# A String theory - no-ghost state - virasoro constraints

1. May 19, 2017

### binbagsss

1. The problem statement, all variables and given/known data

Question (with the following definitions here):

- Consider $L_0|x>=0$ to show that $m^2=\frac{1}{\alpha'}$
- Consider $L_1|x>=0$ to conclude that $1+A-2B=0$

- where $d$ is the dimension of the space $d=\eta^{uv}\eta_{uv}$

For the L1 operator I am able to get the correct expression of $1+A-2B=0$
I am struggling with L0

Any help much appreciated.

2. Relevant equations

$\alpha^u_0={p^u}\sqrt{2 \alpha'}$

$\alpha_{n>0}$ annihilate

$\alpha_{n<0}$ create

$[\alpha_n^u, \alpha_m^v]=n\delta_{n+m}\eta^{uv}$ (*)

where $\eta^{uv}$ is the Minkowski metric

$p^u|k>=k^u|k>$

3. The attempt at a solution

Here is my L0 attempt- Consider $L_0 |x>=0$ to show that $m^{2}=1/\alpha'$

$L_0=(\alpha_0^2+2\sum\limits_{n=1}\alpha_{-n}\alpha_{n}-1)$

So first of all looking at the first term of $|x>$ I need to consider:

$L_0 \alpha_{-1}\alpha_{-1}|k> =(\alpha_0^2+2\alpha_{-1}\alpha_{1}-1)\alpha_{-1}\alpha_{-1}$

Considering the four product operator and using the commutators in the same way as done for $L_1$ I get from this:

$L_0\alpha_{-1}\alpha_{-1}|k> =(\alpha_0^2+4-1)\alpha_{-1}\alpha_{-1}|k>$ (**)

Here's how I got it:(dropped indices in places, but just to give idea, $\eta^{uv}$ the minkowksi metric)
$2\alpha_{-1}\alpha_{1}\alpha_{-1}\alpha_{-1} |k> = 2(\alpha_{-1}(\alpha_{-1}\alpha_1+\eta)\alpha_{-1})|k> = 2(\alpha_{-1}\alpha_{-1}\alpha_1\alpha_{-1}+\eta\alpha_{-1}\alpha_{-1})|k> = 2(\alpha_{-1}\alpha_{-1}(\alpha_{-1}\alpha_{1}+\eta)+\eta\alpha_{-1}\alpha_{-1})|k> =2(\alpha_{-1}\alpha_{-1}(0+\eta|k>)+\eta\alpha_{-1}\alpha_{-1}|k>) = 2(2\alpha_{-1}.\alpha_{-1})$

so from (**) I have:

$L_0\alpha_{-1}\alpha_{-1}|k> =(\alpha_0^2+3)\alpha_{-1}\alpha_{-1}|k>=0$
$=(2\alpha'p^2+3)\alpha_{-1}\alpha_{-1}|k>=0$
$\implies 2\alpha'p^2+3=0$
$\implies 2(-m^2)\alpha'=-3$

So I get $m^{2}=3/\alpha'$ and not $1/\alpha'$ :(

Any help much appreciated ( I see the mass is independent of $A$ and $B$ so I thought I'd deal with the first term before confusing my self to see why these terms vanish)

2. May 24, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.