String theory - no-ghost state - virasoro constraints

[binbagsss]

1. Homework Statement

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. Homework 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)

 

[AndyW]

Thank you for sharing your attempt at solving the problem. I appreciate the effort you have put into it. However, I believe there may be a few errors in your calculations.

Firstly, in your first line, you have written $L_0=\alpha_0^2+2\sum\limits_{n=1}\alpha_{-n}\alpha_{n}-1$. This is incorrect as the correct expression is $L_0=\alpha_0^2+2\sum\limits_{n=1}\alpha_{-n}\alpha_{n}+1$. The reason for this is that the zero mode term $\alpha_0^2$ should have a positive sign and not a negative sign. Additionally, the constant term in the expression should be a positive one, as we will see in the next steps.

Next, you have used the commutation relation $[\alpha_n^u, \alpha_m^v]=n\delta_{n+m}\eta^{uv}$ to calculate $L_0 \alpha_{-1}\alpha_{-1}|k>$. However, this relation holds only for $n, m \neq 0$. For the zero modes, the commutation relation is $[\alpha_0^u, \alpha_0^v]=\eta^{uv}$. Using this, we can write $L_0 \alpha_{-1}\alpha_{-1}|k>=(\alpha_0^2+2\alpha_0.\alpha_0+1)\alpha_{-1}\alpha_{-1}|k>$.

Now, using the definition of the zero mode operators and the momentum operator, we can write $p^u|k>=k^u|k> \implies \alpha_0^u|k>=k^u/\sqrt{2\alpha'}|k>$. Substituting this in the above expression, we get $L_0 \alpha_{-1}\alpha_{-1}|k>= (2\alpha'p^2+3)\alpha_{-1}\alpha_{-1}|k>$. Now, using the fact that $\alpha_{-1}\alpha_{-1}|k>=0$, we get $L_0 \alpha_{-1}\alpha_{-1}|k>=0$.

Therefore, we have