The energy levels of a hydrogen atom are (ignoring fine structures etc.) $E_n=-1/2n^2$, with each energy level labeled by $n\in\bZ^+$, and each energy level has $g_n\ceq n^2$ degeneracy (ignoring spin degeneracy, which merely contributes to an overall factor of the partition function). The partition function is $$\fc Z\beta\ceq\sum_{n=1}^\infty g_n\e^{-\beta E_n} =\sum_{n=1}^\infty n^2\e^{\beta/2n^2},$$ $$(1)$$ which diverges for any $\beta\in\bC$ (of course, normally we can only have $\beta\in\bR$, but the point of saying that it diverges for any complex $\beta$ is that there is no way we can analytically continue the function to get a finite result). Does this mean that statistical mechanics breaks down for this system? Not necessarily. Actually, there are multiple ways we can tackle this divergence.

One should notice that, although this article concentrates on regularizing partition functions and that of the hydrogen atom in particular, all the methods are valid for more general divergent sums.

Here is a sentence that is quoted by many literatures on diverging series, so I want to quote it, too:

It translates to “Divergent series are in general deadly, and it is shameful that anyone dare to base any proof on them.”

The physical answer

A physicist always tell you that one should not be afraid of infinities. Instead, one should look at where the infinity comes out from the seemingly physical model, where there is something sneakily unphysical which ultimately leads to this unphysical divergence. In our case, the divergence comes from high energy levels. It is then a good time to question whether those high energy levels are physical.

There is a radius associated with each energy level in the sense of the Bohr model: $r_n=n^2$. When $r_n\sim L\ceq10^{10}$ (which happens at $n\sim\Lmd\ceq10^5$), the orbit is really microscopic now, and the interaction between the electron and the “box” that contains the whole experimental setup is now having significant effects. Or, if there is not a box at all, we can use the size of the universe instead, which is about $r_n\sim L\ceq10^{36}$ ($\Lmd\ceq10^{18}$). Use the model of particle in a box for energy levels higher than $n=\Lmd$, and we have $$Z=\sum_{n=1}^\Lmd n^2\e^{\beta/2n^2} +\sum_{n_x,n_y,n_z=1}^\infty\fc\exp{-\beta\fr{\p{n_x^2+n_y^2+n_z^2}\pi^2}{2L^2}},$$ where $L$ is the side length of the box (assuming that the box is cubic). If $L$ is very large, we can approximate the second term as a spherically symmetric integral over the first octant to get $L^3\p{2\pi\beta}^{-3/2}$.

For the first term, we need to consider how the magnitude of the summand changes with $n$. The minimum value of the summand is at $n=\sqrt{\beta/2}$. At room temperature, we have $\beta\sim10^3$, so $\sqrt{\beta/2}$ is well between $1$ and $\Lmd$. Therefore, the largest term is either $n=1$ or $n=\Lmd$. The former is $\e^{\beta/2}$, which is of order $10^{217}$, while the latter is $\Lmd^2$, which is of order $10^{36}$ for the case of the size of the universe. We may then be interested in the $n=2$ term $4\e^{\beta/8}$, which is of order $10^{54}$. This is much larger than the $n=\Lmd$ term but much smaller than the $n=1$ term, so it is second largest term in the sum.

An upper bound of the summation is given by replacing every term except the largest term by the second largest term, which gives $$Z<\underbrace{\e^{\beta/2}}_{10^{217}} +\underbrace{\p{\Lmd-1}4\e^{\beta/8}}_{10^{72}}+\underbrace{L^3\p{2\pi\beta}^{-3/2}}_{10^{48}}\approx\e^{\beta/2}.$$ Therefore, the $n=1$ term dominates the entire partition function. This means that the hydrogen atom is extremely likely to be in the ground state (despite the seeming divergence of the partition function). This is intuitive. The probability of the system not being in the ground state is of order $10^{-55}$ for the size of the universe and $10^{-158}$ for a typical macroscopic experiment.

Cutoff regularization

By analyzing the orders of magnitude, we see that we actually do not lose much if we just simply cut off the sum at $n=\Lmd$. This corresponds to a regularization method called the simple cutoff: it replaces the infinite sum by a finite partial sum. This can be generalized a little by considering a more general cutoff function $\chi$ such that $\lim_{x\to0^+}\fc \chi x=1$. Then, an infinite sum $\sum_{n=1}^\infty\fc fn$ can be written as $$\sum_{n=1}^\infty\fc fn=\lim_{\lmd\to0^+}\sum_{n=1}^\infty\fc fn\fc\chi{\lmd n}.$$ The simple cutoff is then the case where $\fc \chi x\ceq\fc\tht{1-x}$ and $\lmd\ceq1/\Lmd$, where $\tht$ is the Heaviside step function. For converging series, this gives the same result as the original sum thanks to the dominated convergence theorem.

