Electron gas is a collection of non-interacting electrons. If these electrons are confined to certain volume (for example, cube of metal), their behavior can be described by the wavefunction which is a solution to the particle in a box problem in quantum mechanics. Allowed energy states for any electron is given by: $$ \epsilon = \frac {h^2} {8mL^2} ({n_x}^2 + {n_y}^2 + {n_z}^2) = \frac {h^2} {8mL^2} n^2 $$

where $n_x$, $n_y$ and $n_z$ are quantum numbers which determine allowed momentum components of the electron:

$$p_x = \frac {hn_x}{2L}, p_y = \frac {hn_y}{2L}, p_z = \frac {hn_z}{2L}$$

Allowed states of the electrons can be visualized in the n-space, where the states can be shown graphically in the $n_x, n_y, n_z$ coordinate system. Every coordinate in the $n$-space actually represents two possible states for certain energy level because there are two electrons with opposite spin states for certain energy level. If we know maximum occupied energy level (Fermi energy), we can calculate maximum quantum number $n_{max}$ from the first equation.

To determine the number of points in the $n$-space for known Fermi energy, we can first count the ''volume'' of the one electron state in the $n$-space (particular $n_x, n_y, n_z$ coordinate). To do so, we note that the difference between two ''neighboring'' quantum numbers is equal to one (quantum numbers are positive integers) which means that the ''volume'' of any cube composed of neighboring points in the n-space is equal to 1. Each cube is enclosed by eight points in the n-space, but since any point is shared by the 8 neighboring cubes, there is actually only one point per each cube since only 1/8 of each point belongs to the individual cube. This means that the ''volume'' of the one electron state in the n-space is equal to 1.

When the ''volume'' per electron state in the $n$-space is known, we can calculate the total number of points in the $n$-space (for certain Fermi energy, $E_f$) by taking the total volume in the $n$-space and dividing it by the ''volume'' per electron state (which is equal to 1, so total number of points equals the total volume).

What I don't understand is that the total volume taken is equal to the 1/8 volume of the sphere with the ''radius'' equal to $n_{max}$: $$V_f = \frac {1} {8}\frac {4 \pi n_{max}^3} {3}$$ According to my understanding, it doesn't really make sense to take the sphere as the total volume because we determined the volume per state as being equal to the volume of the cube with sides having a length 1. Because of this, we should know the total volume of all the cubes for certain $E_f$ and since the volume per state is equal to 1, total number of points is equal to the total volume as mentioned previously.

If this is the case, why is volume of the sphere taken as the paramount when total volume of the cubes should be taken?

  • $\begingroup$ May I know the source this volume calculation is from? $\endgroup$ Jul 30, 2022 at 13:40
  • $\begingroup$ It can be found in Daniel Schroeder: An Introduction to Thermal Physics, chapter 7.3: Degenerate Fermi Gases. $\endgroup$ Jul 30, 2022 at 14:52
  • $\begingroup$ Also, the other source is: youtu.be/z7YGS67GETo $\endgroup$ Jul 30, 2022 at 15:00
  • $\begingroup$ You want to count the number of cubes subject to a constraint. The constraint is that the sum of squares of their co-ordinates is bounded by $r^2$ (related to the Fermi energy). This is the constraint that defines a ball. $\endgroup$ Jul 31, 2022 at 11:01

1 Answer 1


The (1/8) factor appears in the case of 3D n-space because $n_x,n_y,n_z$ are positive integers. This constitutes one of the Octants in the 3D space, as considering the whole sphere would add in redundant values of n.

It makes sense to take the volume of the sphere because its radius determines the value of n of highest occupancy $n_{max}$ that bounds the permissible states. Without knowledge of which are the permissible states, it doesn't make sense to calculate volume. In case we do know the total number of states N, then for a high number of electrons, the volume of a sphere and the number of cubes constituting it will be approximately equal.

  • $\begingroup$ Yes, also the important question is what is in fact the exact volume we are approximating with the sphere? Its volume must be a multiple of the cube volume, but I am really not sure how does such object look in the n-space. To know that we would need to know combination of quantum numbers for each electron and how can we know that? $\endgroup$ Aug 1, 2022 at 10:38
  • $\begingroup$ The combinations that are allowed are the ones that lie within the sphere. The volume we are approximating it with is a sphere with a grainy surface, which are the edges and corners of the cube. In short, sphere made out of cubes. $\endgroup$ Aug 1, 2022 at 11:20
  • $\begingroup$ I think you didn't get my question. We are approximating the object's volume with the sphere, but how does that object look like in the n-space? In another words, what is the object which volume we are approximating with the sphere? Yes, combinations should be inside or on the surface of the sphere, that is clear. $\endgroup$ Aug 1, 2022 at 15:35
  • $\begingroup$ Also, there seem to be a new source of approximation here. There are many combinations which lie on the surface of the sphere, how do we know how many of them are in fact occupied as the volume of the sphere is the same regardless of that number? This is problematic as we use the volume of the sphere to count the number of states and the volume is the same and doesn't depend on how many states with the $E_f$ are actually occupied. $\endgroup$ Aug 1, 2022 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.