The average energy of an n-state system

Question

I have an $n$- state system with energy values $0$, $2 \epsilon$, $3 \epsilon$, $\dots$, $n \epsilon$. I need to find the average energy $\langle E\rangle$.

I proceeded as follows:

The probability that the energy is $s\epsilon$ is

$$\frac{e^{-s\epsilon}}{\sum_{l= 0}^{n} e^{-l \beta \epsilon}}$$

where $\beta \equiv \frac{1}{k_B T}$

The denominator is a geometric series with $n + 1$ terms and ratio $e^{-\beta \epsilon}$

Hence the probability is $\frac{1- e^{-\beta \epsilon}}{1-e^{-(n+1)\beta \epsilon}}e^{-\beta s \epsilon}$

So the expected energy $\langle E \rangle$ is:

$$\frac{1- e^{-\beta \epsilon}}{1-e^{-(n+1)\beta \epsilon}} \sum_{s=0}^{n}s\epsilon e^{-\beta s \epsilon}$$ which equals

$$-\frac{1- e^{-\beta \epsilon}}{1-e^{-(n+1)\beta \epsilon}}\frac{\partial}{\partial \beta} \sum_{s=0}^{n} e^{-\beta s \epsilon}$$

The sum is evaluated as that of a geometric series and differentiated w.r.t $\beta$.

Is this the correct approach to the problem?

Answer 1

Its the right idea, but a little over-complicated. Actually Wikipdia shows explains it nicely:

$$\langle E\rangle =\sum_s E_s P_s =\frac{1}{Z}\sum_sE_s e^{-\beta E_s}=-\frac{1}{Z}\frac{\partial}{\partial \beta}Z=-\frac{\partial \ln Z}{\partial \beta }$$

The difference between your calculation and the one above is that in the one above the partition function is not exlicitly calculated until the end.