x:\eta(x)=0\} }$ or $\displaystyle{ \scriptstyle \{x:\eta(x)=1\} }$.

It is important to distinguish clustering with the concept of cluster. Clusters are defined to be the connected components of \scriptstyle \{x:\eta(x)=0\} or \scriptstyle \{x:\eta(x)=1\}.

= 线性选民模型 =

Model description

This section will be dedicated to one of the basic voter models, the Linear Voter Model.

This section will be dedicated to one of the basic voter models, the Linear Voter Model.

= = = 模型描述 = = 本节将致力于一个基本的选民模型，线性选民模型。

If $\displaystyle{ \scriptstyle p( }$•,•$\displaystyle{ \scriptstyle) }$ be the transition probabilities for an irreducible random walk on $\displaystyle{ \scriptstyle Z^d }$, then:

$\displaystyle{ p(x,y)\geq 0 \quad\text{and} \sum_{y}p(x,y)=1 }$

Then in Linear voter model, the transition rates are linear functions of $\displaystyle{ \scriptstyle \eta }$:

$\displaystyle{ c(x,\eta)= \left\{ \begin{array}{l} \sum_y p(x,y)\eta(y) \quad \text{for all}\quad \eta(x)=0 \\ \sum_y p(x,y)(1-\eta(y)) \quad \text{for all}\quad \eta(x)=1 \\ \end{array} \right. }$

If \scriptstyle p( •,•\scriptstyle) be the transition probabilities for an irreducible random walk on \scriptstyle Z^d , then:

Then in Linear voter model, the transition rates are linear functions of \scriptstyle \eta :

 c(x,\eta)= \left\{
\begin{array}{l}
\end{array} \right.

Or if $\displaystyle{ \scriptstyle \eta_x }$ indicates that a flip happens at $\displaystyle{ \scriptstyle x }$, then transition rates are simply:

$\displaystyle{ \eta\rightarrow\eta_x \quad\text{at rate} \sum_{y:\eta(y)\neq\eta(x)}p(x,y). }$

Or if \scriptstyle \eta_x indicates that a flip happens at \scriptstyle x, then transition rates are simply:

A process of coalescing random walks $\displaystyle{ \scriptstyle A_t\subset Z^d }$ is defined as follows. Here $\displaystyle{ \scriptstyle A_t }$ denotes the set of sites occupied by these random walks at time $\displaystyle{ \scriptstyle t }$. To define $\displaystyle{ \scriptstyle A_t }$, consider several (continuous time) random walks on $\displaystyle{ \scriptstyle Z^d }$ with unit exponential holding times and transition probabilities $\displaystyle{ \scriptstyle p( }$•,•$\displaystyle{ \scriptstyle ) }$, and take them to be independent until two of them meet. At that time, the two that meet coalesce into one particle, which continues to move like a random walk with transition probabilities $\displaystyle{ \scriptstyle p( }$•,•$\displaystyle{ \scriptstyle ) }$ .

A process of coalescing random walks \scriptstyle A_t\subset Z^d is defined as follows. Here \scriptstyle A_t denotes the set of sites occupied by these random walks at time \scriptstyle t . To define \scriptstyle A_t , consider several (continuous time) random walks on \scriptstyle Z^d with unit exponential holding times and transition probabilities \scriptstyle p( •,•\scriptstyle ) , and take them to be independent until two of them meet. At that time, the two that meet coalesce into one particle, which continues to move like a random walk with transition probabilities \scriptstyle p( •,•\scriptstyle ) .

The concept of Duality is essential for analysing the behavior of the voter models. The linear voter models satisfy a very useful form of duality, known as coalescing duality, which is:

$\displaystyle{ P^\eta(\eta_t\equiv 1 \quad\text{on }A)=P^A(\eta(A_t)\equiv 1), }$

where $\displaystyle{ \scriptstyle \eta \in \{0,1\}^{Z^d} }$ is the initial configuration of $\displaystyle{ \scriptstyle \eta_t }$ and $\displaystyle{ \scriptstyle A=\{x\in Z^d, \eta(x)=1\}\subset Z^d }$ is the initial state of the coalescing random walks $\displaystyle{ \scriptstyle A_t }$.

The concept of Duality is essential for analysing the behavior of the voter models. The linear voter models satisfy a very useful form of duality, known as coalescing duality, which is:

where \scriptstyle \eta \in \{0,1\}^{Z^d} is the initial configuration of \scriptstyle \eta_t and \scriptstyle A=\{x\in Z^d, \eta(x)=1\}\subset Z^d is the initial state of the coalescing random walks \scriptstyle A_t.

Limiting behaviors of linear voter models

Let $\displaystyle{ \scriptstyle p(x,y) }$ be the transition probabilities for an irreducible random walk on $\displaystyle{ \scriptstyle Z^d }$ and $\displaystyle{ \scriptstyle p(x,y)=p(0,x-y) }$, then the duality relation for such linear voter models says that $\displaystyle{ \scriptstyle \forall\eta\in S=\{0,1\}^{Z^d} }$

$\displaystyle{ P^{\eta}[\eta_t(x)\neq\eta_t(y)]=P[\eta(X_t)\neq\eta(Y_t)] }$

where $\displaystyle{ \scriptstyle X_t }$ and $\displaystyle{ \scriptstyle Y_t }$ are (continuous time) random walks on $\displaystyle{ \scriptstyle Z^d }$ with $\displaystyle{ \scriptstyle X_0=x }$, $\displaystyle{ \scriptstyle Y_0=y }$, and $\displaystyle{ \scriptstyle \eta(X_t) }$ is the position taken by the random walk at time $\displaystyle{ \scriptstyle t }$. $\displaystyle{ \scriptstyle X_t }$ and $\displaystyle{ \scriptstyle Y_t }$ forms a coalescing random walks described at the end of section 2.1. $\displaystyle{ \scriptstyle X(t)-Y(t) }$ is a symmetrized random walk. If $\displaystyle{ \scriptstyle X(t)-Y(t) }$ is recurrent and $\displaystyle{ \scriptstyle d\leq 2 }$, $\displaystyle{ \scriptstyle X_t }$ and $\displaystyle{ \scriptstyle Y_t }$ will hit eventually with probability 1, and hence

$\displaystyle{ P^{\eta}[\eta_t(x)\neq\eta_t(y)]=P[\eta(X_t)\neq\eta(Y_t)]\leq P[X_t\neq Y_t]\rightarrow 0\quad\text{as}\quad t\to 0 }$

Therefore, the process clusters.

Let \scriptstyle p(x,y) be the transition probabilities for an irreducible random walk on \scriptstyle Z^d and \scriptstyle p(x,y)=p(0,x-y) , then the duality relation for such linear voter models says that \scriptstyle \forall\eta\in S=\{0,1\}^{Z^d}

P^{\eta}[\eta_t(x)\neq\eta_t(y)]=P[\eta(X_t)\neq\eta(Y_t)]

where \scriptstyle X_t and \scriptstyle Y_t are (continuous time) random walks on \scriptstyle Z^d with \scriptstyle X_0=x , \scriptstyle Y_0=y , and \scriptstyle \eta(X_t) is the position taken by the random walk at time \scriptstyle t . \scriptstyle X_t and \scriptstyle Y_t forms a coalescing random walks described at the end of section 2.1. \scriptstyle X(t)-Y(t) is a symmetrized random walk. If \scriptstyle X(t)-Y(t) is recurrent and \scriptstyle d\leq 2 , \scriptstyle X_t and \scriptstyle Y_t will hit eventually with probability 1, and hence

Therefore, the process clusters.

= = = = = = = 设 scriptstyle p (x，y)为 scriptstyle z ^ d 和 scriptstyle p (x，y) = p (0，x-y)上不可约随机游动的转移概率,然后这种线性选民模型的对偶关系说，scriptstyle forall eta in s = {0,1} ^ { z ^ d } : p ^ { eta }[ eta _ t (x) neq eta _ t (y)] = p [ eta (x _ t) neq _ eta (y _ t)]其中 scriptstyle x _ t 和 scriptstyle y _ t 是(连续时间) scriptstyle z ^ d 上的随机游动，并且 scriptstyle x _ 0 = x,scriptstyle y _ 0 = y，scriptstyle eta (x _ t)是当时随机漫步 scriptstyle t 所采取的位置。脚本风格 x _ t 和脚本风格 y _ t 在第2.1节末尾描述了随机游动的结合。Scriptstyle x (t)-y (t)是对称化的随机漫步。如果 scriptstyle x (t)-y (t)是反复出现的，scriptstyle d leq 2，scriptstyle x _ t 和 scriptstyle y _ t 最终将以1的概率出现，因此: p ^ { eta }[ eta _ t (x) neq eta _ t (y)] = p [ eta (x _ t) neq eta (y _ t)] leq p [ x _ t neq _ t ] right tarrow 0 quad text { as } t to 0。

On the other hand, when $\displaystyle{ d\geq 3 }$, the system coexists. It is because for $\displaystyle{ \scriptstyle d\geq 3 }$, $\displaystyle{ \scriptstyle X(t)-Y(t) }$ is transient, thus there is a positive probability that the random walks never hit, and hence for $\displaystyle{ \scriptstyle x\neq y }$

$\displaystyle{ \lim_{t\rightarrow\infty}P[\eta_t(x)\neq\eta_t(y)]=C\lim_{t\rightarrow\infty}P[X_t\neq Y_t]\gt 0 }$

for some constant $\displaystyle{ C }$ corresponding to the initial distribution.

On the other hand, when d\geq 3 , the system coexists. It is because for \scriptstyle d\geq 3 , \scriptstyle X(t)-Y(t) is transient, thus there is a positive probability that the random walks never hit, and hence for \scriptstyle x\neq y

\lim_{t\rightarrow\infty}P[\eta_t(x)\neq\eta_t(y)]=C\lim_{t\rightarrow\infty}P[X_t\neq Y_t]>0

for some constant C corresponding to the initial distribution.

