Daily Archives: Mon, 9 January 2017

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.