Monday, November 14, 2011

A harmonic oscillator potential of natural frequency \omega contains 4 electrons and is in the state of lowest?

A harmonic oscillator potential of natural frequency \omega contains 4 electrons and is in the state of lowest energy.





What is that energy?





What would the energy be if the electrons were replaced by spin-1 particles of the same mass?|||A quantum harmonic oscillator can have energy (n+1/2)h-bar \omega.


Electrons are fermions, and so no two can occupy the same quantum state (Pauli's exclusion principle).


Thus, the lowest energy of four electrons is (1/2)h-bar \omega +(3/2)h-bar \omega+(5/2)h-bar \omega+ (7/2)h-bar \omega=8 h-bar \omega.





On the other hand, spin 1 particles are bosons, and so they can all occupy the lowest energy state.The lowest energy will be 2 h-bar \omega

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