background  and we now proceed in finding the superstring spectrum.

Bosonic sector
--------------

We work in the light-cone gauge $z^+ = p^+ \tau$ with $l_s=1$. The light-cone Lagrangian density of the bosonic $\sigma$-model is given by $$\label{boslagr}
\begin{split}
    \mathscr{L}_{lc}^{B}= &- \frac{1}{4\pi  p^+}\left(\partial^{\alpha}z^i\partial_{\alpha}z^i+
     f^2 \sum_{k=1}^{4}\left(1-\eta_{k}^{2}\right)\left[\left(z^{2k-1}\right)^2+\left(z^{2k}\right)^2\right] \right.  \\
    &+\left. 2f \sum_{k=1}^{4}\left[(C_k-\eta_k)z^{2k-1}\dot{z}^{2k}+(C_k+\eta_k)z^{2k}\dot{z}^{2k-1}\right]\right)\, ,
\end{split}$$ where we have defined $f = \mu  p^+$. The conjugate momenta are computed to be $$\Pi_{2k-1} = \frac{\dot{z}^{2k-1}-f\left(C_k + \eta_k \right) z^{2k}}{2\pi   }\, ,~~~~~
\Pi_{2k} = \frac{\dot{z}^{2k}-f\left(C_k - \eta_k \right) z^{2k-1}}{2\pi   }\, ,$$ and the bosonic light-cone Hamiltonian is given by $$H_{lc}^{B}= \frac{1}{4\pi  p^+}\int_{0}^{2\pi}d\sigma \Bigg[ \dot{z}^i \dot{z}^i+ (z^i)'(z^i)'
  +f^2 \sum_{k=1}^{4}\left(1-\eta_{k}^{2}\right)\left[\left(z^{2k-1}\right)^2+\left(z^{2k}\right)^2\right]\Bigg]\, .$$ In order to solve the equations of motion

$$\begin{aligned}
&\partial^{\alpha}\partial_{\alpha}z^{2k-1}+2f\eta_k \dot{z}^{2k} - f^2 \left(1-\eta_{k}^{2}\right) z^{2k-1}=0\label{moteq1}\, ,\\
&\partial^{\alpha}\partial_{\alpha}z^{2k}-2f\eta_k \dot{z}^{2k-1} - f^2 \left(1-\eta_{k}^{2}\right) z^{2k}=0\label{moteq2}\, ,\end{aligned}$$

it is useful to introduce four complex fields $$X^k = z^{2k-1}+ iz^{2k}\, ,$$ in terms of which the above equations read

$$\begin{aligned}
&\partial^{\alpha}\partial_{\alpha}X^{k}-2 i f\eta_k \dot{X}^{k} - f^2 \left(1-\eta_{k}^{2}\right) X^{k}=0\, ,\label{moteqd1}\\
&\partial^{\alpha}\partial_{\alpha}\bar{X}^{k}+2 i f\eta_k \dot{\bar{X}}^{k} - f^2 \left(1-\eta_{k}^{2}\right) \bar{X}^{k}=0\label{moteqd2}\, .\end{aligned}$$

One can see that a solution of the form $$X^k=e^{-i f \eta_k \tau} Y^k$$ solves if $Y^k$ satisfy the equation $$\partial^{\alpha}\partial_{\alpha}Y^{k} -f^2 Y^k=0\, .$$ Therefore for $Y^k$ and its conjugate $\bar{Y}^k$ we have the following 