I have quite a practical question motivated by physics.

Consider the Riccati equation whose solution gives a quantum-mechanical (QM) analogue of the classical momentum: $$ (p(x))^2 + \dfrac{\hbar}{i}p'(x)= 2 (E-V(x)) \quad. $$

Clearly, in the $\hbar\to0$ limit one obtains the definition of the classical momentum: $$ p_{c}(x)= \sqrt{ E-V(x)}\quad. $$ It's easy to determine what's the Riemann surface on which $p_c(x)$ is defined $-$ a two-dimensional oriented manifold with, typically, a finite number of punctured points (let's limit ourselves with polynomial potentials).

Now, the standard approach in QM is to consider an expansion of $p(x)$ in powers of $\hbar$, which leads to the (Generalised) Bohr-Sommerfeld qantisation: $$ \begin{gathered} p(x) = \sum \limits_{k=0}^\infty \left(\dfrac{\hbar}{i}\right)^k p_k(x)\quad,\\ p_0(x) \equiv p_c(x) \quad. \end{gathered} $$ Here $p_k(x)$ are, of course, assumed to be $\hbar$-independent.

It is now a simple exercise to show that all the $p_k(x)$ live on the same Riemann surface as $p_c(x)$: they all are obtained from $p_c(x)$ recursively by means of algebraic operations and taking derivatives $-$ none of these can drag us out of the Riemann surface.

This result suggests me to make a way stronger statement: namely, that the solutions of the original Riccati equation have to live on the same Riemann surface as $p_c(x)$. Basically, I'm saying that $\hbar\to0$ limit does not change the Riemann surface (well, that's a tricky limit since it turns a differential equation into a trivial equality).

It may be temptingly to say that the last equation (the infinite sum) is exactly what I need. However, those who are familiar with things like asymptotic series know that it's not like that. The reason for this is that $p(x)$ may depend on $\hbar$ in a non-polynomial way, like $\exp(-1/\hbar)$ or $\log (1/\hbar)$ (non-perturbatively, in physical jargon).

So my question is:

Given the Riccati equation (the top one in the question), is it possible to prove that its solutions live on the same Riemann surface as the function $p_c(x)$ defined by the $\hbar\to 0$ limit of the equation?

I would prefer to get the answer which will not rely on employing any (trans-)series expansions of $p(x)$, but would rather be based on a global analysis of the differential equation's Riemann surface (or smth like that).

P.S. If I'm asking something trivial, any references to the relevant textbooks are greatly appreciated.


I'm still very much interested to hear any useful comments/references on the topic, however I've just realised that the answer to my question is negative. Different solutions may live on different Riemann surfaces. Unfortunately, this conclusion relies not on strict mathematical statements, but rather on my knowledge from physics. The quantum momentum $p(x)$ has a first-order pole wherever the wave function $\psi(x)$ has a zero: $$ p(x) =\dfrac{\hbar}{i} \dfrac{1}{\psi(x)} \dfrac{\operatorname{d} \psi(x)}{\operatorname{d} x} $$ These wave function may have arbitrary number of poles on the branch cut of the classical momentum. In the classical limit this sequence of first-order poles coalesces into a branch cut, just like $\int \dfrac{\operatorname{d}x}{x}=\log x$. Which tells us that for non-zero $\hbar$ the Riemann surface is very different from the $\hbar=0$ case.


The answer to the highlighted question is "no". When $V$ is a polynomial, the general solution of the Riccati equation is single valued, it is a meromorphic function in the complex plane. To prove the statement, make the coefficient at $p'$ equal to $1$ by scaling of $x$, and then reduce your Riccati equation to a linear second order equation by setting $p=w'/w$. You obtain a second-order linear equation with polynomial coefficient, so its general solution is entire.

What really happens when $h\to 0$, is that the poles of these meromorphic solutions accumulate to certain ``branch cuts'' of your multivalued classical approximation, and the convergence to this classical approximation happens away from these poles and is not uniform.

I recommend Heading, J. An introduction to phase-integral methods. Methuen & Co., Ltd., London; John Wiley & Sons, Inc., New York 1962 for a short and clear exposition of this topic.

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  • $\begingroup$ Thanks! That's what exactly what I thought by the end, formulated rigorously. $\endgroup$ – mavzolej May 24 '17 at 12:55

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