Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

Virasoro Operators, String Theory/ Quantum Theory/ Ghost state, dimension of s-t watch

    • Thread Starter

    1. The problem statement, all variables and given/known data


    (with the following definitions here in the attachment please)

    - 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

    - 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 itdropped 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)
    Attached Images
Have you ever experienced racism/sexism at uni?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.