Undergrad Can angular momentum be applied to non circular rotations?

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Angular momentum is applicable to all types of motion, not just circular, as demonstrated by a particle moving in a straight line having non-zero angular momentum about any point not on that line. In quantum mechanics (QM), particles lack defined trajectories, and spherical harmonics represent probability distributions rather than specific paths. The use of circular motion in classical physics is primarily for simplification and visualization purposes. Angular momentum arises from the presence of multiple force vectors, regardless of the object's path. Understanding these concepts is crucial for grasping angular momentum in both classical and quantum contexts.
Fruit Water
One of the reasons I've been so stumped about learning about angular momentum in QM, is that in my classical physics class we only applied it to circular motions. Hence, while I am aware that angular momentum is connected to spherical harmonics, the orbital shapes (besides s) isn't really circular motion to me. Can someone clarify this for me?
 
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Fruit Water said:
One of the reasons I've been so stumped about learning about angular momentum in QM, is that in my classical physics class we only applied it to circular motions. Hence, while I am aware that angular momentum is connected to spherical harmonics, the orbital shapes (besides s) isn't really circular motion to me. Can someone clarify this for me?

In classical physics, angular momentum applies to all motion. A partial moving in a straight line has non-zero angular momentum about any point not on that line.

Conservation of linear momentum can be seen as a special case of conservation of angular momentum.

In QM particles do not have defined trajectories such as a circular orbit. The spherical harmonics represent a probability density distribution, not a trajectory.
 
They taught you using a circle since it is the easiest to visualize and grasp the concepts. Just like in entry level physics when they have you calculate aspects of a ball rolling down an incline they do not include atmospheric resistance, surface frictions between the two items, or atomic attraction/repulsion characteristics of the materials. What PeroK stated is spot on, however, to reiterate from a different angle (hehe) think of the vectors of force at play when an object is in motion. If only one vector, no angular momentum, if two or more vectors, there is angular momentum. The path (circle or not) is not a determining factor in if there is angular momentum.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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