If $\displaystyle{ \scriptstyle \tilde{X}(t)=X(t)-Y(t) }$ be a symmetrized random walk, then there are the following theorems:

If \scriptstyle \tilde{X}(t)=X(t)-Y(t) be a symmetrized random walk, then there are the following theorems:

Theorem 2.1

Theorem 2.1

The linear voter model $\displaystyle{ \scriptstyle \eta_t }$ clusters if $\displaystyle{ \scriptstyle \tilde{X}_t }$ is recurrent, and coexists if $\displaystyle{ \scriptstyle \tilde{X}_t }$ is transient. In particular,

The linear voter model \scriptstyle \eta_t clusters if \scriptstyle \tilde{X}_t is recurrent, and coexists if \scriptstyle \tilde{X}_t is transient. In particular,

1. the process clusters if $\displaystyle{ \scriptstyle d=1 }$ and $\displaystyle{ \scriptstyle \sum_x |x|p(0,x)\le \infty }$, or if $\displaystyle{ \scriptstyle d=2 }$ and $\displaystyle{ \scriptstyle \sum_x |x|^2p(0,x)\le\infty }$;
2. the process coexists if $\displaystyle{ \scriptstyle d \geq 3 }$.
1. the process clusters if \scriptstyle d=1 and \scriptstyle \sum_x |x|p(0,x)\le \infty , or if \scriptstyle d=2 and \scriptstyle \sum_x |x|^2p(0,x)\le\infty ;
2. the process coexists if \scriptstyle d \geq 3 .
1. 如果 scriptstyle d = 1和 scriptstyle sum _ x | x | p (0，x) le infty，或者如果 scriptstyle d = 2和 scriptstyle sum _ x | x | ^ 2p (0，x) le infty，进程集群; # 如果 scriptstyle d geq 3，进程共存。

Remarks: To contrast this with the behavior of the threshold voter models that will be discussed in next section, note that whether the linear voter model clusters or coexists depends almost exclusively on the dimension of the set of sites, rather than on the size of the range of interaction.

Remarks: To contrast this with the behavior of the threshold voter models that will be discussed in next section, note that whether the linear voter model clusters or coexists depends almost exclusively on the dimension of the set of sites, rather than on the size of the range of interaction.

Theorem 2.2 Suppose $\displaystyle{ \scriptstyle \mu }$ is any translation spatially ergodic and invariant probability measure on the state space $\displaystyle{ \scriptstyle S=\{0,1\}^{Z^d} }$, then

Theorem 2.2 Suppose \scriptstyle \mu is any translation spatially ergodic and invariant probability measure on the state space \scriptstyle S=\{0,1\}^{Z^d} , then

1. If $\displaystyle{ \scriptstyle \tilde{X}_t }$ is recurrent, then $\displaystyle{ \scriptstyle \mu S(t)\Rightarrow \rho\delta_1+(1-\rho)\delta_0\quad\text{as}\quad t\to\infty }$;
2. If $\displaystyle{ \scriptstyle \tilde{X}_t }$ is transient, then $\displaystyle{ \scriptstyle \mu S(t)\Rightarrow \mu_\rho }$.
1. If \scriptstyle \tilde{X}_t is recurrent, then \scriptstyle \mu S(t)\Rightarrow \rho\delta_1+(1-\rho)\delta_0\quad\text{as}\quad t\to\infty ;
2. If \scriptstyle \tilde{X}_t is transient, then \scriptstyle \mu S(t)\Rightarrow \mu_\rho .

where $\displaystyle{ \scriptstyle \mu S(t) }$ is the distribution of $\displaystyle{ \scriptstyle \eta_t }$; $\displaystyle{ \scriptstyle \Rightarrow }$ means weak convergence, $\displaystyle{ \scriptstyle \mu_{\rho} }$ is a nontrivial extremal invariant measure and $\displaystyle{ \scriptstyle \rho=\mu(\{\eta:\eta(x)=1\}) }$.

where \scriptstyle \mu S(t) is the distribution of \scriptstyle \eta_t ; \scriptstyle \Rightarrow means weak convergence, \scriptstyle \mu_{\rho} is a nontrivial extremal invariant measure and \scriptstyle \rho=\mu(\{\eta:\eta(x)=1\}) .

A special linear voter model

One of the interesting special cases of the linear voter model, known as the basic linear voter model, is that for state space $\displaystyle{ \scriptstyle \{0,1\}^{Z^d} }$:

One of the interesting special cases of the linear voter model, known as the basic linear voter model, is that for state space \scriptstyle \{0,1\}^{Z^d}:

= = = = 一个特殊的线性投票人模型 = = = = 线性投票人模型的一个有趣的特殊情况，被称为基本线性投票人模型，是国家空间的脚本风格{0,1} ^ { z ^ d } :

$\displaystyle{ p(x,y)= \begin{cases} 1/2d & \text{if } |x-y|=1 \text{ and } \eta(x)\neq\eta(y) \\[8pt] 0 & \text{otherwise} \end{cases} }$

So that

$\displaystyle{ \eta_t(x)\to 1-\eta_t(x)\quad\text{at rate}\quad (2d)^{-1}|\{y:|y-x|=1,\eta_t(y)\neq\eta_t(x)\}| }$

In this case, the process clusters if $\displaystyle{ \scriptstyle d\leq 2 }$, while coexists if $\displaystyle{ \scriptstyle d\geq 3 }$. This dichotomy is closely related to the fact that simple random walk on $\displaystyle{ \scriptstyle Z^d }$ is recurrent if $\displaystyle{ \scriptstyle d\leq2 }$ and transient if $\displaystyle{ \scriptstyle d\geq 3 }$.

 p(x,y)= \begin{cases}
1/2d & \text{if } |x-y|=1 \text{ and } \eta(x)\neq\eta(y) \\[8pt]
0 & \text{otherwise}
\end{cases}

So that

\eta_t(x)\to 1-\eta_t(x)\quad\text{at rate}\quad (2d)^{-1}|\{y:|y-x|=1,\eta_t(y)\neq\eta_t(x)\}|

In this case, the process clusters if \scriptstyle d\leq 2 , while coexists if \scriptstyle d\geq 3 . This dichotomy is closely related to the fact that simple random walk on \scriptstyle Z^d is recurrent if \scriptstyle d\leq2 and transient if \scriptstyle d\geq 3 .

P (x，y) = begin { cases }1/2d & text { if } | 我们会找到他的= 1 text { and } eta (x) neq eta (y)[8 pt ]0 & text { otherwise } end { cases }所以: eta _ t (x) to 1-eta _ t (x) quad text { at rate } quad (2d) ^ {-1} | { y: | y-x | = 1，eta _ t (y) neq eta _ t (x)} | 在这种情况下，进程集群如果 scriptstyle d leq 2，而共存如果 scriptstyle d geq 3。这种二分法与这样一个事实密切相关: 如果 scriptstyle d leq2，那么在 scriptstyle z ^ d 上简单的随机漫步是反复出现的; 如果 scriptstyle d geq 3，那么简单的随机漫步是暂时的。

Clusters in one dimension d = 1

For the special case with $\displaystyle{ \scriptstyle d=1 }$, $\displaystyle{ \scriptstyle S=Z^1 }$ and $\displaystyle{ \scriptstyle p(x,x+1)=p(x,x-1)=\frac{1}{2} }$ for each $\displaystyle{ \scriptstyle x }$. From Theorem 2.2, $\displaystyle{ \scriptstyle \mu S(t)\Rightarrow \rho\delta_1+(1-\rho)\delta_0 }$, thus clustering occurs in this case. The aim of this section is to give a more precise description of this clustering.

For the special case with \scriptstyle d=1 , \scriptstyle S=Z^1 and \scriptstyle p(x,x+1)=p(x,x-1)=\frac{1}{2} for each \scriptstyle x . From Theorem 2.2, \scriptstyle \mu S(t)\Rightarrow \rho\delta_1+(1-\rho)\delta_0 , thus clustering occurs in this case. The aim of this section is to give a more precise description of this clustering.

= = = = = 一维集群 d = 1 = = = = = = 对于特殊情况，对于每个 scriptstyle x，scriptstyle s = z ^ 1和 scriptstyle p (x，x + 1) = p (x，x-1) = frac {1}{2}。根据定理2.2，scriptstyle mu s (t) right tarrow rho delta _ 1 + (1-rho) delta _ 0，因此在这种情况下发生了聚类。本节的目的是更精确地描述这种集群。

As mentioned before, clusters of an $\displaystyle{ \scriptstyle \eta }$ are defined to be the connected components of $\displaystyle{ \scriptstyle \{x:\eta(x)=0\} }$ or $\displaystyle{ \scriptstyle \{x:\eta(x)=1\} }$. The mean cluster size for $\displaystyle{ \scriptstyle \eta }$ is defined to be:

$\displaystyle{ C(\eta)=\lim_{n\rightarrow\infty}\frac{2n}{\text{number of clusters in} [-n,n]} }$

provided the limit exists.

As mentioned before, clusters of an \scriptstyle \eta are defined to be the connected components of \scriptstyle \{x:\eta(x)=0\} or \scriptstyle \{x:\eta(x)=1\} . The mean cluster size for \scriptstyle \eta is defined to be:

C(\eta)=\lim_{n\rightarrow\infty}\frac{2n}{\text{number of clusters in} [-n,n]}

provided the limit exists.

Proposition 2.3

Proposition 2.3

Suppose the voter model is with initial distribution $\displaystyle{ \scriptstyle \mu }$ and $\displaystyle{ \scriptstyle \mu }$ is a translation invariant probability measure, then

$\displaystyle{ P\left(C(\eta)=\frac{1}{P[\eta_t(0)\neq \eta_t(1)]}\right)=1. }$

Suppose the voter model is with initial distribution \scriptstyle \mu and \scriptstyle \mu is a translation invariant probability measure, then

P\left(C(\eta)=\frac{1}{P[\eta_t(0)\neq \eta_t(1)]}\right)=1.

Occupation time

Define the occupation time functionals of the basic linear voter model as:

$\displaystyle{ T_t^x=\int_0^t \eta^\rho_s(x)\mathrm{d}s. }$

Define the occupation time functionals of the basic linear voter model as:

T_t^x=\int_0^t \eta^\rho_s(x)\mathrm{d}s.

= = = = 占用时间 = = = 定义基本线性选民模型的占用时间泛函为: t _ t ^ x = int _ 0 ^ t eta ^ rho _ s (x) mathrm { d }。

Theorem 2.4

Theorem 2.4

Assume that for all site x and time t, $\displaystyle{ \scriptstyle P(\eta_t(x)=1)=\rho }$, then as $\displaystyle{ \scriptstyle t\rightarrow \infty }$, $\displaystyle{ \scriptstyle T_t^x/t\rightarrow \rho }$ almost surely if $\displaystyle{ \scriptstyle d\geq 2 }$

Assume that for all site x and time t, \scriptstyle P(\eta_t(x)=1)=\rho, then as \scriptstyle t\rightarrow \infty , \scriptstyle T_t^x/t\rightarrow \rho almost surely if \scriptstyle d\geq 2

proof

proof

By Chebyshev's inequality and the Borel–Cantelli lemma, there is the equation below:

$\displaystyle{ P\left(\frac{\rho}{r}\leq \lim \inf_{t\rightarrow\infty}\frac{T_t}{t}\leq\lim\sup_{t\rightarrow\infty}\frac{T_t}{t}\leq \rho r\right)=1; \quad\forall r\gt 1 }$

The theorem follows when letting $\displaystyle{ \scriptstyle r\searrow 1 }$.

