# Obtaining the number of quanta for a system of harmonic oscillators

So I need to find the entropy of a system made up of two harmonic oscillators having natural frequency $\omega_0$ and $2\omega_0$. The system is said to have a total energy of $E=(n+\frac12)\hbar\omega_0$ where $n$ is an odd integer.

First, naturally I set $n=2m-1$ where $m$ is a natural number which gives $E=(2m-\frac12)\hbar\omega_0$. Now in my notes I have that this is a problem of distributing $m-1$ quanta among two oscillators, and then taking $\Omega=C_N^{M+N-1}$ with $M=m-1$ and $N=2$.

My question is this: How and where did the $m-1$ come from? What is the general approach to obtaining it?

EDIT: I tried setting $E_1+E_2=E$ and then simplifying to get $n_1+2n_2=2(m-1)$. However, I cannot see why $m-1$ is considered the number of quanta as opposed to, say, $2(m-1)$

• It might be helpful to enumerate the possible combinations by hand for $n=1$, 3, and 5. – user8153 May 27 '18 at 20:42
• You mean plug in numbers and count the possible cases for each yes? – Jepsilon May 27 '18 at 20:46
• Yes, that's right. For small values of $n$ that's easy to do, and you might see a pattern emerge. – user8153 May 27 '18 at 20:57