However, although this series may converge for any positive $\lmd$, the limit as $\lmd\to0^+$ may not exist. If it diverges because $\fc fn$ grows too fast (or decays too slowly) as $n\to\infty$, then we should expect that the sum also tends to infinity as $\lmd\to0^+$. Assume that we can characterize this divergence by a Laurent series: $$\sum_{n=1}^\infty\fc fn\fc\chi{\lmd n} =\sum_{k=-\infty}^\infty\gma_k\lmd^k.$$ $$(2)$$ If the $\lmd\to0^+$ limit converge, we would expect $\gma_{k<0}$ to be zero, and then the result is simply $\gma_0$. Therefore, we may also want only $\gma_0$ when the limit does not exist. To pick out $\gma_0$, utilize the residue theorem: $$\sum_{n=1}^\infty\fc fn=\fr1{2\pi\i}\oint\fr{\d\lmd}\lmd \sum_{n=1}^\infty\fc fn\fc\chi{\lmd n},$$ $$(3)$$ where the domain of $\lmd$ is now analytically continued from $\bR^+$ to a deleted neighborhood of $0$. Equation 3 is then a generalized version of Equation 2.

Notice that I have been super slippery in math in the discussion. For example, the Laurent series may not exist at all, and the analytic continuation may not be possible at all; even if they exist, the $\lmd\to0^+$ limit may also be different from $\gma_0$. However, I may claim that we should be able to select smooth enough $\chi$ for all of these to work, and the results will be independent of the choice of $\chi$ as long as Equation 3 works in this form.

Particularly, one can rigorously prove that for $\fc fn\ceq n^k$, the sum obtained by this precedure is $\fc\zeta{-k}$, where $\zeta$ is the Riemann zeta function, as long as $x^k\fc\chi x$ has bounded $\p{k+2}$th derivative and the sum converges. This is proven in an interesting blog article.

Regularizing the hydrogen atom

After saying so much about cutoff regularization in general, what does it say about the partition function of a hydrogen atom? Try multiplying the cutoff function $\fc\chi{\lmd n}$ to the summand in Equation 1: $$Z_\lmd\ceq\sum_{n=1}^\infty n^2\e^{\beta/2n^2}\fc\chi{\lmd n} =\sum_{k=0}^\infty\fr{\p{\beta/2}^k}{k!}\sum_{n=1}^\infty n^{2-2k}\fc\chi{\lmd n} \to\sum_{k=0}^\infty\fr{\p{\beta/2}^k}{k!}\fc\zeta{2k-2},$$ $$(4)$$ where the last step utilizes the result for $\fc fn\ceq n^k$, with which we get rid of the dependence on $\lmd$. The last expression is then identified as $Z$.

Now that we get the expression of $Z$, we can get some useful things. However, this time we cannot simply use the summand divided by $Z$ to get the probability of each energy level because that will break the normalization of the probability distribution. What we can do, however, is to find the expectation value of the energy using $\a E=-\d\ln Z/\d\beta$. On the other hand, we have $\a E\le p_1E_1+\p{1-p_1}E_\infty=-p_1/2$, so the probability $1-p_1$ that the system is not in the ground state is bounded above by $2\a E+1$.

The first check to do is to verify that this result is consistent with the known behavior of the system at cold zero temperature, where the system is almost certainly in the ground state; in other words, $\lim_{\beta\to+\infty}\a E=-1/2$. To get $Z$ for large $\beta$, we notice that $\fc\zeta{+\infty}=1$, so $Z\approx\e^{\beta/2}$, and this leads to $\a E\approx-1/2$ as expected.

Now, we may try to estimate $\a E$ for finite but large $\beta$ (e.g., $\beta=10^3$) and thus give an upper bound for $1-p_1$. We can study the asymptotic behavior of $\a E$ for cold positive temperature. It turns out that $1-p_1\approx3\e^{-3\beta/8}$, which is $10^{-163}$ for $\beta=10^3$. As we can see, without any physical arguments but only with regularization, we get a result that seems sensible and well between the results in the last section for a hydrogen atom confined in a box with a typical macroscopic size or the size of the universe.

