Up to NS fermions

Signed-off-by: Riccardo Finotello <riccardo.finotello@gmail.com>
This commit is contained in:
2020-09-25 17:28:21 +02:00
parent e363da4b90
commit 5f7a0f734f
9 changed files with 984 additions and 240 deletions

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@@ -33,7 +33,7 @@ The parametrisation is such that:
U(\widehat{\vb{n}})
\label{eq:U_props}
\end{eqnarray}
where $\sigma^2$ is the second Pauli matrix, $\widetilde{\vb{n}} = \left( -n^1, n^2, -n^3 \right)$ and $\widehat{\vb{n}} = - \left(\frac{1}{2} -n \right)\, \frac{\vb{n}}{n}$.
where $\sigma^2$ is the second Pauli matrix, $\widetilde{\vb{n}} = \qty( -n^1, n^2, -n^3 )$ and $\widehat{\vb{n}} = - \qty(\frac{1}{2} -n )\, \frac{\vb{n}}{n}$.
The group product of two elements $U(\vb{n} \circ \vb{m} ) = U(\vb{n})\, U(\vb{m})$ has an explicit realisation as:
\begin{equation}
@@ -58,17 +58,17 @@ The group product of two elements $U(\vb{n} \circ \vb{m} ) = U(\vb{n})\, U(\vb{
Let $I = 1,\, 2,\, 3,\, 4$ and define:
\begin{equation}
\tau_I = \left( i\, \1_2,\, \vb{\sigma} \right),
\tau_I = \qty( i\, \1_2,\, \vb{\sigma} ),
\end{equation}
where $\vb{\sigma} = \left( \sigma^1,\, \sigma^2,\, \sigma^3 \right)$ are the Pauli matrices.
where $\vb{\sigma} = \qty( \sigma^1,\, \sigma^2,\, \sigma^3 )$ are the Pauli matrices.
It is possible to show that:
\begin{equation}
\begin{split}
\left( \tau_I \right)^{\dagger}
\qty( \tau_I )^{\dagger}
& =
\eta_{IJ}\, {\tau}^I,
\\
\left( \tau^I \right)^*
\qty( \tau^I )^*
& =
-\sigma_2\, \tau_I\, \sigma_2,
\end{split}
@@ -85,7 +85,7 @@ The following relations are then a natural consequence:
& = &
2\, \eta_{IJ},
\\
\tr(\tau_I \left( \tau_J \right)^{\dagger})
\tr(\tau_I \qty( \tau_J )^{\dagger})
& = &
2\, \delta_{IJ}.
\end{eqnarray}
@@ -99,7 +99,7 @@ We can recover the components using the previous properties:
X^I
=
\frac{1}{2}\, \delta^{IJ}\,
\tr(X_{(s)} \left( \tau_J \right)^{\dagger})
\tr(X_{(s)} \qty( \tau_J )^{\dagger})
=
\frac{1}{2}\, \eta^{IJ}\, \tr(X_{(s)} \tau_J),
\end{equation}
@@ -126,7 +126,7 @@ A rotation in spinor representation is defined as:
\end{equation}
and it is equivalent to:
\begin{equation}
\left( X' \right)^I
\qty( X' )^I
=
\tensor{R}{^I_J}\,
X^J
@@ -137,7 +137,7 @@ through
=
\frac{1}{2}
\tr(
\left( \tau_I \right)^{\dagger}\,
\qty( \tau_I )^{\dagger}\,
U_{L}(\vb{n})\,
\tau_J\,
U_{R}^{\dagger}(\vb{m})
@@ -167,7 +167,7 @@ From the second equation in \eqref{eq:tau_props} and the first equation in \eqre
\end{equation}
Furthermore the direct computation of the determinant of $R$ using the parametrisation~\eqref{eq:su2parametrisation} shows that $\det R = 1$.
Finally the explicit choice of the basis $\tau$ ensures $R$ to be a real matrix which ensures $R \in \SO{4}$.
Since $\left\lbrace U_{L},\, U_{R} \right\rbrace$ and $\left\lbrace -U_{L},\, -U_{R} \right\rbrace$ generate the same \SO{4} matrix then the correct isomorphism takes the form:
Since $\qty{ U_{L},\, U_{R} }$ and $\qty{ -U_{L},\, -U_{R} }$ generate the same \SO{4} matrix then the correct isomorphism takes the form:
\begin{equation}
\SO{4}
\cong

