# Energy of one-electron state $E(k) \sim k^2$ “$\implies$” occupied region in $k$-space indistinguishable from sphere

When we consider a system of $N\gg 1$ one-electrons the occupied region in $k$-space is stated to be indistinguishable from a sphere, since the energy of a one-electron level is directly proportional to $k^2$.

I would rather expect this argument when the energy is proportional to the cube $k^3$ of its wave vector, since a volume in $k$-space goes as $k^3$.

How does $E(k) \sim k^2$ imply that the volume is indistinghuishable from a sphere?

Note. Ashcroft and Mermin say: If it were not spherical it would not be the ground state, for we could then construct a state of lower energy by moving the electrons in levels farthest away from $\mathbf k = 0$ into unoccupied levels closer to the origin.

• NB: A "sphere" is the locus of points in a 3D space all the same distance $r$ from a central point $\vec{p}$. In other words in a strict mathematical parlance a sphere is a 2D surface. The volume enclosed by the sphere is a "ball". See also "circle" and "disk". Physicist are often sloppy with this language, but it is worth remembering it when you are having trouble parsing a formal definition or statement like this. – dmckee --- ex-moderator kitten May 27 '14 at 20:00

It's because energy is a function of only the magnitude of the wave vector that we consider a sphere in k-space, so that all the states on the surface of that sphere have the same energy. The same argument would still work for $E \propto \vert k \vert$ or $k^3$, etc.