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This source http://www.hep.caltech.edu/%7Ephys199/lectures/lect5_6_cas.pdf on page 3 says $E_{0} = \frac{1}{2}\sum_{k}\omega_{k} = \sum_{k}\omega_{k}$ (Natural units are being used and $E_{0}$ is the vacuum energy)

But where did the $\frac{1}{2}$ go? On other derivations of the Casimir effect the $\frac{1}{2}$ is not removed, so why do we remove it here?

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    $\begingroup$ 1. The source does not write $\frac{1}{2}\sum_k\omega_k = \sum_k \omega_k$. It writes $\frac{1}{2}\sum_k\sum_\lambda\omega_k = \sum_k\omega_k$ ($\hbar$ omitted). Did you perhaps just not read that equation carefully enough? 2. To make questions more accessible and guard against link rot, please include all relevant information, such as the explanation of notation or specific terminology used, in your question. $\endgroup$
    – ACuriousMind
    Nov 19, 2022 at 20:18

1 Answer 1

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At the beginning (when there is still 1/2) the sum is also over the polarizations (indicated by $\lambda$). The polarizations of lights are 2, and the frequencies do not change from one polarization to another; so the sum over the two polarizations becomes a multiplication by 2 (i.e. $\omega_{\lambda=1}+\omega_{\lambda=2}=\omega+\omega=2\omega$), and this cancels out the factor 1/2.

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