By Chebyshev's inequality and the Borel–Cantelli lemma, there is the equation below:

P\left(\frac{\rho}{r}\leq \lim \inf_{t\rightarrow\infty}\frac{T_t}{t}\leq\lim\sup_{t\rightarrow\infty}\frac{T_t}{t}\leq \rho r\right)=1; \quad\forall r>1

The theorem follows when letting \scriptstyle r\searrow 1 .

= 阈值选民模型 =

Model description

This section, concentrates on a kind of non-linear voter models, known as the threshold voter model. To define it, let $\displaystyle{ \scriptstyle \mathcal{N} }$ be a neighbourhood of $\displaystyle{ \scriptstyle 0\in Z^d }$ that is obtained by intersecting $\displaystyle{ \scriptstyle Z^d }$ with any compact, convex, symmetric set in $\displaystyle{ \scriptstyle R^d }$; in other word, $\displaystyle{ \scriptstyle \mathcal{N} }$ is assumed to be a finite set that is symmetric with respect to all reflections and irreducible (i.e. the group it generates is $\displaystyle{ \scriptstyle Z^d }$). It can always be assumed that $\displaystyle{ \scriptstyle \mathcal{N} }$ contains all the unit vectors $\displaystyle{ \scriptstyle (1,0,0,\dots,0),\dots,(0,\dots,0,1) }$. For a positive integer $\displaystyle{ \scriptstyle T }$, the threshold voter model with neighbourhood $\displaystyle{ \scriptstyle \mathcal{N} }$ and threshold $\displaystyle{ \scriptstyle T }$ is the one with rate function:

This section, concentrates on a kind of non-linear voter models, known as the threshold voter model. To define it, let \scriptstyle \mathcal{N} be a neighbourhood of \scriptstyle 0\in Z^d that is obtained by intersecting \scriptstyle Z^d with any compact, convex, symmetric set in \scriptstyle R^d ; in other word, \scriptstyle \mathcal{N} is assumed to be a finite set that is symmetric with respect to all reflections and irreducible (i.e. the group it generates is \scriptstyle Z^d ). It can always be assumed that \scriptstyle \mathcal{N} contains all the unit vectors \scriptstyle (1,0,0,\dots,0),\dots,(0,\dots,0,1) . For a positive integer \scriptstyle T , the threshold voter model with neighbourhood \scriptstyle \mathcal{N} and threshold \scriptstyle T is the one with rate function:

= = = 模型描述 = = 本节集中讨论一种非线性选民模型，即阈值选民模型。为了定义它，让 scriptstyle 数学{ n }成为 z ^ d 中 scriptstyle 0的邻域，这个邻域是通过将 scriptstyle z ^ d 与 scriptstyle r ^ d 中的任意紧凑、凸、对称集合相交而得到的; 换句话说，scriptstyle 数学{ n }被假定为对所有反射和不可约的对称的有限集合。它生成的组是 scriptstyle z ^ d)。可以假定 scriptstyle 数学{ n }包含所有单元向量 scriptstyle (1,0,0，点，0)、 dots (0，点，0,1)。对于正整数 scriptstyle t，带有邻里 scriptstyle 数学{ n }和阈值 scriptstyle t 的阈值选民模型是具有 rate 函数的模型:

$\displaystyle{ c(x,\eta)= \left\{ \begin{array}{l} 1 \quad \text{if}\quad |\{y\in x+\mathcal{N}:\eta(y)\neq\eta(x)\}|\geq T \\ 0 \quad \text{otherwise} \\ \end{array} \right. }$
 c(x,\eta)= \left\{
\begin{array}{l}
\end{array} \right.

1 quad text { if } quad | { y in x + mathcal { n } : eta (y) neq eta (x)} | geq t 0 quad text { otherwise } end { array } right.

Simply put, the transition rate of site $\displaystyle{ \scriptstyle x }$ is 1 if the number of sites that do not take the same value is larger or equal to the threshold T. Otherwise, site $\displaystyle{ \scriptstyle x }$ stays at the current status and will not flip.

Simply put, the transition rate of site \scriptstyle x is 1 if the number of sites that do not take the same value is larger or equal to the threshold T. Otherwise, site \scriptstyle x stays at the current status and will not flip.

For example, if $\displaystyle{ \scriptstyle d=1 }$, $\displaystyle{ \scriptstyle \mathcal{N}=\{-1,0,1\} }$ and $\displaystyle{ \scriptstyle T=2 }$, then the configuration $\displaystyle{ \scriptstyle \dots1\quad 1\quad 0\quad 0\quad 1\quad 1\quad 0\quad 0\dots }$ is an absorbing state or a trap for the process.

Limiting behaviors of threshold voter model

If a threshold voter model does not fixate, the process should be expected to will coexist for small threshold and cluster for large threshold, where large and small are interpreted as being relative to the size of the neighbourhood, $\displaystyle{ \scriptstyle |\mathcal{N}| }$. The intuition is that having a small threshold makes it easy for flips to occur, so it is likely that there will be a lot of both 0's and 1's around at all times. The following are three major results:

1. If $\displaystyle{ \scriptstyle T\gt \frac{|\mathcal{N}|-1}{2} }$, then the process fixates in the sense that each site flips only finitely often.
2. If $\displaystyle{ \scriptstyle d=1 }$ and $\displaystyle{ \scriptstyle T=\frac{|\mathcal{N}|-1}{2} }$, then the process clusters.
3. If $\displaystyle{ \scriptstyle T=\theta|\mathcal{N}| }$ with $\displaystyle{ \scriptstyle \theta }$ sufficiently small($\displaystyle{ \scriptstyle \theta\lt \frac{1}{4} }$) and $\displaystyle{ \scriptstyle |\mathcal{N}| }$ sufficiently large, then the process coexists.

If a threshold voter model does not fixate, the process should be expected to will coexist for small threshold and cluster for large threshold, where large and small are interpreted as being relative to the size of the neighbourhood, \scriptstyle |\mathcal{N}| . The intuition is that having a small threshold makes it easy for flips to occur, so it is likely that there will be a lot of both 0's and 1's around at all times. The following are three major results:

1. If \scriptstyle T>\frac{|\mathcal{N}|-1}{2} , then the process fixates in the sense that each site flips only finitely often.
2. If \scriptstyle d=1 and \scriptstyle T=\frac{|\mathcal{N}|-1}{2} , then the process clusters.
3. If \scriptstyle T=\theta|\mathcal{N}| with \scriptstyle \theta sufficiently small(\scriptstyle \theta<\frac{1}{4} ) and \scriptstyle |\mathcal{N}| sufficiently large, then the process coexists.

= = = 限制阈值选民模型的行为 = = = 如果阈值选民模型不固定，那么对于小阈值进程应该共存，对于大阈值进程应该聚类，其中大和小被解释为相对于邻居的大小，scriptstyle | math{ n } | 。我们的直觉是，有一个小的阈值使得翻转很容易发生，所以很可能在任何时候都会有很多0和1。以下是三个主要结果: # 如果 scriptstyle t > frac { | mathcal { n } |-1}{2} ，那么这个进程固定在每个站点只有限次翻转的意义上。# 如果 scriptstyle d = 1和 scriptstyle t = frac { | mathcal { n } |-1}{2} ，那么进程集群。如果 scriptstyle t = theta | mathcal { n } | with scriptstyle theta 充分小(scriptstyle theta < frac {1}{4})和 scriptstyle | mathcal { n } | 足够大，那么进程共存。

Here are two theorems corresponding to properties (1) and (2).

Here are two theorems corresponding to properties (1) and (2).

Theorem 3.1

Theorem 3.1

If $\displaystyle{ \scriptstyle T\gt \frac{|\mathcal{N}|-1}{2} }$, then the process fixates.

If \scriptstyle T>\frac{|\mathcal{N}|-1}{2} , then the process fixates.

Theorem 3.2

Theorem 3.2

The threshold voter model in one dimension ($\displaystyle{ \scriptstyle d=1 }$) with $\displaystyle{ \scriptstyle \mathcal{N}=\{-T,\dots,T\}, T\geq 1 }$, clusters.

The threshold voter model in one dimension (\scriptstyle d=1 ) with \scriptstyle \mathcal{N}=\{-T,\dots,T\}, T\geq 1 , clusters.

proof

proof

The idea of the proof is to construct two sequences of random times $\displaystyle{ \scriptstyle U_n }$, $\displaystyle{ \scriptstyle V_n }$ for $\displaystyle{ \scriptstyle n\geq 1 }$ with the following properties:

The idea of the proof is to construct two sequences of random times \scriptstyle U_n , \scriptstyle V_n for \scriptstyle n\geq 1 with the following properties:

1. $\displaystyle{ \scriptstyle 0=V_0\lt U_1\lt V_1\lt U_2\lt V_2\lt \dots }$,
2. $\displaystyle{ \scriptstyle \{U_{k+1}-V_k,k\geq0\} }$ are i.i.d.with $\displaystyle{ \scriptstyle \mathrm{E}(U_{k+1}-V_k)\lt \infty }$,
3. $\displaystyle{ \scriptstyle \{V_{k}-U_k,k\geq1\} }$ are i.i.d.with $\displaystyle{ \scriptstyle \mathrm{E}(V_{k}-U_k)=\infty }$,
4. the random variables in (b) and (c) are independent of each other,
5. event A=$\displaystyle{ \scriptstyle \{\eta_t(.) }$ is constant on $\displaystyle{ \scriptstyle \{-T,\dots,T\}\} }$, and event A holds for every $\displaystyle{ \scriptstyle t \in \cup_{k=1}^\infty [U_k,V_k] }$.
1. \scriptstyle 0=V_0<U_1<V_1<U_2<V_2<\dots ,
2. \scriptstyle \{U_{k+1}-V_k,k\geq0\} are i.i.d.with \scriptstyle \mathrm{E}(U_{k+1}-V_k)<\infty ,
3. \scriptstyle \{V_{k}-U_k,k\geq1\} are i.i.d.with \scriptstyle \mathrm{E}(V_{k}-U_k)=\infty ,
4. the random variables in (b) and (c) are independent of each other,
5. event A=\scriptstyle \{\eta_t(.) is constant on \scriptstyle \{-T,\dots,T\}\} , and event A holds for every \scriptstyle t \in \cup_{k=1}^\infty [U_k,V_k] .
1. \scriptstyle 0=V_0<U_1<V_1<U_2<V_2<\dots ,
2. \scriptstyle \{U_{k+1}-V_k,k\geq0\} are i.i.d.with \scriptstyle \mathrm{E}(U_{k+1}-V_k)<\infty ,
3. \scriptstyle \{V_{k}-U_k,k\geq1\} are i.i.d.with \scriptstyle \mathrm{E}(V_{k}-U_k)=\infty ,
4. the random variables in (b) and (c) are independent of each other,
5. event A=\scriptstyle \{\eta_t(.)对于 cup _ { k = 1} ^ infty [ u _ k，v _ k ]中的每个 scriptstyle t，事件 a 都是常量。

Once this construction is made, it will follow from renewal theory that

$\displaystyle{ P(A)\geq P(t \in \cup_{k=1}^\infty [U_k,V_k])\to 1 \quad\text{as}\quad t\to\infty }$

Hence,$\displaystyle{ \scriptstyle \lim_{t\rightarrow \infty}P(\eta_t(1)\neq \eta_t(0))=0 }$, so that the process clusters.

