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 $(1)$:


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.


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.


1 Answer 1


I think you're getting confused by dummy indices. Let's look at the crucial step:

$$\left(\sum_{\color{red}{j_1}}e^{-\beta\,{\Large\varepsilon}_{\color{red}{j_1}}}\right)\left(\sum_{\color{blue}{j_2}}e^{-\beta\,{\Large\varepsilon}_{\color{blue}{j_2}}}\right) =Z_{(1)}\times Z_{(1)}.$$

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.

  • $\begingroup$ 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. $\endgroup$
    – BLAZE
    Jan 26, 2017 at 7:03
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 26, 2017 at 16:37

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