# Wave function as a section of a complex line bundle to do QM in polar coordinates

If you want to change the coordinates of a Wave function $$\Psi$$ in 2D QM from cartesian to polar coordinates in a naive way one encounters a problem, namely the (naively defined) radial momentum operator $$P_r$$ is no longer self adjoint: $$(\Phi,P_r\Psi)=\int r\Phi ^*(-i)\partial _r\Psi drd\theta\neq(P_r\Phi,\Psi)$$ According to this lecture https://www.youtube.com/watch?v=C93KzJ7-Es4&t=1630s one can address this issue by interpreting the wave function $$\Psi$$ no longer as a function $$f_\Psi:\mathbb{R^2}→ \mathbb{C}$$ but as a section $$\Psi:\mathbb{R^2}→E$$ on a $$\mathbb{C}$$-line bundle $$E$$ over $$\mathbb{R^2}$$. It is claimed that $$E$$ is also an associated fibre bundle to the frame bundle $$L\mathbb{R^2}$$ over $$\mathbb{R^2}$$ which can be equipped with a connection $$\omega$$. Then the argumentation goes, we can understand a section $$\Psi$$ of an associated $$\mathbb{C}$$-line bundle as a function $$F_\Psi:L\mathbb{R^2}→\mathbb{C}$$ and on such a function we have a covariant derivative $$\text{D}F_\Psi(X)=\text{d}F_\Psi(X^{hor})$$. Now we can pull this back with a local section $$s:U→L\mathbb{R^2}$$ to the base space $$\mathbb{R^2}$$ and get a definition of a covariant derivative for the pulled back function $$s^*F_\Psi:U→\mathbb{C}$$ which is equivalent to the section $$\Psi$$ in the open region $$U$$. With $$T$$ as a vector field on $$\mathbb{R^2}$$ the covariant derivative reads: $$\nabla_Ts^*F_\Psi:=\text{D}s^*F_\Psi(T)=\text{d}s^*F_\Psi+s^*\omega(T)s^*F_\Psi$$ Now we can find a connection $$\omega$$ with which the momentum operator $$P_r$$ defined in terms of a covariant derivative is again a self adjoint operator.

Now my problem with the argumentation: The frame bundle $$L\mathbb{R^2}$$ is trivial as one can easily find a global section of it. Every associated vector bundle of a trivial principal bundle is trivial too. But if $$E$$ is trivial a section $$\Psi$$ can be understood as a function $$f_\Psi:\mathbb{R^2}→\mathbb{C}$$, which is the starting point again. The covariant derivative of such functions is $$\nabla_Tf_\Psi=df_\Psi(T)$$ so that we get again a momentum operator which is not self adjoint. Basically my question now is: What is wrong in my derivation of a covariant derivative for the section $$\Psi$$? Because I don't think both of the formulas are valid and we just choose the one which produces the correct momentum operator.

This is correct, the frame bundle $$L\mathbb{R}^2$$ is trivial, however he quickly states at 16:15 that the domain of the coordinate functions is $$\mathbb{R}^2\backslash\{0\}$$. Punctured $$\mathbb{R}^2$$ appears similarly as an example when introducing De Rham cohomology/K theory precisely because it is no longer homotopically trivial (homotopically it is the circle) - since it doesn't retract to the point we cannot use this argument to say that its supported bundles are trivial.
Quantum mechanically, because the Jacobian in polar/spherical coordinates is not invertible at $$0$$, this gives us a degenerate inner product at that point (especially if we consider multiplication of distributions). In 3d we have to remove an entire line for cylindrical coordinates for a nondegenerate Jacobian! Another example would be toroidal coordinates, where we remove a circle (it retracts to $$S^1\vee S^2$$).
From the point of view of differential geometry, it seems unnatural to just ruin the topology like this with a "change of coordinates" - but in QM this is incredibly natural as we want to solve for the electron wavefunctions of an atom with a spherically symmetric nucleus. So, one can explore different topologies coming from modified coordinates/topology on $$\mathbb{R}^n$$ by instead studying the symmetries of particular PDEs appearing in quantum mechanics. Olver's book on the Applications of Lie Groups to Differential Equations would be in line with this approach.