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Strings - Visaro operators - basically commutator algebra

  1. May 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Question:

    (With the following definitions here: definitions.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##
    - 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>##?

    Many thanks in advance
     
    Last edited: May 11, 2017
  2. jcsd
  3. May 16, 2017 #2
    bump. many thanks in advance.
     
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