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.