如果你也在 怎样代写弦论string theory这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。弦论string theory在物理学中是一个理论框架，其中粒子物理学中的点状粒子被称为弦的一维物体取代。弦理论描述了这些弦如何在空间传播并相互作用。在大于弦的距离尺度上，弦看起来就像一个普通的粒子，其质量、电荷和其他属性由弦的振动状态决定。在弦理论中，弦的许多振动状态之一对应于引力子，一种携带引力的量子力学粒子。因此，弦理论是一种量子引力的理论。
弦论string theory是一个广泛而多样的学科，它试图解决基础物理学的一些深层次问题。弦理论为数学物理学贡献了许多进展，这些进展被应用于黑洞物理学、早期宇宙宇宙学、核物理学和凝聚态物理学中的各种问题，它也刺激了纯数学的一些重大发展。由于弦理论有可能提供对引力和粒子物理学的统一描述，它是万物理论的候选者，是描述所有基本力量和物质形式的独立数学模型。尽管在这些问题上做了很多工作，但目前还不知道弦理论在多大程度上描述了现实世界，也不知道该理论在选择其细节方面允许多大的自由度。
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物理代写|弦论代写string theory代考|超引力代写supergravity|Canonical quantization
The worldsheet theory can be quantized by replacing Poisson brackets by commutators:
$$
[,]{P . B} \longrightarrow i[,] . $$ We obtain the fundamental commutation relations $$ \begin{aligned} {\left[P^{\mu}(\tau, \sigma), P^{\nu}\left(\tau, \sigma^{\prime}\right)\right] } &=0 \ {\left[X^{\mu}(\tau, \sigma), X^{\nu}\left(\tau, \sigma^{\prime}\right)\right] } &=0 \ {\left[P^{\mu}(\tau, \sigma), X^{\nu}\left(\tau, \sigma^{\prime}\right)\right] } &=-i \eta^{\mu \nu} \delta\left(\sigma-\sigma^{\prime}\right) . \end{aligned} $$ Equivalently, we obtain from (2.56), (2.57) and (2.58), the commutation relations $$ \left[\alpha{n}^{\mu}, x^{\nu}\right]=\left[\alpha_{n}^{\mu}, p^{\nu}\right]=\left[\tilde{\alpha}{n}^{\mu}, x^{\nu}\right]=\left[\tilde{\alpha}{n}^{\mu}, p^{\nu}\right]=0
$$
$$
\begin{gathered}
{\left[\alpha_{n}^{\mu}, \alpha_{n^{\prime}}^{\nu}\right]=\left[\tilde{\alpha}{n}^{\mu}, \tilde{\alpha}{n^{\prime}}^{\nu}\right]=n \delta_{n+n^{\prime}, 0} \eta^{\mu \nu}, \quad\left[\alpha_{n}^{\mu}, \quad \tilde{\alpha}{n^{\prime}}^{\nu}\right]=0} \ {\left[p^{\mu}, x^{\nu}\right]=-i \eta^{\mu \nu} .} \end{gathered} $$ As before $$ \alpha{0}^{\mu}=\tilde{\alpha}{0}^{\mu}=\frac{1}{2} l{s} p^{\mu} \text { (closed), } \quad \alpha_{0}^{\mu}=l_{s} p^{\mu} \text { (open) }
$$
Define
$$
a_{m}^{\mu}=\frac{1}{\sqrt{m}} \alpha_{m}^{\mu}, \quad a_{m}^{\mu+}=\frac{1}{\sqrt{m}} \alpha_{-m}^{\mu}, \quad m>0 .
$$
Then we have
$$
\left[a_{m}^{\mu}, a_{n}^{\nu+}\right]=\left[\tilde{a}{m}^{\mu}, \tilde{a}{n}^{\nu+}\right]=\eta^{\mu \nu} \delta_{m, n}, \quad m, n>0
$$
This is the algebra of raising and lowering operators for quantum mechanical harmonic oscillators. The ground state $|0\rangle$ is defined by the condition
$$
a_{m}^{\mu}|0\rangle=0, \quad m>0 .
$$
As it turns out, this does not determine the state of the string completely, since the oscillators can be in their ground state, but the string center of mass can still have a non-zero momentum $k$. A state with this property will be denoted by $|0, k\rangle$. The momentum carried by the state $|\phi\rangle$ defined by
$$
|\phi\rangle=a_{m_{1}}^{\mu_{1}+} a_{m_{2}}^{\mu_{2}+} \ldots a_{m_{n}}^{\mu_{n}+}|0, k\rangle
$$
is $k$ since
$$
p^{\mu}|\phi\rangle=k^{\mu}|\phi\rangle .
