Simple harmonic oscillators-Quantum mechanics

Jenkz
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Homework Statement


An ion in a harmonic ion trap sees a potential which is effectively that of a simple harmonic
oscillator. It has a natural oscillation frequency given by v = 1 MHz. Ignoring any internal
excitations, it is known to be in a superposition of the n = 0, 1 and 2 SHO energy states.
A measurement is then made and it is found to be in the n = 2 level.

a)What is the energy of the ion after the measurement has been made?

The Attempt at a Solution


Why is the answer E_n = (2n+1)/2 \hbar\omega

I do not understand the (2n+1) / 2

Thanks!
 
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Hi,
The average energy in the nth state (or in the phonon picture: number of phonons in a mode associated with frequency \omega) for a single harmonic oscillator is given by:
E_n=\frac{2n+1}{2}\hbar\omega=(n+\frac{1}{2})\hbar\omega.
where
h\nu=\hbar\omega.
 
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