Although the asymptotic behavior at cold temperature ($\beta\to+\infty$) looks good, its behavior is very wrong at some regimes. At some temperature, the monoticity of $\a E$ reverts, and then it gets even lower than the ground state energy $-1/2$ and heads all the way to $-\infty$ at some finite temperature.This is clearly unphysical. This suggests that it is wrong to use the regularized result.

Plots

Here is a plot that shows how $\a E$ starts to decrease with temperature at some point and becomes even lower than the ground state energy:

Here is a plot that shows how $\a E$ goes to infinity at different temperatures:

Here are also plots for $Z$ and $\d Z/\d\beta$, if you are curious:

The two vertical asymptotes of $\a E$ corresponds to the two zeros of $Z$, which are $\beta=0$ and $\beta=1.0721$. It also has a zero, correponding to the zero of $\d Z/\d\beta$ at $\beta=0.5530$. The point where $\a E=-1/2$ is $\beta=11.2486$, and the point where $\a E$ has a local maximum is $\beta=13.8021$.

Another aspect where we can see that this result is wrong is that, if we look at the hot negative temperature limit $\beta\to-\infty$, although we have $\a E\to0=\sup_nE_n$ as expected, it is approaching from the wrong side. In fact, because $Z>0$ while $\d Z/\d\beta<0$ for $\beta<0$, we have $\a E>0$ for $\beta<0$, exceeding the supremum of the energy levels, which is unphysical.

Zeta function regularization

For a series $\sum_n\fc fn$, if it diverges, we can instead consider $\sum_n\fc fn^{-s}$ for some $s$ whose real part is big enough for the series to converge. Then, we can try to analytically continue to $s=-1$ to get a finite result for the original series. This is called the zeta function regularization.

A famous example is $\fc fn\ceq n$, which gives $\sum_nn=\fc\zeta{-1}=-1/12$. Generally, for $\fc fn\ceq n^k$, we have $\sum_nn^k=\fc\zeta{-k}$. This is the same as the result for the simple cutoff regularization. This raises the question of whether the results obtained from those two methods are necessary the same whenever they both exist. I do not have a rigorous proof, but a strong argument is that both of them are the result of some analytic continuation, so they should be the same by the uniqueness of analytic continuation.

We can check this with the hydrogen atom. We have, for $s>1/2$ and $\beta$ real, $$Z_s\ceq\sum_{n=1}^\infty n^{-2s}\e^{-s\beta/2n^2} =\sum_{k=0}^\infty\fr{\p{-s\beta/2}^k}{k!}\sum_{n=1}^\infty n^{-2s-2k} =\sum_{k=0}^\infty\fr{\p{-s\beta/2}^k}{k!}\fc\zeta{2s+2k}.$$ Analytically continue this result to $s=-1$, and then this gives the same result as Equation 4. The rest will be the same as the last section.

Can we trust this result?

However, can we trust this result, though? Everything is becoming fishy. Probabilities are no longer well-defined because how we normally derive them using $Z$ is causing a divergent sum of probabilities and thus invalid. Yet somehow we are trying to estimate the bound of the probability of the system not being in the ground state and getting an expected result. You must have been feeling uncomfortable about this.

The first thing to ask is what we mean by “the expectation value” when the probability distribution is not even well-defined. If it means nothing physical, can we still trust its expression? The simple answer is no.

As we already see, although the result at cold temperature is sensible, the result at some regimes is clearly unphysical. We can also see similar problems with other systems. Consider the system that has energy levels $E_n=\ln n$ (with no degeneracies). We can easily get $Z=\sum_nn^{-\beta}=\fc\zeta\beta$, and thus there is a absi at $\beta=1$. For $\beta>1$, $Z$ converges, and everything looks good. For $\beta<1$, the system is so hot that $Z$ diverges. Previous arguments suggest that, in this region, the regularized $Z$ is still $\fc\zeta\beta$. However, we then have $\a E<0$ in this region, which is lower than the ground state energy. This clearly should not be trusted.

In another aspect, we should note that since the estimation for $1-p_1$ does not depend on the size of the box confining the hydrogen atom, its rough agreement with the result in the last section should be considered a coincidence.

