# Inaccuracies of an energy level's degeneracy

I know that the inaccuracy is negligible but I'm trying to understand how it can be considered negligible in more detail.

The formula for the discrete energy levels in a box with equal dimensions $$L$$ is: $$E = \frac{h^2(n_x^2+n_y^2+n_z^2)}{8mL^2}$$ Where the $$n$$'s vary in integers. The degeneracy of an energy level is the number of ways the n-values can be rearranged, being a maximum of $$3!=6$$.

However, in the continuous approach it is approximated as the surface of an 8th of a sphere in n-dimensions with radius $$\sqrt{n_x^2+n_y^2+n_z^2}$$ which can be way more than the maximum of 6.

I have some specific questions regarding this approximation.

1. Can this be considered as counting in the degeneracy of other energy levels along with it?

2. If yes, is that what compensates for the fact that the n-sphere surface contains non-integer $$n$$-coordinates? Or is there another compensation for that?

3. I understand that the larger the $$n$$-sphere surface (i.e. the corresponding energy level is large), the smaller the factorial difference is with the true number of degeneracy of the covered energy levels. What about the absolute difference? Does this stay the same, increase or decrease?

4. If the $$n$$-sphere surface is very large (i.e. the corresponding energy level is large) does this mean that you're counting in the degeneracy of other energy levels that are further away from your chosen energy level?

There are way more than 3! possibilities in general. For instance, for $$n_x^2+n_y^2+n_z^2=69$$, there is $$(8,2,1)$$ and permutations thereof, and $$(7,4,2)$$ and permutations thereof, so $$12$$ possible combinations in total. The number of such "accidental" additional degeneracies increases as you increase the sum $$n_x^2+n_y^2+n_z^2$$, hence the continuous limit.

(By simple inspection, $$n_x^2+n_y^2+n_z^2=66,69,74,77\ldots$$ have $$12$$ degenerate states. To give you a sense of how this increases there are at least $$30$$ states with $$n_x^2+n_y^2+n_z^2=206,209,230,266,269,321,350,381,389,298,413,414,437,458,474$$ $$,486,494,497,$$ $$506,509,530,545,546$$ and $$621$$. Reversing this $$144$$ combinations of $$(n_x,n_y,n_z$$) that will produce $$349\le n_x^2+n_y^2+n_z^2\le 358$$.)

• Thanks for your correction, this helps me a bit. Do you happen to have an algorithm or link that spits out the number of integer degenerate states when inputting a value for $n_x^2+n_y^2+n_z^2$? One other question, does this mean that the volume of a spherical shell in n-dimensions with thickness $d\bigg(\sqrt{n_x^2+n_y^2+n_z^2}\bigg)$ would be more or less equal to the number of degenerate states as $n_x^2+n_y^2+n_z^2$ increases? If yes, would this be factorially closer or in absolute terms?
– Phy
Apr 16, 2020 at 18:10
• well you want to think of $n=\sqrt{n_x^2+n_y^2+n_z^2}$ as the length of a vector in "n-space". So in spherical you'd have $n^2 dn \sin\theta d\theta d\phi$ as you would have $r^2 dr \sin\theta\,d\theta\,d\phi$ in spherical. It's not hard to just produce a simple loop to count the number of times $n^2$ occurs given $(n_x,n_y,n_z)$. Apr 16, 2020 at 18:27
• I'm sorry but I'm completely unfamiliar with producing loops or algorithms to find the number of solutions. I have tried searching but couldn't find any online version of it either. And yes, I'd want to calculate an 8th of an n-shell volume which would be equal to $\frac{\pi n^2}{2} \cdot dn$. Therefore, would the outcome of this formula approach the number of degenerate states closer factorially or absolutely as $n$ increases?
– Phy
Apr 16, 2020 at 18:59

I think what is misleading here is focusing in the number of states corresponding to a specific energy, which is zero in the continuous limit. It is better instead to think of the number of states in an energy shell or even in a sphere (i.e. with energies less than the specified energy). In this case, in addition to counting degeneracy, one has to consider the states with different quantum numbers, but very close energies.

Let me note that "inaccuracies" are not always negligeable. In physics one often makes approximations, i.e. neglecting the parameters/effects whose magnitude is small, as compared to other parameters/effects. You always must have something to compare with in order to justify the approximation and calculate by how much you err. Thus, in the case of the infinite square well, one would assume that the frequency of the driving field, or temperature, or the level broadening is much bigger than the energy level spacing, to justified the transition to the continuous limit. This is a very bad approximation, since in the infinite square well the level spacing grows with $$n_x, n_y, n_z$$, unlike in any realistic potential.

The context might be however not the infinite square well, but using the periodic boundary conditions to calculate the energy density per unit volume. In thus case the discretization is an artificial tool in order to do calculations for the density of states that is already continuous. Therefore what I said in the beginning about the states with different $$n_x, n_y, n_z$$ but close in energy, will be reinforced in the end by sending $$L$$ to infinity!

• Thanks for your explanation. There is a big issue I have when talking about a shell in n-dimensions. I would deduce that the true number of quantum states within an infinitely thin n-shell would be achieved by counting the number of n-grid lattice points that is within/on the n-shell. But when one calculates it using n-shell volume, thus by multiplying the surface of the shell by an infinitely small thickness $0.00000...00001$, I would deduce that this thickness would make the outcome way less than the number of lattice points. What am I reasoning wrong here?
– Phy
Apr 17, 2020 at 12:56
• I think that talking about an infinitely thin shell doesn't make much sense here - this is a kind of coarse-graining procedure, where you have always deal with the volumes containing lots of points. Otherwise, one shouldn't be doing such an approximation. Apr 17, 2020 at 12:59
• I'm trying to comprehend how the number of lattice points within the shell would be approximately the same as its volume but I don't know how.
– Phy
Apr 17, 2020 at 16:02
• There is exactly one state for every combination $n_x, n_y, n_z$, that is one state occupies volume 1, and their density is uniform, i.e. volume in $n$-space is literally the number of states in it. Then the number states in any shell of radius $n$ and thickness $dn$ is $4\pi n^2dn/8$. This can be converted to the energy units using that $n^2=n_x^2+n_y^2+n_z^2$. Apr 17, 2020 at 17:50
• Ok, this has helped me quite a bit. So does this mean that I can not compare the outcome of $4\pi n^2 dn/8$ to the amount of integer solutions for a particular $n^2=n_x^2+n_y^2+n_z^2$? I.e. they will never be the same purely because the approach is discrete vs continuous?
– Phy
Apr 17, 2020 at 18:39