Why particle in a bended box seeemd to be resilient?

Suppose a particle in a box confined in a 1 dimensional segment or length $$L$$. However, let's put the box on 2 D surface. Further, let's bend the box into a semi circle, the radius was thus $$\displaystyle \frac{L}{\pi}$$.

With the particle confined in the segment, $$i\hbar \partial_t=-\frac{\hbar^2}{2m} (\partial_x^2+\partial_y^2) =-\frac{\hbar^2}{2m R^2} \partial_\theta^2$$ (Cite: Polar coordinates, Differential Calculus , it's actually just a Jacobin.) This reduced the question back to a one dimensional harmonic oscillator. The solution was thus $$k=\frac{n\pi}{\pi} =n$$ $$E_n=\hbar \omega_n=\frac{n^2\pi^2 \hbar^2 }{2mR^2 \pi^2 }=\frac{n^2 \hbar^2 }{2mR^2 }$$ with wave equation $$\psi_n(\theta) =\begin{cases} \sqrt\frac{2}{\pi} \sin(k_n (\theta +\frac{\pi}{2})) \text{ for n even}\\ \sqrt\frac{2}{\pi} \cos(k_n (\theta +\frac{\pi}{2})) \text{ for n odd} \end{cases} =\begin{cases} \sqrt\frac{2}{\pi} \sin(n \theta +\frac{\pi}{2} n) \text{ for n even}\\ \sqrt\frac{2}{\pi} \cos(n \theta +\frac{\pi}{2} n) \text{ for n odd} \end{cases}$$

Using geometry $$r=\theta R=\theta \frac{L}{\pi}\rightarrow \theta =\frac{\pi r}{L}$$ where $$r$$ was the arc length or the "1 dimensional unbended position". The solution was thus $$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}$$ with the wave solution (notice that $$d\theta =\frac{\pi}{L} dr$$ the normalization factor) $$\psi_n(r) =\begin{cases} \sqrt\frac{2}{L} \sin(n \frac{\pi r}{L} +\frac{\pi}{2} n) \text{ for n even}\\ \sqrt\frac{2}{L} \cos(n \frac{\pi r}{L} +\frac{\pi}{2} n) \text{ for n odd} \end{cases}$$

The particle did not "see" bending, and the energy remained the same! (A Related post could be found here.)

Question 1: Was this just a particular coincidence for 1 D, with $$\partial_x^2+\partial_y^2=\frac{1}{R^2}\partial_\theta^2$$? Will higher dimensional system be different as their coordinate system was different?

Question 1b: For example, what would happen to 2 dimensional box, as torsion seemed to be involved?

Question 2: Why the energy and the wave equation remained the same? Since there's bending, should there be some angular contribution, or some extra leveling, because the fact that it's being bent?

In consideration of perpetuation theory

Question 3 In adiabatic approximation, the ground states went to ground states. However, in this case, the new system did not "see" the bending. Does that mean the wave equation would remain in the same energy states, even though the process may be abrupt,(as long as it's being confined within the infinite potential wall with equal curvature across the segment)?

The reason the particle doesn't feel the bend is that when you converted the Hamiltonian to polar coordinates, you threw away the $$\partial_r$$,$$\partial^2_r$$ terms, which reduced your problem to purely 1D. The idea of 'bending' only makes sense when you can also move in the radial direction. For instance, during uniform circular motion, centripetal acceleration comes from the velocity vector changing direction in the 2D plane, but if you only considered one dimension, the velocity vector is not changing at all. In 1D, the only two directions are clockwise and counter-clockwise.