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Is it mathematically valid to simplify the expression $$\left ( \bar \Psi \right)^2 \left(1, \ -1\right)$$ to 1 if $$\Psi = \begin{pmatrix} \cos(x) & 0 \\ 0 & i \sin(x) \end{pmatrix}$$ (where $\bar \Psi$ is the complex conjugate of $\Psi$)? Since \begin{align} \left( \bar \Psi \right)^2 \left(1,\ -1\right) &= \bar \Psi \bar \Psi \left(1,\ -1\right) &= \bar \Psi \Psi \\ &= 1 \end{align}

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  • $\begingroup$ Use mathjax. Please do not post images of math. $\endgroup$ – DanielSank Nov 19 '19 at 9:01
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You have $$\bar\Psi = \begin{pmatrix} \cos(x) & 0 \\ 0 & -i \sin(x) \end{pmatrix}.$$ Then, $$ \bar\Psi\bar\Psi= \begin{pmatrix} \cos^2(x) & 0 \\ 0 & -\sin^2(x) \end{pmatrix}. $$ This yields $$ \bar\Psi\bar\Psi\begin{pmatrix}1 \\ -1\end{pmatrix}= \begin{pmatrix} \cos^2(x) & 0 \\ 0 & -\sin^2(x) \end{pmatrix}\begin{pmatrix}1 \\ -1\end{pmatrix}= \begin{pmatrix}\cos^2(x) \\ \sin^2(x)\end{pmatrix}. $$

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  • $\begingroup$ Oops! I meant the second term to be a row vector! And in that case, it would be simplified to one (right?), sorry for the mistake. $\endgroup$ – Matrix001 Nov 19 '19 at 10:27
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    $\begingroup$ @Matrix001 Please edit your question to fix your error that you mentioned, otherwise the question isn't really answerable until we know what you are asking. $\endgroup$ – tpg2114 Nov 19 '19 at 10:59

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