A Problem

Revisit a problem

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}}

Standard Distribution Approach:
The system can be viewed as a set of discrete states where the state $n$ (having $n$ open links) has energy $E_n = n\epsilon$. According to the Boltzmann distribution, the probability of being in state $n$ is:

$$P_n = \frac{e^{-\beta n \epsilon}}{Z}$$

For an infinite chain ($N \to \infty$), the partition function $Z$ is a geometric series:

$$Z = \sum_{n=0}^{\infty} (e^{-\beta \epsilon})^n = \frac{1}{1 - e^{-\beta \epsilon}}$$

The mean number of open links $\langle n \rangle$ is then simply the expected value of this geometric distribution:

$$\langle n \rangle = \frac{\sum n e^{-\beta n \epsilon}}{\sum e^{-\beta n \epsilon}} = \frac{e^{-\beta \epsilon}}{1 - e^{-\beta \epsilon}} = \frac{1}{e^{\beta \epsilon} - 1}$$

Another Approach:

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.

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

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}

Post-Exam Reflection & Correction

Upon reviewing the derivation with rigorous statistical mechanics, I realized there are two important nuances that bridge my answer with the exact solution:

1. Mathematical Correction on Summation
In the calculation of $\langle n \rangle$, there was a minor algebraic error. The summation of the geometric-like series should be: $$\sum_{n=0}^{\infty} n P_1^n = \frac{P_1}{(1-P_1)^2}$$ Substituting this into the mean value formula: $$\langle n \rangle = (1-P_1) \sum n P_1^n = (1-P_1) \frac{P_1}{(1-P_1)^2} = \frac{P_1}{1-P_1}$$ My original derivation missed the $P_1$ factor in the numerator (yielding $1/(1-P_1)$), which would incorrectly suggest $\langle n \rangle \to 1$ even at absolute zero temperature.

2. Physical Definition of Probability
My model defined $P_1 = \frac{e^{-\beta \epsilon}}{1+e^{-\beta \epsilon}}$, which effectively normalizes each link as an independent system. However, in the exact Zipper model, the statistical weight is simply the Boltzmann factor $x = e^{-\beta \epsilon}$.

If we use the corrected mean formula $\langle n \rangle = \frac{P_1}{1-P_1}$ but substitute the exact weight $P_1 \to x = e^{-\beta \epsilon}$, we get: $$\langle n \rangle = \frac{e^{-\beta \epsilon}}{1-e^{-\beta \epsilon}} = \frac{1}{e^{\beta \epsilon}-1}$$ This perfectly reproduces the standard solution without any approximation approximations. The "independent trial" intuition is valid if the probability weight is defined strictly via the partition function.