Absence of negative-norm states in old covariant quantization of the bosonic string This question is about the discussion about the absence of negative-norm states in the old covariant quantization of the bosonic string as presented e.g. in Becker Becker & Schwarz (BBS). Their discussion is in page 44 and the following ones.
They start by defining spurious states as states $|\psi\rangle$ obeying:
$$(L_0-a)|\psi\rangle=0,\quad\text{and},\quad  \langle \phi|\psi\rangle=0,\quad \forall \ \text{physical states $|\phi\rangle$}.\tag{2.106}$$
Afterwards they state that any such spurious state can be written in the form $$|\psi\rangle=L_{-1}|\chi_1\rangle+L_{-2}|\chi_2\rangle\tag{2.108}$$
where the level one and level two spurious states $|\chi_1\rangle$ and $|\chi_2\rangle$ obey the condition $$(L_0-a+n)|\chi_n\rangle=0,\quad n=1,2.$$
After that they observe that if $|\psi\rangle$ is both spurious and physical it has zero norm since it must be orthogonal to itself. After these preliminaries they do two things:

*

*They first consider the spurious state $L_{-1}|\chi_1\rangle$. They demand it to be physical (and hence have zero norm) and find that for this to be so we must have $a =1$.


*They then consider spurious states of the form $|\psi\rangle=(L_{-2}+\gamma L_{-1}^2)|\tilde{\chi}\rangle$. They further say that this will be spurious if $(L_0+1)|\tilde{\chi}\rangle=L_m|\tilde{\chi}\rangle=0$. Afterwards they impose the physical state condition and find that the condition $L_1|\psi\rangle=0$ imposes $\gamma=\frac{3}{2}$ and that $L_2|\psi\rangle=0$ imposes $D = 26$.
After that they claim that: "The zero-norm spurious states are unphysical. The fact that they are spurious ensures that they decouple from all physical processes. In fact, all negative-norm states decouple, and all physical states have positive norm. Thus, the complete physical spectrum is free of negative norm states when the two conditions $a=1$ and $D=26$ are satisfied, as is proved in the next section".
In the next section they turn to light-cone gauge quantization (which is manifestly free of negative-norm states) and find $a =1$ and $D = 26$ by other means.
But the question here is about this old covariant quantization method. To be honest what they did is completely obscure. Why doing the two steps (1) and (2) above should free us from the negative-norm states? Why should we be demanding that $L_{-1}|\chi_1\rangle$ and $(L_{-2}+\gamma L_{-1}^2)|\tilde{\chi}\rangle$ be physical in order to eliminate the negative-norm states?
I do understand that the light-cone gauge quantization confirms that for $a = 1$ and $D = 26$ these negative-norm states are absent, but I want to understand, in a self-contained way, all of this discussion in old covariant quantization that I have outlined above.
 A: In the Polyakov action, when $g_{ab}$ was variant, its EOM was $T^{ab}=0$. Afterward, we fixed the gauge, this equation is lost. However, we still want to keep this equation. The minimal requirement is that the entries of $T^{ab}$ are zero for physical states
$$\langle \psi|T^{ab}|\psi'\rangle=0.$$
For this to be true, remember that $L_n$'s are the mode expansion of $T$ on the worldsheet, the minimal requirement is that
$$L_0|\psi\rangle=a|\psi\rangle\quad\textrm{and}\quad L_{n>0}|\psi\rangle=0.$$
These $L_n$'s obey the Virasoro algebra
$$[L_n,L_m]=(n-m)L_{n+m}+\frac{D}{12}n(n-1)(n+1)\delta_{n+m,0}.$$
If a state $|\chi\rangle$ can be written in the form of $L_{-n}|\psi\rangle$, it is orthogonal to all physical states
$$\langle\psi'|L_{-n}|\psi\rangle=(\langle\psi|L_{n}|\psi'\rangle)^*=0.$$
So these states are called spurious. It is reasonable that two physical states which only differ by a spurious state are equivalent. And of course, the spurious state which is the difference between the two physical states is a physical state. So we define states which are both spurious and physical null. The Hilbert space is physical states moded out by null states.
First consider $L_{-1}|\chi_1\rangle$. Here $\chi_1$ obeys $L_0|\chi_1\rangle=(a-1)|\chi_1\rangle.$
We want this state to be null so that we have the same number of level one states as lightcone quantization. This is to say
$$L_1L_{-1}|\chi_1\rangle=2L_0|\chi_1\rangle=2(a-1)|\chi_1\rangle$$
should be zero. So $a=1$.
Then consider $(L_{-2}+\gamma L_{-1}^2)|\chi_2\rangle$. Restricting $a=1$ and using Virasoro algebra for some times, we find that $L_{-1}(L_{-2}+\gamma L_{-1}^2)|\chi_2\rangle=0$ and $L_{-2}(L_{-2}+\gamma L_{-1}^2)|\chi_2\rangle=0$ respectively give
$$3-2\gamma=0\quad\textrm{ and }\quad\frac{D}{2}-4-6\gamma=0.$$
So $D=26$.
