Number of States in a 1D Simple Harmonic Oscillator

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Ang Han Wei
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Homework Statement


A system is made of N 1D simple harmonic oscillators. Show that the number of states with total energy E is given by [itex]\Omega[/itex](E) = [itex]\frac{(M+N-1)!}{(M!)(N-1)!}[/itex]


Homework Equations


Each particle has energy ε = [itex]\overline{h}[/itex][itex]\omega[/itex](n + [itex]\frac{1}{2}[/itex]), n = 0, 1

Total energy is given by E = [[itex]\frac{N}{2}[/itex] + M][itex]\overline{h}[/itex][itex]\omega[/itex], M is an integer

The Attempt at a Solution



If the entire system is in the ground state, n = 0 for all particles and the energy will be [itex]\frac{N}{2}[/itex][itex]\overline{h}[/itex][itex]\omega[/itex]

So M must be the number of excited particles having the value of n = 1

Hence, the total number of states should be the number of ways that I can choose M particles to be "excited" out of N particles.
[itex]\Omega[/itex] = [itex]\frac{N!}{(M!)(N-M)!}[/itex] but that is not the case.

I am not sure where I have gone wrong.
 
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Your mistake is in assuming the excited oscillators have to be in the n=1 state. For example, say you had three oscillators. If two are in the n=1 state and one is in the n=0 state, the system has the same energy as one with one oscillator in the n=2 state and two oscillators in the n=0 state.
 
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