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Absolute Square

Also known as the squared Norm. The absolute square of a Complex Number $z$ is written $\vert z\vert^2$ and is defined as

\begin{displaymath}
\vert z\vert^2 \equiv zz^*,
\end{displaymath} (1)

where $z^*$ denotes the Complex Conjugate of $z$. For a Real Number, (1) simplifies to
\begin{displaymath}
\vert z\vert^2=z^2.
\end{displaymath} (2)

If the Complex Number is written $z=x+iy$, then the absolute square can be written
\begin{displaymath}
\vert x+iy\vert^2=x^2+y^2.
\end{displaymath} (3)


An important identity involving the absolute square is given by

$\displaystyle \vert a\pm b e^{-i \delta}\vert^2$ $\textstyle =$ $\displaystyle (a\pm b e^{-i \delta})(a\pm b e^{i \delta})$  
  $\textstyle =$ $\displaystyle a^2+b^2 \pm a b (e^{i \delta} + e^{-i \delta})$  
  $\textstyle =$ $\displaystyle a^2+b^2\pm 2 a b \cos\delta.$ (4)

If $a=1$, then (4) becomes
$\displaystyle \vert 1\pm b e^{-i\delta}\vert^2$ $\textstyle =$ $\displaystyle 1+b^2\pm 2b\cos\delta$  
  $\textstyle =$ $\displaystyle 1+b^2\pm 2b[1-2\sin^2({\textstyle{1\over 2}}\delta)]$  
  $\textstyle =$ $\displaystyle 1\pm 2b+b^2\mp 4b\sin^2({\textstyle{1\over 2}}\delta)$  
  $\textstyle =$ $\displaystyle (1\pm b)^2\mp 4b\sin^2({\textstyle{1\over 2}}\delta).$ (5)

If $a=1$, and $b=1$, then
\begin{displaymath}
\vert 1- e^{-i\delta}\vert^2 = (1-1)^2+4\cdot 1\sin^2({\textstyle{1\over 2}}\delta)= 4\sin^2({\textstyle{1\over 2}}\delta).
\end{displaymath} (6)

Finally,
$\displaystyle \vert e^{i\phi_1}+e^{i\phi_2}\vert^2$ $\textstyle =$ $\displaystyle (e^{i\phi_1}+e^{i\phi_2})(e^{-i\phi_1}+e^{-i\phi_2})$  
  $\textstyle =$ $\displaystyle 2+e^{i(\phi_2-\phi_1)}+e^{-i(\phi_2-\phi_1)}$  
  $\textstyle =$ $\displaystyle 2+2\cos(\phi_2-\phi_1) = 2[1+\cos(\phi_2-\phi_1)]$  
  $\textstyle =$ $\displaystyle 4\cos^2(\phi_2-\phi_1).$ (7)



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© 1996-9 Eric W. Weisstein
1999-05-25