# Adiabatic Pertubation in an Infinite Potential Well

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