Posted on Categories:Quantum field theory, 物理代写, 量子场论

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|PHYS4124

avatest.orgt™量子场论Quantum field theory代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.org™， 最高质量的量子场论Quantum field theory作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此量子场论Quantum field theory作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Group theory

A group is a set of elements $\left{g_i\right}$ and a rule $g_i \times g_j=g_k$ which tells how each pair of elements is multiplied to get a third. The rule defines the group, independent of any particular way to write the group elements down as matrices. More precisely, the mathematical definition requires the rule to be associative $\left(g_i \times g_j\right) \times g_k=g_i \times\left(g_j \times g_k\right)$, there to be an identity element for which $\mathbb{1} \times g_i=g_i \times \mathbb{1}=g_i$, and for the group elements to have inverses, $g_i^{-1} \times g_i=1$. A representation is a particular embedding of these $g_i$ into operators that act on a vector space. For finite-dimensional representations, this means an embedding of the $g_i$ into matrices. Often we talk about the vectors on which the matrices act as being the representation, but technically the matrix embedding is the representation. Any group has the trivial representation $r: g_i \rightarrow 1$. A representation in which each group element gets its own matrix is called a faithful representation.

Recall that the Lorentz group is the set of rotations and boosts that preserve the Minkowski metric: $\Lambda^T g \Lambda=g$. The $\Lambda$ matrices in this equation are in the 4-vector representation under which
$$X_\mu \rightarrow \Lambda_{\mu \nu} X_\nu$$
Examples of Lorentz transformations are rotations around the $x, y$ or $z$ axes:
$$\left(\begin{array}{llll} 1 & & & \ & 1 & & \ & & \cos \theta_x & \sin \theta_x \ & & -\sin \theta_x & \cos \theta_x \end{array}\right),\left(\begin{array}{cccc} 1 & & & \ & \cos \theta_y & & -\sin \theta_y \ & \sin \theta_y & & \cos \theta_y \end{array}\right),\left(\begin{array}{cccc} 1 & & & \ & \cos \theta_z & \sin \theta_z & \ & -\sin \theta_z & \cos \theta_z & \ & & & 1 \end{array}\right)$$
and boosts in the $x, y$ or $z$ directions:
$$\left(\begin{array}{ccccc} \cosh \beta_x & \sinh \beta_x & & \ \sinh \beta_x & \cosh \beta_x & & \ & & 1 & \ & & & 1 \end{array}\right),\left(\begin{array}{llll} \cosh \beta_y & & \sinh \beta_y & \ & 1 & & \ \sinh \beta_y & & \cosh \beta_y & \ & & & 1 \end{array}\right),\left(\begin{array}{llll} \cosh \beta_z & & & \sinh \beta_z \ & 1 & & \ & & 1 & \ \sinh \beta_z & & & \cosh \beta_z \end{array}\right) .$$
These matrices give an embedding of elements of the Lorentz group into a set of matrices. That is, they describe one particular representation of the Lorentz group (the 4-vector representation). We would now like to find all the representations.

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|General representations of the Lorentz group

The irreducible representations of the Lorentz group can be constructed from irreducible representations of $\mathrm{SU}(2)$. To see how this works, we start with the rotation generators $J_i$ and the boost generators $K_j$. You can think of them as the matrices in Eq. (10.14), which is a particular representation, but the algebraic properties in Eqs. (10.17) to (10.19) are representation independent.
Now take the linear combinations
$$J_i^{+} \equiv \frac{1}{2}\left(J_i+i K_i\right), \quad J_i^{-} \equiv \frac{1}{2}\left(J_i-i K_i\right)$$
which satisfy
\begin{aligned} & {\left[J_i^{+}, J_j^{+}\right]=i \epsilon_{i j k} J_k^{+},} \ & {\left[J_i^{-}, J_j^{-}\right]=i \epsilon_{i j k} J_k^{-},} \ & {\left[J_i^{+}, J_j^{-}\right]=0 .} \end{aligned}
These commutation relations indicate that the Lie algebra for the Lorentz group has two commuting subalgebras. The algebra generated by $J_i^{+}$(or $J_i^{-}$) is the 3D rotation algebra, which has multiple names, $\operatorname{so}(3)=\operatorname{sl}(2, \mathbb{R})=\operatorname{so}(1,1)=\operatorname{su}(2)$, due to multiple Lie groups having the same algebra. So we have shown that
$$\operatorname{so}(1,3)=\operatorname{su}(2) \oplus \operatorname{su}(2) .$$
Thus, representations of $\mathrm{su}(2) \oplus \mathrm{su}(2)$ will determine representations of the Lorentz group.
The decomposition $\mathrm{so}(1,3)=\mathrm{su}(2) \oplus \mathrm{su}(2)$ makes studying the irreducible representations very easy. We already know from quantum mechanics what the representations of $\operatorname{su}(2)$ are, since $\mathrm{su}(2)=3$ ) is the algebra of Pauli matrices, which generates the 3D rotation group $\mathrm{SO}(3)$. Each irreducible representation of $\mathrm{su}(2)$ is characterized by a halfinteger $j$. The representation acts on a vector space with $2 j+1$ basis elements (see Problem 10.2). It follows that irreducible representations of the Lorentz group are characterized by two half-integers: $A$ and $B$. The $(A, B)$ representation has $(2 A+1)(2 B+1)$ degrees of freedom.