Another thing to note is that the result of the regularizations depend on whether we “flatten” the energy levels. We can “flatten” all the energy levels: pretend no degeneracies exist. For example, suppose a system with $g_n\ceq n$ and $E_n\ceq n$. However, we can rewrite the same system as $E_n\ceq1,2,2,3,3,3,\ldots$ (or equivalently $E_n\ceq\floor{\sqrt{2n}+1/2}$) ¹, with no degeneracies. This “re-grouping” of the energy levels can affect the result of regularizations and whether a zeta function regularization exists. For an immediate example, if we flatten the energy levels of the hydrogen atom, the zeta function regularization does not exist. Another simple example is that, for a system with $E_n=\mrm{const}$, we can essentially re-group the all-degenerate states to have any positive integer sequence $g_n$ to get very arbitrary results for the partition function.

Abscissa of convergence

Forget about the hydrogen atom, and let us consider a general system with (ever-increasing) energy levels $E_n$ and degeneracies $g_n$. For a given system, there is an abscissa of convergence $\beta_\mrm c$, below (hotter than) which the partition function diverges. In other words, $Z\ceq\sum_ng_n\e^{-\beta E_n}$ converges for $\Re\beta>\beta_\mrm c$ and diverges for $\Re\beta<\beta_\mrm c$. For most physical systems, we have $\beta_\mrm c=0$, meaning that it can have any positive temperature, which sounds sensible. The hydrogen atom has $\beta_\mrm c=+\infty$, and a two-level system has $\beta_\mrm c=-\infty$. A system with $E_n=\ln n$ and no degeneracy has $\beta_\mrm c=1$.

The term “abscissa of convergence” is borrowed from the study of general Dirichlet series. The form of $Z$ is indeed very much like a general Dirichlet series, but a general Dirichlet series requires $E_\infty=+\infty$, which is not true for the hydrogen atom. However, the existence of an abscissa of convergence is still true for the more general case.

What does it mean physically to have an abscissa of convergence $\beta_\mrm c$? First, if $\beta_\mrm c=-\infty$, then the system is well behaved at any temperature, which is good and does not need further care.

If $\v{\beta_\mrm c}<\infty$, normally one should say the system cannot reach a certain temperature: the system can never be in equilibrium with a heat bath hotter than $\beta_\mrm c$. Thermodynamically, one can say that the system needs to absorb an infinite amount of heat to reach this temperature. One can see this easily by considering any sensible system, which has $\beta_\mrm c=0$: for $\beta$ to go below zero means to make the temperature hotter than infinity, which of course needs an infinite amount of heat intuitively. One may want to see whether it is possible to regularize $Z$ to get a finite result for $\Re\beta<\beta_\mrm c$. A valid claim to make is that, if $Z$ can be analytically continued to the half real axis to the left of $\beta_\mrm c$, then any sensible regularization of $Z$ there will give the same result as the analytic continuation. Actually, the analytic continuation is exactly the zeta function regularization if there is no degeneracy (or regarding degenerate states as different energy levels). However, it is possible that the analytic continuation does not exist. There may be a branch cut or a natural boundary. For example, if $E_n\ceq\ln p_n$ with no degeneracy, where $p_n$ is the $n$th prime number, then $Z$ is the prime zeta function, which has a natural boundary at $\Re\beta=0$. Even if such a regularization exists, it should be questioned whether it is physical.

If $\beta_\mrm c=+\infty$, then the system is not well behaved at any temperature. This is the case for the hydrogen atom. Physically, this means that the system cannot be in equilibrium with a heat bath at any temperature. The problem with regularization is the same as the case with $\v{\beta_\mrm c}<\infty$.

In a previous article about statistical ensembles, when I defined the partition function, I briefly mentioned that it is only defined for those intensive variables ($\beta$ in the context of this article) such that the partition function converges. I did not talk about what to do with the partition function when it diverges, but what that article implied is that it is simply undefined and that no physical meaning should be assigned to it in principle. The existence of an abscissa of convergence tells us that there is a “hottest possible temperature” for any given system. The hydrogen atom is symply the case where the hottest possible temperature coincides with the coldest possible temperature (which is the absolute zero). For most sensible systems, the hottest possible temperature is just the positive hot limit. For systems such as $E_n\ceq\ln n$, the hottest possible temperature is a finite positive temperature, which is at $3.16\times10^5\,\mrm K$, resulting from $\beta_\mrm c=1$. This can be conterintuitive at first, but one should realize that it is not essentially different from the more common case of $\beta_\mrm c=0$.

Theorem. The conformal map $\fc wz$ transforms the trajectory with energy $-B$ in potential $\fc Uz\ceq A\v{\d w/\d z}^2$ into the trajectory with energy $-A$ in potential $\fc Vw\ceq B\v{\d z/\d w}^2$.

