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Here is the key difference between Cartesian and polar coordinates: In Cartesian coordinates, the basis vector is fixed in space, hence the second term is always zero and we simply need to differentiate the component. However, as one can imagine, in polar coordinates, the unit vector $\hat{r}$ is actually changing direction as the particle move ...
Find out the components in the polar coordinates using vector/tensor transformation rules. My answer: From the coordinate transformation we have, \begin{equation} \begin{gathered} dx=\cos\theta dr-r\sin\theta d\theta,\\ dy=\sin\theta dr+r\cos\theta d\theta.
Notice in step 4 that when we simplify x and y, we obtain the Cartesian coordinate.Therefore, the Cartesian coordinate (-5,-2) correspond to the polar coordinate (5.39, 201.8 degrees). Practice ...
At the end of the day, a simple recipe is to just transform the Schrödinger equation in Cartesian coordinates as a partial differential equation without any interpretation of the new terms as momenta with respect to the new coordinates. Let me illustrate my point below.
I'm following along with these notes, and at a certain point it talks about change of basis to go from polar to Cartesian coordinates and vice versa. It gives the following relations: It gives the following relations:
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If we choose ($\hat r,\hat\phi$) as reference frame -(that is, if we move with it) then the Cartesian will be changing direction. More generally, the very notion of motion cannot be manipulated without the choice of a frame. theQman, the two couples $(\hat x,\hat x)$ and $(\hat r,\hat\phi)$ are precisely built as basis vectors.
However, the velocity vector is the same vector wether you write it using the spherical coordinates or Cartesian coordinates. In the Cartesian coordinate system, the velocity is given by: $$\vec{v} = v_x \hat{e_x} + v_y \hat{e_y} +v_z \hat{e_z}$$
The angular coordinate in a polar coordinate set can never equal a Cartesian (rectangular) coordinate, simply due to its different dimension (different unit). Thinking correctly of the angular coordinate as a number-of-degrees also gives you the impression of a curved axis and basis vector rather than a straight vector.
to 1st order. If the uncertainty is the same magnitude as the coordinate, then this is not a good estimator...of course, if this is the case, the 1-sigma circle is going to include or be close to the origin, in which case all the polar coordinates are going to be rightfully uncertain.