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I watched some derivations of the density of states in a square box of length L with potential $V=0$ for the points inside the box and $V=\infty$ outside the box. Using separation of variable one can get $\frac{2mE}{\hbar^2} = k_x^2+k_y^2+k_z^2$ with $k_j = \frac{\pi}{L}n_j$ and $n_j$ an integer. Defining $k^2 = k_x^2+k_y^2+k_z^2$ and assuming $k$ is very big we can calculate how many different wave functions there are with a $k$ having a length between $k$ and $k+dx$ by taking the volume of the shell of a sphere with radius $k$, which is $4\pi k^2dk$ and divide this by the volume $(\frac{\pi}{L})^3$. We must multiply the result by $2$ since each combination of $k_j$ can have two different spins. Summing up I would have said that the density of states is: $g(k)dk = 2\frac{4\pi k^2dk}{(\frac{\pi}{L})^3}$

The thing is the solution says that we only consider the positive $n_j$ so we should multiply the whole thing with $1/8$. I don't understand why wave function with negative $n_j$ should not be taken into account since for example a wave function with $n_x=1, n_y=2, n_z=3$ does not have the same "direction" as a wave with $n_x=-1, n_y=2, n_z=3$ so I don't get why we shouldn't count both.

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  • $\begingroup$ One way of understanding this is that you simply define k_x, k_y and k_z to be the (positive) square root of (2mE_i)/(h-bar)^2, where E_i is the energy associated with the i-th component. Another way of thinking about it is that changing the sign of the wave vector components only possibly changes the sign of the wave function, meaning that they represent the same physical states. $\endgroup$
    – User3141
    Commented Apr 18, 2021 at 11:42
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    $\begingroup$ @FelisSuper Please post answers as answers, not as comments. $\endgroup$
    – BioPhysicist
    Commented Apr 18, 2021 at 12:04

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The energy eigenstates take on the form of products of sines/cosines (depending on how you set up the coordinate system). The arguments of these functions will be proportional to $n_i$ for $i=x,y,z$, and so negative values of these quantum numbers will only change the eigenfunctions by a factor of $1$ or $-1$, thus not changing the actual physical state.

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  • $\begingroup$ But why aren't $\psi(x)$ and -$\psi(x)$ considered as different wave functions? I know they have the same absolute value, but they behave differently, for example it is not the same when you have a superposition of another wave with $\psi(x)$ or with -$\psi(x)$. I don't see why the Pauli exclusion's principle would forbid having an electron in the sate $\psi(x)$ and the other in the sate -$\psi(x)$ $\endgroup$
    – Nicolas Schmid
    Commented Apr 18, 2021 at 13:21
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    $\begingroup$ @NicolasSchmid Yes, $\psi_1+\psi_2$ is different than $\psi_1-\psi_2$, but that doesn't mean $\psi_2$ and $-\psi_2$ are different states. This has nothing to do with the Pauli exclusion principle, as this is true for even single-particle states. $\endgroup$
    – BioPhysicist
    Commented Apr 18, 2021 at 14:25
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You only need to consider positive values as the energy does not depend on the sign of any of the components of $\vec k$. You then have counted only 1/8 of all possibilities and must multiply by 8.

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  • $\begingroup$ I downvoted because your last sentence is wrong, and because your explanation with the energy doesn't help me. $\endgroup$
    – Nicolas Schmid
    Commented Apr 18, 2021 at 13:30
  • $\begingroup$ It might be right but without arguing why we can not have two states with the same energy, it doesn't explain anything. $\endgroup$
    – Nicolas Schmid
    Commented Apr 18, 2021 at 13:37
  • $\begingroup$ @my2cts Your answer is wrong. If you consider the case of a free particle in a box with periodic boundary conditions, the Hamiltonian is again quadratic in the components of $\vec k$ but we have to count as independent states those corresponding to positive and negative components. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Apr 22, 2021 at 5:47
  • $\begingroup$ @my2cts I also add that proposing to close the question after being downvoted does not look like fair behavior. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Apr 22, 2021 at 5:49
  • $\begingroup$ @GiorgioP Of course these are independent states. Since there is 8 fold degeneracy they are counted by multiplying with the factor 8. $\endgroup$
    – my2cts
    Commented Apr 22, 2021 at 7:01

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