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In statistical mechanics we are going to derive the grand potential for $N$ quantum harmonic oscillators which are in my case not the same (see equivalent Schwabl's Statistical Mechanics p.33 for $N$ equally oscillators). For this the Hamiltonian

$$H = \sum_{j=1}^{N} \hbar \omega\Big( a_{j}^{\dagger}a_{j} + \frac{1}{2}\Big)$$

is given. Now the grand potential $\Omega(E)$ can be calculated via

$$\Omega(E) = \sum_{n_{1}=0}^{N} \cdot\cdot\cdot \sum_{n_{n}=0}^{N} \delta\Bigg(E- \hbar \omega \sum_{j=1}^{N}\Big(n_{j}+\frac{1}{2}\Big)\Bigg)$$

The derivation above is was given by a solution from our excercise sheet. My question is now: Why is each sum the same, and why we are not summing over $\sum_{n_{1}=0}^{N} \cdot\cdot\cdot \sum_{n_{n}=n}^{N}$.

If anybody would have a clear way (understandable way) how to solve these sums, I would be very happy.

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