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.


**UPDATE**


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.



Riemann surface from Riccati equation
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.