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Group theory

$$X_\mu \rightarrow \Lambda_{\mu \nu} X_\nu$$

$$\left(\begin{array}{llll} 1 & & & \ & 1 & & \ & & \cos \theta_x & \sin \theta_x \ & & -\sin \theta_x & \cos \theta_x \end{array}\right),\left(\begin{array}{cccc} 1 & & & \ & \cos \theta_y & & -\sin \theta_y \ & \sin \theta_y & & \cos \theta_y \end{array}\right),\left(\begin{array}{cccc} 1 & & & \ & \cos \theta_z & \sin \theta_z & \ & -\sin \theta_z & \cos \theta_z & \ & & & 1 \end{array}\right)$$

$$\left(\begin{array}{ccccc} \cosh \beta_x & \sinh \beta_x & & \ \sinh \beta_x & \cosh \beta_x & & \ & & 1 & \ & & & 1 \end{array}\right),\left(\begin{array}{llll} \cosh \beta_y & & \sinh \beta_y & \ & 1 & & \ \sinh \beta_y & & \cosh \beta_y & \ & & & 1 \end{array}\right),\left(\begin{array}{llll} \cosh \beta_z & & & \sinh \beta_z \ & 1 & & \ & & 1 & \ \sinh \beta_z & & & \cosh \beta_z \end{array}\right) .$$

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|General representations of the Lorentz group

$$J_i^{+} \equiv \frac{1}{2}\left(J_i+i K_i\right), \quad J_i^{-} \equiv \frac{1}{2}\left(J_i-i K_i\right)$$

\begin{aligned} & {\left[J_i^{+}, J_j^{+}\right]=i \epsilon_{i j k} J_k^{+},} \ & {\left[J_i^{-}, J_j^{-}\right]=i \epsilon_{i j k} J_k^{-},} \ & {\left[J_i^{+}, J_j^{-}\right]=0 .} \end{aligned}

$$\operatorname{so}(1,3)=\operatorname{su}(2) \oplus \operatorname{su}(2) .$$

## 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:Quantum field theory, 物理代写, 量子场论

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|PHYS5125

avatest.orgt™量子场论Quantum field theory代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.org™， 最高质量的量子场论Quantum field theory作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此量子场论Quantum field theory作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Feynman rules for scalar QED

Expanding out the scalar QED Lagrangian we find
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2-\phi^{\star}\left(\square+m^2\right) \phi-i e A_\mu\left[\phi^{\star}\left(\partial_\mu \phi\right)-\left(\partial_\mu \phi^{\star}\right) \phi\right]+e^2 A_\mu^2|\phi|^2 .$$
We can read off the Feynman rules from the Lagrangian. The complex scalar propagator is
$$=\frac{i}{p^2-m^2+i \varepsilon} \text {. }$$
This propagator is the Fourier transform of $\left\langle 0\left|\phi^{\star}(x) \phi(0)\right| 0\right\rangle$ in the free theory. It propagates both $\phi$ and $\phi^{\star}$, that is both particles and antiparticles at the same time – they cannot be disentangled.
The photon propagator was calculated in Section 8.5:
$$\sim m \sim=\frac{-i}{p^2+i \varepsilon}\left[g_{\mu \nu}-(1-\xi) \frac{p_\mu p_\nu}{p^2}\right],$$
where $\xi$ parametrizes a set of covariant gauges.

