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Homework Help: QFT T-duality, Massless vector states

  1. Dec 10, 2017 #1
    1. The problem statement, all variables and given/known data

    Part C) Please:

    massless.png
    2. Relevant equations

    above,below

    3. The attempt at a solution

    so I think I understand the background of these expressions well enough, very briefly, changing the manifold from ## R^n ## to a cylindrical one- ##R^{(n-1)}^{+1}## we need to cater for winding modes, the momentum and winding momentum for the circular dimension can not take arbitrary values and are quantified, ##n,m \in Z##

    And importantly, the level-matching constraint is no longer required to hold and instead replaced by the second equation in c) .

    For the combinations I get:

    a) ##n=m=0 ## ##N=\bar{N}=1##
    b) ##n=m=1=N## ##\bar{N}=0##
    c) ##n=2## ##m=0=N=\bar{N}##
    d) ##m=2## ##n=N=\bar{N}=0##

    I am completely stuck on which of these combinations transforms as a vector. The only notes relevant to it I seem to have is the following attached, (bit underlined in pink):

    everytableineverylongue.png

    Is this referring to the ladder operator carrying a transverse index? or the state |p> ?

    So out of the combinations above I have:

    a) would require both a ## \alpha^j ## and a ## \bar{\alpha^j} ##
    b) would require just a ## \alpha^j ##
    c) & d) would require no ladder operators.

    Is the above relevant/needed at all or not, for what transforms as a vector or what doesn't, what defintion am I needing to go by here?

    Many thanks in advance.
     
    Last edited by a moderator: Dec 10, 2017
  2. jcsd
  3. Dec 15, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Dec 19, 2017 #3
    bump please, thank you so much.
     
  5. May 19, 2018 #4
    ok from my notes and just following the idea i believe i can get the correct answer as to which states transform as a vector and which don't. However, I don;t understand why

    Both, my lecture notes and David Tong's , lecture notes seem to just jump into these ideas of SO(D-2) being a transverse group- I understand the reduction of D to D-2 via light-cone gauge, I have no idea why this is a 'transverse group' and why this implies any states with a transverse index must transform as a vector. (Nor the irreducible decomposition into a graviton blah blah etc).

    In T-duality I understand the expression for the mass, as observed by an observer in the compactified dimension, so the momentum-mass relation rather than summing over the momentum of each dimension, sums over all of those of R^n and not this compactified dimension. This compactified dimension has winding momentum + the same momentum term as the other modes of the other dimensions in R^n, however this time it is quantized, and so as usual we use L_0^L - L_0^R =0 to get the mass expression (where the L and R denote the left moving modes of the string and the right , for a closed string), the \alpha_0 as usual $\approx p $ , $p$ the momentum,and so \alpha_0^L.\alpha_0^L-\alpha_0^L.\alpha_0^R gives m^2+(momentum terms from the X^{25} dimension). The result is the mass depends on the N operators, as before, but also the integers n and m via the relationship given in the OP, and is invariant under R \to 1\2\pi R - T-duality. So massless combinations are given by various sets of integers (N^L,N^R,n,m) . Now I have two questions looking at this:

    1) The N^L and N^R are operators and so the 'integers' they take is an eigenvalue of the state. e.g \alpha_{-1}^i^L\alpha_{-1}^i^R |0>gives arise to eigenvalues 1 for both N^L and N^R. (via commutator relations of a three product of \alpha operators) , but, how are n and m specified in this respect with respect to the chosen state \alpha_{-1}^i^L\alpha_{-1}^i^R |0>. Do we just merely state these values with a state? surely it can be specified within the state notation, this wouldn't really make sense to me to state them separately.

    2) As said above, my lecture notes immediately simply state that any state with one transverse index transforms as a vector since the dimensions associated to i form the rotation group (D-2). Now let me start with the notation u=0,1,2....24,25, D=26, now in the light cone let me denote 0 by + and 1 by -, the + \alpha operators are set to zero via freedom left over and then one can show that the \alpha^- can be expressed in terms of the remaining (oscillatory)dimensions, so now let me denote 2,3...25 as i. Now let me turn to a compactified dimension, i=2,3...24 only.

    Ok so my notes conclude, the state used in my question 1, having two transverse indexes, transforms as a tensor rather than a vector. This would make sense if I understand why a transverse index implies the state must transform as a vector.

    My notes also say:

    \alpha^25^R.\alpha_{-1}^i^L|0> transforms as a vector, only having one transverse index

    and

    \alpha^25^L \alpha^25^R |0> transforms as a scalar, having no transverse index.


    So basically just to ask why the X^25 associated to this compactified dimension are not part of the transverse rotation group, is it to do with the nature of the underlying space or not- i.e. circular compared to R^n. Could one say it occupies a group itself, with only one element itself, and why does this group transform as a scalar rather than a vector- probably a really stupid but if such a group only has one element, as it does here, can it only transform as a scalar, does there need to be more than one element for it to transform as a vector.

    (So say we introduced another, identical in its properties, circular dimension X^26, then would \alpha^25^L_{-1}\alpha^25^R_{-1}|0> transform as a scalar or a vector?)

    Many thanks
     
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