# Partition function for two weakly or strongly interacting independent systems

the partition function for two weakly interacting systems is$$Z(a+b)=\sum_i\sum_je^{-(\epsilon_i+\epsilon_j)/\tau}=Z(a)Z(b)$$

I don't quite understand why we need to assume weakly interaction? Furthermore, what will happen to the partition function (or the Boltzmann factor) for two strongly interacting systems?

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• Just think about what's the meaning of $\varepsilon_i$ and $\varepsilon_j$. – Kite.Y Dec 27 '17 at 17:49

The key is: when you add something (e.g. another system $B$ to system $A$), whether if such a thing will alter the total number of states of combined systems $\Omega(A+B)$ from simply multiplying $\Omega(A)\Omega(B)$. Strongly interacting systems always alter it, as energy can transform to other unexpected form of energy, such as radiation when you only expect energy from spins.

When the total number of states is altered, $Z(a+b)=Z(a)+Z(b)$ does not hold.

Consider there is a box with one (distinguishable) particle - the particle can go to the left side and the right side (it has energy of left $E_{L}$ and right $E_{R}$), so we have $Z=\exp[-E_{L}/\tau]+\exp[-E_{R}/\tau]$.

Now we add another (distinguishable) particle into the box, then by counting, we have: $$Z_{total}=\big(\exp[-E_{L}/\tau]+\exp[-E_{R}/\tau]\big)^2$$

Same as suggested by $$Z(\text{particle 1 + particle 2})=Z(\text{particle 1})*Z(\text{particle 2})$$

However, suppose now we cut the box into half, and each half box with one particle. So the number of state of each box is one: $$\Omega(A)=\Omega(B)=1$$

As for its Partition function:

$$Z(A)=\exp[-E_{L}/\tau] \ ;\ Z(B)=\exp[-E_{R}/\tau]$$

And at last, we combine these two half boxes, and remove the wall between them. then by counting we have: $$\Omega(A+B)=4$$ (This combined box is identical to the the box that hasn't been cut.)

but that is not same as $\Omega(A)*\Omega(B)=1$.

The partition function $Z_{total}=\big(\exp[-E_{L}/\tau]+\exp[-E_{R}/\tau]\big)^2$ - it differs to $Z(A)*Z(B)$.