This result is pretty amazing in that it reveals a quite implicit duality between the two potentials, and it looks very symmetric as written.

This theorem, as I know of, was first introduced in the appendix of V. I. Arnold’s book Huygens and Barrow, Newton and Hooke. Part of this article is already covered in the relevant part of the book.

Power-law central-force potentials

Before I show the proof of it, let me first introduce it by a much more well-known example.

As we all know, Bertrand’s theorem states that the only two types of central-force potentials where all bound orbits are closed are $U\propto r^{-1}$ (the Kepler problem) and $U\propto r^2$ (the harmonic oscillator). How the two potentials are special among all sorts of different central-force potentials makes people wonder if there is any connection between them. Fortunately, there is one, and it is obvious once we notice that the complex squaring transforms any center-at-origin ellipses into focus-at-origin ellipses. Inspired by this, it is easy to see that trajectories in the Kepler problem can be transformed into trajectories of harmonic oscillators under complex squaring.

You may ask, how can we notice complex squaring does the said transformation on ellipses? The observation is noticing the simple algebra $$\p{z+\fr1z}^2=z^2+\fr1{z^2}+2,$$ which means that the Joukowski transform $z\mapsto z+1/z$ of a unit circle simply translates under complex squaring. We can then try to generalize this to circles of other radii, whose Joukowski transformations are just ellipses! (If you remember, this is the second time Joukowski transformation appears in my blog. The first time was here.)

Then, are the Kepler problem and the harmonic oscillator the only two central-force potentials whose trajectories can be transformed into each other by a complex function? The answer is no. In fact, for any trajectory in almost any power-law central-force potential, we can take some power of it to get a trajectory in another power-law central-force potential.

This result can be summarized as follows. Taking the $\p{\alp/2+1}$th power of a trajectory with energy $E$ in the potential $U=ar^\alp$ ($\alp\ne-2$) gives a trajectory with energy $F$ in the potential $V=br^\beta$, where $$\p{\alp+2}\p{\beta+2}=4,\quad b=-\fr14\p{\alp+2}^2E,\quad F=-\fr14\p{\alp+2}^2a.$$ To prove this, we just need to reparameterize the transformed trajectory in a new time coordinate $\tau$ defined as $\d\tau=\v z^\alp\,\d t$, where $z$ is the complex position of the original trajectory. Then, by some calculation and utilizing the energy conservation, we can show that the parameter equation in terms of the new time coordinate satisfy the equation of motion we expect. I will not show the details here because they would be redundant once I prove the more general case using the same methods.

Corollaries and applications

There is an interesting special case, which is $\alp=-2$. There is no potential that is dual to $U\propto r^{-2}$. Another interesting case is $\alp=-4$, which is dual to itself ($\beta=-4$). It kind of means that the coefficient in the potential is “interchangeable” with the energy, and the trajectories can be derived from each other by taking the complex reciprocal.

We can get some interesting results with $a=0$, which is just the case of a free particle, whose trajectories are all straight lines. Since in this case we necessary have $F=0$, we can say that the zero-energy trajectory in any power-law potential is related to a straight line by a power. From this result, we can derive some interesting corollaries. For example, the zero-energy trajectory in the Kepler problem is a parabola (square of a straight line), which is well-known. The zero-energy trajectory in $U\propto-r^{-4}$ is a circle passing through the origin (reciprocal of a straight line), which is a pretty interesting not-so-well-known result.

Another interesting result is that, the deflection angle of an incident zero-energy particle scattered by the potential $U\propto-r^\alp$ is $\tht$ under paraxial limit, if $$\alp=\fr{2\vphi}{\pi-\vphi},\quad\vphi=\pm\tht-2k\pi,\quad k\in\bN.$$ This result can be easily derived by using the conformal transform of the real line (actually, a straight line that approaches the real line). The crucial part here is that $k$ cannot take negative integers because we need $\alp>-2$. The reason is that, when $\alp\le-2$, paraxial zero-energy particles are bound to sink into the origin, and thus no scattering actually happens. This small pitfall indicates that the trajectory in the dual potential is not a two-side infinite straight line, either, in that limit, in contrast to being seemingly a free particle.

Some straightforward proofs

Let’s go back to the theorem I stated at the beginning of this article.

