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## 物理代写|粒子物理代写Particle Physics代考|C P T symmetry

The Dirac equation is by construction invariant under space inversions, or parity transformations $(\mathcal{P})$. We just showed the existence of a second symmetry transformation, that of charge conjugation $(\mathcal{C})$. We wish to investigate here the consequences of time reversal $\mathcal{T}$, with $(\mathcal{T} x)^\mu=(-t, \boldsymbol{x})$. We can easily check that $\gamma^0 \gamma^5 \Psi^C(\mathcal{T} x)$, which is an anti-unitary operation, also satisfies the Dirac equation. It is interesting to consider the product of these three symmetries, namely the product of ${ }^5$
$$\begin{gathered} \mathcal{P}: \Psi(x) \rightarrow \mathrm{i} \gamma^0 \Psi(\mathcal{P} x) \ \mathcal{C}: \Psi(x) \rightarrow \Psi^C(x)=C \gamma^0 \Psi^(x) \end{gathered}$$ and $$\mathcal{T}: \Psi(x) \rightarrow \gamma^0 \gamma^5 \Psi^C(\mathcal{T} x)$$ where $(\mathcal{P} x)^\mu=(t,-\boldsymbol{x})$ denotes space inversion. We can verify that the product $C P T$ of the three operators taken in any order is an invariance of the theory. Indeed, start with \begin{aligned} P C T \Psi(x) & =\mathrm{i} \gamma^0\left(C \gamma^0\left[\gamma^0 \gamma^5 C \gamma^0 \Psi^(\mathcal{P} \mathcal{T} x)\right]^*\right) \ & =\mathrm{i} \gamma^0\left(C \gamma^0\left(\gamma^0 \gamma^5 C^{-1} \gamma^0 \Psi(-x)\right)\right) \ & =\mathrm{i} \gamma^5 \Psi(-x) \end{aligned}
and ask the following question: Given a wave function $\Psi(x)$, which satisfies the Dirac equation in an external electromagnetic field $A_\mu(x)$, can we find the equation satisfied by its $C P T$-transformed wave function $\mathrm{i} \gamma^5 \Psi(-x)$ ? The answer is simple:
\begin{aligned} 0 & =\left[\mathrm{i} \not \partial_x-e \not A(x)-m\right] \Psi(x) \ & =\mathrm{i} \gamma^5\left[\mathrm{i} \not \partial_x-e \not A(x)-m\right] \Psi(x) \ & =\left[-\mathrm{i} \not \partial_x+e \not A(x)-m\right] \mathrm{i} \gamma^5 \Psi(x) \ & =\left[\mathrm{i} \not \partial_x+e \not A(-x)-m\right] C P T \Psi(x) \end{aligned}
where, in the last step, we have performed the change of variable $x^\mu \rightarrow-x^\mu$.

## 物理代写|粒子物理代写Particle Physics代考|Chirality

In section 7.3.6 we introduced the so-called “standard” representation, which separates the components of a Dirac spinor that vanish in the $c \rightarrow \infty$ non-relativistic limit. At the other end, at the ultra-relativistic limit where the mass is negligible, we expect, at least in the absence of any external potential, the Dirac equation to split into a pair of Weyl equations. We shall introduce a formal way to demonstrate this fact.

We define two orthogonal projectors $P_L$ and $P_R, L$ and $R$ standing for “left” and “right” respectively, by
$$P_L=\frac{1+\gamma^5}{2}, \quad P_R=\frac{1-\gamma^5}{2}$$
They are Hermitian operators and satisfy the projection, orthogonality and completeness relations: $P_{L, R}^2=P_{L, R}, P_L P_R=P_R P_L=0$ and $P_L+P_R=1$. With the help of these projectors we define the “left” and “right” components of a general Dirac spinor $\Psi(x)$ by
$$\Psi_{L, R}(x)=P_{L, R} \Psi(x), \quad \Psi(x)=\Psi_L(x)+\Psi_R(x)$$
In the Weyl basis we used in equation (7.38), in which $\gamma^5$ is diagonal, $P_L$ and $P_R$ project onto the $\xi$ and $\eta$ spinors, respectively. Using $\Psi_L$ and $\Psi_R$, the free Dirac action (7.61) becomes
$$S=\int \mathrm{d}^4 x\left[\bar{\Psi}_L \mathrm{i} \not \partial \Psi_L+\bar{\Psi}_R \mathrm{i} \not \partial \Psi_R-m\left(\bar{\Psi}_L \Psi_R+\bar{\Psi}_R \Psi_L\right)\right]$$

