5.3: Типові приклади

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Page ID: 63553

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Приклад 1

\[\begin{array}{cc} {K' = HKH,}&{H = \exp (\frac{\mu}{2} \hat{h} \cdot \vec{\sigma})}\\ {\vec{k} = \vec{k}_{\parallel}+\vec{k}_{\perp}}&{\vec{k}_{\parallel} = (\vec{k} \cdot \hat{h}) \hat{h}} \end{array}\]

Використовуючи (6a) і (7b):

\[\begin{array}{cc} {\vec{k}_{\parallel} \cdot \vec{\sigma} H = H \vec{k}_{\parallel} \cdot \vec{\sigma},}&{\vec{k}_{\perp} \cdot \vec{\sigma} H = H^{-1} \vec{k}_{\perp} \cdot \vec{\sigma}}\\ {}&{\vec{k}'_{\parallel} = \vec{k}_{\parallel} = k \hat{h}} \end{array}\]

\[\begin{array}{c} {(k'_{0}+\vec{k}'_{\parallel} \cdot \vec{\sigma}) = H^{2} (k_{0}+\vec{k}_{\parallel} \cdot \vec{\sigma})}\\ {(\cosh \mu+\sinh \mu \hat{h} \cdot \vec{\sigma})(k_{0}+\vec{k}_{\parallel} \cdot \vec{\sigma})} \end{array}\]

\[\begin{array}{c} {k'_{0} = k_{0} \cosh \mu+ k \sinh \mu}\\ {k' = k_{0} \sinh \mu+ k \cosh \mu} \end{array}\]

Приклад 2

\[\begin{array}{cc} {K' = UKU^{-1},}&{U = \exp (-i \frac{\phi}{2} \hat{u} \cdot \vec{\sigma})}\\ {\vec{k} = \vec{k}_{\parallel}+\vec{k}_{\perp}}&{\vec{k}_{\parallel} = (\vec{k} \cdot \hat{u}) \hat{u}} \end{array}\]

\[\begin{array}{cc} {\vec{k}_{\parallel} \cdot \vec{\sigma} U^{-1} = U^{-1} \vec{k}_{\parallel} \cdot \vec{\sigma},}&{\vec{k}_{\perp} \cdot \vec{\sigma} U^{-1} = U \vec{k}_{\perp} \cdot \vec{\sigma}}\\ {}&{\vec{k'}_{\parallel} = \vec{k}_{\parallel}} \end{array}\]

\[\begin{array}{c} {\vec{k}'_{\perp} \cdot \vec{\sigma} = (\cos \frac{\phi}{2} 1-i \sin \frac{\phi}{2} \hat{u} \cdot \vec{\sigma})^{2} \vec{k}_{\perp} \cdot \vec{\sigma}}\\ {= (\cos \phi 1-i \sin \phi \hat{u} \cdot \vec{\sigma}) \vec{k}_{\perp} \cdot \vec{\sigma}} \end{array}\]

\[\begin{array}{c} {\vec{k}'_{\perp} \cdot \vec{\sigma} = \cos \phi \vec{k}_{\perp}+\sin \phi \hat{u} \times \vec{k}_{\perp}} \end{array}\]