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Strings, Virasoro Operators & constraints, mass of state

  1. May 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Question:

    (With the following definitions here: vira.png

    - 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:

    ##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
     
  2. jcsd
  3. May 19, 2017 #2

    fresh_42

    Staff: Mentor

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