I'm working on a problem of my quantum mechanics homework set.

The problem is as follows: A particle is in the ground state of infinite potential well (the well is determined by the region 0 < x < L) At t=0, the wall at x=L starts to move adiabatically accordantly to the expression $$L(t) = (2-e^{-\frac{t}{\tau}})L$$

Then the problem asks for the work absorbed by the wall at an instant t.

Attempt of a solution:

We know from the work-energy theorem that the work done by the particle is equal to variation of its kinetic energy: $$W = \Delta K$$

Inside the well, the particle acts like a free particle (no potential), and so the kinetic energy will be given by the energy of the ground state of the infinite potential well: $$E = \frac{\pi^2 \hbar^2}{2mL(t)}$$

Then $\Delta K$ will be $$\Delta K = K(t) - K(0) = \frac{\pi^2 \hbar^2}{2mL^2}\left(\frac{1}{(2-e^{-\frac{t}{\tau}})^2}-1\right)$$

The correct answer is the negative of what I found:

$$W = \frac{\pi^2 \hbar^2}{2mL^2}\left(1-\frac{1}{(2-e^{-\frac{t}{\tau}})^2}\right)$$

My question is: Should not the work absorbed by the wall be equal to the work done by the particle? Or the work done by the particle is the inverse of what I used?

Thank you all


2 Answers 2


The particle and the wall do the work of the opposite sign, but equal magnitude. One can think of it mechanistically as exerting the forces that are equal, but in the opposite directions (aka Newton's third law). One could also think of it in terms of an ideal gas, where one distinguishes the "work done by the gas" and the "work done on the gas" - both equal, but having the opposite signs.


Reading this topic in Liboff's book I found another argument: As the length of the well increases, its energy will decrease. The difference between the initial and final energy must be the energy absorbed by the moving wall.


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