Before I post the question: I'm not looking for someone to solve it for me. I just want to give my solution and ask if what I'm doing is correct. I feel like I've attempted to solve the problem correctly (used the correct formulas etc) but the solutions I'm getting are a little hairy. Meaning, they don't look to neat.
A system possess three energy levels $E_1 = \epsilon$, $E_2 = 2\epsilon$ and $E_3 = 3\epsilon$, with degeneracies $g(E_1)=1$, $g(E_2)=2$ and $g(E_3)=1$.
(a) Find the partition function $Z$ of the system.
(b) Find the total energy $E$ of the system from the partition function.
(c) Find the heat capacity $C$ of the system
(a) $$Z = \sum_{r}g(E_r)e^{-\beta E_r}$$
where $\beta = \dfrac{1}{kT}$. Using the formula and plugging in the values gives the following:
$$Z= 1+2e^{-2\epsilon \beta} + e^{-3\epsilon \beta}$$
(b)The total energy is simply the expected value or ensemble average
$$\langle E \, \rangle=\sum_{r}E_rP_r = -\dfrac{1}{z}\dfrac{\partial z}{\partial \beta}= – \dfrac{\partial \ln z}{\partial \beta}$$
Just differentiating $\ln Z$ w.r.t $\beta$ I got
$$\langle E \, \rangle= \dfrac{-4\epsilon e^{-2\epsilon \beta}-3\epsilon e^{-3\epsilon \beta}}{1+2e^{-2\epsilon \beta}+ e^{-3\epsilon \beta}}$$
This doesn't look too bad, insofar as it looks like it could be a legitimate value for the total energy.
It's when I do the calculation for (c). I get a very messy answer and that's what brings me back to part (b) and thinking that it's incorrect.
For part (c) I used the relationship between the heat capacity and total energy of the system. That is
$$C_v=\dfrac{\partial \langle E \, \rangle }{\partial T}$$
I got a messy fraction. I might be missing some algebra trick though so I'll post the answer anyway.
$$C_v= -\dfrac{8\epsilon^2 e^{-\frac{2\epsilon}{kT}}-11\epsilon^2 e^{-\frac{5\epsilon}{kT}}}{kT^2(1+2e^{-2\epsilon \beta}+ e^{-3\epsilon \beta^2})^2}$$
Edit: $E_1 = 0$ this is the level of the zero energy scale.
Best Answer
If you have put the reference of energy in $E_1=0$, then the partition function should read $Z=1+2e^{-\epsilon\beta}+e^{-2\epsilon\beta}$. You can draw the levels in a diagram to see it clearly.
Original --------------------------- New energy reference $E_1=0$
$E_3=3\epsilon$ ---------------------- $E_3=2\epsilon$
$E_2=2\epsilon$ ============ $E_2=\epsilon$
$E_1=\epsilon$ ---------------------- $E_1=0$
You are interested in the difference of energy between the levels. Between 1 and 2 there is $\epsilon$ energy. You must keep this difference whatever point of reference you set.
With this, I obtain (using Mathematica, I hope these are correct):
$$\langle E\rangle = \frac{2\epsilon}{1+e^{\epsilon\beta}}$$
(To do this Mathematica simplfied multplying by $e^{-\epsilon\beta}$ both numerator and denominator).
And finally:
$$C_T=\frac{2\epsilon^2\beta^2k_be^{-\epsilon\beta}}{(1+e^{\epsilon\beta})^2}=\frac{\epsilon^2\beta^2k}{1+\cosh(\epsilon\beta)}$$