$$
物理代写|弦论代写string theory代考|超引力代写supergravity|Virasoro algebra
We start by computing the classical Poisson bracket of two Virasoro generators, viz $\left[L_{m}, L_{n}\right]{P . B}$. This Poisson bracket involves four equal terms, we find $$ \left[L{m}, L_{n}\right]{P . B}=i \sum{l}(m-l) \alpha_{m+n-l}^{\mu} \alpha_{l \mu} .
$$
Equivalently,
$$
\left[L_{m}, L_{n}\right]{P . B}=-i \sum{l}(n-l) \alpha_{m+n-l}^{\mu} \alpha_{l \mu}
$$
Thus
$$
\left[L_{m}, L_{n}\right]{P . B}=\frac{i}{2} \sum{l}(m-n) \alpha_{m+n-l}^{\mu} \alpha_{l \mu}=i(m-n) L_{m+n}
$$
This is called the Virasoro algebra.
This algebra is due to the fact that our gauge choice $h^{\alpha \beta}=\eta^{\alpha \beta}$ did not fully fix the reparametrization invariance. Indeed, any combined reparametrization $\xi^{\alpha}$ and Weyl scaling $\Lambda$ for which
$$
\partial^{\alpha} \xi^{\beta}+\partial^{\beta} \xi^{\alpha}=\Lambda \eta^{\alpha \beta}
$$
preserves the gauge choice. This can be seen as follows. Recall that
$$
\delta h_{\alpha \beta}=h_{\alpha \beta}^{\prime}-h_{\alpha \beta}=-\partial_{\beta} \xi^{\delta} h_{\alpha \delta}-\partial_{\alpha} \xi^{\delta} \cdot h_{\delta \beta}-\xi^{\delta} \partial_{\delta} h_{\alpha \beta} .
$$
If $h=\eta$, then
$$
\delta h_{\alpha \beta}=h_{\alpha \beta}^{\prime}-\eta_{\alpha \beta}=-\partial_{\beta} \xi_{\alpha}-\partial_{\alpha} \xi_{\beta} .
$$
This can be undone by the Weyl scaling
$$
\delta h_{\alpha \beta}=2 \omega h_{\alpha \beta}=2 \omega \eta_{\alpha \beta}, \quad 2 \omega=\Lambda .
$$
The condition $(2.115)$ is solved by
$$
\xi^{\pm}=\xi^{0} \pm \xi^{1}, \quad \xi^{+}=\xi^{+}\left(\sigma^{+}\right), \quad \xi^{-}=\xi^{-}\left(\sigma^{-}\right) .
$$
弦论超引力代写
物理代写|弦论代写STRING THEORY代考|超引力代写SUPERGRAVITY|CANONICAL QUANTIZATION
世界表理论可以通过用交换子替换泊松括号来量化：
$$
[,]{P . B} \longrightarrow i[,] . $$ We obtain the fundamental commutation relations $$ \begin{aligned} {\left[P^{\mu}(\tau, \sigma), P^{\nu}\left(\tau, \sigma^{\prime}\right)\right] } &=0 \ {\left[X^{\mu}(\tau, \sigma), X^{\nu}\left(\tau, \sigma^{\prime}\right)\right] } &=0 \ {\left[P^{\mu}(\tau, \sigma), X^{\nu}\left(\tau, \sigma^{\prime}\right)\right] } &=-i \eta^{\mu \nu} \delta\left(\sigma-\sigma^{\prime}\right) . \end{aligned} $$ Equivalently, we obtain from (2.56), (2.57) and (2.58), the commutation relations $$ \left[\alpha{n}^{\mu}, x^{\nu}\right]=\left[\alpha_{n}^{\mu}, p^{\nu}\right]=\left[\tilde{\alpha}{n}^{\mu}, x^{\nu}\right]=\left[\tilde{\alpha}{n}^{\mu}, p^{\nu}\right]=0
$$
$$
\begin{gathered}
{\left[\alpha_{n}^{\mu}, \alpha_{n^{\prime}}^{\nu}\right]=\left[\tilde{\alpha}{n}^{\mu}, \tilde{\alpha}{n^{\prime}}^{\nu}\right]=n \delta_{n+n^{\prime}, 0} \eta^{\mu \nu}, \quad\left[\alpha_{n}^{\mu}, \quad \tilde{\alpha}{n^{\prime}}^{\nu}\right]=0} \ {\left[p^{\mu}, x^{\nu}\right]=-i \eta^{\mu \nu} .} \end{gathered} $$ As before $$ \alpha{0}^{\mu}=\tilde{\alpha}{0}^{\mu}=\frac{1}{2} l{s} p^{\mu} \text { (closed), } \quad \alpha_{0}^{\mu}=l_{s} p^{\mu} \text { (open) }
$$
Define
$$
a_{m}^{\mu}=\frac{1}{\sqrt{m}} \alpha_{m}^{\mu}, \quad a_{m}^{\mu+}=\frac{1}{\sqrt{m}} \alpha_{-m}^{\mu}, \quad m>0 .
