Calculating Helmholtz energy (incl degeneracy)

[guest1234]

Hi

I'm trying to calculate Helmholtz free energy from definition [itex]\frac{\partial A}{\partial T}=\frac{\partial}{\partial T}(kT\mathrm{ln}\; Z)[/itex]. First
[itex]\langle E\rangle=\sum\limits_i p_i E_i=\sum\limits_i E_i\frac{g_i e^{-\beta E_i}}{Z}=\frac{1}{Z}\sum\limits_i E_i g_i e^{-\beta E_i}=-\frac{1}{Z}\sum\limits_i \frac{\partial \left(g_i e^{-\beta E_i}\right)}{\partial\beta}=-\frac{1}{Z}\frac{\partial \left(\sum\limits_i g_i e^{-\beta E_i}\right)}{\partial\beta}=-\frac{1}{Z}\frac{\partial Z}{\partial\beta}=-\frac{\partial\left(\mathrm{ln}\;Z\right)}{\partial\beta}[/itex]

Therefore the expected value of internal energy in term of partial derivatives is the same as a non-degenerate one. On the other hand, from the definition of entropy

[itex]S=-k\sum\limits_{i}p_i \mathrm{ln}\;p_i=-k\sum\limits_i \frac{g_i e^{-\beta E_i}}{Z}\mathrm{ln}\left(g_i e^{-\beta E_i }Z^{-1}\right)=-\frac{k}{Z}\sum\limits_i g_i e^{-\beta E_i}(\mathrm{ln}\; g_i -\beta E_i - \mathrm{ln}\;Z)=[/itex]

[itex]=\frac{k}{Z}\sum\limits_i\beta E_i g_i e^{-\beta E_i}+\frac{k}{Z}\sum\limits_i\beta \mathrm{ln}\;Z g_i e^{-\beta E_i}-\frac{k}{Z}\sum\limits_i \mathrm{ln}\;g_i g_i e^{-\beta E_i}=k\beta\sum\limits_i\frac{E_i g_i e^{-\beta E_i}}{Z}+\frac{k\mathrm{ln}\;Z}{Z}\sum\limits_i\frac{g_i e^{-\beta E_i}}{Z}-\frac{k}{Z}\sum\limits_i \mathrm{ln}\;g_i g_i e^{-\beta E_i}=[/itex]

[itex]=k\beta\langle E \rangle+k\mathrm{ln}\; Z-\frac{k}{Z}\sum\limits_i g_i e^{-\beta E_i}\mathrm{ln}\;g_i[/itex]

The last term really bugs me, as we cannot use [itex]\frac{\partial A}{\partial T}=\frac{\partial}{\partial T}(kT\mathrm{ln}\; Z)=k\mathrm{ln}\;Z+kT\frac{\partial}{\partial T}\mathrm{ln}\; Z[/itex] anymore, because

[itex]
\frac{\partial (\mathrm{ln}\; Z)}{\partial T}=\frac{1}{Z}\frac{\partial Z}{\partial T}=\frac{1}{Z}\sum\limits_i g_i \frac{\partial}{\partial T}e^{-\frac{E_i}{kT}}=\frac{1}{Z}\sum\limits_i g_i \frac{E_i}{kT^2}e^{-\beta E_i}=\frac{1}{kT^2}\sum\limits_i \frac{g_i E_i e^{-\beta E_i}}{Z}=\frac{\beta\langle E\rangle}{T}
[/itex]

Am I missing something or is [itex]A=\langle E\rangle-TS[/itex] only (general) way to calculate Helmholtz free energy?

 

[eljose79]

Thank you for your question. From your calculations, it seems like you are on the right track towards calculating the Helmholtz free energy. However, there are a few things that I would like to clarify.

Firstly, the definition of Helmholtz free energy is A = U - TS, where U is the internal energy and S is the entropy. This means that A is not equal to <E> - TS, but rather A = <E> - TS + PV, where P is the pressure and V is the volume. So the last term in your calculation should actually be -PV, and not -kln(g_i).

Secondly, the expected value of the internal energy can be calculated using the partial derivative method, as you have shown. However, for the entropy, it is more common to use the relation S = -k ln(Z), which is derived from the definition of entropy as S = k ln(W), where W is the number of microstates. This is because it is easier to calculate the partition function Z than the number of microstates W.

Finally, I would like to address your concern about the last term in the calculation. You are correct that we cannot use the formula <E> = -kT (∂ ln(Z)/∂T) anymore, as this only applies to non-degenerate systems. For degenerate systems, we need to use the formula <E> = -kT (∂ ln(Z)/∂T) + PV, which takes into account the degeneracy of energy levels.

I hope this helps clarify your doubts. Keep up the good work in your calculations!