Is Pressure isotropic in an Ideal Gas in QM? I have made some calculations which to my surprise show that the pressure is anisotropic in an ideal gas. I don't know if it is correct.
The calculation goes as follows:
Basically I modelled the ideal gas as N identitcal particles in a non cubical box. The energy is proportional to (ignoring the constant factor of mass and Planck's constant): $E_{total}\propto\sum_{i,j}^{N,3}(\frac{n_{ij}}{l_j})^2$
where i is the index of the ith particle and j is the index for the jth dimension (i.e x,y,z) and $n_{i,j}$ is the quantum number associated with the ith particle in the jth dimension.
If I now calculate the pressure by taking the derivative wrt to one dimension i.e $l_k$ and dividing by the area $l_m\times l_n$ where m,n are orthogonal directions to $l_k$ then I will get  an expression which depends on $l_k$. However if $l_k$ is anisotropic i.e the lengths of the box are not all equal then the pressure will also be anisotropic.
Where is the catch?
 A: If the gas were truly non-interacting, and you were able to do the expansion in such a way that the particles did not jump between energy levels in the process, then your conclusion would be correct.
However, in a real gas there are interaction processes which rapidly transfer energy between different particles and between different degrees of freedom (here, the degrees of freedom are the kinetic energies in the $x$, $y$, and $z$ directions). In fact, such processes are a prerequisite for the fundamental postulate of statistical mechanics to hold, and this postulate is necessary to derive the usual equations of state, including the fact that pressure is isotropic. (I say `necessary'... there are other approaches to axiomatising statistical mechanics, but the fundamental postulate is a common starting point.)
Intuitively you can think of it in this way. The pressure in a particular direction is proportional to the average kinetic energy of particles in that direction. If it is possible to expand one dimension of the box (say the $x$-direction) without causing a shift in energy levels then the average kinetic energy in that direction will decrease, causing an anisotropic decrease in pressure. However, the interaction processes work so as to divide the total energy equally between different degrees of freedom (this is a consequence of the equipartition theorem, which is a good rule-of-thumb as long as the average energy spacing between levels is much less than $k_B T$). Hence during expansion the interactions will transfer energy from the $y$ and $z$ directions to the $x$ direction, ensuring the average kinetic energies (and thus the pressures) remain the same in all directions.
