177 lines
4.6 KiB
TeX
177 lines
4.6 KiB
TeX
In this appendix we explain the conventions used for \SU{2} and show the details of the isomorphism between \SO{4} and a class of equivalence of $\SU{2} \times \SU{2}$.
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\subsection{Conventions}
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We parameterise \SU{2} matrices $U$ with a vector $\vec{n} \in \R^3$ such that:
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\begin{equation}
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U(\vec{n})
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=
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\cos(2 \pi n)\, \1_2
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+
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i\, \frac{\vec{n} \cdot \vec{\sigma}}{n}\, \sin(2 \pi n),
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\label{eq:su2parametrisation}
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\end{equation}
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where $n = \norm{\vec{n}}$ and $0 \le n \le \frac{1}{2}$.
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We also identify all $\vec{n}$ when $n=\frac{1}{2}$ since in this case $U(\vec{n})= -\1_2$.
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The parametrisation is such that:
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\begin{eqnarray}
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U^*(\vec{n})
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& = &
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\sigma^2\, U(\vec{n})\, \sigma^2
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=
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U(\widetilde{\vec{n}}),
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\\
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U^{\dagger}(\vec{n})
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& = &
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U^T(\widetilde{\vec{n}})
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=
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U(-\vec{n}),
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\\
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-U(\vec{n})
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& = &
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U(\widehat{\vec{n}})
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\label{eq:U_props}
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\end{eqnarray}
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where $\sigma^2$ is the second Pauli matrix, $\widetilde{\vec{n}} = \qty( -n^1, n^2, -n^3 )$ and $\widehat{\vec{n}} = - \qty(\frac{1}{2} -n )\, \frac{\vec{n}}{n}$.
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The group product of two elements $U(\vec{n} \circ \vec{m} ) = U(\vec{n})\, U(\vec{m})$ has an explicit realisation as:
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\begin{equation}
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\begin{split}
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\cos(2 \pi \norm{\vec{n} \circ \vec{m}})
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& =
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\cos(2 \pi n)\, \cos(2 \pi m)
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-
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\sin(2 \pi n)\, \sin(2\pi m)\, \frac{\vec{n} \cdot \vec{m}}{n\, m},
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\\
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\sin(2 \pi \norm{\vec{n} \circ \vec{m}})\,
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\frac{\vec{n} \circ \vec{m}}{\norm{\vec{n} \circ \vec{m}}}
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& =
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\cos(2 \pi n)\, \sin(2\pi m)\, \frac{\vec{m}}{m}
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+
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\sin(2 \pi n)\, \cos(2\pi m)\, \frac{\vec{n}}{n}.
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\end{split}
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\label{eq:product_in_SU2}
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\end{equation}
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\subsection{The Isomorphism}
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Let $I = 1,\, 2,\, 3,\, 4$ and define:
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\begin{equation}
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\tau_I = \qty( i\, \1_2,\, \vec{\sigma} ),
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\end{equation}
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where $\vec{\sigma} = \qty( \sigma^1,\, \sigma^2,\, \sigma^3 )$ are the Pauli matrices.
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It is possible to show that:
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\begin{equation}
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\begin{split}
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\qty( \tau_I )^{\dagger}
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& =
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\eta_{IJ}\, {\tau}^I,
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\\
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\qty( \tau^I )^*
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& =
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-\sigma_2\, \tau_I\, \sigma_2,
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\end{split}
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\label{eq:tau_props}
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\end{equation}
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where $\eta_{IJ} = \mathrm{diag}(-1,1,1,1)$.
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The following relations are then a natural consequence:
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\begin{eqnarray}
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\tr(\tau_I)
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& = &
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2\, i\, \delta_{I1},
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\\
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\tr(\tau_I \tau_J)
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& = &
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2\, \eta_{IJ},
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\\
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\tr(\tau_I \qty( \tau_J )^{\dagger})
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& = &
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2\, \delta_{IJ}.
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\end{eqnarray}
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Now consider a vector in the spinor representation:
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\begin{equation}
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X_{(s)} = X^I\, \tau_I.
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\end{equation}
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We can recover the components using the previous properties:
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\begin{equation}
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X^I
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=
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\frac{1}{2}\, \delta^{IJ}\,
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\tr(X_{(s)} \qty( \tau_J )^{\dagger})
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=
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\frac{1}{2}\, \eta^{IJ}\, \tr(X_{(s)} \tau_J),
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\end{equation}
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where the trace acts on the space of the $\tau$ matrices.
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If the vector $X^I$ is real, using~\eqref{eq:tau_props} we have:
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\begin{equation}
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\begin{split}
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X_{(s)}^{\dagger}
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& =
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X^I\, \eta_{IJ}\, \tau^J
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=
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\frac{1}{2} \tr(X_{(s)} \tau_I)\, \tau^I,
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\\
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X_{(s)}^*
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& =
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- \sigma_2\, X_{(s)}\, \sigma_2.
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\end{split}
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\label{eq:X_dagger}
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\end{equation}
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A rotation in spinor representation is defined as:
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\begin{equation}
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X'_{(s)} = U_{L}(\vec{n})\, X_{(s)}\, U_{R}^{\dagger}(\vec{m})
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\end{equation}
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and it is equivalent to:
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\begin{equation}
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\qty( X' )^I
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=
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\tensor{R}{^I_J}\,
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X^J
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\end{equation}
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through
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\begin{equation}
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R_{IJ}
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=
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\frac{1}{2}
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\tr(
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\qty( \tau_I )^{\dagger}\,
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U_{L}(\vec{n})\,
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\tau_J\,
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U_{R}^{\dagger}(\vec{m})
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).
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\end{equation}
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The matrix $R$ is the $4$-dimensional rotation matrix we are looking for since:
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\begin{equation}
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\tr(X'_{(s)}\, (X')^{\dagger}_{(s)})
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=
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\tr(X_{(s)}\, X^{\dagger}_{(s)})
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\qquad
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\Rightarrow
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\qquad
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\finitesum{K}{1}{4} R_{IK} R^*_{JK} = \delta_{I \,J}.
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\end{equation}
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From the second equation in \eqref{eq:tau_props} and the first equation in \eqref{eq:U_props} we then get the reality condition on $R$:
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\begin{equation}
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R_{NM}
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=
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\frac{1}{2}\, \eta_{NI}\, \eta_{MJ}\,
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\tr(\tau_I ^{\dagger}\, U_{R}\, \tau_J\, U_{L}^{\dagger})
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=
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\frac{1}{2}
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\tr(\tau_N\, U_{R}\, \tau_M^\dagger\, U_{L}^{\dagger})
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=
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R_{NM}^*.
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\end{equation}
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Furthermore the direct computation of the determinant of $R$ using the parametrisation~\eqref{eq:su2parametrisation} shows that $\det R = 1$.
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Finally the explicit choice of the basis $\tau$ ensures $R$ to be a real matrix which ensures $R \in \SO{4}$.
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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:
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\begin{equation}
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\SO{4}
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\cong
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\frac{\SU{2} \times \SU{2}}{\Z_2}.
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\end{equation}
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