# Confusion over the justification for factorizing the partition function for weakly interacting molecules

Consider a single particle system with states $\psi_j$ and corresponding energy levels ${\Large\varepsilon}_j$. The partition function for this single particle system is $$Z_{(1)}=\sum_je^{-\beta\,{\Large\varepsilon}_j}$$ where $$\beta=\frac{1}{k_B\,T}$$ with $T$ as the thermodynamic temperature and $k_B$ is Boltzmann's constant.

Now consider a system of two identical weakly interacting distinguishable particles. The microstates are now $\psi=\psi_{j_1}\,\psi_{j_2}$ for every combination of the two single-particle states and the energies are $${\Large\varepsilon}_{j_1j_2}={\Large\varepsilon}_{j_1}+{\Large\varepsilon}_{j_2}\tag{1}$$

The partition function for this two-particle system is

\begin{align}Z_{(2)}&=\sum_{j_1}\sum_{j_2}e^{-\beta\,{\Large\varepsilon}_{j_1j_2}}\\&=\sum_{j_1}\sum_{j_2}e^{-\beta\,({\Large\varepsilon}_{j_1}+{\Large\varepsilon}_{j_2})}\\&=\sum_{j_1}\sum_{j_2}e^{-\beta\,{\Large\varepsilon}_{j_1}}e^{-\beta\,{\Large\varepsilon}_{j_2}}\\&=\left(\sum_{\color{red}{j_1}}e^{-\beta\,{\Large\varepsilon}_{\color{red}{j_1}}}\right)\left(\sum_{\color{blue}{j_2}}e^{-\beta\,{\Large\varepsilon}_{\color{blue}{j_2}}}\right)\tag{2}\\&=Z_{(1)}\times Z_{(1)}\tag{3}\\&={Z_{(1)}}^2\end{align}

Where in going from equation $(2)$ to $(3)$ we used the fact that $j=\color{red}{j_1}=\color{blue}{j_2}$; since it is just a dummy index of notation. Mathematically I understand completely all of the above.

But physically and intuitively I cannot understand why $Z_{(2)}={Z_{(1)}}^2$.

Looking at equation $(1)$:

$${\Large\varepsilon}_{j_1j_2}={\Large\varepsilon}_{j_1}+{\Large\varepsilon}_{j_2}$$

by my logic $Z_{(2)}={Z_{(1)}}^2$ if and only if ${\Large\varepsilon}_{j_1}={\Large\varepsilon}_{j_2}$.

But I already know that in general ${\Large\varepsilon}_{j_1}\ne{\Large\varepsilon}_{j_2}$

My same confusion applies when we factorize the partition formula for composite systems that are weakly interacting. For example, consider a single diatomic molecule in a gas. The molecule has translational, vibrational and rotational components to its motion which can be treated as independent of each other giving states with the total energy

\begin{align}{\Large\varepsilon}&={\Large\varepsilon}_{\mathrm{translational}}+{\Large\varepsilon}_{\mathrm{vibrational}}+{\Large\varepsilon}_{\mathrm{rotational}}\\&={\Large\varepsilon}_{\mathrm{trans}}+{\Large\varepsilon}_{\mathrm{vib}}+{\Large\varepsilon}_{\mathrm{rot}}\qquad\qquad\qquad\qquad\text{(for simplicity)}\end{align}

The partition function then factorizes as before and we end up with $$Z=\sum_{j_{\mathrm{trans}}}\sum_{j_{\mathrm{vib}}}\sum_{j_{\mathrm{rot}}}e^{-\beta\,\left({\Large\varepsilon}_{{j}_{\mathrm{trans}}}+{\Large\varepsilon}_{{j}_{\mathrm{vib}}}+{\Large\varepsilon}_{{j}_{\mathrm{rot}}}\right)}=Z_{\mathrm{trans}}\times Z_{\mathrm{vib}}\times Z_{\mathrm{rot}}$$

and by my logic this must mean that ${\Large\varepsilon}_{\mathrm{trans}}={\Large\varepsilon}_{\mathrm{vib}}={\Large\varepsilon}_{\mathrm{rot}}$

Clearly I am missing the point and do not fully understand the physics that is taking place here.

I would be most grateful if someone could give me some hints or an explanation that will dispel my confusion.

## EDIT:

One answer addresses the fact that the index of summations will not be equal outside the summation. This part I acknowledge and understand. But since $j_1\ne j_2$ outside the summation then this must mean that $${\Large\varepsilon}_{j_1}\ne{\Large\varepsilon}_{j_2}$$ so how can we have $$Z_{(1)}\times Z_{(1)}=\sum_je^{-\beta\,{\Large\varepsilon}_j}\sum_je^{-\beta\,{\Large\varepsilon}_j}$$ when the energies are different $({\Large\varepsilon}_{j_1}\ne{\Large\varepsilon}_{j_2})$?

It's almost as if the summation has taken away information about the energies.

$$\left(\sum_{\color{red}{j_1}}e^{-\beta\,{\Large\varepsilon}_{\color{red}{j_1}}}\right)\left(\sum_{\color{blue}{j_2}}e^{-\beta\,{\Large\varepsilon}_{\color{blue}{j_2}}}\right) =Z_{(1)}\times Z_{(1)}.$$
In the first product, it's legal to rename $j_1$ to $j$. In the second product, it's also legal to rename $j_2$ to $j$. However, that does not mean that $j_1 = j_2$! Dummy variables have no meaning outside of the sums/integrals they appear in, so it doesn't even make sense to set different dummy variables equal. Even though we might call $j_1$ and $j_2$ by the same letter, the fact remains that we are still summing over all configurations of the two molecules, even ones where $j_1 \neq j_2$.
As a simpler example, let's consider dummy variables in the integral $$\left( \int_0^1 x \, dx \right) \left( \int_0^1 y \, dy \right).$$ We can rename both $x$ and $y$ to $z$, giving $$\left( \int_0^1 z \, dz \right) \left( \int_0^1 z \, dz \right).$$ However, this doesn't tell us that $x = y = z$. Such a statement doesn't even make sense. If you know programming, there's never a point where more than one of $x$, $y$ or $z$ are "in scope" to do a comparison; they are local variables.
• Thank you for your answer (+1); I understand what you are saying, but I just cannot understand how $\left(\sum_{\color{red}{j_1}}e^{-\beta\,{\Large\varepsilon}_{\color{red}{j_1}}}\right)\left(\sum_{\color{blue}{j_2}}e^{-\beta\,{\Large\varepsilon}_{\color{blue}{j_2}}}\right) =Z_{(1)}\times Z_{(1)}$ when the energies are different $({\Large\varepsilon}_{j_1}\ne{\Large\varepsilon}_{j_2})$. I have updated my answer accordingly. It's like I'm searching for a 'physics' reasoning instead of a mathematical one. Could you please take a look at my edit? Many thanks. – BLAZE Jan 26 '17 at 7:03
• The partition function is a sum over states, it does not make reference to a particular microstate, so in your comment (and edit) it is not obvious to what the indices $j_1$ and $j_2$ refer. As mentioned in the above answer, $j_1$ and $j_2$ have no meaning outside of the summations, they are just labels. – ComptonScattering Jan 26 '17 at 16:37