Proof. Consider a new time coordinate $\tau$ defined as $\d\tau=\v{\d w/\d z}^2\,\d t$. Then, the motion of $w$ satisfies $$\begin{align*} m\fr{\d^2w}{\d\tau^2} &=m\fr{\d t}{\d\tau}\fr{\d}{\d t}\p{\fr{\d t}{\d\tau}\fr{\d w}{\d t}}\\ &=m\v{\fr{\d z}{\d w}}^2\fr{\d}{\d t}\p{\v{\fr{\d z}{\d w}}^2\fr{\d w}{\d z}\fr{\d z}{\d t}}\\ &=m\fr{\d z}{\d w}\p{\fr{\d z}{\d w}}^*\p{\p{\fr{\d^2z}{\d w^2}\fr{\d w}{\d z}\fr{\d z}{\d t}}^*\fr{\d z}{\d t} +\p{\fr{\d z}{\d w}}^*\fr{\d^2 z}{\d t^2}}. \end{align*}$$ Here we need to substitute $\d^2 z/\d t^2$ by the equation of motion for $z$. By computing the real and imaginary parts separately, we can derive that for any holomorphic function $f$, the gradient of $\v f^2$ expressed as a complex number is $\nabla\v f^2=2\p{\d f/\d z}^*f$. Therefore, the equation of motion for $z$ is $$m\fr{\d^2z}{\d t^2}=-2A\fr{\d w}{\d z}\p{\fr{\d^2w}{\d z^2}}^*.$$ According to series reversion, we have $\d^2 w/\d z^2=-\p{\d w/\d z}^3\d^2 z/\d w^2$. Therefore, the equation of motion for $z$ can also be written as $$m\fr{\d^2z}{\d t^2}=2A\v{\fr{\d w}{\d z}}^2\p{\fr{\d w}{\d z}}^{*2}\p{\fr{\d^2 z}{\d w^2}}^*.$$ Substitute this, and we have $$m\fr{\d^2w}{\d\tau^2}=\fr{\d z}{\d w}\p{\fr{\d^2z}{\d w^2}}^* \p{m\v{\fr{\d z}{\d t}}^2+2A\v{\fr{\d w}{\d z}}^2}.$$ Substitute the energy conservation of the motion of $z$: $$\fr12m\v{\fr{\d z}{\d t}}^2+A\v{\fr{\d w}{\d z}}^2=-B,$$ and we have $$m\fr{\d^2w}{\d\tau^2}=-2B\fr{\d z}{\d w}\p{\fr{\d^2z}{\d w^2}}^*,$$ which is the equation of motion for $w$ that we expect.

To get the energy of the motion of $w$, we calculate $$\begin{align*} \fr12m\v{\fr{\d w}{\d\tau}}^2+B\v{\fr{\d z}{\d w}}^2 &=\fr12m\v{\fr{\d w}{\d z}\fr{\d z}{\d t}\fr{\d t}{\d\tau}}^2+B\v{\fr{\d z}{\d w}}^2\\ &=\v{\fr{\d w}{\d z}}^2\p{-B-A\v{\fr{\d w}{\d z}}^2}\v{\fr{\d z}{\d w}}^4+B\v{\fr{\d z}{\d w}}^2\\ &=-A, \end{align*}$$ which is the energy conservation of the motion of $w$ in the potential $V$ that we expect. $\square$

Noticing that we are only interested in the trajectory, we can just use Maupertuis’ principle to get a simpler proof.

Proof. $$\mcal S_0=\int\v{\d z}\sqrt{2m\p{-B-A\v{\fr{\d w}{\d z}}^2}}=\int\v{\d w}\sqrt{2m\p{-A-B\v{\fr{\d z}{\d w}}^2}}.$$ The abbreviated action is then exactly the same for the motion of $z$ and the motion of $w$. Therefore, by Maupertuis’ principle, for any physical trajectory of $z$, the trajectory of $w$ is also physical. $\square$

Details worth noting

Invertibility of the conformal map

There are two different definitions of a conformal transformation in two dimensions. One is that a function defined on an open subset of $\bC$ is conformal iff it is holomorphic and its derivative is nowhere zero. The other is that a function is conformal iff it is biholomorphic (is bijective and has a holomorphic inverse).

You may think here I have adopted the second definition because when I say $\fc Vw\ceq B\v{\d z/\d w}^2$, I am implicitly assuming that I can take the inverse of $\fc wz$ to get the function $\fc zw$ and then take the derivative of it. However, if that is the case, an immediate problem is that then the duality between the Kepler problem and the harmonic oscillator, from which I introduced the more general result in the first place, would not be actually covered by the “more general” result. This is because $z\mapsto z^2$ is not biholomorphic (because it is not injective).

