Revisit a problem

Revisiting Question 2: A Biophysics Perspective on the Zipper Model

This is a biophysics problem from my graduate course. It serves as a fascinating application of the Boltzmann distribution to a 1D chain (Zipper Model).

The Exam Question:
A zipper has $N$ links, each of which can be in a closed state with energy $0$ or an open state with energy $\epsilon$. The zipper can only unzip from one end. This means link $n$ can only be open if all previous links ($1, 2, \dots, n-1$) are already open.

(a) Find the probability $P_n$ that exactly $n$ links are open.
(b) Find the mean number of open links $\langle n \rangle$ in the limit of large $N$.

During the exam, rather than following the standard partition function approach, I constructed a statistical model mixing Binomial and Boltzmann distributions. While the resulting expression was slightly different from the canonical solution, I found they converged well in the physical limit.

For every individual link, the probability that it is open is defined as:

$$P_1=\frac{e^{-\beta \epsilon} }{1+e^{-\beta \epsilon}}=\frac{e^{-\beta \epsilon} (1- e^{-\beta \epsilon})}{(1+e^{-\beta \epsilon})(1- e^{-\beta \epsilon})}=\frac{e^{-\beta \epsilon}(1-e^{-\beta \epsilon}) }{1-e^{-2\beta \epsilon}}$$

Since the probability $P_1$ is small in the high-energy limit ($\epsilon \gg kT$), we have: $$P_1 \approx e^{-\beta \epsilon}$$ The condition that "at least one link is open" is equivalent to the first link being open. Thus, the answer for question (a) is $P_1$, which matches the solution provided in class.

Derivation Part 1

If the chain is sufficiently long, the opening of subsequent links can be treated as independent trials. If the first link is open, the probability of the second link opening is identical to the first. Therefore, the calculated probability that only (exactly) two links are open is:

$$P_2'=P_1^2(1-P_1)=\left(\frac{e^{-\beta \epsilon} }{1+e^{-\beta \epsilon}}\right)^2(1-P_1)=\frac{e^{-2 \beta \epsilon} }{ (1 + e^{ -2 \beta \epsilon } + 2 e^{-\beta \epsilon})(1+e^{-\beta \epsilon}) } $$

For comparison, the probability calculated from the canonical partition function $Z$ is:

$$P_2=\frac{e^{-2\beta \epsilon}}{Z}=\frac{e^{-2\beta \epsilon}(1-e^{-\beta \epsilon}) (1+e^{-\beta \epsilon})}{1+e^{-\beta \epsilon}}=\frac{e^{-2\beta \epsilon}(1-e^{-2\beta \epsilon}) }{1+e^{-\beta \epsilon}}$$

Observation: The two expressions for $P_2$ converge to the same limit when the energy $\epsilon$ is small. This suggests that my statistical model, which blends Binomial and Boltzmann logic, is physically valid under these constraints.

Part (b): Mean Number of Open Links

Exam Question 2b

Based on the logic in (a), the probability that at least $n$ links are open is: $$P_n = P_1^n$$ Consequently, the probability that exactly $n$ links are open is:

$$P_n'=P_1^n(1-P_1)$$

The mean number of open links $\langle n \rangle$ is then calculated as:

$$\langle n \rangle = \frac{\sum nP_1^n(1-P_1)}{\sum P_1^n(1-P_1)} = \frac{\sum nP_1^n}{\sum P_1^n}$$
Through algebraic derivation for a large $N$, I obtained:
$$\langle n \rangle = \frac{1}{1-P_1}-\frac{NP_1^{N+1}}{1-P_1^N} \approx \frac{1}{1-P_1} = 1+e^{-\beta \epsilon}$$

Rearranging this form to compare with the standard result:

$$\langle n \rangle = 1+e^{- \beta \epsilon}=\frac{(1+e^{-\beta \epsilon})(1-e^{-\beta \epsilon})}{1-e^{-\beta \epsilon}}=\frac{1-e^{-2\beta \epsilon} }{1-e^{-\beta \epsilon}} \approx \frac {1 }{1-e^{-\beta \epsilon}}=\frac{e^{\beta \epsilon} }{e^{\beta \epsilon}-1}$$
In the limit where $e^{\beta \epsilon} \approx 1$, we obtain the classical result:
$$\langle n \rangle = \frac{1 }{e^{\beta \epsilon}-1}$$

Conclusion: Although my approach was unconventional, I found that the result is consistent with the standard Boltzmann distribution.