Consider free electron gas in a rectangular solid with each side length $L_x$, $L_y$, $L_z$.
"Griffiths <Introduction to Quantum Mechanics>" use infinite square barrier as fixed boundary condition. e.i. $\psi(0)=\psi(L_{x,y,z})=0$, this lead to quantum numbers positive integers $n_{x,y,z}=1,2, ...$ corresponding to standing wave solutions $\psi=\sqrt{\frac{8}{V}}sin(\frac{n_x\pi}{L_x})sin(\frac{n_y\pi}{L_y})sin(\frac{n_z\pi}{L_z})$ with $E=\frac{\hbar^2k^2}{2m}, k_{x,y,z}=\frac{n\pi}{L_{x,y,z}}$.
But "Ashcroft&Mermin <Solid State Physics>" use periodic boundary conditions. e.i. $\psi(x,y,z)=\psi(x,y,z+L_z);\psi(x,y,z)=\psi(x+L_x,y,z);\psi(x,y,z)=\psi(x,y+L_y,z)$. Thus this time the quantum numbers are just integers $n_{x,y,z}=0,\pm1,\pm2, ...$ corresponding to running wave solutions $\psi=\sqrt{\frac{1}{V}}e^{ikr}$ with $E=\frac{\hbar^2k^2}{2m}, k_{x,y,z}=\frac{2n\pi}{L_{x,y,z}}$.
These two boundary conditions give very different energy levels. For instance, the ground state wavefunction of fixed boundary condition is $\psi_{111}$ since n cannot be zero. But for periodic boundary conditons we can have state like $\psi_{100}, \psi_{110}$ since n can be zero. I am confused by the difference in these two books. So what is the real ground state of a electron gas model? Or in what condition should we use these two different boundary?