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@@ -15,9 +15,9 @@ In the main text we set
where $\cL(\vb{n}_{\vb{\infty}}) \in \SU{2}$.
The previous equation implies
\begin{equation}
\left( D\, \rM_{\vb{\infty}}\, D^{-1} \right)^\dagger
\qty( D\, \rM_{\vb{\infty}}\, D^{-1} )^\dagger
=
\left( D\, \rM_{\vb{\infty}}\, D^{-1} \right)^{-1},
\qty( D\, \rM_{\vb{\infty}}\, D^{-1} )^{-1},
\end{equation}
which can be rewritten as
\begin{equation}
@@ -147,7 +147,7 @@ which is satisfied by:
k'_{ab} \in \Z,
\end{split}
\end{equation}
where $p^{(L)},\, q^{(L)} \in \left\lbrace 0, 1 \right\rbrace$.
where $p^{(L)},\, q^{(L)} \in \qty{ 0, 1 }$.
Notice that changing the value of $p^{(L)}$ corresponds to swapping $a$ and $b$: since the hypergeometric function is symmetric in those parameters we can fix $p^{(L)}=0$.
Redefining $k'$ we can always set $q^{(L)}=0$.
We therefore have:
@@ -170,16 +170,16 @@ We cannot thus choose a vanishing $k_{\delta^{(L)}_{\vb{\infty}}}$ but we have t
We find a third relation by considering the entry
\begin{equation}
\Im\left(
\Im\qty(
e^{+2\pi i \delta_{\vb{\infty}}^{(L)}}\,
D^{(L)}\,
\rM_{\vb{\infty}}^{(L)}\,
\left( D^{(L)} \right)^{-1}
\right)_{11}
\qty( D^{(L)} )^{-1}
)_{11}
=
\Im\left(
\Im\qty(
\cL(n_{\vb{\infty}})
\right)_{11}.
)_{11}.
\end{equation}
Using
\begin{equation}
@@ -221,7 +221,7 @@ We then write
\qquad
k_{abc}\in \Z,
\end{equation}
with $f^{(L)} \in \left\lbrace 0, 1 \right\rbrace$.
with $f^{(L)} \in \qty{ 0, 1 }$.
The request
\begin{equation}
A
@@ -284,10 +284,10 @@ So far we can summarise the results in
$K^{(L)}$ is finally determined from
\begin{equation}
\left( D^{(L)}\, \rM_{\vb{\infty}}\, \left( D^{(L)} \right)^{-1} \right)_{21}
\qty( D^{(L)}\, \rM_{\vb{\infty}}\, \qty( D^{(L)} )^{-1} )_{21}
=
e^{-2\pi i \delta_{\vb{\infty}}^{(L)}}\,
\left( \cL(n_{\vb{\infty}}) \right)_{21},
\qty( \cL(n_{\vb{\infty}}) )_{21},
\label{eq:fixing_K_21}
\end{equation}
and get:
@@ -310,7 +310,7 @@ We check the consistency condition \eqref{eq:K_consistency_condition} using~\eqr
The result is
\begin{equation}
\begin{split}
\left( K^{(L)} \right)^{-1}
\qty( K^{(L)} )^{-1}
& =
\frac{(-1)^{m_a + m_b + m_c}}{2 \pi^2}\,
\cG(1 - a^{(L)},\, 1 - b^{(L)},\, 2 - c^{(L)})\,
@@ -341,10 +341,10 @@ We can then rewrite~\eqref{eq:cos_n1} as
It is then possible to verify that the sum of the left and right hand sides of~\eqref{eq:n12+n22} and the last equation are equal to $1$.
The same consistency check can also be performed by computing $K^{(L)}$ from
\begin{equation}
\left( D^{(L)}\, \rM_{\vb{\infty}}\, \left( D^{(L)} \right)^{-1} \right)_{12}
\qty( D^{(L)}\, \rM_{\vb{\infty}}\, \qty( D^{(L)} )^{-1} )_{12}
=
e^{-2\pi i \delta_{\vb{\infty}}^{(L)}}\,
\left( \cL(n_{\vb{\infty}}) \right)_{12},
\qty( \cL(n_{\vb{\infty}}) )_{12},
\end{equation}
instead of \eqref{eq:fixing_K_21}.