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Why is $\omega = \sqrt{K/m}$ valid for a quantum oscillator?

I'm working in the 3rd edition of Modern Physics by Serway, Moses, and Moyer. In 6.6, it talks about a quantum oscillator. I don't fully understand how the definition of frequency works. Now, we ...

nphirning

asked Aug 12, 2014 at 19:26

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Why is $\omega = \sqrt{K/m}$ valid for a quantum oscillator?

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