## 物理代写|粒子物理代写Particle Physics代考|C P T symmetry

$$\left.\mathcal{P}: \Psi(x) \rightarrow \mathrm{i} \gamma^0 \Psi(\mathcal{P} x) \mathcal{C}: \Psi(x) \rightarrow \Psi^C(x)=C \gamma^0 \Psi^{(} x\right)$$

$$\mathcal{T}: \Psi(x) \rightarrow \gamma^0 \gamma^5 \Psi^C(\mathcal{T} x)$$

$$\left.P C T \Psi(x)=\mathrm{i} \gamma^0\left(C \gamma^0\left[\gamma^0 \gamma^5 C \gamma^0 \Psi^{(\mathcal{P} T} x\right)\right]^*\right) \quad=\mathrm{i} \gamma^0\left(C \gamma^0\left(\gamma^0 \gamma^5 C^{-1} \gamma^0 \Psi(-x)\right)\right)=\mathrm{i} \gamma^5 \Psi(-x)$$

$$0=\left[\mathrm{i}, \partial_x-e A(x)-m\right] \Psi(x) \quad=\mathrm{i} \gamma^5\left[\mathrm{i} \partial \partial_x-e A(x)-m\right] \Psi(x)=\left[-\mathrm{i} \partial \partial_x+e A(x)-m\right] \mathrm{i} \gamma^5 \Psi(x) \quad=\left[\mathrm{i}, \partial_x+e A(-x)-m\right] C P T \Psi(x)$$

## 物理代写|粒子物理代写Particle Physics代考|Chirality

$$P_L=\frac{1+\gamma^5}{2}, \quad P_R=\frac{1-\gamma^5}{2}$$

$$\Psi_{L, R}(x)=P_{L, R} \Psi(x), \quad \Psi(x)=\Psi_L(x)+\Psi_R(x)$$

$$S=\int \mathrm{d}^4 x\left[\bar{\Psi}_L \mathrm{i} \not \partial \Psi_L+\bar{\Psi}_R \mathrm{i} \not \partial \Psi_R-m\left(\bar{\Psi}_L \Psi_R+\bar{\Psi}_R \Psi_L\right)\right]$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 物理代写|粒子物理代写Particle Physics代考|The Klein-Gordon Equation

We start with the simplest Lorentz covariant wave equation, namely the Klein-Gordon equation. From our study of the Lorentz group, we expect it a priori to be relevant to the description of the quantum mechanics of a spinless particle. Since the wave function in quantum mechanics is complex valued, it is natural to start with the Klein-Gordon equation (7.1) for a complex function $\Phi(x)$
$$\left(\frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\Delta+m^2\right) \Phi(t, \boldsymbol{x})=0$$
It is straightforward to verify that in the non relativistic limit, $c \rightarrow+\infty$, the KleinGordon equation reduces to the Schrödinger equation. To show this it is enough to extract the mass from the energy dependence of $\Phi(x)$. This is obtained by parametrising $\Phi$ in the following way:
$$\Phi(t, \boldsymbol{x})=\exp \left(-\frac{\mathrm{i} m c^2}{\hbar} t\right) \Psi(t, \boldsymbol{x})$$

Then, the Klein-Gordon equation becomes
$$\left(\frac{1}{c^2}\left(\frac{\partial}{\partial t}\right)^2-\frac{\mathrm{i}}{\hbar} 2 m \frac{\partial}{\partial t}-\Delta\right) \Psi(x)=0 .$$
In the non-relativistic approximation, $c \rightarrow+\infty$, the first term can be neglected, and we recover the free Schrödinger equation
$$\mathrm{i} \hbar \frac{\partial}{\partial t} \Psi(x)=-\frac{\hbar^2}{2 m} \Delta \Psi(x)$$
So, at first sight, the Klein-Gordon equation seems to have the right properties to give the relativistic generalisation of the Schrödinger equation. We suspect though that this could not be right because, if it were that simple, Schrödinger would have written directly this more general relativistic equation. Let us show that, indeed, interpreting $\Phi(x)$ as the particle wave function, does lead to physical inconsistencies.