Once this construction is made, it will follow from renewal theory that

Hence,\scriptstyle \lim_{t\rightarrow \infty}P(\eta_t(1)\neq \eta_t(0))=0 , so that the process clusters.

Remarks: (a) Threshold models in higher dimensions do not necessarily cluster if $\displaystyle{ \scriptstyle T=\frac{|\mathcal{N}|-1}{2} }$. For example, take $\displaystyle{ \scriptstyle d=2,T=2 }$ and $\displaystyle{ \scriptstyle \mathcal{N}=\{(0,0),(0,1),(1,0),(0,-1),(-1,0)\} }$. If $\displaystyle{ \scriptstyle \eta }$ is constant on alternating vertical infinite strips, that is for all $\displaystyle{ \scriptstyle i,j }$:

$\displaystyle{ \eta(4i,j)=\eta(4i+1,j)=1,\quad \eta(4i+2,j)=\eta(4i+3,j)=0 }$

then no transition ever occur, and the process fixates.

Remarks: (a) Threshold models in higher dimensions do not necessarily cluster if \scriptstyle T=\frac{|\mathcal{N}|-1}{2} . For example, take \scriptstyle d=2,T=2 and \scriptstyle \mathcal{N}=\{(0,0),(0,1),(1,0),(0,-1),(-1,0)\} . If \scriptstyle \eta is constant on alternating vertical infinite strips, that is for all \scriptstyle i,j :

then no transition ever occur, and the process fixates.

(b) Under the assumption of Theorem 3.2, the process does not fixate. To see this, consider the initial configuration $\displaystyle{ \scriptstyle \dots 0 0 0 1 1 1 \dots }$, in which infinitely many zeros are followed by infinitely many ones. Then only the zero and one at the boundary can flip, so that the configuration will always look the same except that the boundary will move like a simple symmetric random walk. The fact that this random walk is recurrent implies that every site flips infinitely often.

(b) Under the assumption of Theorem 3.2, the process does not fixate. To see this, consider the initial configuration \scriptstyle \dots 0 0 0 1 1 1 \dots , in which infinitely many zeros are followed by infinitely many ones. Then only the zero and one at the boundary can flip, so that the configuration will always look the same except that the boundary will move like a simple symmetric random walk. The fact that this random walk is recurrent implies that every site flips infinitely often.

(b)在定理3.2的假设下，过程不固定。要看到这一点，考虑初始配置 scriptstyle dots 0011 dots，其中无限多个零后面跟着无限多个1。然后，只有边界处的0和1可以翻转，这样配置看起来总是一样的，除了边界会像一个简单的对称随机游走一样移动。事实上，这种随机漫步是反复出现的，这意味着每个站点都会无限频繁地翻转。

Property 3 indicates that the threshold voter model is quite different from the linear voter model, in that coexistence occurs even in one dimension, provided that the neighbourhood is not too small. The threshold model has a drift toward the "local minority", which is not present in the linear case.

Property 3 indicates that the threshold voter model is quite different from the linear voter model, in that coexistence occurs even in one dimension, provided that the neighbourhood is not too small. The threshold model has a drift toward the "local minority", which is not present in the linear case.

Most proofs of coexistence for threshold voter models are based on comparisons with hybrid model known as the threshold contact process with parameter $\displaystyle{ \scriptstyle \lambda\gt 0 }$. This is the process on $\displaystyle{ \scriptstyle [0,1]^{Z^d} }$ with flip rates:

$\displaystyle{ c(x,\eta)= \left\{ \begin{array}{l} \lambda \quad \text{if}\quad\eta(x)=0\quad \text{and}|\{y\in x+\mathcal{N}:\eta(y)=1\}|\geq T; \\ 1 \quad \text{if}\quad \eta(x)=1;\\ 0 \quad \text{otherwise} \end{array}\right. }$

Most proofs of coexistence for threshold voter models are based on comparisons with hybrid model known as the threshold contact process with parameter \scriptstyle \lambda>0 . This is the process on \scriptstyle [0,1]^{Z^d} with flip rates:

c(x,\eta)= \left\{
\begin{array}{l}

\end{array}\right.

Proposition 3.3

Proposition 3.3

3.3号提案

For any $\displaystyle{ \scriptstyle d, \mathcal{N} }$ and $\displaystyle{ \scriptstyle T }$, if the threshold contact process with $\displaystyle{ \scriptstyle \lambda=1 }$ has a nontrivial invariant measure, then the threshold voter model coexists.

For any \scriptstyle d, \mathcal{N} and \scriptstyle T , if the threshold contact process with \scriptstyle \lambda=1 has a nontrivial invariant measure, then the threshold voter model coexists.

Model with threshold T = 1

= = = 具有阈值 t = 1 = = = 的模型

The case that $\displaystyle{ \scriptstyle T=1 }$ is of particular interest because it is the only case in which it is known exactly which models coexist and which models cluster.

The case that \scriptstyle T=1 is of particular interest because it is the only case in which it is known exactly which models coexist and which models cluster.

Scriptstyle t = 1的情况特别有趣，因为它是唯一知道哪些模型共存和哪些模型集群的情况。

In particular, there is interest in a kind of Threshold T=1 model with $\displaystyle{ \scriptstyle c(x,\eta) }$ that is given by:

$\displaystyle{ c(x,\eta)= \left\{ \begin{array}{l} 1 \quad\text{if exists one}\quad y \quad\text{with}\quad |x-y|\leq N \quad\text{and}\quad \eta(x)\neq\eta(y) \\ 0 \quad \text{otherwise}\\ \end{array} \right. }$

In particular, there is interest in a kind of Threshold T=1 model with \scriptstyle c(x,\eta) that is given by:

 c(x,\eta)= \left\{
\begin{array}{l}
\end{array} \right.

$\displaystyle{ \scriptstyle N }$ can be interpreted as the radius of the neighbourhood $\displaystyle{ \scriptstyle \mathcal{N} }$; $\displaystyle{ \scriptstyle N }$ determines the size of the neighbourhood (i.e., if $\displaystyle{ \scriptstyle \mathcal{N}_1=\{-2,-1,0,1,2\} }$, then $\displaystyle{ \scriptstyle N_1=2 }$; while for $\displaystyle{ \scriptstyle \mathcal{N}_2=\{(0,0),(0,1),(1,0),(0,-1),(-1,0)\} }$, the corresponding $\displaystyle{ \scriptstyle N_2=1 }$).

\scriptstyle N can be interpreted as the radius of the neighbourhood \scriptstyle \mathcal{N} ; \scriptstyle N determines the size of the neighbourhood (i.e., if \scriptstyle \mathcal{N}_1=\{-2,-1,0,1,2\} , then \scriptstyle N_1=2 ; while for \scriptstyle \mathcal{N}_2=\{(0,0),(0,1),(1,0),(0,-1),(-1,0)\} , the corresponding \scriptstyle N_2=1 ).

Scriptstyle n 可以解释为邻域 scriptstyle 数学{ n }的半径; scriptstyle n 确定邻域的大小(即，如果 scriptstyle 数学{ n }1 = {-2,-1,0,1,2} ，那么 scriptstyle n _ 1 = 2; 而对于 scriptstyle 数学{ n } _ 2 = {(0,0) ，(0,1) ，(1,0) ，(0,-1) ，(- 1,0)} ，相应的 scriptstyle n _ 2 = 1)。

By Theorem 3.2, the model with $\displaystyle{ \scriptstyle d=1 }$ and $\displaystyle{ \scriptstyle \mathcal{N}=\{-1,0,1\} }$ clusters. The following theorem indicates that for all other choices of $\displaystyle{ \scriptstyle d }$ and $\displaystyle{ \scriptstyle \mathcal{N} }$, the model coexists.

By Theorem 3.2, the model with \scriptstyle d=1 and \scriptstyle \mathcal{N}=\{-1,0,1\} clusters. The following theorem indicates that for all other choices of \scriptstyle d and \scriptstyle \mathcal{N} , the model coexists.

Theorem 3.4

Theorem 3.4

Suppose that $\displaystyle{ \scriptstyle N\geq 1 }$, but $\displaystyle{ \scriptstyle (N,d)\neq(1,1) }$. Then the threshold model on $\displaystyle{ \scriptstyle Z^d }$ with parameter $\displaystyle{ \scriptstyle N }$ coexists.

Suppose that \scriptstyle N\geq 1 , but \scriptstyle (N,d)\neq(1,1) . Then the threshold model on \scriptstyle Z^d with parameter \scriptstyle N coexists.

The proof of this theorem is given in a paper named "Coexistence in threshold voter models" by Thomas M. Liggett.

The proof of this theorem is given in a paper named "Coexistence in threshold voter models" by Thomas M. Liggett.

= 参见 =

• Probabilistic Cellular Automata
• sequential dynamical system
• contact process

• 概率细胞自动机
• 顺序动力系统
• 接触过程

= 笔记 =

1. Holley, Richard A.; Liggett, Thomas M. (1975). "Ergodic Theorems for Weakly Interacting Infinite Systems and the Voter Model". The Annals of Probability (in English). 3 (4): 643–663. doi:10.1214/aop/1176996306. ISSN 0091-1798.

= 参考文献 =

• Thomas M. Liggett, "Stochastic Interacting Systems: Contact, Voter and Exclusion Processes", Springer-Verlag, 1999.

