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### AuthorTopic: The energy levels of the particle on a ring.  (Read 2621 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### Twickel

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##### The energy levels of the particle on a ring.
« on: February 20, 2012, 05:21:05 AM »

Having trouble as to why for Sin j cannot equal 0. Since Sin (0) = 0 and Sin (2pi)= 0, doesn't this fit the boundary condition of the particle on a ring, ( the wave is cyclic)

Was wondering if this answer is correct.

Why can the particle on a ring have an energy level of 0?

The particle on a ring can have an energy state of 0, corresponding to j=0, since the only requirements for the particle on a ring are to have a integer  ( the wavefunctions must repeat on successive cycles. Meaning they must have the same values at 0 and 360 degrees.) number of waves around the ring. The sine wave function is not defined at j= 0 however the cosine wave function is defined at j=0/ This ground state has the particle motionless around the ring  with a constant zero amplitude. J= 0 is not degenerate since the sine function is not defined for 0.

Thanks.
« Last Edit: February 20, 2012, 06:09:19 AM by Twickel »
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