## 物理代写|粒子物理代写Particle Physics代考|The Dirac Equation

Dirac derived his equation in 1928 as a relativistic generalisation of Schrödinger’s equation for the electron. Schrödinger’s equation is first order in time derivatives, but second order in derivatives with respect to the spatial coordinates. It does have a conserved probability current with positive definite density. We just saw that this property is lost in the obvious relativistic extension, the Klein-Gordon equation, which has second order derivatives with respect to both time and space. It is this difference, which motivated Dirac to look for an equation with first order derivatives. For a single scalar function there is no such non-trivial Lorentz covariant first order differential equation, so Dirac assumed a multi-component wave function and looked for a matrix equation. In doing so, he discovered the spinorial representations of the Lorentz group. In the notation we used in section 7.3 the equation reads
$$(\mathrm{i} \not \partial-m) \Psi(x)=0$$
As we have shown already, this equation is Lorentz covariant, provided the $\gamma$ matrices satisfy the Clifford algebra relation (7.41). It is obtained from the Lagrangian density given in equation (7.60).

## 物理代写|粒子物理代写Particle Physics代考|The Klein-Gordon Equation

$$\left(\frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\Delta+m^2\right) \Phi(t, \boldsymbol{x})=0$$

$$\Phi(t, \boldsymbol{x})=\exp \left(-\frac{\mathrm{i} m c^2}{\hbar} t\right) \Psi(t, \boldsymbol{x})$$

$$\left(\frac{1}{c^2}\left(\frac{\partial}{\partial t}\right)^2-\frac{\mathrm{i}}{\hbar} 2 m \frac{\partial}{\partial t}-\Delta\right) \Psi(x)=0 .$$

$$\mathrm{i} \hbar \frac{\partial}{\partial t} \Psi(x)=-\frac{\hbar^2}{2 m} \Delta \Psi(x)$$

## 物理代写|粒子物理代写Particle Physics代考|The Dirac Equation

$$(\mathrm{i} \partial \partial-m) \Psi(x)=0$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Particle Physics, 物理代写, 粒子物理

## avatest™帮您通过考试

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## 物理代写|粒子物理代写Particle Physics代考|Lagrangian, Hamiltonian and Green functions

The Lagrangian formulation of the Dirac equation presents some subtleties, because the latter is a first order differential equation. Recall the results we obtained for the case of a complex scalar field. The Lagrangian density (7.27) depends on the fields $\phi$ and $\phi^*$ as well as their first derivatives. This is consistent with the fact that the equations of motion are second order differential equations and we must assign initial values for the fields and their first derivatives. When we vary with respect to the field $\phi$ in order to obtain the Euler-Lagrange equations, we must also take into account the variation of $\partial \phi$. For the Dirac equation, however, which is a first order equation, we can vary only with respect to $\psi$ and $\bar{\psi}$, not to their derivatives. This implies in turn that the standard way to obtain the Hamiltonian through a Legendre transformation should be reformulated. In this section we want to present the rules which will make it possible for us to use the Lagrangian and Hamiltonian formalisms, without attempting a mathematically rigorous justification.

We choose the Lagrangian density corresponding to the Dirac equation in the form
$$\mathcal{L}{\mathrm{D}}=\frac{\mathrm{i}}{2}\left(\bar{\psi} \gamma^\mu \partial\mu \psi-\partial_\mu \bar{\psi} \gamma^\mu \psi\right)-m \bar{\psi} \psi$$
This is justified by the fact that we obtain the Dirac equations for $\psi$ and $\bar{\psi}$ as a consequence of the stationarity requirement of the action $S=\int \mathcal{L}{\mathrm{D}} d^4 x$ under independent variations of $\bar{\psi}$ and $\psi$, respectively. Because of the linear dependence of $\mathcal{L}{\mathrm{D}}$ on $\psi$ or $\bar{\psi}$, the action has neither a minimum nor a maximum. Thus, the overall sign of the action can be chosen at will. Provided that the field vanishes at infinity, we can rewrite the action as
$$S=\int \mathrm{d}^4 x(\mathrm{i} \bar{\psi} \not \partial \psi-m \bar{\psi} \psi)$$