• 托马斯 · m · 利格特,”随机相互作用系统: 接触、选民和排除过程”，Springer-Verlag，1999。

Category:Stochastic models Category:Lattice models

This page was moved from wikipedia:en:Voter model. Its edit history can be viewed at 投票者模型/edithistory


text:
# Funding for Solar Geoengineering from 2008 to 2018

By Ella Necheles, Lizzie Burns, Amy Chang, and David Keith.

### Our Project

As the visibility of solar geoengineering research grows, we thought it would be useful to provide a publicly accessible record of the solar geoengineering projects that have been funded over the past ten years.

How one defines a solar geoengineering “project,” however, is not straightforward. With a diverse range of efforts having taken place, there are numerous approaches that one could take.

For our purposes, we used the following definition to focus our scope. We realize, however, that this does not encompass every solar geoengineering effort that has been undertaken to date. Furthermore, we recognize that many solar geoengineering efforts have not even been funded, but rather have been conducted voluntarily. As one researcher explained: “Nobody supports me for GeoMIP. I do all of that on nights and weekends.”

Therefore, while the criteria below provided a scope for our project, it did not capture every effort that has been dedicated to solar geoengineering to date. For our project:

### Highlights

Several interesting patterns emerged from this data set. However, given the data set’s small size and incomplete nature (including the missing volunteer work), we do not believe these patterns represent a thorough analysis of the trends in solar geoengineering research and advocacy. We merely highlight them below in case they prove useful.

Overall, the total amount of global funding supporting solar geoengineering research and advocacy has been rather minimal, particularly when compared to the total amount of funding that has supported other climate related research and advocacy efforts since 2008. That said, solar geoengineering funding has increased gradually over the last decade (except for 2015 and 2016, when there was a slight decline). For example, in 2008, there was a little more than $1 million in solar geoengineering funding, and in 2018 there was a little more than$8 million in funding.

Looking closer at the geographic distribution, the US, UK, and Germany have hosted the majority of solar geoengineering projects over the past decade. In Germany, funding has supported numerous projects since 2012 and 2013, and that funding has stayed rather constant over time; in the US, there has been a gradual increase in funding from 2008 to 2016, and then a spike in 2016 (largely due to Harvard’s Solar Geoengineering Research Program and the Carnegie Climate Geoengineering Governance Initiative); and in the UK, we see an opposite trend, where there was some funding towards the beginning to middle of the period in question, but then a decline in recent years.

Various other countries, such as China and Japan, also host large geoengineering projects. Still, there are regions that have had little to no engagement in solar geoengineering research or advocacy. For example, South America and Africa both lack countries with significant solar geoengineering programs. Considering the large impacts that solar geoengineering could have on regions within these continents (both in terms of benefits and risks), it is essential that these regions are engaged moving forward. The DECIMALS Fund was launched in 2018 to serve this purpose, and aims to fund 7-8 projects in developing nations through 2020.

In addition to geographic trends, we also analyzed the funding sources. Over the decade in question, government funding swelled (reaching its peak in 2014) before decreasing. Private funding remained constant before increasing in 2016 (again, largely due to Harvard’s Solar Geoengineering Research Program and the Carnegie Climate Geoengineering Governance Initiative). And mixed sources funding (public and private) remained such a small percentage of overall funds that its pattern is difficult to trace.

When we looked closer at funding sources by location, we noticed several differences. In Europe and Asia, the government provided the majority of funding, while in the US, the philanthropic sector provided a greater amount of funding. The following table shows the approximate funding amounts by location and funding type between 2008 and 2018:

Lastly, over the last ten years, funding support for interdisciplinary projects (focusing on both natural and social sciences) has increased rather steadily over the decade. Likewise, funding support for purely social science work has increased over time, except for 2015 and 2016, when it experienced a decline. In comparison, funding support for purely natural science research has gradually increased and then decreased over time.

### Thank You

We would like to thank the many individuals who shared information about their projects. If you notice any errors or omissions, please email Selena Wallace (swallace [at] seas.harvard.edu).

Ella Necheles is a sophomore at Harvard College. Lizzie Burns is the Program Director for Harvard’s Solar Geoengineering Research Program. Amy Chang is the Coordinator for Environmental Research at Harvard’s Solar Geoengineering Research Program. David Keith is the Gordon McKay Professor of Applied Physics at the Harvard John A. Paulson School of Engineering and Applied Sciences, and Professor of Public Policy at the Harvard Kennedy School.

This post was updated on August 11, 2019 to reflect new information learned.

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text:
# Make an operator to be hermitian

1. Jul 16, 2014

### masoodsa

I have an operator which isnot Hermitian is there any way to make it hermitian ?

2. Jul 16, 2014

### ChrisVer

add its hermitian conjugate and redefine it.

3. Jul 16, 2014

### bhobba

Any operator can be written of the form H = A + iB where A and B are Hermitian.

In fact one could base QM on any operator using that result via the obvious extension to the Born Rule E(H) = E(A) + i E(B)

Interestingly in proving the Born rule via Gleason that's what one does.

So why does one stick with Hermitan operators? The reason is the properties of continous transformations between states and symmetry. Wigners Theroem implies such must be Unitary:
http://en.wikipedia.org/wiki/Wigner's_theorem

If U is such a transformation dependant on t (not necessarily time) and we consider Δt small. U(Δt) = 1 + D Δt. UU(bar) = 1 + DΔt + D(bar)Δt = 1 or D = -D(bar) which means D is pure imaginary. If we want to associate D with an observation, which from symmetry considerations you want to do because it leads to the same equations as classical physics (see Ballentine chapter 3), then its expectation would be imaginary as well which would be a rather strange thing. So what we do is define divide D by i and use that operator which is of course real.

Thanks
Bill


text:
# New to pointers

This topic is 2144 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hey guys!

I'm not new to programming, but it's been many years since I really used C++, and I don't think we even touched on pointers(it was in highschool). Anyways, I have some working code, but since I'm not a pointer expert I wanted to make sure that I was doing this correctly.

Currently, I'm working on a small prototype for a game to kind of help me get some basic ideas down as well as help me get back into the way C++ does things. I'm using SFML and my question really applies to how I have my game screens set up. In each game screen, it holds a pointer to the RenderWindow so that when Draw is called the RenderWindow would not have to be passed before each frame is drawn.

Since the RenderWindow will have its own destructor called all I do is set the pointer to 0. Is that all I really have to do?

Here's some code just to be certain.

This is not all my code, just the code in question. Like I said, I wanted to make sure that I'm doing this correctly. Everything compiles and it runs fine!

tl;dr

Is this an acceptable use of pointers?

 class BasicGameScreen { private: //Window used for Rendering sf::RenderWindow *MainApp; public: void Initialize(sf::RenderWindow &App) { MainApp = &App; } void :Draw() { //drawing whatever MainApp->Draw(Background); MainApp->Draw(Sprite); } ~BasicGameScreen() { MainApp = 0; } }  Edited by Jebbles

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No problems with that really. You might like to consider passing the MainApp pointer into BasicGameScreen's constructor so you avoid having a BasicGameScreen ever exist without the pointer initialised - as is if you called Draw() before Initialize() by mistake, you'd crash in a nasty way.

The other approach to consider is to pass a reference to MainApp to the Draw() method as a paramter. I generally prefer this when possible as it reduces the amount of bookeeping within classes and makes it easier to re-use a class in the future.

The other option would be to store a reference to MainApp rather than a pointer, initialised again in the constructor but having reference members of classes is a tricky business and can have implications for storing instances of your class in standard containers etc.

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No problems with that really. You might like to consider passing the MainApp pointer into BasicGameScreen's constructor so you avoid having a BasicGameScreen ever exist without the pointer initialised - as is if you called Draw() before Initialize() by mistake, you'd crash in a nasty way.

The other approach to consider is to pass a reference to MainApp to the Draw() method as a paramter. I generally prefer this when possible as it reduces the amount of bookeeping within classes and makes it easier to re-use a class in the future.

The other option would be to store a reference to MainApp rather than a pointer, initialised again in the constructor but having reference members of classes is a tricky business and can have implications for storing instances of your class in standard containers etc.

If in your case MainApp pointer can never be null in a BasicGameScreen instance you need to use a reference as that makes that clear in the semantics of the class. A reference has to be initililased at the time to class instance is created, which means that whatever it aliases needs to exist as long as this class exists.

If you can't guarantee either of these conditions use a pointer. If you can't guarantee condition two and you are using a reference you can get a reference that's pointing at random memory which the runtime will always try to interpret as the definition of the class being referenced.

Having pointer or reference parameters can mean problems with any type of copy construction, that's the problems you face with STL containers anyway. If a shallow copy is fine, this is the copy constructor C++ will generate for you, there should be no trouble. However when a shallow copy isn't fine you will have to define your own assignment and copy constructors to do a deep copy Edited by NightCreature83

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Having pointer or reference parameters can mean problems with any type of copy construction, that's the problems you face with STL containers anyway. If a shallow copy is fine, this is the copy constructor C++ will generate for you, there should be no trouble. However when a shallow copy isn't fine you will have to define your own assignment and copy constructors to do a deep copy

How do you suggest defining a default constructor or an assignment operator for a class containing a reference?

The simple rule of "if it cannot be null, use a pointer" applies to function parameters but cannot be universally applied to class members as well.

[source]
class xxx
{
public:
xxx(int &a) : a(a) { }
xxx(const xxx &b) : a(b.a) { }

int &a;
};

void f()
{
int z;

std::vector<xxx> a;
a.push_back(xxx(z));
}
[/source]

Won't compile (with GCC anyway). A class containing a pointer member can easily be made to work however. Edited by Aardvajk

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Isn't it just a warning you are getting about z not being inialised as MSVC2010 compiles that code fine.

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Hmm, checked again, definatly an error from GCC but it is complaining that the default assignment operator is not sufficient, which admittedly seems a bit odd.

Not sure if GCC or VS is correct here as far as standard is concerned. I thought a class containing a reference member was non-copyable but thinking about it now, I'm not sure why.

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With C++03, in order for a type to be usable in a standard library container, it needs to have both a copy constructor and an assignment operator. With C++11, the situation is a little more complex. Only some of the operations require an assignment operator to be available. Some functions only need the type to be move constructable, copy constructable, etc. Since MSVC 2010 is more or less a C++11 compiler, it doesn't freak out if there's a missing assignment operator where other compilers might complain. If you change your function to:
 void f() { int z; std::vector<xxx> a; a.push_back(xxx(z)); std::vector<xxx> b; b = a; }
it should start complaining about the container assignment since container assignment requires that assignment of the contained type be allowed.

