Riemann surface
matlab
无需担心！我们的量子场论代写专家团队将专业地解决您在量子场论学习中遇到的各种挑战。我们拥有广泛的专业知识和丰富的经验，可以协助您完成高水平的作业和论文，确保您在学习道路上顺利前行！
以下是一些我们可以帮助您解决的问题：
量子力学基础：涵盖量子力学的基本概念、波函数、测量和算符等。
量子场的正则形式：研究量子场的正则量子化方法，如场算符、场正则对易关系等。
相对论性量子场论：介绍相对论性量子场论的基本原理，如洛仑兹不变性、自由场等。
量子电动力学：研究电磁场与带电粒子的相互作用，如量子电动力学的基本框架和计算方法。
量子色动力学：探索强相互作用的量子场论，如夸克和胶子的相互作用和强子的产生与衰变。
量子场论中的计算方法：介绍量子场论中的计算技巧和近似方法，如费曼图、重整化等。
量子场论在凝聚态物理中的应用：探索量子场论在凝聚态物理领域中的应用，如拓扑相、量子液体等。
无论您面临的量子场论问题是什么，我们都会竭尽全力提供专业的帮助，确保您的学习之旅顺利无阻！
Problems. Solution 1.1. Recall
$$
\left{\alpha^j, \alpha^k\right}_{a b}=2 \delta^{j k} \delta_{a b}, \quad\left{\alpha^j, \beta\right}_{a b}=0 \quad\left(\beta^2\right){a b}=\delta{a b}
$$
Suppose $\beta|\psi\rangle=\lambda|\psi\rangle$
$$
\beta^2|\psi\rangle=\lambda^2|\psi\rangle=1|\psi\rangle=|\psi\rangle \lambda^2=1 \text { or } \lambda= \pm 1
$$
Use hint. $\operatorname{Tr} \alpha_1^2 \beta=-\operatorname{Tr} \alpha_1 \beta \alpha_1=-\operatorname{Tr} \alpha_1^2 \beta \Longrightarrow \operatorname{Tr} \alpha_1^2 \beta=0$
$$
\left{\alpha^1, \alpha^1\right}_{a b}=2\left(\alpha^1\right){a b}^2=2 \delta{a b}
$$
so $\left(\alpha^1\right)^2=1 \Longrightarrow \operatorname{Tr} \beta=0$
Since $\operatorname{Tr} \beta=0$ and $\lambda_\beta= \pm 1$ and $\operatorname{Tr} \beta=\sum \lambda_\beta$, then $\beta$ must be even-dimensional for a set of \pm 1 ‘s to sum up and cancel each other.
Likewise,
$$
\begin{gathered}
\alpha_i|\psi\rangle=\lambda|\psi\rangle \
\alpha_i^2|\psi\rangle=\lambda \alpha_i|\psi\rangle=\lambda^2|\psi\rangle=|\psi\rangle \quad \Longrightarrow \lambda= \pm 1 \
\operatorname{Tr} \alpha_1^2 \alpha_j=-\operatorname{Tr} \alpha_1 \alpha_j \alpha_1=-\operatorname{Tr} \alpha_1^2 \alpha_j \Longrightarrow \operatorname{Tr} \alpha_1^2 \alpha_j=0 \text { so } \operatorname{Tr}\left(\alpha_j\right)=0
\end{gathered}
$$
So then $\alpha^j$ even dimensional as well.
