这是一份liverpool利物浦大学MATH423的成功案例

$$
\partial^{2} X^{\mu}=\partial^{\alpha} \partial_{\alpha} X^{\mu}=0 .
$$
Note that when imposing the conformal gauge on the Polyakov action the equation of motion for $h_{\alpha \beta}$, i.e., $T_{\alpha \beta}=0$, becomes a constraint, which has to be imposed on the solutions of (11).

The general solution is a superposition of left- and right-moving waves,
$$
X^{\mu}(\sigma)=X_{L}^{\mu}\left(\sigma^{+}\right)+X_{R}^{\mu}\left(\sigma^{-}\right) .
$$
However, we also have to specify boundary conditions at the ends of the string. One possible choice are periodic boundary conditions,
$$
X^{\mu}\left(\sigma^{0}, \sigma^{1}+\pi\right)=X^{\mu}\left(\sigma^{0}, \sigma^{1}\right) .
$$

MATH423 COURSE NOTES ：

$$
X^{\mu}(\sigma)=x^{\mu}+\left(2 \alpha^{\prime}\right) p^{\mu} \sigma^{0}+\mathrm{i} \sqrt{2 \alpha^{\prime}} \sum_{n \neq 0} \frac{\alpha_{n}^{\mu}}{n} e^{-\mathrm{i} n \sigma^{0}} \cos \left(n \sigma^{1}\right)
$$
There is, however, a second possible choice of boundary conditions for open strings, namely Dirichlet boundary conditions. Here the ends of the string are kept fixed:
$$
\left.X^{\mu}\right|{\sigma^{1}=0}=x{(1)}^{\mu},\left.\quad X^{\mu}\right|{\sigma^{1}=\pi}=x{(2)}^{\mu}
$$
With these boundary conditions the solution takes the form
$$
X^{\mu}(\sigma)=x_{(1)}^{\mu}+\left(x_{(2)}^{\mu}-x_{(1)}^{\mu}\right) \frac{\sigma^{1}}{\pi}+\mathrm{i} \sqrt{2 \alpha^{\prime}} \sum_{n \neq 0} \frac{\alpha_{n}^{\mu}}{n} e^{-\mathrm{in} \sigma^{0}} \sin \left(n \sigma^{1}\right)
$$