Some of the interactions that connect $A_\mu$ to $\phi$ and $\phi^{\star}$ have derivatives in them, which will give momentum factors in the Feynman rules. To see which momentum factors we get, look back at the quantized fields:
\begin{aligned} \phi(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p e^{-i p x}+b_p^{\dagger} e^{i p x}\right), \ \phi^{\star}(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p^{\dagger} e^{i p x}+b_p e^{-i p x}\right) . \end{aligned}

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|External states

Now we know the vertex factors and propagators for the photon and the complex scalar field. The only thing left in the Feynman rules is how to handle external states. For a scalar field, this is easy – we just get a factor 1 . That is because a complex scalar field is just two real scalar fields, so we just take the real scalar field result. The only thing left is external photons.
For external photons, recall that the photon field is
$$A_\mu(x)=\int \frac{d^3 k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_k}} \sum_{i=1}^2\left(\epsilon_\mu^i(k) a_{k, i} e^{-i k x}+\epsilon_\mu^{i \star}(k) a_{k, i}^{\dagger} e^{i k x}\right) .$$
As far as free states are concerned, which is all we need for $S$-matrix elements, the photon is just a bunch of scalar fields integrated against some polarization vectors $\epsilon_\mu^i(k)$. Recall that external states with photons have momenta and polarizations, $|k, \epsilon\rangle$, so that $\left\langle 0\left|A_\mu(x)\right| k, \epsilon_i\right\rangle=\epsilon_\mu^i(k) e^{-i k x}$. This leads to LSZ being modified only by adding a factor of the photon polarization for each external state: $\epsilon_\mu$ if it is incoming and $\epsilon_\mu^{\star}$ if it is outgoing.

For example, consider the following diagram:
where $k^\mu=p_1^\mu+p_2^\mu$. The first polarization $\epsilon_\mu^1$ is the polarization of the photon labeled with $p_\mu^1$. It gets contracted with the momenta $p_2^\mu+k^\mu$ which come from the $-i e A_\mu\left[\phi^{\star}\left(\partial_\mu \phi\right)\right.$ $\left.-\left(\partial_\mu \phi^{\star}\right) \phi\right]$ vertex. The other polarization, $\epsilon_\mu^4$, is the polarization of the photon labeled with $p_\mu^4$ and contracts with the second vertex.

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Feynman rules for scalar QED

$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2-\phi^{\star}\left(\square+m^2\right) \phi-i e A_\mu\left[\phi^{\star}\left(\partial_\mu \phi\right)-\left(\partial_\mu \phi^{\star}\right) \phi\right]+e^2 A_\mu^2|\phi|^2 .$$

$$=\frac{i}{p^2-m^2+i \varepsilon} \text {. }$$

$$\sim m \sim=\frac{-i}{p^2+i \varepsilon}\left[g_{\mu \nu}-(1-\xi) \frac{p_\mu p_\nu}{p^2}\right],$$

\begin{aligned} \phi(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p e^{-i p x}+b_p^{\dagger} e^{i p x}\right), \ \phi^{\star}(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p^{\dagger} e^{i p x}+b_p e^{-i p x}\right) . \end{aligned}

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|External states

$$A_\mu(x)=\int \frac{d^3 k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_k}} \sum_{i=1}^2\left(\epsilon_\mu^i(k) a_{k, i} e^{-i k x}+\epsilon_\mu^{i \star}(k) a_{k, i}^{\dagger} e^{i k x}\right) .$$

$$\left(\partial_\mu+i e A_\mu\right) \phi \rightarrow\left(\partial_\mu+i e A_\mu+i \partial_\mu \alpha\right) e^{-i \alpha(x)} \phi=e^{-i \alpha(x)}\left(\partial_\mu+i e A_\mu\right) \phi .$$

$$D_\mu \phi \equiv\left(\partial_\mu+i e A_\mu\right) \phi \rightarrow e^{-i \alpha(x)} D_\mu \phi,$$

$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2+\left(D_\mu \phi\right)^{\star}\left(D_\mu \phi\right)-m^2 \phi^{\star} \phi$$

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Gauge symmetries and conserved currents

$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2+\partial_\mu \phi^{\star} \partial_\mu \phi+i e A_\mu\left(\phi \partial_\mu \phi^{\star}-\phi^{\star} \partial_\mu \phi\right)+e^2 A_\mu^2 \phi^{\star} \phi-m^2 \phi^{\star} \phi .$$

\begin{aligned} \left(\square+m^2\right) \phi & =-2 i e A_\mu \partial_\mu \phi+e^2 A_\mu^2 \phi, \ \left(\square+m^2\right) \phi^{\star} & =2 i e A_\mu \partial_\mu \phi^{\star}+e^2 A_\mu^2 \phi^{\star} . \end{aligned}

## 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。