Any nice interpretation of differentiating $j(j+1)$ to get $2j+1$?

Question

If $j$ is a continuous variable, then differentiating the function $f(j)=j(j+1)$ with respect to $j$ gives $f'(j)=2j+1$. Of course I've chosen the letter to evoke quantum mechanical angular momentum, in which case for integer or half-integer values of $j$ we can interpret these two expressions as the eigenvalue of the squared angular momentum and the multiplicity of the angular momentum.

Is there any nice interpretation of this? As I was looking at the automatically generated list of "Questions that may already have your answer," I came across a comment that asks exactly the same question.

One reason to believe that it has no very special interpretation is that since the actual variable is discrete, the derivative $f'$ would really represent an approximation to a divided difference, and the relevant difference for a unit change in $j$ does not necessarily equal the derivative unless you evaluate the derivative at the correct place.

It seems to me that there is a second reason not to expect anything special here, which is that the correspondence doesn't seem to work except in three dimensions. For a rotor in $d$ dimensions, the eigenvalue of the squared angular momentum operator is $j(j+d-2)$. I don't know what the multiplicity of states is in general, but I suppose it's a polynomial of order $d-2$. E.g., for $d=2$, the multiplicity is 2 ($m=\pm j$), which doesn't equal the derivative of $j(j+d-2)=j^2$. On the other hand, I guess it's possible that there is a nice interpretation, and the nice interpretation tells us that there's something special about three dimensions.

Related: What is known about the hydrogen atom in $d$ spatial dimensions?

@AccidentalFourierTransform: the multiplicity above is the number of micro-states Ω, whose derivative is the density of micro-states You have a function and its derivative. Isn't the function you're talking about differentiating the multiplicity, whereas in the topic of this question the multiplicity is what you get after differentiation? Or maybe I'm just not understanding you. — , May 26 '17 at 00:58

Prahar · Answer 1 · 2017-05-28T22:46:01.753

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The number of spherical harmonics of weight $\ell$ on $S^d$ is given by $$ N(d,\ell) = \frac{d+2\ell-1}{d-1} {d+\ell-2\choose \ell} $$ Further, the eigenvalue of this spherical harmonic is $-\Delta(d,\ell) = \ell ( \ell + d - 1 )$.

In particular, note that for $d=2$ (which is the case to consider for 3 dimensional angular momenta, we find $$ N(2,\ell)= 2\ell+1~, \qquad -\Delta(2,\ell) = \ell(\ell+1)~. $$ In this special case, it is true that $-\partial_\ell \Delta(2,\ell) = N(2,\ell)$ as you have observed. I did note the following generalization to higher dimensions, $d\geq3$, $$ \partial_\ell^{d-3} N(d,\ell) = - \Delta(d,\ell) + \frac{d^2}{4} - \frac{7d}{12} + \frac{1}{2}~. $$ I do not see any real property of interest however. I believe the formula is a pure coincidence.

PS - This is one case where I would love to be proven wrong!

edited May 28 '17 at 22:46

answered May 28 '17 at 21:36

Prahar

25,924

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This reads to me as a rewording of the argument I made in the fourth paragraph of the question. There are some added details, but most of those added details were already worked out in the comment by AccidentalFourierTransform. The reason I posted this as a question and offered a bounty was that I was hoping someone could come at this from a different angle. – May 29 '17 at 17:21
Perhaps interesting as well: if we use a finite difference instead of a derivative, we have $\Delta^d_\ell N=0$, $\Delta^{d-1}\ell N=2$, $\Delta^{d-2}\ell N=2 d+2 \ell-3$, $\Delta^{d-3}_\ell N=(d+\ell-2)^2$, etc. – AccidentalFourierTransform May 30 '17 at 17:26

Any nice interpretation of differentiating $j(j+1)$ to get $2j+1$?

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