How can a reversible adiabatic expansion not increase entropy? In the second stage of the Carnot cycle, a gas is thermally insulated and allowed to expand and do work on the piston. 
I understand the reason people give is that because entropy is $\,dS = \,dQ/T$ then a reversible adiabatic expansion doesn't change the entropy because there is no heat energy flowing in or out of the system. 
But quantum mechanically (particle in a box), we know that increasing the volume causes an increase in the number of available energy levels i.e. microstates, and since entropy is essentially a measure of the number of microstates, how can this process not cause the entropy to increase? 
 A: The available volume goes up, but the momenta of the particles go down. Since the number of available states is proportional to phase space volume, which includes both momentum and position factors, the total number stays constant. You can do an explicit calculation to show this.
However, it's tricky to see why, classically, these two effects must cancel exactly. It's much more clear in quantum mechanics, where the reasoning goes like this:


*

*For a sufficiently slow process, the adiabatic theorem applies, so that initial microstates are mapped one-to-one to final microstates.

*The entropy is proportional to classical phase space volume, which is proportional to the number of quantum microstates.


Thus the entropy doesn't change.
A: The entropy of a gas does not simply depend on the number of ways to arrange the particles that make up the gas but also on the number of ways of distributing the available energy between those particles. In an adiabatic expansion there is no heat transfer but he gas does do work on its surroundings. This reduces the internal, and so reduces then number of ways that energy be distributed between the particles of the gas. 
A: Basics
Suppose we study a fixed volume:


*

*Reversible means that no entropy is produced inside the volume.

*Adiabatic means that no entropy is transported in or out of the
    volume.

*Isentropic means that entropy inside the volume stays
    constant.


If you have 1 and 2 you also have 3. But it is also possible to have an isentropic change of state where entropy is produced inside the volume. (when exactly the produced entropy leaves the volume) 
.
Part 2

we know that increasing the volume causes an increase in the number of
  available energy levels i.e. microstates, ..., how can this process
  not cause the entropy to increase?

For the change in internal energy of a gas we have the eqn:
$dU=C_v \, dT=T  \, ds - p  \, dV$
So entropy stays constant if temperature decreases as volume increases in a certain matter:
$\frac{dT}{dV}=-p/C_v$