## 物理代写|粒子物理代写Particle Physics代考|The plane wave solutions

We have shown that the solutions of the Dirac equation are solutions of the KleinGordon equation $\left(\square+m^2\right) \psi=0$ as well. Consequently, a plane wave solution $\psi(x) \sim$ $\exp (-\mathrm{i} k \cdot x)$ of the Dirac equation has to satisfy the condition $k^2=k_0^2-\boldsymbol{k}^2=m^2$, which means that its energy $k_0$ can have either sign. We shall be interested in the full set of plane wave solutions of both positive and negative energies, since only their union forms a basis. We fix the zero component of the wave vector $k^\mu$ to $k_0=+\sqrt{\boldsymbol{k}^2+m^2} \equiv E_k$. Then, we denote the positive energy solution of wave vector $\boldsymbol{k}$ by
$$\psi^{(+)}(x)=\mathrm{e}^{-\mathrm{i} k \cdot x} u(\boldsymbol{k})$$
and the negative energy one by
$$\psi^{(-)}(x)=\mathrm{e}^{\mathrm{i} k \cdot x} v(\boldsymbol{k})$$
where $u$ and $v$ are four-component spinors, whose components are labelled $u_r$ and $v_r$, $r=\mathbf{1}, \ldots, 4$
From $\left(\mathrm{i} \gamma^\mu \partial_\mu-m\right) \psi^{( \pm)}(x)=0$, we obtain
$$(\not k-m) u(\boldsymbol{k})=0 \text { and }(\not k+m) v(\boldsymbol{k})=0$$
Let us choose the $\gamma$ matrices in the standard representation. For $k=0$, the equations (7.72) simplify to
$$\left(\gamma^0-1\right) u(\mathbf{0})=0 \quad \text { and }\left(\gamma^0+1\right) v(\mathbf{0})=0$$
and lead to $u_3=u_4=v_1=v_2=0$. A possible basis of the solutions is
$$\hat{u}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 1 \ 0 \ 0 \ 0 \end{array}\right), \quad \hat{u}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 1 \ 0 \ 0 \end{array}\right), \quad \hat{v}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 1 \ 0 \end{array}\right), \quad \hat{v}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 0 \ 1 \end{array}\right)$$

## 物理代写|粒子物理代写Particle Physics代考|Lagrangian, Hamiltonian and Green functions

$$\mathcal{L D}=\frac{\mathrm{i}}{2}\left(\bar{\psi} \gamma^\mu \partial \mu \psi-\partial_\mu \bar{\psi} \gamma^\mu \psi\right)-m \bar{\psi} \psi$$

$$S=\int \mathrm{d}^4 x(\mathrm{i} \bar{\psi} \partial \partial-m \bar{\psi} \psi)$$

## 物理代写|粒子物理代写Particle Physics代考|The plane wave solutions

$$\psi^{(+)}(x)=\mathrm{e}^{-\mathrm{i} k \cdot x} u(\boldsymbol{k})$$

$$\psi^{(-)}(x)=\mathrm{e}^{\mathrm{i} k \cdot x} v(\boldsymbol{k})$$

$$(k-m) u(\boldsymbol{k})=0 \text { and }(k+m) v(\boldsymbol{k})=0$$

$$\left(\gamma^0-1\right) u(\mathbf{0})=0 \quad \text { and }\left(\gamma^0+1\right) v(\mathbf{0})=0$$

$$\hat{u}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 1 \ 0 \ 0 \ 0 \end{array}\right), \quad \hat{u}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 1 \ 0 \ 0 \end{array}\right), \quad \hat{v}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 1 \ 0 \end{array}\right), \quad \hat{v}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 0 \ 1 \end{array}\right)$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。