Continuing with fermions

Signed-off-by: Riccardo Finotello <riccardo.finotello@gmail.com>
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2020-09-30 14:43:55 +02:00
parent c47dd39f25
commit 2358ce8ab8
2 changed files with 58 additions and 50 deletions

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@@ -2188,57 +2188,65 @@ From the usual definition of the stress-energy tensor in terms of the Virasoro g
\end{split}
\end{equation}
We already hinted to the fact that the vacua state involved are not in general \SL{2}{R} invariant.
In particular we can see that that the excited vacua \eexcvacket is a primary field
\begin{equation}
\begin{split}
L_{(\rE)\, k} \eexcvacket
& =
0,
\qquad
k > 0,
\\
L_{(\rE)\, 0} \eexcvacket
& =
\frac{\rE^2}{2} \eexcvacket,
\end{split}
\end{equation}
with non trivial conformal dimensions $\Delta\qty( \eexcvacket ) = \frac{\rE^2}{2}$.
This operator is an excited spin field $\rS_{\rE_{(t)}}\qty( x )$ inserted at $x = 0$.
Its equivalent expression using bosonisation is:
\begin{equation}
\rS_{\rE}\qty( x ) = e^{i \rE \phi( x )},
\end{equation}
where $\phi$ is such that
\begin{equation}
\left\langle \phi( z ) \phi( w ) \right\rangle
=
-\frac{1}{( z - w)^2}.
\end{equation}
The minimal conformal dimension is achieved for $n_{\rE}=n_{\brE}=0$, i.e.\ $\Delta\qty( \twsvacket ) = \frac{\epsilon^2}{8}$, and identifies a plain spin field.
We can further check this idea by showing that the conformal dimensions are consistent.
Using~\eqref{eq:usual-twisted-fermion-conformal-twisted} we get:
\begin{equation}
\begin{split}
L_{(\rE)\, 0} \twsvacket
& =
L_0\,
\qty(%
b^{*\, ( \brE )}_0\,
b^{*\, ( \brE )}_{-1}\,
\dots\,
b^{*\, ( \brE )}_{2-n_{\rE}}
\eexcvacket
)
\\
& =
\left[%
\finitesum{n}{1}{n_{\rE}}
\qty( n - \frac{\rE + 1}{2} )
+
\frac{\rE^2}{2}
\right]
\twsvacket
=
\frac{\epsilon^2}{8} \twsvacket.
\end{split}
\end{equation}
%%% TODO %%%
Looking back at the analysis of the excited and twisted vacua, we already
hinted to the fact that they are not in general \SL{2}{R} invariant.
In particular we can see that that the excited vacua $\eexcvacket$
is a primary field
\begin{equation}
L_{(\rE) k > 0} \eexcvacket =0,
\qquad
L_{(\rE) 0} \eexcvacket = \frac{\rE^2}{2} \eexcvacket
,
\end{equation}
with non trivial conformal dimensions
$\Delta\qty( \eexcvacket ) = \frac{\rE^2}{2}$.
This operator is an excited spin field $\rS_{\rE_{(t)}}\qty( x )$
inserted at $x=0$ whose bosonized expression is given by
\begin{equation}
\rS_{\rE}\qty( x ) = e^{i \rE \phi( x )},
\end{equation}
where $\phi$ is such that
\begin{equation}
\left\langle \phi( z ) \phi( w ) \right\rangle = -\frac{1}{( z -
w)^2}
.
\end{equation}
In fact the minimal conformal dimension is achieved
for $n_{\rE}=n_{\brE}=0$, i.e.
$
\Delta\qty( \twsvacket ) = \frac{\epsilon^2}{8}
$
and we know this is the basic spin field.
We can further check this idea
by showing that the conformal dimensions are consistent.
Using \eqref{eq:usual-twisted-fermion-conformal-twisted} we get
\begin{equation}
\begin{split}
L_{(\rE) 0}\twsvacket
& =
L_0
\qty( b^{*\, ( \brE )}_0 b^{*\, ( \brE )}_{-1} \dots b^{*\, ( \brE )}_{2-n_{\rE}} \eexcvacket)
\\
& =
\left[ \sum\limits^{n_{\rE}}_{n = 1} ( n - \frac{\rE + 1}{2} )
+\frac{\rE^2}{2}
\right] \twsvacket
= +\frac{1}{8} \epsilon^2\twsvacket .
\end{split}
\end{equation}
\subsection{Generic Case With Defects}
We will now apply the same procedure to the generic case of one complex

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@@ -7,7 +7,7 @@
\usepackage{import}
\author{Riccardo Finotello}
\title{Theoretical and Computational Aspects of String Theory and Their Phenomenological Implications}
\title{Theoretical and Computational Aspects for Phenomenology in String Theory}
\advisor{Igor Pesando}
\institution{Università degli Studi di Torino}
\school{Scuola di Dottorato}