# Why is the $k$-space in multiples of $2\pi/L$?

So when you find the solution to the Schrödinger equation you get that the wave function can have $$k=n\pi/L$$, $$n=1, 2,3 \dots$$

The problem I have is that when calculating the density of states of a fermi gas, you use the $$k$$-space.

In this space, you use $$k = \frac{2\pi}{L}n$$, $$n=\dots; -2,-1, 0, +1, +2, \dots$$.

Why don't these equations match as they are both Schrödinger solution and using boundary condition with sin functions?

• So when you find the solution to the Schrödinger equation you get that the wave function can have $k=n\pi/L$ Certainly this is not always true – Aaron Stevens Apr 16 at 19:00
• Could you explain how or where you obtained the second expression for k? – Crimson Apr 16 at 19:07
• When you take the boundary conditions in a box of traveling wave and not a standing wave. Then you put the different values of k in k-space and get the equation. – hi hi Apr 17 at 14:28

Speaking about Schrödinger's equation, without specifying the potential and which boundary conditions are present, is equivalent to say almost nothing.

The first case, $$k=\pi n/L$$, corresponds to the solutions of the Schrödinger's equation with zero potential, in the interval $$[0,L]$$, provided hard wall boundary conditions are used, i.e. the wavefunction must vanish at the boundary points $$0$$ and $$L$$.

The second case, $$k=2 \pi n/L$$, corresponds to the solutions of the Schrödinger's equation with zero potential, in the interval $$[0,L]$$, provided periodic boundary conditions are used, i.e. the wavefunction must satisfy the conditions $$\psi(0)=\psi(L)$$ and $$\psi^{\prime}(0)=\psi^{\prime}(L)$$.

When calculating the density of states, one exploits the fact that for large systems the density of states gets independent on the exact boundary conditions and on the precise shape of the boundary surface, so the density of states evaluated with periodic boundary conditions is used as a convenient way of doing calculations.

First of all, I would notice that while eigenfunctions on the case of hard wall boundary conditions can be written in term of $$sin$$ functions, this is not the case with periodic boundary conditions where the eigenstates are of the form $$e^{i k x}$$.

The density of states in energy per unit of length is the same in the two cases as discussed in the first answer to a related question.

Notice however that such equality, i.e. the independence of the density of states on the boundaries for large systems is a much more general result.

• why can you se the periodic boundary conditions, as you are missing out many energies levels? – hi hi Apr 17 at 13:02
• I concentrated my attention on the title of your question. Now I see your second formula which is wrong with periodic boundary conditions (p.b.c). There is no square, no absolute value in the expression for $k$, in the case of p.b.c.. It is true that you are not writing it corresponds to a system with p.b.c, but it is also true that the only way to have a traveling wave is to use them. If the second formula is not coming from a system with p.b.c., please edit your post to specify which system you are considering. – GiorgioP Apr 17 at 18:11
• Yes but you can get to the same final solution of the density of states or number of states in the fermi sphere using for equations for k. So my question is, why can you use both p.b.c. and hard wall boundary conditions to arrive at the same solution, where the particles are in a box for example? – hi hi Apr 18 at 12:39
• The short answer is what I wrote above (independence of the density of states on boundaries when the size is large enough). For a detailed answer, it depends on the degree of rigor one requires. – GiorgioP Apr 18 at 14:36
• Surely the way you've written it, the density of states differs by a factor of 2 for all system sizes, no? – Jahan Claes Apr 18 at 19:35