$$
\begin{gathered}
\int d^3 x V_{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}} a_{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle=\sum{j=1}^n( \pm 1)^{j-1} \int d^3 x V_{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} \delta_{\mathbf{x}-\mathbf{x}j} \prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle= \ =\sum{j=1}^n( \pm 1)^{j-1} V_{\mathbf{x}j-\mathbf{y}} a{\mathbf{x}j}^{\dagger} a{\mathbf{y}}^{\dagger} a_{\mathbf{y}} \prod_{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle \end{gathered} $$ Now $$ a{\mathbf{y}} \prod_{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle=\sum{k \neq j}^n \delta_{\mathbf{x}k-\mathbf{y}}(-1)^m \prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}|0\rangle $$ So then $$ \begin{gathered} \int d^3 x d^3 y V{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}} a_{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle=\int d^3 y \sum{j=1}^n( \pm 1)^{j-1} V_{\mathbf{x}j}-\mathbf{y}{\mathbf{x}j}^{\dagger} a{\mathbf{y}}^{\dagger} \sum_{k \neq j}^n \delta_{\mathbf{x}k-y}(-1)^m \prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}|0\rangle= \ =\sum{j=1}^n \sum_{k \neq j}^n V_{\mathbf{x}j-\mathbf{x}_k}( \pm 1)^{j-1+m} a{\mathbf{x}j}^{\dagger} a{\mathbf{x}k}^{\dagger} \prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}|0\rangle \end{gathered} $$ To bring $a{\mathbf{x}k}^{\dagger}$ back in its ordered place with $\prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}$, then $m$ (anti)commutations must occur (we really are repeating what we did before, exactly, again). So there’s another factor of $(-1)^m$. Likewise for $a{\mathbf{x}j}^{\dagger}$, to be back in its ordered place, another factor of $(-1)^{j-1}$ results. $$ \Longrightarrow \sum{j=1}^n \sum_{k \neq j}^n V\left(\mathbf{x}j-\mathbf{x}_k\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}_n}^{\dagger}|0\rangle
$$
Now $\sum_{j=1}^n \sum_{k \neq j}^n$ counts each unique pair twice so
$$
\begin{gathered}
\sum_{j=1}^n \sum_{k \neq j}^n V\left(\mathbf{x}j-\mathbf{x}_k\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle=2 \sum{j=1}^n \sum_{k=1}^{j-1} V\left(\mathbf{x}j-\mathbf{x}_k\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle \ \Longrightarrow \frac{1}{2} \int d^3 x d^3 y V(\mathbf{x}-\mathbf{y}) a{\mathbf{x}{\mathbf{y}}^{\dagger}}^{\dagger} a{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}} \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}1, \ldots, \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle= \ =\sum{j=1}^n \sum_{k=1}^{j-1} V\left(\mathbf{x}j-\mathbf{x}_k\right) \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}_1, \ldots \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}_n}^{\dagger}|0\rangle \
\Longrightarrow i \hbar \frac{\partial}{\partial t}|\psi, t\rangle=H|\psi, t\rangle \text { is satisfied. }
\end{gathered}
$$
Also recall the commutation and anticommutation rules:
$$
\begin{aligned}
{\left[a(\mathbf{x}), a\left(\mathbf{x}^{\prime}\right)\right] } & =0 \
{\left[a^{\dagger}(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right] } & =0 \
{\left[a(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right] } & =\delta^3(\mathbf{x}-\mathbf{x})
\end{aligned}
$$
$$
\begin{aligned}
\left{a(\mathbf{x}), a\left(\mathbf{x}^{\prime}\right)\right} & =0 \
\left{a^{\dagger}(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right} & =0 \
\left{a(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right} & =\delta^3\left(\mathbf{x}-\mathbf{x}^{\prime}\right)
\end{aligned}
$$
Using some shorthand notation,
$$
