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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.