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With C++03, in order for a type to be usable in a standard library container, it needs to have both a copy constructor and an assignment operator. With C++11, the situation is a little more complex. Only some of the operations require an assignment operator to be available. Some functions only need the type to be move constructable, copy constructable, etc. Since MSVC 2010 is more or less a C++11 compiler, it doesn't freak out if there's a missing assignment operator where other compilers might complain.

Interesting, thanks SiCrane.

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# History of delta barrier in quantum mechanics

I'm interested in finding something out about the history of the problem of the delta potential barrier in quantum mechanics.

Which was the first study to propose this problem, and perhaps any particular motivation for it?

I am aware of some of the usual motivations for using it, but any examples that come to mind are appreciated.

• Ed, can you be more explicit about what to you want? References to published work? Points of view? The problem was exploited in the nineties as a mine of examples of "supersymmetry" in quantum mechanics, as well as "scale invariance".
– arivero
Aug 19 '15 at 15:28
• Thanks @arivero. I am interested in references in which a potential V(x)=A delta(x) with A a constant is first used inside a non-relativistic Schrödinger Equation. It could be a textbook or an article (or a set of either) where the delta barrier is first presented. I know that there is a text by S. Flügge that treats the delta potential in 1971. This is the oldest I've found, but it seems recent. I don't have access to this book so wonder what else could be out there. Does that clarify better? Thanks Aug 19 '15 at 15:49
• Ed, I see. Indeed the plain delta, and its cousin the dirac comb, is traditional textbook material. If you are about this, and not modern uses, I reluctantly agree that an historian has better opportunity to pinpoint the origin of the problem.
– arivero
Aug 19 '15 at 15:57
• note that the $\delta$ potential can appear also as solid state physics (the comb) and as mathematics of operators (theory of self-adjoint extensions).
– arivero
Aug 19 '15 at 15:59
• @arivero. Thanks. I guess the super-users have spoken infinite wisdom and feel this should be in this history exchange. In any case, tracking further down, I found a reference from I. R. Lapidus in American Journal of Physics 37, 930 (1969) explicitly solving the problem, and referring to it as standard. So once again, there is going to be something older. Aug 20 '15 at 9:04

After looking for a while this is what I have found, although it does not mean that perhaps earlier references do not exist.

In the review article by M. Belloni and R. W. Robinett,

"The infinite well and Dirac delta function potentials as pedagogical, mathematical and physical models in quantum mechanics", PHYSICS REPORTS-REVIEW SECTION OF PHYSICS LETTERS Volume: 540 Issue: 2 Pages: 25-122 Published: JUL 10 2014,

there is a section in which the delta potential is discussed at some length. They seem to find the first references to the use of the delta barrier in the classic article by R. de L. Kronig and W. G. Penney

"Quantum Mechanics of Electrons in Crystal Lattices" Proc. R. Soc. Lond. Ser. A Volume: 130 Pages: 499-513 Published: 1931.

There, the delta potential is introduced as the limit of zero width but constant area of square potential barriers.

Morse and Feshbach's treatise "Methods of Theoretical Physics" in v. 2 (published in 1953) discusses the single delta potential problem (as a well). Belloni and Robinett above make a note of this.

Intriguingly, I have never used references prior to the eighties. The oldest I used were:

R.G. Newton Inverse scattering by a local impurity in a periodic potential in one dimension, JMP 24 (1983)

P. Seba, The generalized point interaction in one dimension, Cz J Phys B 36:667 (1986)

M. Carreau; Four-parameter point-interaction in 1D-quantum systems, J Phys A: Math Gen, v 26 (1993)

There was a book on the topic in 1988, by Albeverio et al, that instead closing the topic, caused some subsequent work due to issues on the denomination of the different solutions. While the plain "delta barrier" was clear to everyone, some different barriers were expected to appear when using the $\delta '$ distribution, and it was controversial if the book had choosen the right solution.

Mathematically the question is about self-adjoint extensions of hermitian operators, and I'd expect it to appear in textbooks addressing this issue.


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# Quantum Computing for Everyone - Part I: Classical vs. Quantum Computing

, 16 Jun 2018
An introduction to quantum computing and the differences between classical and quantum computers.
 Part II: Quantum Gates

## Series Overview

Quantum computing can be a difficult subject to grasp initially. Not because the math, which is only slightly more complicated than high school algebra, but because some of the concepts are counterintuitive based on our daily experiences. Therefore, this article has been broken up into four parts starting with the difference between classical and quantum computers and ending with a demonstration of running an algorithm on a real quantum processor. If you already have some knowledge of quantum computing, you may be able to skip one or more of the parts. However, if you are new to this subject, I recommend going through the articles in order as each part builds on the concepts from the previous parts. Below is a quick overview of the articles in this series:

1. Part I - Introduction to Quantum Computing (this part): This part will go over the difference between classical computers and quantum computers, explaining how quantum computers are able to achieve the power that they have. The qubit, the fundamental unit of information (analogous to a bit), will also be introduced
2. Part II - Quantum Gates: This part will introduce the concept of quantum gates and how they operate on qubits to change their states. A description of the most common gates is also provided, along with some examples worked out in detail to show how to figure out the final state of a qubit being operated on by a quantum gate. As this section makes heavy use of matrix algebra, a review of matrices is provided at the start of the article (including matrix multiplication, transposing matrices, and getting the complex conjugate of a matrix).
3. Part III - Quantum Circuits and OpenQASM: The third part of this series will go over quantum circuits and how they are represented. Some example circuits will be constructed for specific problems. In addition, OpenQASM will be introduced, which is an intermediate programming language used to describe quantum circuits. Using OpenQASM, two programs will be written that will be simulated, and ran on a real quantum processor, in Part IV.
4. Part IV - IBM Quantum Experience: This part will introduce the IBM Quantum Experience platform. It will introduce the composer, which allows the building of quantum circuits graphically, as well as the OpenQASM editor. The OpenQASM programs created in Part III will also be simulated on this platform as well as ran on a real quantum processor.
5. Part V - IBM Quantum Experience API (planned): This part will introduce the IBM Quantum Experince RESTful API which allows an OpenQASM program to be submitted and either simulated, or run on the actual quantum processor, with the results returned in JSON format.
6. Part VI - Simulating Quantum Circuits (planned): This part will show how to simulate a quantum circuit by parsing, and interpreting, an OpenQASM program.

The fifth and sixth parts are currently in the planning stages. If/When they are completed, the links above, and at the end of Part IV will be updated. If there are any other parts that you think should be added, or would like to see, please feel free to leave a comment. I may not respond to all comments, but I definitely read all of them.

Quantum computing is a broad subject which fills large textbooks. In an attempt to condense the information, some details will be left out and there will be concepts presented without any proof for which I ask the reader to simply accept it. However, I will try to give enough information that for the problems that are not fully worked out, and just an answer provided, that the reader should be able to solve themselves (if they are so inclined). If there is anything that you felt was left out of this series and should be included, or something that needs further clarification, feel free to leave a comment on the articles page. I will do my best to accommodate everyone, but please remember that it is impossible to cover everything. When going through this series, if you ever feel lost or just can't understand why some of the things in quantum mechanics happen the way they do, just remember the following quote from famous physicist Richard Feynman:

I think I can safely say that nobody understands quantum mechanics.
-Richard Feynman, The Character of Physical Law (MIT Press: Cambridge, Massachusetts, 1995)

## Introduction

Ever since quantum computers were proposed in the 1980's (from people like Yuri Manin and Richard Feynman), slow but steady progress has been made. These advances were only accessible to quantum physicists in dedicated labs with state-of-the-art equipment. However, that all changed in May 2016 when IBM announced that it would be making a 5 qubit quantum computer available over the internet for members of the public to access and run experiments on. This is the first time that a quantum computer was made available to the general public and it has the potential to change the face of programming.

Before being able to design experiments to either run or simulate, we need to understand what makes a quantum computer different from a classical computer (such as the one you are likely reading this article on) and how quantum computers work. The math and physics behind quantum computing can be very complex and difficult to understand. However, this article will try to keep things as simple as possible and avoid as much math/physics as possible. Unfortunately, this cannot be avoided completely, so as much detail as possible with be given and some math examples worked in an attempt to make it as understandable as possible.

## Quantum vs. Classical Computers

All computers today (laptops, notebooks, even cell phones) are all classical computers. They all function by using transistors as switches and operating on electrical current. When the switch is open, and no electricity is flowing, it represents a 0 bit. If the switch is closed, and electricity is flowing, it represents a 1 bit. Everything a computer is able to do, is done by operating on one or more bits at a time very quickly. At any time while the computer is running, it can be in one of (2^n) possible states, where (n) is the total number of bits, ranging from (00..0) to (11..1).

Quantum computers also have bits, called qubits (pronounced CUE-bits). The difference between a qubit and a bit is that while a bit can be either 0 or 1, a qubit can be either 0, 1, or a mixture of 0 AND 1 at the same time. When the qubit is a mixture of a 0 and 1, this is known as a superposition. This allows the quantum computer to exists in exponentially many logical states simultaneously.

By Original reproduction: Jbw2, SVG version: WhiteTimberwolf - adapted from Nanocomputers and Swarm Intelligence (page 157) ISTE, Waldner, JB (2007), ISBN 2746215160. Reproduced with author's authorization, CC BY 3.0, Link

This image is a visual interpretation of a superposition state. Image each ball is an electron who's spin can either be up (repsented by an up arrow or (\vert 1 \rangle)) or down (repsented by a down arrow or (\vert 0 \rangle)). If the spin is down it represents a binary 0 and if the spin is up it represents a binary 1. Assuming the least significant bit is to the right, the first four electrons represent 5 in binary.

However, in the second case we can see that the electron in the least significant space is in a superposition of both up and down at the same time. Once a measurement is performed, this superposition will collapse and the electron's spin will either be up or down. If the spin is measured to be up, then the second set of electrons also represents 5 in binary. If the spin is measured to be down though, then the second case represents 4 in binary.

This is part of where quantum computers get their power. By being able to have objects in multiple states at the same time.

Along with exploiting superposition, quantum computers also take advantage of another unusal feature of quantum mechanics called entanglement. When individual qubits become entangled, they can no longer be described individually; they can only be described as part of the system as a whole. There is also a correlation between entangled particles even if they are separated by immense distances. Measuring one of the qubits immediately affects the other entangled particle.

Both of these features, superposition and entanglement, have no classically analog. They are both counter-intuitive (an object existing in more than one state at the same time and separated objects have an effect on each other with nothing connecting them), but they are what give a quantum computer its power.

