# Level of String Fields

I know that the Tachyon state $$c_1|0;k\rangle$$, where $$|0;k\rangle = e^{ikX}|0\rangle$$, has eigenvalue $$-1$$ under $$L_0$$ operator. Using $$L_0 = \alpha'p^2 + \sum_{n\ge1}\alpha_{-n}\cdot \alpha_n$$ it is easy to see. Then if I define tachyon states as the zero level of the string states, the other $$L_0$$ eigenvector states have its levels given by its eigenvalue plus one, that is, the eigenvalue of the operator $$L_0+1$$. But I dont understand for example why $$c_{-1}|0\rangle$$ or $$b_{-2}c_1|0\rangle$$ has level 2, or why $$c_{0}|0\rangle$$ has leval one. To me it would have level $$-1$$ since $$L_0$$ should comute with ghost modes. It is a simple question but I don't know how to answer.

Your expression of $$L_0$$ includes only the matter operator, you are missing the ghost part: $$L_0^{gh} = \sum_{n \in \mathbb Z} n \, b_{-n} c_n$$ This is why you had the impression that $$L_0$$ commutes with the ghosts, but it's not the case.
Moreover, the state $$c_1 |0; k\rangle$$ has not eigenvalue $$-1$$ but: $$L_0 c_1 |0; k\rangle = (\alpha' k^2 - 1) c_1 |0; k\rangle$$ You have forgotten that $$\mathrm{e}^{i k \cdot X}$$ has eigenvalue $$k$$ for the momentum operator. This is compatible with the fact that the on-shell condition $$L_0 = 0$$ should give the mass-shell condition for the tachyon: $$k^2 = - m^2$$, with $$m^2 = - 1 / \alpha'$$.
Moreover, the level operator $$\widehat L_0$$ is defined as $$L_0$$ without the zero-mode matter operator (which is the momentum operator) $$L_0 := \alpha' k^2 + \widehat L_0$$ and can be written as (for $$d = 26$$ dimensions, flat Minkowski background) $$\widehat L_0 = N_X + N_b + N_c \in \mathbb N$$ where $$N_X, N_b, N_c$$ are the matter, $$b$$-ghost and $$c$$-ghost level operators: $$N_X = \sum_{n > 0} n N_{X,n}, \qquad N_b = \sum_{n > 0} n N_{b,n}, \qquad N_c = \sum_{n > 0} n N_{c,n}$$ and $$N_{X,n}, N_{b,n}, N_{c,n}$$ are the matter, $$b$$-ghost and $$c$$-ghost number operators: $$N_{X,n} = \frac{1}{n} \, \alpha_{-n} \cdot \alpha{-n}, \qquad N_{b,n} = b_{-n} c_n, \qquad N_{c,n} = c_{-n} b_n.$$ This is equivalent to the expression you wrote for the matter, plus the one I wrote for the ghosts, but written in a nice form. Indeed, you can see directly that each number operator counts the number of excitations in a given mode, and then the contribution to the level operator is weighted by the frequency of the mode.
Ok, I think I can use Equation (2.6.24) on Polchinski to write: $$[L_0,c_n] = -nc_n \ ,\ \ \ \ \ \ \ [L_0,b_n] = -nb_n$$. Since the ghosts are holomorphic tensor fields. Then, I was wrong, the $$L_0$$ does commutes with the ghost modes. Now, I have $$L_0c_n|k\rangle = c_nL_0|k\rangle -nc_nL_0c_1|k\rangle = -nc_n|k\rangle$$ and $$L_0b_n|k\rangle = b_nL_0|k\rangle -nb_nL_0c_1|k\rangle = -nb_n|k\rangle$$ Now it is clear that $$c_n|k\rangle$$ is on level $$1-n$$. Samething about $$b_n|k\rangle$$. The levels of $$b_nc_m|k\rangle$$ also follows: $$L_0b_nc_m|k\rangle = b_nL_0c_m|k\rangle - n b_n c_m|k\rangle$$ $$= -mb_nc_m|k\rangle - n b_n c_m|k\rangle = -(m+n)b_nc_m|k\rangle$$ Therefore, $$b_nc_m|k\rangle$$ has level $$1-n-m$$.