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Showing functions are eigenfunctions of angular momentum.
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[QUOTE="Simon Bridge, post: 4988134, member: 367532"] [SIZE=4]Put more precisely: since phi is the angle the total angular momentum makes with the z axis, then any wavefunction with no phi dependence will also have no component of angular momentum along the z axis. Hence, zero eigenvalue. Notice the emphasis on physics (well, geometry) rather than mathematics (calculus) - yes it comes out zero because of the maths, but the Universe does not care about what our calculations yield. The maths is just a model - it is trying to describe Nature. Look for the truth in Nature. Similarly ... [SIZE=4]The "z axis" does not have to be vertical - you are not dealing with gravity here - it is determined by the orientation of the apparatus doing the measuring. It can as easily (and more usually) be along the direction of motion or the direction of an applied field. It's just a common label for an "axis of interest". You should abandon ideas about the orientation of the Cartesian axes. The phi [i]direction[/i] is not mentioned in the problem ... it is not actually a direction since knowing phi does not tell you where to look, but describes a set of infinitely many directions from the z axis. It is important to distinguish between a direction (component of a vector) and the dependence that the magnitude of the component has. i.e. ##\vec B = (0, kx^2, 0)## has a y direction that depends on the x component of position.[/SIZE] [/SIZE] [/QUOTE]
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Showing functions are eigenfunctions of angular momentum.
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