H|\psi, t\rangle=\left(\int d^3 x a_{\mathbf{x}}^{\dagger}\left(\frac{-\hbar^2}{2 m} \nabla^2+U_{\mathbf{x}}\right) a_{\mathbf{x}}+\frac{1}{2} \int d^3 x d^3 y V_{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}}\right) \cdot \int d^3 x_1 \ldots d^3 x_n \psi\left(\mathbf{x}1, \ldots, \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger} \mid 0 $$ Now $$ a{\mathbf{x}} a_{\mathbf{x}1}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}=\left(( \pm 1) a{\mathbf{x}1}^{\dagger} a{\mathbf{x}}+\delta_{\mathbf{x}-\mathbf{x}1}\right) a{\mathbf{x}2}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}=( \pm 1) a{\mathbf{x}1}^{\dagger}\left(( \pm 1) a{\mathbf{x}2}^{\dagger} a{\mathbf{x}}+\delta_{\mathbf{x}-\mathbf{x}2}\right) a{\mathbf{x}3}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}+\delta{\mathbf{x}-\mathbf{x}1} a{\mathbf{x}2}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger} $$ If this term is applied to $|0\rangle$, note $a_x|0\rangle=0$. $$ a{\mathbf{x}} a_{\mathbf{x}1}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}|0\rangle=\sum{j=1}^n( \pm 1)^{j-1} \delta_{\mathbf{x}-\mathbf{x}j} \prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle $$ For the kinetic and self-potential terms of $H|\psi, t\rangle$, $$ \int d^3 x a{\mathbf{x}}^{\dagger}\left(\frac{-\hbar^2}{2 m} \nabla^2+U_{\mathbf{x}}\right) \int d^3 x_1 \ldots d^3 x_n \psi\left(\mathbf{x}1, \ldots, \mathbf{x}_n, t\right) \sum{j=1}^n( \pm 1)^{j-1} \delta_{\mathbf{x}-\mathbf{x}j} \prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle $$ Evaluate $\int d^3 x$, in particular, with $\delta{\mathbf{x}-\mathbf{x}j}$ $$ \sum{j=1}^n a_{\mathbf{x}j}^{\dagger}\left(\frac{-\hbar^2}{2 m} \nabla_j^2+U{\mathbf{x}j}\right) \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}_1, \ldots, \mathbf{x}_n, t\right)( \pm 1)^{j-1} \prod{k \neq j}^n a_{x_k}^{\dagger}|0\rangle
$$
To (anti)commmute $a_{\mathbf{x}j}^{\dagger}$ over to its ordered place in $\prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}( \pm 1)^{j-1}, j-1$ (anti)commutations must take place. $$ \Longrightarrow \sum{j=1}^n\left(-\frac{\hbar^2}{2 m} \nabla_j^2+U_{\mathbf{x}j}\right) \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}_1, \ldots, \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle $$ Now, on $a{\mathbf{x}} a_{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}_n}^{\dagger}|0\rangle$.
Problem 2.2. Verify that eq. (2.14) follows from $U(\Lambda)^{-1} U\left(\Lambda^{\prime}\right) U(\Lambda)=U\left(\Lambda^{-1} \Lambda^{\prime} \Lambda\right)$.
Solution 2.2. $\Lambda^{-1} \Lambda^{\prime} \Lambda=\Lambda^{-1}\left(1+\delta \omega^{\prime}\right) \Lambda=1+\Lambda^{-1} \delta \omega^{\prime} \Lambda$
$$
\begin{gathered}
U\left(\Lambda^{-1} \Lambda^{\prime} \Lambda\right)=1+\frac{i}{2}\left(\Lambda^{-1} \delta \omega^{\prime} \Lambda\right){\rho \sigma} M^{\rho \sigma}=1+\frac{i}{2}\left(\Lambda^{-1}\right)\rho^a\left(\delta \omega^{\prime}\right){a b}(\Lambda)\sigma^b M^{\rho \sigma} \
U\left(\Lambda^{\prime}\right)=U\left(1+\delta \omega^{\prime}\right)=1+\frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu} \
U(\Lambda)^{-1} U\left(\Lambda^{\prime}\right) U(\Lambda)=U(\Lambda)^{-1} U(\Lambda)+\frac{i}{2} U(\Lambda)^{-1} \delta \omega_{\mu \nu} M^{\mu \nu} U(\Lambda)
\end{gathered}
$$
Using $U(\Lambda)^{-1} U\left(\Lambda^{\prime}\right) U(\Lambda)=U\left(\Lambda^{-1} \Lambda^{\prime} \Lambda\right)$,
$$
\Longrightarrow U\left(\Lambda^{-1}\right) \delta \omega_{\mu \nu} M^{\mu \nu} U(\Lambda)=\delta \omega_{\mu \nu} U\left(\Lambda^{-1}\right) M^{\mu \nu} U(\Lambda)=\delta \omega_{\mu \nu}\left(\Lambda^{-1}\right)\rho^\mu(\Lambda)\sigma^\nu M^{\rho \sigma}
$$
Note that $\delta \omega_{\mu \nu}$ is just an entry in a rank-2 tensor, a number, and for arbitrary Lorentz transformation, each a uniquelly determined quantity $\delta \omega_{\mu \nu}$. (6 of them for antisymmetric $\delta \omega_{\mu \nu}$ ).
$$
\Longrightarrow U\left(\Lambda^{-1}\right) M^{\mu \nu} U(\Lambda)=\left(\Lambda^{-1}\right)\rho^\mu(\Lambda)\sigma^\nu M^{\rho \sigma}
$$
E-mail: help-assignment@gmail.com 微信:shuxuejun
help-assignment™是一个服务全球中国留学生的专业代写公司
专注提供稳定可靠的北美、澳洲、英国代写服务
专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务