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Bosonic Strings and their Verma modules

  1. Apr 1, 2015 #1
    Hi there!

    I have som troubles with representation theory.
    It is obvious that bosonic strings fields $X^{\mu}$ has zero conformal dimension $h=0$. But when one builds Verma module (open string for example) highest weight state has the following definition
    $$
    L_0 \vert h \rangle = 1 \vert h \rangle
    $$

    All descendants states have even higher grading with respect to $L_0$. This shift is necessary for the absence or spurious states (negative norm).

    How Verma module with $h=1$ coonected to open bosonic string with $h=0$?
     
  2. jcsd
  3. Apr 1, 2015 #2
    If Virasoro algebra has not central charge, Verma modules with $h=1$ and $h=0$ are in some sense equivalent
    $$
    \vert 1 \rangle = L_+ \vert 0 \rangle,
    $$
    where
    $$
    L_0 \vert 0\rangle =0 \;\; L_0 \vert 1 \rangle=-\vert 1 \rangle
    $$
    Applying lowering operators $L_-$
    $$
    L_- L_+ \vert 0\rangle = (L_- L_+ - 2L_0) \vert 0 \rangle =0
    $$
    State $\vert 1 \rangle$ is highest weight state from this simple exercise. But when Virasoro algebra has non-trivial central extension state $\vert 0 \rangle$ will apper in spectrum
     
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