Why is the pressure in quantum infinite wells the derivative of total energy, not potential?

In a classical-mechanics setup, we define the potential energy as being the negative of the work done by the force when a particle from point $A$ to point $B$. If an external agency moves a particle from point $A$ to point $B$ without changing its kinetic energy $T$, then the net work done on the particle must be zero, and so the external work done must be equal to the change in the potential energy. This then implies that the magnitude of the external force is equal to $partial U/partial a$.

The part of this argument that fails in the quantum-mechanical context is that we can't guarantee that the kinetic energy of the particle is constant throughout the process. In other words, $langle hat{H} rangle = langle hat{V} rangle + langle hat{T} rangle$, and in changing the parameter $a$ we inevitably end up changing $langle hat{T} rangle$ as well as $langle hat{V} rangle$. The particle in a box is an extreme example of this; if we change the size of the box, then $langle hat{V} rangle = 0$ at all times, and the only dependence on of the energy on the size of the box comes from $langle hat{T} rangle$.

A closer analogy to the quantum-mechanical situation would be the particles in a monatomic ideal gas. In that case, the work we do on the system goes into the kinetic energy of the gas molecules, not into increasing their potential energy.

In a thermodynamic context, the pressure of a system is given by (minus) the derivative of the internal energy with respect to the volume, while keeping the system thermally isolated:

$$p = -left(frac{partial E}{partial V}right)_S$$

In other words, to compress the system by an infinitesimal amount $dV$, you need to supply energy equal to $dE = - p dV$ in the form of mechanical work. Replacing $E$ with $left<hat Hright>$ provides a semiclassical prediction of the (time-averaged) pressure exerted by the particle on the walls of the box - though of course, this is just a rough idea which ignores the subtle but crucial issues of thermodynamic equilibrium and quantum statistical mechanics.

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The reason your intuition is wrong is that $mathbf F = -nabla V$ is the force experienced by a particle moving in some potential field. Note that the derivative is taken with respect to spatial coordinates of the particle in question, not the size of the box in which it lives.

If we can replace the mean values of the functions by the functions of the mean values ; $$dot{p}_0=langle dot{P}rangle=-leftlangle frac{partial H}{partial X}rightrangleapprox -left.frac{partial H}{partial X}right|_{(X=x_0,P=p_0)}=-frac{partial mathcal{H}(x_0,p_0)}{partial x_0}$$

Classical physics can be seen as a good approximation whenever it is good approximation to replace the mean of the functions $frac{partial H}{partial P}, -frac{partial H}{partial X},$ and $Omega (X,P)$ by the functions of the mean. This is turn requires that the fluctuations about the mean have to be small. (exact if there is no fluctuations). The term $langle frac{partial H}{partial X} rangle = langle frac{dV}{dX}rangle =langle V'(X)rangle$ by $V'(X=x_0)$. To see when this is a good approximation, let us expand $V'$ in taylor series around $x_0$. We will work in coordinate basis

$$V'(x)=V'(x_0)+(x-x_0)V''(x_0)+frac{1}{2}(x-x_0)^2V'''(x_0)+cdots$$

Let us now take the mean of both sides. The first term on the right-hand side which alone we keep in our approximation, corresponds to classical force at $x_0$ and thus reproduces Newton's law.The second vanishes in all cases, since the mean of $x-x_0$ does. The succeeding terms, which are corrections to the classical approximation, represent the fact that unlike the classical particle, which responds only to the force $F=-V'$ at $x_0$, the quantum particle responds to the force at neighboring points as well (Note that these terms is zero if the potential is at the most quadratic in the variable $x$).