Total number of orbitals inside a sphere in $k$ space In kittel's book on solid state physics it says that for the volume element  $(2\pi/L)^3$ there is one distinct triplet of quantum numbers $k_x,k_y,k_z$. Thus in the sphere of volume $4\pi k_F^3/3$ the total number of orbitals is $$\frac{4\pi k_F^3/3}{(2\pi/L)^3}.2$$where the factor 2 comes from two allowed values of the spin quantum number $m_s$ for each value of $k$
my questions are

*

*when $k$ can be $0$ or $(2\pi/L)$ shouldn't there be $2$ distinct set of quantum numbers possible in the volume element $(2\pi/L)^3$


*why did we divide the volume of fermi sphere with the volume element  $(2\pi/L)^3$ to get the total number of orbitals


*isn't number of orbitals independent of spin quantum number . As far as i know the number of orbitals in a given subshell depends only on magnetic quantum number, So why did we multiply by $2$.
 A: Assuming a cubic sample of side $L$ and periodic boundary conditions.

*

*The wavevector $\mathrm{k}$ can assume any value on the grid
\begin{equation}
\mathrm{k} = (k_x, k_y, k_z) = \left(\frac{2\pi}{L}n_x, \frac{2\pi}{L}n_y, \frac{2\pi}{L}n_z\right)
\end{equation}
where $n_x, n_y, n_z$ are integers. The Fermi momentum is the maximum momentum allowed for particles when the sample is at zero temperature. From the relation above you can see that on average a state occupies on average a volume $\left(\frac{2\pi}{L}\right)^3$ in the momentum space. See also the picture below (in 2D).

$k$-points in a two-dimensional space. The area per point, highlighted by the red dashed square, is $(2\pi / L)^2$. If the space were $d$-dimensional, the generalized volume would be $(2\pi / L)^d$ per allowed point." />


*If the Fermi volume is spherical, i.e. $V_F = \frac{4}{3}\pi k_F^3$, and way larger than the volume occupied by a state, i.e. $\left(\frac{2\pi}{L}\right)^3$, a good approximation for the total number of states is simply
\begin{equation}
N = \frac{\frac{4}{3}\pi k_F^3}{\left(\frac{2\pi}{L}\right)^3}
\end{equation}
If I consider that my particles (Fermions) have spin $s$, each momentum states can host $(2s+1)$ particles (spin-multiplicity), so one must correct the total number of states as
\begin{equation}
\mathcal{N} = (2s+1)N=(2s+1)\frac{\frac{4}{3}\pi k_F^3}{\left(\frac{2\pi}{L}\right)^3}
\end{equation}
If I consider spin-1/2 particles, like electrons, the multiplicity factor is $2$.

*The term orbitals is a little outdated, but, as I just mentioned, each orbital hosts 2 electrons with opposite spin orientation.

A: Let's first be clear about some basics then we go on to questions:
The periodic boundary condition requires that any wave in the sample $e^{ikr}$ have the same value for a position $r$ as it has for $r+L$ (after one round around the circle). This imposes quantization of $k$
$$k=\frac{2\pi n}{L} \ \ \ n\ \text{is integer}.$$
In three dimensions,
$$\mathbf{k}=\frac{2\pi }{L}(n_1,n_2,n_3)$$

The above picture taken from Blundell shows what I said in a two-dimensional case.
So here each $\mathbf{k}$ point now occupies a volume of $(2\pi /L)^3$. Because of this discretization of values of $\mathbf{k}$, whenever we have to sum over all possible $\mathbf{k}$ we obtain
$$\sum_\mathbf{k}\rightarrow \frac{L^3}{(2\pi)^3}\int \mathbf{dk}$$

At $T=0$ for electrons,
$$N=2\times\frac{L^3}{(2\pi)^3} \int \mathbf{dk} \theta(E_F-\epsilon(\mathbf{k}))=2\times\frac{L^3}{(2\pi)^3} \int^{|k|<k_F} \mathbf{dk} =2\frac{L^3}{(2\pi)^3}\left(\frac{4}{3}\pi k_F^3\right)$$
where the prefactor of $2$ accounts for the two possible spin states each possible wavevector $\mathbf{k}$.


Problem ($1$)

As stated earlier,
$$\mathbf{k}=\frac{2\pi }{L}(n_1,n_2,n_3)$$
As you can see from the figure as we go up the density of states increases which allows us to use integrals.
What you are stating in question is wrong.

Problem ($2$)

This has been answered above. The word orbitals are not right in this situation we are dealing with free electron gas. What we have found is the number of electrons not the number of orbital or something.

Problem ($3$)

The free electron gas is not bound to atoms so we are not in concert with a magnetic quantum number. There can be two spin-state correspondings to a partial wave function.

Edit: If we focus on the magnitude of the wave vector given by $k$. Allowed states with a wave vector whose magnitude lies between $k$ and $k+dk$ given by the density of states:
$$g(k)dk=\frac{\text{volume in k-space of a spherical shell}}{\text{volume in k-space occupied per allowed state}}$$
$$g(k)dk=\frac{4\pi k^2dk}{(2\pi /L)^3}\propto k^2dk$$
That's what I meant by increasing the density of state.
