Recent content by Keita

Thank you for your suggestion. I also appreciate your note on Zee’s equation(17) on page 72.

Thank you. I appreciate your suggestion.

Sorry for the delay of my response. I appreciate your explanation. Still, I would like to stand to my argument. In the original x-y coordinate, $$ z = \frac{1}{2}ax^2 + cxy + \frac{1}{2}by^2 = \frac{1}{2} \vec{x}^{T} M \vec{x}. $$ $$ M = \begin{pmatrix} a & c \\ c & b \end{pmatrix}. $$ (pages...

Thank you for your answer. I understood your suggestion correctly. Now, let me show you my argument. In the original coordinate, we have the following. $$ z \sim \frac{1}{2} \vec {x}^{T} M\vec {x} (1)$$ (Top of page 84) In the rotated coordinate, we have the following. $$ z \sim \frac{1}{2}...

##R^T## Thank you for your suggestion. Let me confirm I understood your explanation correctly. Does your suggestion mean the following? $$ z \sim \frac{1}{2} \vec{x}^{T} M\vec{x} $$ (Top of page 84) $$ z \sim \frac{1}{2} \left(R\vec{x}\right) ^{T} M\left(R\vec{x}\right) =...

In p. 84, Zee says “In the new coordinates, M is replaced by M’ = R[-1]MR.” However, I figure out M is replaced by M’ = RMR[-1]. Why is M replaced by M’ = R[-1]MR?

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