I Angular Momentum Hydrogen Atom Problem: Physically Explained When L=0

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In quantum mechanics, the angular momentum for a hydrogen atom is described by the equation L = √(l(l+1))ħ. When L = 0, it indicates that l must also be 0, resulting in a wave function that is spherically symmetric and lacks angular dependence. This spherically symmetric condition is not exclusive to hydrogen; it also applies to other potential energy systems, such as the 3D harmonic oscillator and spherical potential wells. In these cases, energy eigenfunctions with L = 0 maintain the same spherical symmetry. Understanding these properties is crucial for analyzing quantum systems with similar characteristics.
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In quantum mechanics hydrogen atom problem ##L=\sqrt{l(l+1)}\hbar##. What that means physically when ##L=0##?
 
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L = 0 is only possible when l = 0 which means that the wave function has no angular dependence and spherically symmetric.
 
And also for a particle moving in some potential energy other than the hydrogenic ##V\propto 1/r##, the energy eigenfunctions with ##L=0## are spherically symmetric. These systems include the 3D harmonic oscillator ##V(x,y,z) = C(x^2 + y^2 + z^2 )## and the spherical potential well where ##V(r) = 0## if ##r<R## and ##V(r) = V_0## when ##r\geq R##.
 

Thread 'Angular Momentum Vector and Its Magnitude'

For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...
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