Number of states in the free electron gas

Question

Considering the free electron gas model and the representation of stationary states in the k-space, the book I'm reading (Griffith's Introduction to Quantum Mechanics) says that "each intersection point represents a distinct (one-particle) stationary state, and each block represents a state". I also read other similar questions in which someone stated that each state is shared by 8 blocks, so it gives a contribution of $1/8$ to each block, but I still can't figure out why this is true (each vertex is shared by 4 blocks, isn't it?). What am I doing wrong?

Answer 1

I suppose OP is concerned with a density of states, that's what the literature is about.

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If you are considering the center of the box of gas at the origin so that it's bounded by the planes $x=\pm L/2\cdots$. The wave function in this case given by $$\psi(x,y,z)=\frac{1}{\sqrt{V}}e^{i\mathbf{k}\cdot \mathbf{r}}$$ The periodic boundary conditions can be applied $$\psi\left(\frac{L}{2},y,z\right)=\psi\left(-\frac{L}{2},y,z\right)$$ which implies that $$k_x=\frac{2\pi n_x}{L}$$ similarly two other boundary conditions implies $$k_i=\frac{2n_i\pi}{L}, \ \ \ \ \ i=x,y,z$$

In this case, the spacing between the allowed states is $2\pi/L$ so the volume per allowed state $(2\pi/L)^3$.

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On the other hand, side if you consider the corner of the box at the origin and the whole box in the first coordinate, you would find $$k_i=\frac{n_i\pi}{L}\ \ \ \ \ \ i=x,y,z$$ In this case the spacing between the states is $\pi/L$ and the volume per allowed state is $(\pi/L)^3$. But when taking the volume of the shell we need to consider the first quadrant thus $$\frac{1}{8}\times 4\pi k^2dk$$ This is the reason for taking factor $1/8$.