$$
Then we have
$$
\left[a_{m}^{\mu}, a_{n}^{\nu+}\right]=\left[\tilde{a}{m}^{\mu}, \tilde{a}{n}^{\nu+}\right]=\eta^{\mu \nu} \delta_{m, n}, \quad m, n>0
$$
This is the algebra of raising and lowering operators for quantum mechanical harmonic oscillators. The ground state $|0\rangle$ is defined by the condition
$$
a_{m}^{\mu}|0\rangle=0, \quad m>0 .
$$
As it turns out, this does not determine the state of the string completely, since the oscillators can be in their ground state, but the string center of mass can still have a non-zero momentum $k$. A state with this property will be denoted by $|0, k\rangle$. The momentum carried by the state $|\phi\rangle$ defined by
$$
|\phi\rangle=a_{m_{1}}^{\mu_{1}+} a_{m_{2}}^{\mu_{2}+} \ldots a_{m_{n}}^{\mu_{n}+}|0, k\rangle
$$
is $k$ since
$$
p^{\mu}|\phi\rangle=k^{\mu}|\phi\rangle .
$$
物理代写|弦论代写STRING THEORY代考|超引力代写SUPERGRAVITY|VIRASORO ALGEBRA
我们首先计算两个 Virasoro 生成器的经典泊松括号，即 viz $\left[L_{m}, L_{n}\right]{P . B}$. This Poisson bracket involves four equal terms, we find $$ \left[L{m}, L_{n}\right]{P . B}=i \sum{l}(m-l) \alpha_{m+n-l}^{\mu} \alpha_{l \mu} .
$$
Equivalently,
$$
\left[L_{m}, L_{n}\right]{P . B}=-i \sum{l}(n-l) \alpha_{m+n-l}^{\mu} \alpha_{l \mu}
$$
Thus
$$
\left[L_{m}, L_{n}\right]{P . B}=\frac{i}{2} \sum{l}(m-n) \alpha_{m+n-l}^{\mu} \alpha_{l \mu}=i(m-n) L_{m+n}
$$
这称为 Virasoro 代数。
这个代数是由于我们的规范选择H一种b=这一种b没有完全修复重新参数化的不变性。事实上，任何组合的重新参数化X一种和外尔缩放Λ为此∂一种Xb+∂bX一种=Λ这一种b
$$
\partial^{\alpha} \xi^{\beta}+\partial^{\beta} \xi^{\alpha}=\Lambda \eta^{\alpha \beta}
$$
保留仪表选择。这可以看如下。回想起那个
$$
\delta h_{\alpha \beta}=h_{\alpha \beta}^{\prime}-h_{\alpha \beta}=-\partial_{\beta} \xi^{\delta} h_{\alpha \delta}-\partial_{\alpha} \xi^{\delta} \cdot h_{\delta \beta}-\xi^{\delta} \partial_{\delta} h_{\alpha \beta} .
$$
If $h=\eta$, then
$$
\delta h_{\alpha \beta}=h_{\alpha \beta}^{\prime}-\eta_{\alpha \beta}=-\partial_{\beta} \xi_{\alpha}-\partial_{\alpha} \xi_{\beta} .
$$
This can be undone by the Weyl scaling
$$
\delta h_{\alpha \beta}=2 \omega h_{\alpha \beta}=2 \omega \eta_{\alpha \beta}, \quad 2 \omega=\Lambda .
$$
The condition $(2.115)$ is solved by
$$
\xi^{\pm}=\xi^{0} \pm \xi^{1}, \quad \xi^{+}=\xi^{+}\left(\sigma^{+}\right), \quad \xi^{-}=\xi^{-}\left(\sigma^{-}\right) .
$$
物理代写|弦论代写string theory代考|超引力代写supergravity 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
