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I know very little about quantum physics, but I just can't wrap my head around the fact that non-spherical orbitals appear to be "oriented", constituting a set of fixed "frame of reference axes".

Does it really work like that? Are there for a single atom given directions in which the orbitals "protrude"? And if so, what are they based on? If I understand it right, the nucleus is seemingly spherical - symmetrical with no unique axis of symmetry. What would then decide the orientation of the orbitals in space (with respect to the surroundings)?

Edit: I understand any orientation works to satisfy the wave equation. I was more curious why the atom would choose one direction, instead of any of the others. Is that fundamentally random? And is this setup retained further, or can it change at any time?

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marked as duplicate by Robin Ekman, Rob Jeffries, ZeroTheHero, Jon Custer, Yashas May 2 '17 at 3:19

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    $\begingroup$ Possible duplicate of Where do symmetries in atomic orbitals come from? $\endgroup$ – Robin Ekman May 1 '17 at 17:01
  • $\begingroup$ See also physics.stackexchange.com/q/174957 $\endgroup$ – Robin Ekman May 1 '17 at 17:01
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    $\begingroup$ Why does the earth orbit in a plane when the Sun's gravitational field is spherically symmetrical? $\endgroup$ – By Symmetry May 1 '17 at 17:04
  • $\begingroup$ Thank you for the link, skimming through the answers, I reckon part of my question is answered - any "orientation" of the axes will be the solution to the wave equation. But I was mainly unsure as to why would, upon observation, the atom choose one direction, instead of any of the others. Is that fundamentally random? And is this setup retained further, or can it change at any time? $\endgroup$ – Necrophagist May 1 '17 at 17:12
  • $\begingroup$ Well, the real answer to that is that what you have been taught about "collapse of the wavefunction" is a caricature and not very helpful. My usual recommendation is this lecture with Sidney Coleman. But in brief: you need to treat also the measurement system as a quantum system and realize that it has to interact with the atom, and this interaction means that their states will be entangled. Suppose that the initial state is something like $|\psi\rangle |0\rangle$ where this means that the atom is in the state $|\psi\rangle$ and your detector $\endgroup$ – Robin Ekman May 1 '17 at 17:22
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This is a fairly tricky concept, and I see research physicists stumbling with (analogues of) this problem with a somewhat alarming frequency. When you say

the nucleus is seemingly spherical

what that really means is that the electrons interact with a spherical object (essentially a point Coulomb charge at the nuclear centre) and therefore their dynamics must be invariant under any rotation about this point. However:

  • This does not mean that all the solutions to those dynamics must be spherically symmetric.
  • What it does mean is that for every non-symmetric solution $S_1$ of those dynamics that points along some direction $\hat n_1$ and every other direction $\hat n_2$, there must exist a separate, equally valid, yet distinct, solution $S_2$ that points along $\hat n_2$.

Thus, it's perfectly possible for an atom to exist in an anisotropic state, like, say, the $2p_z$ state of hydrogen. The only thing that symmetry requires is that we have analogous $2p_{\hat{n}}$ states along any given axis $\hat n$ as possibilities, which is obviously true.


That still doesn't answer, of course, the question of which way a given atom in a $2p$ state will point, but the answer to that is simply that it depends on the situation, and you need to specify more information to say anything useful.

If all you know is that a given atom is in a $2p$ state but you don't have any more information, then all you can say is that the electron is in a mixed state of all possible orientations, which is isotropic and can be written down as $$ \hat{\rho}_{2p,\text{ isotropic}} = \sum_{j=x,y,z}|2p_j⟩⟨2p_j| = \frac{1}{4\pi} \int |2p_{\hat{n}}⟩⟨2p_{\hat{n}}| \mathrm d \Omega_{\hat{n}}. $$ This means that the electron is not in a pure state and therefore you cannot assign it a wavefunction, which is standard fare when you have not specified enough information to pin down a single state.

In practice, however, when you work with $2p$ states, or other anisotropic orbitals, you've normally prepared them yourself, by e.g. stimulating a transition up from the $1s$ state with a linearly polarized laser, or some other anisotropic pumping method. In those cases, the orientation will be fixed by the pumping mechanism, i.e. the symmetry gets broken by your preparation apparatus, which selects one out of the many possible orientations for the state.

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