# Infinite square well – force to decrease well width

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.2, Exercise 5.2.4.

One way of comparing the classical and quantum pictures of a particle in an infinite square well is to calculate the force exerted on the walls by the particle. If a particle is in state ${\left|n\right\rangle }$, its energy is

$\displaystyle E_{n}=\frac{\left(n\pi\hbar\right)^{2}}{2mL^{2}} \ \ \ \ \ (1)$

If the particle remains in this state as the walls are slowly pushed in, so that ${L}$ slowly decreases, then its energy ${E_{n}}$ will increase, meaning that work is done on the system. The force is the change in energy per unit distance, so the force required is

$\displaystyle F=-\frac{\partial E_{n}}{\partial L}=\frac{\left(n\pi\hbar\right)^{2}}{mL^{3}} \ \ \ \ \ (2)$

If we treat the system classically, then a particle with energy ${E_{n}}$ between the walls is effectively a free particle in this region (since the potential ${V=0}$ there), so all its energy is kinetic. That is

 $\displaystyle E_{n}$ $\displaystyle =$ $\displaystyle \frac{1}{2}mv^{2}\ \ \ \ \ (3)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2E_{n}}{m}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n\pi\hbar}{mL} \ \ \ \ \ (5)$

The classical particle bounces elastically between the two walls, which means its velocity is exactly reversed at each collision. The momentum transfer in such a collision is

$\displaystyle \Delta p=2mv=\frac{2n\pi\hbar}{L} \ \ \ \ \ (6)$

The time between successive collisions on the same wall is

$\displaystyle \Delta t=\frac{2L}{v}=\frac{2mL^{2}}{n\pi\hbar} \ \ \ \ \ (7)$

Thus the average force exerted on one wall is

$\displaystyle \bar{F}=\frac{\Delta p}{\Delta t}=\frac{\left(n\pi\hbar\right)^{2}}{mL^{3}} \ \ \ \ \ (8)$

Comparing with 2, we see that the quantum and classical forces in this case are the same.