Then, why did this never become a problem when we were studying the duality between the Kepler problem and the harmonic oscillator? All we have talked about is how we can derive a trajectory in the Kepler problem by squaring the trajectory of a harmonic oscillator, but we have not discussed about how we can reverse this process, as an essential part of the duality. You may think the reverse of the process would be totally natural given how symmetric our theorem is regarding the two potentials. However, the reverse is not actually well-defined since the inverse of squaring, i.e., taking the square root, is not a single-valued function. Nevertheless, it is still well-defined in some sense: starting with whichever branch we like, tracing one point on the trajectory of the Kepler problem, and moving it along this trajectory for two cycles, we will end up with a trajectory of the harmonic oscillator if we take the square root of the position and ensure we always choose the branch so that the mapping is continuously done.

What about other power-law central potentials? In those cases, we have non-closed trajectories, so we cannot just move along the trajectory for two cycles. For example, if we take $w=z^3$, then the potential would be $U=9A\v z^4$. For any non-closed trajectory, we can uniquely map it to a trajectory of the potential $V=B\v w^{-4/3}/9$. However, we cannot uniquely do the reverse mapping. There would be three different trajectories in the potential $U$ that can be mapped to the same trajectory in $V$, and we can in turn map the trajectory in $V$ to any of the three trajectories in $U$ depending on which branch we choose.

Therefore, to generalize this for more general potentials, we can use similar arguments. Because $z\mapsto w$ has non-zero derivative everywhere in our considered region, it is everywhere locally invertible by the Lagrange inversion theorem. We can then bijectively map the trajectories in the two dual potentials locally for every small (and finite) segment and then patch them together to get the global correspondence between the two trajectories. This mapping may not be well-defined globally, but the trajectories can still be considered dual to each other. If the potential also becomes multi-valued due to the mapping $w\mapsto z$ being multi-valued, then we should imagine this situation like this: at some point, the potential may be different when the particle visit here for the second time. This case does not happen if we only look at power-law potentials, but it does happen for more general cases.

What makes this sense of duality weaker is that one trajectory can be dual to multiple different trajectories. A case worth noting is that sometimes one trajectory can be mapped to infinitely many different trajectories. This happens when the trajectory runs around a logarithmic branch point. However, we can gain the sense of duality back if we can also consider the case where $z\mapsto w$ is multi-valued. The notion of conformal transformation is now too limited to cover this case, a better notion is a global analytic function, which generalizes the notion of analytic function to allow for multiple branches.

Requirements for the potential

Not any potential can be expressed as $A\v{\d w/\d z}^2$. How can we determine whether a potential can be expressed in this form?

Theorem. A continuous potential $U$ can be expressed in the form of $A\v{\d w/\d z}^2$ (where $\fc wz$ is a conformal transformation) iff one of the following conditions is met:

• $U$ is zero everywhere, or
• $\ln\v U$ is a harmonic function on the domain of $U$.

Proof. First, prove the necessity.

An obvious requirement is that the potential must be positive everywhere or negative everywhere (or zero everywhere, but that is trivial). The sign is determined by the sign of $A$. Therefore, without loss of generality, we can assume $A=1$ because we can always absorb a factor of $\sqrt{\v A}$ into $w$ and adjust the overall sign of $U$ accordingly.

We can decompose $\p{\d w/\d z}^2$ in the polar form $$\p{\d w/\d z}^2=\v{\d w/\d z}^2\e^{\i\vphi}=U\e^{\i\vphi},$$ where $\vphi$ is a real function of $z$. Applying the Cauchy–Riemann equations to $\p{\d w/\d z}^2$ gives $$\i\partial_x\p{\fr{\d w}{\d z}}^2=\partial_y\p{\fr{\d w}{\d z}}^2 \implies\i\p{\e^{\i\vphi}\partial_xU+\i U\e^{\i\vphi}\partial_x\vphi} =\e^{\i\vphi}\partial_yU+\i U\e^{\i\vphi}\partial_y\vphi.$$ Equate the real and imaginary parts, and we have $$\begin{cases}U\partial_x\vphi=-\partial_yU,\\U\partial_y\vphi=\partial_xU.\end{cases}$$ Use the symmetry of second derivatives on $\vphi$, and we have $$\partial_x\partial_y\vphi-\partial_y\partial_x\vphi=0 \implies\partial_x\fr{\partial_xU}U+\partial_y\fr{\partial_yU}U=0.$$ In the language of vector analysis, this is just $\nabla^2\ln U=0$.