## The Qubit

Now that we know qubits can exist in more than one state at the same time, we need some notation to represent this so we can work with them. Similar to how a classical bit can be represented by either a 0 or a 1, qubits can be represented as (|0\rangle) (known as the "ground state") or as (|1\rangle) (known as the "excited state"). This notation is known as Dirac's bra-ket notation. The (|0\rangle) (or (|1\rangle)) is known as a "ket" and represents a two-dimensional column vector over the complex numbers ((\mathbb{C}^2)):

\begin{aligned} |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{aligned}

The "bra" is written as (\langle0|) (or (\langle1|)) and represents a two-dimensional row vector over (\mathbb{C}^2):

\begin{aligned} \langle0| = \begin{pmatrix} 1 & 0 \end{pmatrix} \\ \langle1| = \begin{pmatrix} 0 & 1 \end{pmatrix} \end{aligned}

However, we also need to have a way to represent a qubit when it is in a superposition (i.e. a mixture) of both (|0\rangle) and (|1\rangle). We do this by written a linear combination of the two states:

\begin{aligned} a|0\rangle + b|1\rangle \end{aligned}

Where (a) and (b) can be positive, negative, fractional, or even a complex number. If we take the square of the absolute value of (a) and (b) (that is (\lvert a \rvert^2) and (\lvert b \rvert^2)) we get the probability of measuring either a (|0\rangle) or (|1\rangle), respectively. This also means that

\begin{aligned} \lvert a \rvert^2 + \lvert b \rvert^2 = 1 \end{aligned}

That is to say the probability of measuring a (|0\rangle) plus the probability of measuring a (|1\rangle) is (100\%) (i.e. we must end up measuring either a (|0\rangle) or (|1\rangle), those are the only possibilities).

So, if the qubit was in a superposition where it had a (50/50) chance of being (|0\rangle) or (|1\rangle) it would be represented as (\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle). In matrix form it would be:

\begin{aligned} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \end{aligned}

Why is the value of (a) and (b) both equal to (\frac{1}{\sqrt{2}}) and not (\frac{1}{2}) if there is a (50/50) chance of (|0\rangle) or (|1\rangle) being the result? Remember that (\lvert a \rvert^2 + \lvert b \rvert^2) must equal (1). If (a) and (b) were both (\frac{1}{2}), then:

\begin{aligned} \lvert \frac{1}{2} \rvert^2 + \lvert \frac{1}{2} \rvert^2 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} (\neq 1) \end{aligned}

However, with (a) and (b) both equal to (\frac{1}{\sqrt{2}}):

\begin{aligned} \lvert \frac{1}{\sqrt{2}} \rvert^2 + \lvert \frac{1}{\sqrt{2}} \rvert^2 = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1 \end{aligned}

## Visualizing a Qubit

Superpositions of qubits is not something that can be easily visualized as there is nothing similar in our daily lives. In order to make it easier, a tool has been developed called a Bloch Sphere. A Bloch Sphere is exactly what it sounds like; it is a sphere with a radius of 1 and any possible state that a qubit can be in corresponds with a point on the surface of the sphere.

A Bloch Sphere looks like this:

By Glosser.ca - Own work, CC BY-SA 3.0, Link

With (|\psi\rangle) representing the state of the qubit. With this representation, we can also describe the state of the qubit using angles and the following equation:

\begin{aligned} |\psi\rangle = \cos(\frac{\theta}{2})|0\rangle + e^{i\phi}\sin(\frac{\theta}{2})|1\rangle = \cos(\frac{\theta}{2})|0\rangle + (\cos \phi + i\sin\phi)\sin(\frac{\theta}{2})|1\rangle \end{aligned}

Where (0\leq\theta\leq\pi) and (0\leq\phi<2\pi). This notation, and representation, will help when we discuss quantum gates and how they change a qubit's state (i.e. changing the location of the qubit on the Bloch Sphere).

The (e^{i\phi}) value is known as a phase factor. The phase factor does not have any physical meaning in of itself. That means (|\psi\rangle) and (e^{i\phi}|\psi\rangle) are the same. However, when measuring two quantum states that are interacting, the differences in the phase factor can sometimes be measured and have important consequences.

## Conclusion

This wraps up Part I of this series. We have seen the difference between a classical and quantum computer, introduced the qubit, and how to represent/visualize a qubit for the purpose of quantum computing. In Part II we will look at how qubits can be manipulated and used to perform calculations.

## Acknowledgements

I wanted to thank Dominick Marciano Sr. for their review and feedback on this series. Also, a special thank you must be given to Sean Ewington, without who this article series would not have been possible.

## References

The following are a list of links for further information. Most, if not all, are contained within the article but are included here for easy reference.

## History

June 26, 2017: First version

 Part II: Quantum Gates

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I started programming around 10 years old. C/C++ was the first language I learned and then moved on to Assembly and finally VB/VB.NET which I have stuck with all these years. However, I have kept myself open to new languages and currently learning to work with Rust.

Besides computer programming, I have also learned microchip programming, starting with the Microchip PIC line and now using Raspberry Pi when it is convenient. I also have experience with HTML/ASP.NET/JavaScript/JQuery and built several websites for one of the largest environmental companies in NJ that I use to work for. These included their main website, and e-commerce site for the environmental products they sold, and a website for their training division which managed students and test grades. I wrote several large applications for them as well (e.g. a database front-end, a PDA application for data entry, and a program for analyzing data from an X-ray fluorescence machine and writing a report with recommendations based on the data).

Currently I work for a MSP that serves a good portion of the charter schools in NYC as well as a few small business and NPO's in NY, CT, and MA. Besides providing troubleshooting help, I develop applications which help streamline our work or our clients.

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# How to plot two columns of single DataFrame on Y axis

I have two data frames (Action, Comedy). Action contains two columns (year, rating) ratings columns contains average rating with respect to year. The Comedy data frame contains the same two columns with different mean values.

I merged both data frames into a total_year data frame.

The output of total_year:

Now, I would like to plot total_year on a line graph in which the X axis should contain the year column and the Y axis should contain both the action and the comedy columns.

I can plot only 1 column at a time on Y axis using following code:

total_year[-15:].plot(x='year', y='action' ,figsize=(10,5), grid=True)

How can I plot both columns on the Y axis?

I took this photo from google just to let you know guys I want to draw graph in this way:

• i got answer of this Question here Dec 12 '17 at 14:58

Feeding your column names into the y values argument as a list works for me like so:

total_year[-15:].plot(x='year', y=['action', 'comedy'], figsize=(10,5), grid=True)

Using something like the answer at this link is better and gives you way more control over the labels and whatnot: adding lines with plt.plot()

Using Seaborn:

Need to melt down columns so that they can be treated as hue in plotting

    import seaborn as sns
df_melted = df.melt("year",var_name="Action&Comedy",value_name="Rating")

    sns.relplot(data=df_melt, x="year", y="Rating", hue="Action&Comedy",kind="line", height=4, aspect=.7)

• Is there a way that doesn't involve melting? In my case, I have one column for the hue (which algorithm I'm running), one column for the score (which I'd like plotted on the Y axis) and a third column for the optimal possible score (which I'd like added as another line). How could I do this? Dec 29 '21 at 18:47


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gmat prep 2 question : Integrated Reasoning (IR)
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# gmat prep 2 question

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12 Jun 2012, 12:09
Loan x has a principal of 10000x and a yearly simple interest rate of 4%. loan Y has a principal of $10000y and a yearly simple interest of 8%. Loans x and y will be consolidated to form loan Z with a principal of (10000x + 10000y) and a yearly simple interest rate of r%, where r= 4x + 8y / (x+y). In the table, select a value for x and a value for y corresponding to a yearly simple interest rate of 5% for the consolidated load. make only two selections, value x y 21 32 51 64 81 96 Dont know how to solve this question without T and E . Can someone explain the shortcut Intern Affiliations: The Princeton Review Joined: 14 Feb 2012 Posts: 8 Location: United States Followers: 0 Kudos [?]: 24 [6] , given: 0 Re: gmat prep 2 question [#permalink] ### Show Tags 12 Jun 2012, 15:45 6 This post received KUDOS 1 This post was BOOKMARKED What are the correct answers? It seems to me that the answers are x=96 and y=32. Here's the quick math: r is supposed to equal 5. You're given the formula for calculating r. r = (4x + 8y) / (x + y). Is that the correct formula? The original looks like the x+y is only under the 8y, not under the entire 4x + 8y. Anyway, let's just assume that the formula I wrote is correct. Just substitute 5 in for r and solve for x in terms of y. 5 = (4x + 8y) / (x + y). Multiply both sides by (x+y) 5x + 5y = 4x + 8y. Get x's on one side and y's on the other. x = 3y. Look for answer choices in which one is triple another. Only 32 and 96 work. Senior Manager Affiliations: UWC Joined: 09 May 2012 Posts: 399 Location: Canada GMAT 1: 620 Q42 V33 GMAT 2: 680 Q44 V38 GPA: 3.43 WE: Engineering (Entertainment and Sports) Followers: 28 Kudos [?]: 1112 [1] , given: 100 Re: gmat prep 2 question [#permalink] ### Show Tags 13 Jun 2012, 01:41 1 This post received KUDOS 1 This post was BOOKMARKED Here's my take: 4x+8y/(x+y) = 5 4x+8y = 5x + 5y x = 3y Choices: 21 32 51 64 81 96 Check lowest 3 choices: 21*3 = 63 (not in lower three choices; eliminated) 32*3 = 96 (present in lower three; RIGHT ANSWER) Thus: x = 96 y = 32 (oops: error corrected!) Last edited by macjas on 13 Jun 2012, 16:54, edited 2 times in total. Intern Affiliations: The Princeton Review Joined: 14 Feb 2012 Posts: 8 Location: United States Followers: 0 Kudos [?]: 24 [0], given: 0 Re: gmat prep 2 question [#permalink] ### Show Tags 13 Jun 2012, 13:13 macjas wrote: Here's my take: 4x+8y/(x+y) = 5 4x+8y = 5x + 5y x = 3y Choices: 21 32 51 64 81 96 Check lowest 3 choices: 21*3 = 63 (not in lower three choices; eliminated) 32*3 = 96 (present in lower three; RIGHT ANSWER) Thus: x = 32 y = 96 Other way around, right? If x= 3y, x=96 and y =32. Intern Joined: 05 Dec 2011 Posts: 47 Followers: 0 Kudos [?]: 1 [0], given: 18 Re: gmat prep 2 question [#permalink] ### Show Tags 01 Jul 2012, 23:47 Hey macjas then why this line is provided in the question "Loan x has a principal of 10000x and a yearly simple interest rate of 4%. loan Y has a principal of$10000y and a yearly simple interest of 8% " is of no use . Then why this line has been provided in the question ??just to confuse us !!

