Been struggling with this concept. Or is this just one of those things in quantum mechanics which attempting to understand is futile? Guess we can see it as the electron just linearly oscillating back and forth but it doesn't feel right.
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"the electron just linearly oscillating back and forth". How would that work? An s-orbital electron's position state is spherically symmetric. – PM 2Ring Nov 06 '21 at 21:25
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It does oscillate radially back and forth—but, to satisfy Heisenberg uncertainty principle, it does it as a radial standing wave. – Ruslan Nov 06 '21 at 21:26
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It still has spin angular momentum – Lewis Miller Nov 06 '21 at 21:44
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1"is this just one of those things in quantum mechanics which attempting to understand is futile?" Remember: quantum physics is the deeper theory, and classical physics is only an approximation. We don't try to understand deeper theories in terms of approximate theories. We do the opposite: we use the deeper theories to understand why/when the approximate theories work. Are you really asking why the relationship between energy and angular momentum is the way it is for an orbiting object in classical physics (the approximate theory), even though it's not always that way in quantum physics? – Chiral Anomaly Nov 06 '21 at 22:21
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Classically a particle without angular momentum will have a radial orbit. This is the closest classical picture to an s-orbital. The orbital is spherically symmetric because the orientation of the orbital is undetermined
my2cts
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Classically, there is nothing like orbitals. There are orbits. Orbital was a name introduced in 1932 by Mulliken as a short form of one-electron orbital wave function. – GiorgioP-DoomsdayClockIsAt-90 Nov 06 '21 at 22:47
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It is possible to understand the situation but the first thing to have clear in mind is that classical intuition based on trajectories becomes useless in quantum mechanics (QM). In no way the quantum description allows speaking about an electron linearly oscillating back and forth. Oscillations refer to a trajectory, while QM is about the expectation values of dynamic observables and there is no trajectory.
Each cartesian angular momentum component has the same form, in terms of operators, like its classical counterpart. For instance: $$ L_z = xp_y-y p_x. $$ In the position representation of the Schrödinger picture, $p_x=-i \hbar \frac{\partial{}}{\partial{x}}$, and similarly for $y$ and $z$.
Classically, each cartesian component of the angular momentum can be interpreted geometrically as the area of the parallelogram corresponding to the position vector and the momentum vector. Therefore the two aligned vectors (oscillations forth and back) correspond to zero angular momentum.
The zero eigenvalues of the quantum version have a different origin. A spherically symmetric wavefunction ($\psi(r)$) is such that $$ \frac{\partial{\psi}}{\partial{x}}=\frac{x}{r}\psi, $$ and again similarly for $y$ and $z$. Therefore, it is always an eigenfunction of each component of the angular momentum with zero eigenvalue.
It is clear that notwithstanding the same expression of the angular momentum in terms of position and momentum, the interpretation of zero angular momentum is quite different in the two cases.
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See the answer by my2cts. If you would like to build some intuition about how quantum orbits are related to classical orbits, basic handwavy arguments about how the classical orbits would behave goes a long way in seeing "why" the probability densities of the hydrogen wavefunctions look the way they do.
For example, because any nonzero angular momentum causes a classical to never reach the origin, the wavefunctions corresponding to $\ell\neq0$ orbits vanish at $r=0$. If $\ell=n-1$, the angular momentum is the "most it could be" given its energy (alternatively, the energy is minimized for the given angular momentum), so the orbit is circular, and its wavefunction will be pretty localized to the radius of a classical circular orbit.
Nicholas Rui
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