Considering the case where $U$ is negative everywhere, we have that $\ln\v U$ is a harmonic function.

Then, prove the sufficiency.

The case where $U$ is zero everywhere is trivial. Otherwise, because $\ln\v U$ is defined everywhere on the domain of $U$, we must have $U$ is non-zero everywhere. Because $U$ is continuous, we have $U$ is either positive everywhere or negative everywhere.

Without loss of generality, assume $U$ is positive everywhere. Let $\vphi$ be the harmonic conjugate of $\ln U$. Then, $\ln U+\i\vphi$ is a holomorphic function. We can then define $$\fr{\d w}{\d z}=\sqrt U\e^{\i\vphi/2},$$ which is also a holomorphic function. $\square$

From now on, we will call this requirement on $U$ as being log-harmonic for obvious resons.

We should notice that whether $U$ is log-harmonic does not respect that any potential can have an additive constant and still be essentially the same potential. An immediate example is that a function that is positive everywhere may be negative somewhere if we add a constant to it. We may then want to ask whether $U$ can be log-harmonic if we allow it to be added an additive constant. It is easy to do this: we can just apply the same test to $U+C$, and see if there is some $C$ that makes it work. To illustrate, solve the equation $\nabla^2\ln\v{U+C}=0$ for $C$, and then see whether it is a constant over the whole complex plane.

A property of log-harmonic functions is that the product of two log-harmonic functions is also log-harmonic.

Trajectories that run out of the domain

Trajectories often run out of the domain of the potential. For example, in the discussions about power-law potentials before, though not emphasized, the origin is outside the domain of the potential because it is either a pole or a zero of $\d w/\d z$ (except the trivial case where $w$ is simply proportional to $z$). Another example that is rather overlooked is that unbound trajectories go to infinity while infinity is often not in the domain of the potential, either.

What need to take care of is that, when the trajectories run out of the domain, the trajectory is cut off there, and the rest of the trajectory is never considered (even if it may come back to the domain again later). Take the Kepler problem ane the harmonic oscillator as an example. If a trajectory of the harmonic oscillator passes through the origin, which is outside the domain, the trajectory degrades from a closed ellipse to a segment. If you take the square of a segment passing through the origin, you will get a broken line folded into itself, which looks like a particle in the Coulomb field may sink into the origin and then goes back along the exact path it came along. This would confusing if it were physical.

Arbitrariness in the construction of the conformal map

The construction of $z\mapsto w$ is not unique for a given $U$.

Rotation and translation

First, we can observe that the substitution $w\to w'\ceq w\e^{\i\tht}+w_0$ does not change $\v{\d w/\d z}$ (nor thus $U$). The real number $\tht$ is a function of $z$ in principle, but if we want $w$ to be holomorphic on a connected region, then $\tht$ must be a constant (except the trivial case where $w=0$).

The dual trajectory does change, though, but the dual potential $V$ is also changed, too. Because $\v{\d z/\d w'}=\v{\d z/\d w}$, we have $$\fc{V'}{w'}=\fc Vw=\fc V{\p{w'-w_0}\e^{-\i\tht}}.$$ Therefore, the dual trajectory and the dual potential are also rotated and translated by the same amount.

Scaling

Before introducing scaling, I need to add some words about the unit systems. In the above discussions, I have never mentioned what units or dimensions do $z,w,A,B$ have. The natural way of thinking is to let $z,w$ have the dimension of length and let $A,B$ have the dimension of energy. However, this is not the only way of thinking. We will later see that the $z$-space and the $w$-space can have totally different dimensions.

The dimensions or units of variables in a physical formula can be totally different from what they were originally intended to be. For example, when a particle is rotating, its motion needs to satisfy $\dot{\mbf r}=\bs\omg\times\mbf r$, where $\bs\omg$ is the angular velocity. However, although $\mbf r$ has the dimension of length when it is first introduced, this formula is satisfied by any rotating vectors. A typical example is that the angular momentum changes according to this formula when a rigid body is doing precession. For another example, in classical mechanics and general relativity, the coordinates used to describe the motion of a particle are often not in the dimension of length, but have all sorts of dimensions. For another example that is less well-known, just because the Berry connection has the same gauge transformation as the electromagnetic potential, a bunch of formulas that are useful in electromagnetic theory can be applied to the Berry connection to define all sorts of interesting quantities with rich physical implications. The units of Berry connection are, however, very unimportant because they are literally arbitrary.