这是一份imperial帝国理工大学 GG14作业代写的成功案例
but
$$
\begin{aligned}
&F_{n}=\frac{1}{\sqrt{2} \alpha^{\prime}} \cdot \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{j n \sigma} J_{++} d \sigma \
&G_{r}=\frac{1}{\sqrt{2} \alpha^{\prime}} \cdot \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{j r \sigma} J_{++} d \sigma
\end{aligned}
$$
and
$$
\begin{aligned}
\bar{F}{n} &=\frac{1}{\sqrt{2} \alpha^{\prime}} \cdot \frac{1}{2 \pi} \int{-\pi}^{\pi} e^{-i n \sigma} J_{–} d \sigma \
\bar{G}{r} &=\frac{1}{\sqrt{2} \alpha^{\prime}} \cdot \frac{1}{2 \pi} \int{-\pi}^{\pi} e^{-i r \sigma} J_{–} d \sigma
\end{aligned}
$$
G102 COURSE NOTES ：
where
$$
\alpha_{0}^{\mu}=-i \sqrt{2 \alpha^{\prime}} \partial^{\mu} .
$$
Consequently, the constraint $F_{0}|\phi\rangle_{\epsilon}=0$ implies that
$$
\not \lambda=0, \sqrt{\alpha^{\prime}} \partial \psi_{\mu}^{1}=\psi_{\mu}^{2},-\sqrt{\alpha^{\prime}} \not \psi_{\mu}^{2}=\psi_{\mu}^{1}, \ldots,
$$
while $F_{1}|\phi\rangle_{\epsilon}=0$ implies
$$
-\sqrt{2 \alpha^{\prime}} \partial^{\mu} \psi_{\mu}^{2}+\frac{\Gamma^{\mu}}{\sqrt{2}} \psi_{\mu}^{1}=0 .
$$