H
Size: a a a
2021 June 12
NM
Может выравнивание убрать?
NM
Зачем = под subset
H
убрал выравнивание, не помогло
H
H
H
\begin{eqnarray}
&& \mathcal{L} \supset \frac{g^2}{4} \left[ h_1^2 + h_2^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ h_1^2 + h_2^2 \right] Z_{\mu} Z^{\mu} =
\nonumber \\
&& \frac{g^2}{4} \left[ (h^0)^2 + (H^0)^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ (h^0)^2 + (H^0)^2 \right] Z_{\mu} Z^{\mu}.
\end{eqnarray}
&& \mathcal{L} \supset \frac{g^2}{4} \left[ h_1^2 + h_2^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ h_1^2 + h_2^2 \right] Z_{\mu} Z^{\mu} =
\nonumber \\
&& \frac{g^2}{4} \left[ (h^0)^2 + (H^0)^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ (h^0)^2 + (H^0)^2 \right] Z_{\mu} Z^{\mu}.
\end{eqnarray}
H
изначально было так
\begin{eqnarray}
\mathcal{L} &\supset & \frac{g^2}{4} \left[ h_1^2 + h_2^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ h_1^2 + h_2^2 \right] Z_{\mu} Z^{\mu} =
\nonumber \\
&=& \frac{g^2}{4} \left[ (h^0)^2 + (H^0)^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ (h^0)^2 + (H^0)^2 \right] Z_{\mu} Z^{\mu}.
\end{eqnarray}
\begin{eqnarray}
\mathcal{L} &\supset & \frac{g^2}{4} \left[ h_1^2 + h_2^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ h_1^2 + h_2^2 \right] Z_{\mu} Z^{\mu} =
\nonumber \\
&=& \frac{g^2}{4} \left[ (h^0)^2 + (H^0)^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ (h^0)^2 + (H^0)^2 \right] Z_{\mu} Z^{\mu}.
\end{eqnarray}
R
Переслано от Rikudo
\begin{align}
\mathcal{L} &\supset \frac{g^2}{4} \left[ h_1^2 + h_2^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ h_1^2 + h_2^2 \right] Z_{\mu} Z^{\mu} = {} \nonumber \\
{}&= \frac{g^2}{4} \left[ (h^0)^2 + (H^0)^2 \right] W^+_{\mu} W^{- \mu}
+ \frac{(g^2 + g^{\prime 2})}{8} \left[ (h^0)^2 + (H^0)^2 \right] Z_{\mu} Z^{\mu}.
\end{align}H
Да, спасибо, получилос!)
2021 June 14