$m_l=0$ corresponds to s-orbital and the values associated are: $$\lambda=2\pi r/m_l\to\infty$$ It means a wave of of infinite wavelength has a constant height at all values of $\phi$(azimuthal angle). $$J_z=\pm hr/\lambda\to0\text{ also }J_z=\pm pr$$ which implies zero linear and angular momentum as $r\ne0$. Now what does zero angular and linear momentum signify/mean?(I think the electron is at rest but it cannot be)

  • $\begingroup$ Wher did you get this formula for $\lambda$? do you mean the wavelenght of absorbed/emitted photons? $\endgroup$ – Antonio Ragagnin Jul 29 '14 at 8:41
  • $\begingroup$ $\lambda=h/p$ and $J_z=\pm hr/\lambda$ $\endgroup$ – RE60K Jul 29 '14 at 9:08

The plane wave – the wave function for a particle moving in a clear direction and with a sharp momentum, $\vec p$, is $$ \psi (\vec x) = C \exp (i\vec p\cdot \vec x / \hbar ) $$ where $\exp(ia)=\cos a + i \sin a$. However, the dependence of an $s$-wave wave function on the angular direction is trivial (constant), $$\psi_s (r,\theta,\phi) = \psi(r). $$ There is no dependence. You may imagine this wave function to be a superposition of some waves moving in all directions of the space, and backwards.

Because the wave function doesn't depend on the angles, in particular, it doesn't depend on $\phi$, you may imagine that it is a wave with an infinitely long wavelength in the angular directions. Note that $\cos 0\phi = 1$. The shorter the wavelength, the higher momentum, and vice versa – and it sort of holds for the angular momentum and the wavelength measured in spherical or azimuthal coordinates and in radians, too.

The zero angular momentum means that there is no rotation. If such a (slowly moving) particle (atom) is absorbed by a gyroscope (near the axis), the rate of the gyroscope rotation won't change.

  • $\begingroup$ so why doesn't such a slowly moving particle get absorbed/atracted into the nucleus $\endgroup$ – RE60K Jul 29 '14 at 9:10
  • $\begingroup$ First, it is only not moving in the angular directions; the average value of the momentum in the radial dimension is nonzero. Second, this $|p_r|$ has to be nonzero, much like the average $|r|$, so the particle can't quite stop and can't collapse to the nucleus, due to the uncertainty principle $\Delta p \cdot \Delta x \geq \hbar/2$ which prohibits $r=0$ much like $|\vec p|=0$ etc. The electron finds a compromise between minimizing the kinetic energy (small $p$ is good) and minimizing potential energy (small $r$ is nice), and the sum, total energy is minimized for Bohr-radius-like radius. $\endgroup$ – Luboš Motl Jul 29 '14 at 11:37
  • $\begingroup$ So it means the electron can move only towards and away from the nucleus in a radial direction where $p_z=0; p_x,p_y\ne0\implies J_z=0; J_x,J_y\ne0$. So is this mean the electron will vibrate in radial direction at mean position $r_0=0.529 A^o$.Also how can x and y components be measured when we are trying to acheive 100% accuracy in z component? $\endgroup$ – RE60K Jul 29 '14 at 15:26
  • $\begingroup$ Yes, the electron in an atom is "vibrating" in the radial direction around the average value that is what you write and that is of the same order for other states of the same atom or other atoms, too. The $x,y,z$ components of the position commute with each other, and so do the 3 components of the momentum. The 3 components of the angular momentum do not commute with each other, but the commutator of two of them is the third one, so the states with $J_x=J_y=J_z=0$ are an "exception" for which the commutator is effectively zero so the $s$-states are eigenstates of all three components. $\endgroup$ – Luboš Motl Jul 29 '14 at 16:22

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