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Display information for equation id:math.1762.29 on revision:1762

* Page found: Homöopolare chemische Bindung des Wasserstoffmoleküls (eq math.1762.29)

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TeX (original user input): 

\begin{align}
& \left| \begin{matrix}
D-{{E}^{(1)}} & A-{{E}^{(1)}}|T{{|}^{2}}  \\
A-{{E}^{(1)}}|T{{|}^{2}} & D-{{E}^{(1)}}  \\
\end{matrix} \right|=0 \\
& {{\left( D-{{E}^{(1)}} \right)}^{2}}-{{\left( A-{{E}^{(1)}}|T{{|}^{2}} \right)}^{2}}=0={{E}^{(1)}}^{2}\left( 1-|T{{|}^{4}} \right)-2{{E}^{(1)}}\left( D-|T{{|}^{2}} \right)+{{D}^{2}}-{{A}^{2}} \\
\end{align}

TeX (checked): 

{\begin{aligned}&\left|{\begin{matrix}D-{{E}^{(1)}}&A-{{E}^{(1)}}|T{{|}^{2}}\\A-{{E}^{(1)}}|T{{|}^{2}}&D-{{E}^{(1)}}\\\end{matrix}}\right|=0\\&{{\left(D-{{E}^{(1)}}\right)}^{2}}-{{\left(A-{{E}^{(1)}}|T{{|}^{2}}\right)}^{2}}=0={{E}^{(1)}}^{2}\left(1-|T{{|}^{4}}\right)-2{{E}^{(1)}}\left(D-|T{{|}^{2}}\right)+{{D}^{2}}-{{A}^{2}}\\\end{aligned}}

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|D−E(1)A−E(1)|T|2A−E(1)|T|2D−E(1)|=0(D−E(1))2−(A−E(1)|T|2)2=0=E(1)2(1−|T|4)−2E(1)(D−|T|2)+D2−A2
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