Strings, Virasoro Operators&constraints, commutator algrebra

[binbagsss]

Homework Statement

[/B]
Question:

(With the following definitions here: View attachment 203368 )

- 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.

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>##

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 .

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)
Many thanks in advance

 

[mark726]

,

Scientist

Dear Scientist,

Thank you for sharing your attempts on this problem. I can see that you have put a lot of effort into understanding the operators and their commutation relations. However, I believe there may be some misunderstandings that are causing you to get incorrect results.

Firstly, for the L1 operator, you have correctly shown that the expression ##1+A-2B=0## holds. However, your method of using the commutation relations to move the operators around may not be the most efficient approach. Instead, you could use the definition of the L1 operator and directly evaluate its action on the state |x>. This would give you:

##L_1|x> = \frac{1}{2}(\sum\limits_{n=-\infty}^{n=\infty} \alpha^u_{1-n} \alpha^v_{n} \eta_{uv})|x> = \frac{1}{2}(\alpha^u_0 \alpha^v_1 \eta_{uv} + \alpha^u_{-1} \alpha^v_2 \eta_{uv} + \alpha^u_{-2} \alpha^v_3 \eta_{uv} + \dots)|x>##

Using the commutation relations, you can simplify this expression to:

##L_1|x> = \frac{1}{2}(\alpha^u_0 \alpha^v_1 \eta_{uv} - \alpha^u_{-1} \alpha^v_1 \eta_{uv})|x> = \frac{1}{2}(\alpha^u_0 - \alpha^u_{-1})(\alpha^v_1 \eta_{uv})|x>##

Since ##\alpha^u_0## and ##\alpha^u_{-1}## commute, you can rewrite this as:

##L_1|x> = \frac{1}{2}(\alpha^u_0^2 - \alpha^u_{-1}\alpha^u_0)(\alpha^v_1 \eta_{uv})|x>##

Using the definition of the L0 operator, you can see that ##\alpha^u_0^2 = \frac{1}{2}(\alpha^u_0 \alpha^u_0 + \alpha^u_{-1}\alpha^