Saltar al contenido principal

# 5.2: Invarianza de Lorentz y multiplicación bilateral

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Para matrices hermitianas:$$K^{\dagger} = K, \bar{K} = \tilde{K}$$ y lo mismo para R. ¿Por qué multiplicación bilateral? Eliminar factores no físicos indicados como$$\underbrace{}$$.

$\begin{array}{c} {\begin{pmatrix} {e^{(\mu-i \phi)/2}}&{0}\\ {0}&{e^{-(\mu-i \phi)/2}} \end{pmatrix} \begin{pmatrix} {k_{0}+k_{3}}&{k_{1}-i k_{2}}\\ {k_{1}+i k_{2}}&{k_{2}-k_{3}} \end{pmatrix} =}\\ {\begin{pmatrix} {e^{\mu/2}(k_{0}+\mu_{3}) \underbrace{e^{-i \phi/2}}}&{\underbrace{e^{\mu/2}}(k_{1}-i k_{2}) e^{-i \phi/2}}\\ {\underbrace{e^{-\mu/2}}(k_{1}+i k_{2}) e^{-i \phi/2}}&{e^{\mu/2}(k_{0}-k_{3}) \underbrace{e^{i \phi/2}}} \end{pmatrix} \times}\\ {\begin{pmatrix} {k_{0}+k_{3}}&{k_{1}-ik_{2}}\\ {k_{1}+ik_{2}}&{k_{2}-k_{3}} \end{pmatrix} \begin{pmatrix} {e^{(\mu+i \phi)/2}}&{0}\\ {0}&{e^{-(\mu+i \phi)/2}} \end{pmatrix} =}\\ {\begin{pmatrix} {e^{\mu/2}(k_{0}+\mu_{3}) \underbrace{e^{i \phi/2}}}&{\underbrace{e^{-\mu/2}}(k_{1}-i k_{2}) e^{-i \phi/2}}\\ {\underbrace{e^{-\mu/2}}(k_{1}+i k_{2}) e^{i \phi/2}}&{e^{\mu/2}(k_{0}-k_{3}) \underbrace{e^{-i \phi/2}}} \end{pmatrix} \times}\\ {\begin{pmatrix} {e^{(\mu-i \phi)/2}}&{0}\\ {0}&{e^{-(\mu-i \phi)/2}} \end{pmatrix} \begin{pmatrix} {k_{0}+k_{3}}&{k_{1}-i k_{2}}\\ {k_{1}+i k_{2}}&{k_{2}-k_{3}} \end{pmatrix} \begin{pmatrix} {e^{(\mu+i \phi)/2}}&{0}\\ {0}&{e^{-(\mu+i \phi)/2}} \end{pmatrix} =}\\ {\begin{pmatrix} {e^{\mu/2}(k_{0}+\mu_{3})}&{e^{-i \phi/2} (k_{1}-i k_{2})}\\ {e^{i \phi/2} (k_{1}+i k_{2})}&{e^{\mu/2}(k_{0}-k_{3})} \end{pmatrix}} \end{array}$

O bien, en forma de$$4 \times 4$$ matriz:

$\begin{array}{c} {\begin{pmatrix} {k_{1}'}\\ {k_{2}'}\\ {k_{3}'}\\ {k_{0}'} \end{pmatrix} = \begin{pmatrix} {\cos \phi}&{-\sin \phi}&{0}&{0}\\ {\sin \phi}&{\cos \phi}&{0}&{0}\\ {0}&{0}&{\cosh \mu}&{\sinh \mu}\\ {0}&{0}&{\sinh \mu}&{\cosh \mu} \end{pmatrix} \begin{pmatrix} {k_{1}'}\\ {k_{2}'}\\ {k_{3}'}\\ {k_{0}'} \end{pmatrix}} \end{array}$

Rotación circular alrededor del eje z por$$\phi$$ y rotación hiperbólica a lo largo del mismo asix por el ángulo ehiper bólico$$\mu$$: Lorentz de cuatro tornillos:$$\mathcal{L}(\phi, \hat{z}, \mu)$$. Estas transformaciones forman un grupo abeliano.

En el álgebra de Pauli se mantiene la simplicidad formal de estas relaciones incluso para direcciones axiales arbitrarias. Sin duda, la obtención de resultados explícitos de los productos bilaterales puede llegar a ser algo más cumber. Sin embargo, los resultados vectoriales estándar se pueden extraer fácilmente.

This page titled 5.2: Invarianza de Lorentz y multiplicación bilateral is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by László Tisza (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.