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.


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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.


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