# Strings - Visaro operators - basically commutator algebra

1. May 11, 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 $1+A-2B=0$
- Consider $L_2|x>=0$ to conclude that $d-4A-2B=0$

- 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 and L2

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

My working for L1 is here:

Operating on the three terms in turn:

where $L_1 = \frac{1}{2}(\sum\limits^n_{n=-\infty} \alpha^u_{1-n} \alpha^v_{n} \eta_{uv})$ and so using (*) the only alpha operators that do not commute are the negatives of each other,

so from $L_1$ for the first term we need to look at $\alpha_0\alpha_1=\alpha_1\alpha_0$

for the second term we look at $\alpha_{-1}\alpha_2$ etc and where $\alpha_0$ commutes with all.

$L_1 \alpha^u_{-1} \alpha^v_{-1} \eta_{uv} |k>=2\alpha^u_0 \alpha^v_{-1}\eta_{uv} |k>$ [1]

where this has came from considering the four product of alpha operators that we need to look at : $\alpha_0 \alpha_1 \alpha_{-1} \alpha_{-1}$ applying the commutator relation (*) twice to move $\alpha_1$ to the right which annihilates.

In a similar way I get:
$L_1 \alpha^u_{0} \alpha^v_{-2} \eta_{uv} |k>=2\alpha^u_0 \alpha^v_{-1}\eta_{uv} |k>$ [2]
$L_1 \alpha^u_{0} \alpha^v_{0} \alpha^a_{-1} \alpha^b_{-1} \eta_{ab} |k>=2\alpha_0.\alpha_0 \alpha_0^u\alpha^v_{-1}\eta_{uv} |k>$ [3]

So putting [1] , [2] and [3] together:

$L_1|x>=(2\alpha_0 + 2A\alpha_0 + 2B\alpha_0(\alpha_0.\alpha_0))\alpha_{-1}|k>=0$

$(2+2A+2B\alpha_0.\alpha_0)\alpha_0\alpha_{-1}|k>=0$

$\sqrt{2\alpha'}(2)(1+A+\alpha_0.\alpha_0 B)p^u\alpha_{-1}|k>=0$

$(1+A+B\alpha_0.\alpha_0)k^u|k'>=0$

where I have defined $|k'>=\alpha_{-1}|k>$ .I'm not sure I completely understand the $\alpha_{-1}|k>$ here, I think this works because the whole expression vanishes for eigenstate $\alpha_{-1}|k>$ and since $\alpha_0$ commutes with all we can move this all the way to right , can someone correct me if this is wrong please?

$p^2 = -m^2$ and (from using the mass result deduced from $L_0$ which I am stuck on, see below ) $m^2=\frac{1}{\alpha'}$
so $\alpha_0.\alpha_0=p^22 \alpha'=-m^2 2 \alpha'$
$m^2=1/\alpha' \implies \alpha_0.\alpha_0=-2$

Therefore we have:

$\implies (1+A-2B)k^u|k>=0$
$\implies (1+A-2B)=0$

I expected something similar is needed for L0 and L2 .

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)

Here are my ideas for L2:

The relevant four products to consider are:

$\alpha_1\alpha_1\alpha_{-1}\alpha_{-1}$ , $\alpha_0\alpha_0\alpha_2\alpha_{-2}$, $\alpha_0\alpha_0\alpha_1\alpha_1\alpha_{-1}\alpha_{-1}$

from the first, second and third term of $|x>$ respectively.

I am confused how this is going to work out, similar to the logic used for L1 I use (*) to move annihilaters to the right, however my end result is going to be something of the form:

$p^u \alpha_{-1}+Ap^u\alpha_{-2}+Bp^u\alpha_{-1}|k>$ i.e. there is no common eigenstate so how can any conclusions be made? unlike L1 where we had $\alpha_{-1} |k>$ common,

or is it the case that

$\alpha_{-2} |k>= \alpha_{-1}\alpha_{-1}|k>$?