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01 Jul 2012, 23:51
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debabrata44 wrote:
Hey macjas then why this line is provided in the question "Loan x has a principal of 10000x and a yearly simple interest rate of 4%. loan Y has a principal of \$10000y and a yearly simple interest of 8% " is of no use . Then why this line has been provided in the question ??just to confuse us !!

Exactly! It's of no use! This is one of the fundamental ways in which IR differs from the Q/V section of the GMAT. IR is supposed to reflect 'real world' data; you need to figure out whether a data piece of data is relevant or not.
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02 Jul 2012, 01:12
thanks a lot macjas !!
Re: gmat prep 2 question   [#permalink] 02 Jul 2012, 01:12
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text:
# Pass key value pair as value with pgfkeys

I want to pass a key=value pair as the value of another key=value pair. The problem is, that the equal sign gets interpreted by pgfkeys and thus I need to encapsulate it using curly brackets, which in turn prevent pgfkeys to interpret the equal sign (surprise).

The following minimal example doesn't make sense, but it illustrates what I'm trying to do. In reality I try to pass on the key=value pair to another command, which takes key=value pairs.

\documentclass[a4paper]{scrartcl}

\usepackage[utf8]{inputenc}
\usepackage{pgfkeys}

\begin{document}

\newcommand{\mycmd}[1]{%
\pgfkeys{/mykey/.initial=hello}%
\pgfkeys{/mykey2/.initial=world}%
\pgfkeys{#1}%
\pgfkeys{\pgfkeysvalueof{/mykey2}}%
\pgfkeysvalueof{/mykey}%
\pgfkeysvalueof{/mykey2}%
}

\mycmd{mykey2={mykey=test}}.

\end{document}

The error I get is of course: "I do not know the key '\mykey=test' ...". How can I solve this?

Edit:

Thanks to percusse I now use a style for what I'm doing. In the example below mycmdA gives the error 'I do not know the key '/fill/', while mycmdB works. Does anybody know why?

\documentclass[a4paper]{scrartcl}

\usepackage[utf8]{inputenc}
\usepackage{pgfkeys}

\begin{document}

This is a test

\newcommand{\mycmdA}[1]{%
\pgfkeys{/mystyle/.style={##1}}%
\pgfkeys{#1}%
}

\newcommand{\mycmdB}[1]{%
\pgfkeys{/mystyle/.style={}}%
\pgfkeys{/mystyle/.code=\pgfkeys{/style/.style={##1}}}
\pgfkeys{#1}%
}

\mycmdA{mystyle={fill=red}}
\mycmdB{mystyle={fill=red}}

\end{document}

• I think that the first line in \mycmdA is equivalent to\pgfkeys{/fill=red}, which will fail because the key /fill has not been defined. I don't understand what you are trying to achieve but perhaps you want \pgfkeys{/mystyle/.code=\pgfkeys{/mystyle/##1}} or \pgfkeys{/mystyle/.cd, /mystyle/.style={##1}} ? The later will still fail because /mystyle/fill has not been set but at least the error is in the right pgfkeys "directory" – Andrew May 4 '17 at 11:48

You don't need to reinvent the wheel. Use pgfkeys machinery for /.style and /.code handlers just as in the TikZ pictures. If mykey2 is going to be a key setter then it wouldn't be fun to also have a value. That part of the key mechanism is fairly nontrivial. But this will print test (expansion is not necessary here but just in case you need it)

\pgfkeys{mykey/.initial=hello}%
%\pgfkeys{mykey2/.initial=world}%
\pgfkeys{mykey2/.code={\expandafter\pgfkeys\expandafter{#1}}}% Expand once to remove braces
\pgfkeys{mykey2={mykey=test}}%
\pgfkeysvalueof{/mykey}%

you can instead make mykey2 a /.style and pass options to it as a key holder. Then whenever you execute mykey2 it will execute all of the stored key values (/.append style is needed for adding without resetting).

• Thanks, but how would I use /.style. Sorry, I'm very new to pgfkeys. – Thomas Leitz May 3 '17 at 11:36
• More specifically: What does "execute mykey2" mean in this context? I also read this in the pgf documentation, but I don't know what it means. – Thomas Leitz May 3 '17 at 12:11


text:
# Analytic function on complement of closed unit disk onto open unit disk

Does there exist an analytic function whose domain is the complement of the closed unit disk and whose range is the open unit disk?

I am studying John B. Conway's book Functions of One Complex Variable for my own amusement. One of the exercises following Chapter III, Section 3, asks whether the open unit disk can be mapped conformally onto the punctured open unit disk. I solved that one (it's true), and then asked myself whether the converse were true as well. I immediately realized that that question is equivalent to whether the complement of the closed unit disk can be mapped conformally onto the open unit disk. Then I realized that I didn't even know the answer to the simpler question asked above.

Take $$g(z)$$ any Blaschke product of degree two that is not $$cz^2$$; then $$g$$ maps the punctured unit disc holomorphically onto the unit disc since $$g(0)=g(w)$$ for some $$w \ne 0$$. Then $$f(z)=g(1/z)$$ will map the exterior of the closed unit disc holomorphically onto the open unit disc.
For example $$g(z)=\frac{(1-2z)^2}{(2-z)^2}$$ is one such Blaschke product, giving
$$f(z)=\frac{(z-2)^2}{(2z-1)^2}$$
It is considerably more difficult (though doable) to find a conformal such $$f$$ (in other words with the additional condition that $$f'(z) \ne 0, |z|>1$$)


text:
# How can I normalize this chemical analysis to a mineral with 3 cations?

$\ce{FeO}$ = 70 wt%

$\ce{TiO2}$ = 30 wt%

The molecular weight given of $\ce{FeO}$ = 72, and the mw of $\ce{TiO2}$ = 80.

The first step I did was divide each wt% by their molecular weights to get a molecular proportion:

$$\ce{FeO}: {70 \over 72} = 0.972$$

$$\ce{TiO_2}: {30 \over 80} = 0.375$$

Then, I multiplied each proportion by the number of oxygens in each chemical formula

$$0.972 \times 1 \ \rm{oxygen} = 0.972$$ $$0.375 \times 2 \ \rm{oxygens} = 0.75$$

The next step is where I get really confused. I think I'm supposed to add the two numbers above together, and divide that number into a number that is supposedly the cations assumed of the mineral? (As in 5 for plagioclase, 4 for pyroxene, 3 for olivine, etc) But I do not understand how I'm supposed to gather which number to use based on only the data given. If anyone can point me in the right direction it would be greatly appreciated.

The first step I did was divide each wt% by their molecular weights to get a molecular proportion:

Actually that's not what you did. What you did is how much moles of each oxide you have assuming 100 grams of mineral. But that's less important now.

You are mostly correct in your steps. Eventually you have 0.972 Fe, 0.375 Ti, and 0.972 + 2×0.375 = 1.722 O. The (non normalised) mineral formula is then: Fe0.972Ti0.375O1.722.

You can easily check that it's correct using charge balance:

0.972×2 + 0.375×4 - 1.722×2 = 0

The question now is how to normalise it. Since you were asking about 3 cations, you need to multiply everything by 3/(0.972 + 0.375), giving you:

Fe2.165Ti0.835O3.835

But that's also ugly. Usually mineral formulas are normalised to round anion numbers, so let's say you multiply it all by 5/1.722 you get:

Fe2.822Ti1.089O5

which is slightly prettier.

Without having any more information on the mineral as to what's it supposed to be, you can only guess. This is what we do in EMPA analyses - we know what mineral it is and we calculate the formula according to that. If we don't, we guess and try to figure out which one it is, and then calculate according to that.

I don't know much about the question, but it could be that not all iron in the mineral is Fe2+. Some of it may be Fe3+, but analysing both is not an easy task so most iron in minerals is reported as either all FeO or all Fe2O3 regardless of what oxidation state it is actually in. So it's possible that you have (FeO + Fe2O3) = 70 wt%. If you play enough with that, you might end up with a better looking formula. Again, I don't know the question and I don't know if that is what you're supposed to do in the question.

• Ok thank you! The question in my book literally was what I wrote, they didn't give anymore info. I just should've known that normalize to three cations meant use 3/(0.972 + 0.375) as the normalization factor. – Jay Apr 20 '15 at 21:16


text:
Question
Figure 21.70 In this circuit, assume the currents through $R_1$, $R_2$ and $R_3$ are $I_1$, $I_2$ and $I_3$ respectively and all are flowing in the clockwise direction.
1. Find the equation obtained by applying Kirchhoff’s junction rule at point A.
2. Find the equations obtained by applying Kirchhoff’s loop rule in the upper and lower loops.
3. Assume $R_1 = R_2 = 6 \Omega$, $R_3 = 12 \Omega$, $r_1 = r_2 = 0 \Omega$, $\xi_1 = 6 \textrm{ V}$ and $\xi_2 = 4 \textrm{ V}$. Calculate $I_1$, $I_2$ and $I_3$.
4. For the situation in which $\xi_2$ is replaced by a closed switch, repeat parts (a) and (b). Using the values for $R_1$, $R_2$, $R_3$, $r_1$ and $\xi_1$ from part (c) calculate the currents through the three resistors.
5. For the circuit in part (d) calculate the output power of the voltage source and across all the resistors. Examine if energy is conserved in the circuit.
6. A student implemented the circuit of part (d) in the lab and measured the current though one of the resistors as 0.19 A. According to the results calculated in part (d) identify the resistor(s). Justify any difference in measured and calculated value.
Question Image
1. $I_1 + I_3 = I_2$
2. see video
3. $I_1 = \dfrac{8}{15}$, $I_2 = \dfrac{7}{15}$, $I_3 = \dfrac{-1}{15}$
4. $I_1 = \dfrac{2}{5}$, $I_2 = \dfrac{3}{5}$, $I_3 = \dfrac{1}{5}$
5. $P_{total} = \dfrac{18}{5} \textrm{ W}$, $P_{R_1} = \dfrac{24}{25} \textrm{ W}$, $P_{R_2} = \dfrac{54}{25} \textrm{ W}$, $P_{R_3} = \dfrac{12}{25} \textrm{ W}$
Yes, energy is conserved.
6. The resistor is $R_3$.
The ammeter introduces additional resistance, $r$, in series with $R_3$, thereby reducing the current.
Solution Video


text:
# Order of shells formed in a massive star

In addition to the question with the picture here, why do shells don't form the same order as the alpha process?

image credit: Gemini Observatory/NSF/C.Aspin