# Regarding density of states, not counting of kx, ky, kz of its negative wavefunctions [closed]

In the density of states, we divide the equation by 8 as we remove the count from kx, ky, kz in negative direction. why is that? I understand that it's kinda intuitive, but I don't follow the intuition

## 1 Answer

For a free particle in a box of volume $$L^3$$, the eigenfunction will be: $$\psi_{n_x,n_y,n_z}(x, y, z)=\sqrt{\frac{8}{L}} \sin(\pi n_x x/L) \sin(\pi n_y y/L) \sin(\pi n_z y/L).$$

I can take the same eigenstate, with a negative set of quantum number (I could also just flip the sign of one or two of them, the argument would be the same): $$\psi_{-n_x,-n_y,-n_z}(x, y, z)=-\sqrt{\frac{8}{L}} \sin(\pi n_x x/L) \sin(\pi n_y y/L) \sin(\pi n_z y/L).$$

But, in this case wave functions with $$(n_x, n_y,n_z)$$ represent the same physical state as wave functions with $$(\pm n_x, \pm n_y, \pm n_z)$$ because they are just the same functions with a different sign. Thus, they are not distinct and since the density of state is really counting the number of states on which the system can be found, if two wavefunctions represent the same physical state, you don't count them twice, just once.

On a more mathematical side, you only want linearly independent wavefunction to form a basis, and this is what you count: the number of linearly independent states. Since $$\psi_{\vec n} = -\psi_{-\vec n}$$, they are not linearly independent.

Edit: Here is an interesting thing (in 1D for notation ease). If instead of considering hard boundary conditions with $$\psi(0) = \psi(L) = 0$$ you consider periodic boundary conditions: $$\psi'(x) = \psi'(x + L)$$, your eigenstate has the form: $$\psi'_{n}(x)\sim\exp\left(\frac{2n\pi ix}{L}\right)$$ and you can remark that: $$\psi'_{n}(x) \neq \alpha\psi'_{-n}(x)$$. They are genuinely different states. So here, we should not divide by 8 the density of states.

Everything is fine because in the case of hard boundary condition: you have $$k= n \pi/L$$ with $$n\in \mathbb{N}$$ while with periodic boundary conditions you have $$k= 2n \pi/L$$ with $$n\in \mathbb{Z}$$. So in the case of a periodic boundary condition, you would have a factor $$V/(2\pi)^3 = \dfrac{1}{8}V/\pi^3$$ instead of a factor $$V/\pi^3$$. The factor $$1/8$$ appears automatically and you don't have to divide, manually at the end by 8 to restrain your spheres to positive quantum number. The same thing happens also for the energy, see this PSE post.

Edit 2: By "indistinguishable" he really means linearly independent wavefunctions (wavefunction1 $$\neq \alpha$$ wavefunction2 ) and not wavefunctions with the exact same probability density distribution $$|\psi(x)|^2$$. Because $$|\psi'_{-n}(x)|^2 = |\psi'_{n}(x)|^2\forall x$$ for the periodic boundary codition wavefunction but the two states are genuinely different one that both need to be counted in the density of states.

• Thanks for the very comprehensive answer. I believe that I have gained knowledge about it very well. Commented Nov 11, 2023 at 21:26