# BRST-quantization of open string at level $N=1$

I have a question regarding an equation in Polchinski's "String Theory, Volume 1, An introduction to the bosonic string". The equation is (4.3.27) on p.135.

This section is about the brst-cohomology of the open bosonic string. My question in particular is about the $N=1$ level and the derivation of the cohomology.

By calculating $Q_B |\psi_1\rangle$ and eliminating all terms containing $c_0$ and $b_n$ where $n \geq 0$ since these annihilate the states, I end up with the following equation:

$$0 = Q_B |\psi_1\rangle = (c_{-1}L^{(m)}_{1} + c_{1}L^{(m)}_{-1})(e_\mu\alpha^\mu_{-1}+\beta b_{-1} + \gamma c_{-1})|0;\boldsymbol{k}\rangle$$

which by further multiplication simplifies to:

$$0 = (e_\mu c_{-1}L^{(m)}_{1}\alpha^\mu_{-1}+ \beta c_{-1}L^{(m)}_{1} b_{-1} + e_\mu c_{1}L^{(m)}_{-1} \alpha^\mu_{-1} + \beta c_{1}L^{(m)}_{-1} b_{-1} + \gamma c_{1}L^{(m)}_{-1} c_{-1})|0;\boldsymbol{k}\rangle$$

where $c_{-1}c_{-1}=0$ was used.

In Polchinski eq (4.3.27), this turns out to be:

$$0 = \sqrt{2\alpha'}(k^\mu e_\mu c_{-1} + \beta k_\mu \alpha_{-1}^\mu)|0;\boldsymbol{k}\rangle . \tag{4.3.27}$$

I don't quite see how the Virasoro operators and the ghost operators act on the state to produce this result, other than that it is probably the first and fourth term that gives the result with the others being zero. I've been trying to figure it out, but I haven't been able to, would someone be able to enlighten me?

The matter sector and the ghost sector commutes. The $L_1$ and $c_1$ annihilate the $|0;k\rangle$ state. Together with the commutation relations $\{c_n,b_m\}=\delta_{n+m}$ and $\left[\alpha_n^\mu,\alpha_m^\nu\right]=n\delta_{m+n}\eta^{\mu\nu}$ you have the sufficient information to know that the second, third and firth terms are zero.

If you does not understand why, see bellow:

First, note that $\left[L_0,\,L_1\right]=-L_1$. You can see that by the Virasoro Algebra:

$$\left[L_m,\,L_n\right]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m)\delta_{m+n}$$

This means that if $L_0|\psi\rangle = c|\psi\rangle$, then $L_0\left(L_1|\psi\rangle\right)=(c-1)\left(L_1|\psi\rangle\right)$, i.e. the $L_0$ decrease the level of the state by one. Now, in your case, $|\psi\rangle=|0;\,k\rangle$, a state in the N=0 level, so $L_1$ should annihilate this state since there is no $N=-1$ level. You can check this in a more explicit method by work out the $\alpha^{\mu}_m\,$-operators if you want.

Now, in the ghost sector, the ground state is $|\downarrow\,\rangle$, annihilated by $c_m$, $b_m$ for $m>0$, and $b_0$. In particular, $c_1|0;k\rangle=0$. You are going to use this later.

So, in the equation:

$$0 = \left(e_\mu c_{-1}L^{(m)}_{1}\alpha^\mu_{-1}+ \beta c_{-1}L^{(m)}_{1} b_{-1} + e_\mu c_{1}L^{(m)}_{-1} \alpha^\mu_{-1} + \beta c_{1}L^{(m)}_{-1} b_{-1} + \gamma c_{1}L^{(m)}_{-1} c_{-1}\right)|0;k\rangle$$

the second and third operators annihilate the $|0;k\rangle$ state. This is so because the matter sector commutes with the ghost sector, e.g. $L_1^{(m)}$ commutes with the $bc$-ghosts. Furthermore, $c_1$ and $c_{-1}$ anti-commute, so $c_1c_{-1}|0;k\rangle=-c_{-1}c_{1}|0;k\rangle=0$, and the firth term annihilate the state as well.

Now the equation reduces to:

$$0 = \left(e_\mu c_{-1}L^{(m)}_{1}\alpha^\mu_{-1} + \beta c_{1}L^{(m)}_{-1} b_{-1} \right)|0;k\rangle$$

Using $L_{\pm1}=(2\alpha')^{1/2}p\cdot\alpha_{\pm1}+\,...$ and we are done. The $...$ are operators that annihilate $|0,k\rangle$ and commutes with $\alpha_{\pm 1}$.

$$0 = (2\alpha')^{1/2}\left(e_\mu c_{-1}(p\cdot\alpha_{+1})\alpha^\mu_{-1} + \beta c_{1}(p\cdot\alpha_{-1}) b_{-1} \right)|0;k\rangle$$

using the commutation relations $[\alpha_{+1}^\mu,\,\alpha_{-1}^\nu]=\eta^{\mu\nu}$ and $\{b_{-1},c_1\}=1$:

$$0 = (2\alpha')^{1/2}\left(c_{-1}(p\cdot e) + \beta (p\cdot\alpha_{-1})\right)|0;k\rangle$$

Precisely the equation in Polchinski eq (4.3.27).