Solution#

We can think about a particle on every node. The first particle is fixed at the origin. Every subsequent particle has 2 choices (up or down) relative to the previous one.

The total number of microstates $\mathcal{W}$ is: $\mathcal{W} = \underbrace{1}_{\text{Fixed}} \times \underbrace{2 \times 2 \times \dots \times 2}_{N} = 2^N$

Solution#

Let $\sigma_i = y_i - y_{i-1} \in \{+1, -1\}$ be the “sign” of each step. Let $n_+$ be the number of steps up and $n_-$ be the number of steps down.

The total deflection is: $y_N = \sum_{i=1}^N \sigma_i = n_+ - n_- = y$

We also know the total number of steps is $N$: $n_+ + n_- = N \implies n_- = N - n_+$

Substituting this into the deflection equation: $y = n_+ - (N - n_+) = 2n_+ - N$ $\implies n_+ = \frac{y + N}{2}$

The number of ways to choose $n_+$ “up” steps out of $N$ total steps is given by the binomial coefficient: $\mathcal{W}(y) = \binom{N}{n_+} = \binom{N}{\frac{y+N}{2}}$

The probability of finding such a state is $P(y) = \frac{\mathcal{W}(y)}{\mathcal{W}} = \frac{1}{2^N} \binom{N}{n_+}$.

Solution: The Partition Function Trick#

We define a “partition function” $Z(\beta)$ for the ensemble: $Z(\beta) = \sum_y e^{\beta y} \mathcal{W}(y)$

Note that at $\beta = 0$: $Z(0) = \sum_y \mathcal{W}(y) = 2^N$

The typical deflection can be expressed using derivatives of $Z(\beta)$: $\langle y_N^2 \rangle = \frac{\sum_y y^2 \mathcal{W}(y)}{\sum_y \mathcal{W}(y)} = \left. \frac{1}{Z(0)} \frac{\partial^2 Z}{\partial \beta^2} \right|_{\beta=0}$

Calculating $Z(\beta)$#

Substituting our expression for $y = 2n_+ - N$: $Z(\beta) = \sum_{n_+=0}^N e^{\beta(2n_+ - N)} \binom{N}{n_+} = \sum_{n_+=0}^N (e^\beta)^{n_+} (e^{-\beta})^{N-n_+} \binom{N}{n_+}$

Using the binomial identity $(a+b)^N = \sum \binom{N}{k} a^k b^{N-k}$ with $a=e^\beta$ and $b=e^{-\beta}$: $Z(\beta) = (e^\beta + e^{-\beta})^N = (2 \cosh \beta)^N = 2^N \cosh^N \beta$

Finding the Deflection#

First derivative: $Z'(\beta) = 2^N \cdot N \cosh^{N-1} \beta \cdot \sinh \beta$

Second derivative: $Z''(\beta) = 2^N N \left[ (N-1) \cosh^{N-2} \beta \sinh^2 \beta + \cosh^{N-1} \beta \cosh \beta \right]$ $Z''(0) = 2^N N [ (N-1)(1)(0) + (1)(1) ] = 2^N \cdot N$

Finally: $\langle y_N^2 \rangle = \frac{Z''(0)}{Z(0)} = \frac{2^N \cdot N}{2^N} = N$

This result makes physical sense: for a random walk of $N$ steps where each step is $\pm 1$, the variance (typical deflection squared) scales linearly with the number of steps $N$.

TIP

Full Reference: You can download the original problem set with full diagrams here: Download Directed Polymer PDF