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.
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
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.
If you want more details, I can recommend having a look at the draft of my book, chapters 7 and 8 where I gave a lot of details on expressions for the CFTs.
$\begingroup$
$\endgroup$
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 $1-n-m$.