Question

Given a particle on a 1-D random walk with some drift velocity νd = ΔxdΔt, the position in at some time step j is given by xj = xj − 1 + kj L+ Δxdx where L is the step length and kj ∈ {−1,+1} with equal probability. I tried calculating the mean square displacement using ⟨x^2N⟩ = ⟨(xN − 1 + kNL + Δxd)^2⟩ = ⟨x^2N − 1⟩ + 2Δxd ⟨xN − 1⟩ + L^2 + Δx^2d. Solving this recurrence gives that the mean squared displacement is exponential with respect to time, although intuitively, and when I graph it, it should be quadratic with respect to time. Where am I going wrong?

The easiest approach is to transform into coordinates that comove with the drift, i.e.˜xj = xj + Δxdj. In those coordinates˜xj, the problem reduces to a standard (fixed-step) random walk. Nevertheless, your approach is also fine. You likely made an error when solving the recurrence relation. Note that the recurrence relation consists of two equations,⟨xj⟩ = ⟨xj−1⟩ + Δxd, ⟨x^2j⟩ = ⟨x^2j − 1⟩ + 2Δxd ⟨xj−1⟩ + L^2 + Δx^2d.
Answer With initial condition at the origin x = 0, and independent is, the mean squared displacement at step reads : E{x^2n} = n^2Δx^2d + nL^2.E{x2n}=n2Δx2d+nL2.

Answer 1

The error in the mean square displacement calculation for a 1-D random walk with drift is likely due to a mistake in solving the recurrence relation. Displacement in physics is known to depend on the square of the time when acceleration is present, and for a random walk with drift, it should be proportional to time, not exponential.

Step-by-step explanation:

The misunderstanding in your calculation for the mean square displacement in a 1-D random walk with drift seems to be in how the recurrence relation was solved.

Typically, for classical mechanics, when acceleration is present, the displacement of an object is expected to depend on the square of the time elapsed (t2).

This is reflected in kinematic equations like x = xo + vot + 1/2at2, where displacement can be quadratic in time due to the acceleration term.

In the case of your random walk with drift, the mean squared displacement should be proportional to the number of steps (or time), not exponential in time.

The correct expression would be the sum of the drift displacement squared plus the random walk component, resulting in E{x2n} = n2Δx2d + nL2, assuming statistical independence of steps and an initial condition at the origin.