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How to distinguish the $p_x$, $p_y$, $p_z$ orbitals since one could choose the $x-$, $y-$ and $z-$axes arbitrary?

How could the can the wavefunctions that describe the probability of the presence of electrons in the associated orbitals have a physical meaning? One could decide to make any arbitrary tiny 3D rotations so that the resulting orbitals would no more privilege some specific directions?

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    $\begingroup$ It’s not just the names that are arbitrary. The orientation in space of the orthonormal basis $\hat x, \hat y, \hat z$ is arbitrary as well. $\endgroup$ – G. Smith Aug 13 '20 at 16:47
  • $\begingroup$ @G. Smith : and so what is the conclusion ? we could permut any of them, so how experimentally one could identify which one is which one ? $\endgroup$ – Mathieu Krisztian Aug 13 '20 at 17:41
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It looks like you think that $p_x$, $p_y$ and $p_z$ have some special absolute meaning. They don't. A single physical state $|\psi\rangle$ could be described as $|p_x\rangle$ in one reference frame, as $|p_z\rangle$ in another, and as $\frac1{\sqrt2}(|p_x\rangle+|p_z\rangle)$ in yet another frame, with all these frames being related by rotations.

These states become special when external electric field is applied along a coordinate axis. Let $\vec E$ be aligned with $\vec{e}_x$. Then, since the electric field splits energy levels of these $p$ states (Stark effect), you'll be able to distinguish $|p_x\rangle$ from $|p_y\rangle$ or from $|p_z\rangle$ by noting the difference in e.g. frequencies of electromagnetic emissions.

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  • $\begingroup$ excellent : thank you. Ok I get the point : so you mean that the axes are absolute when one has an additional field, as the example of the E field that is in a given direction. Is it the key point ? $\endgroup$ – Mathieu Krisztian Aug 22 '20 at 18:36
  • $\begingroup$ @MathieuKrisztian yes. And in isotropic space all directions are equivalent, so you can choose axes arbitrarily without changing the result. $\endgroup$ – Ruslan Aug 22 '20 at 18:37
  • $\begingroup$ excellent. I wait 24 hours in order to give the bounty. $\endgroup$ – Mathieu Krisztian Aug 22 '20 at 18:38
  • $\begingroup$ All the examples of "one and the same" I could find are of the form "x and y are one and the same thing", not "one and the same thing is x and y". I don't think you're using the phrase idiomatically. $\endgroup$ – Acccumulation Aug 22 '20 at 19:49
  • $\begingroup$ @Acccumulation I've changed the wording to avoid the wrong usage of the phrase. $\endgroup$ – Ruslan Aug 22 '20 at 20:24
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The eigenstates of angular momentum, in particular $L^2$ and an arbitrary $L_z$, are the spherical harmonics:

$$ Y_l^m(\theta, \phi)$$

For fixed $l$ (the $L^2$ quantum number), the $2l + 1$ functions are eigenstates of $L_z$ with magnetic quantum number $-l \le m \le l$. They constitute a representation of the rotation group SO(3). That means they are closed under rotation.

If you have $p_z$ and rotate your coordinate system, it's now a mixture of $p_{z'}, p_{x'}, p_{y'}$, but there are no terms of the form $l\ne 1$. Of course, the energies are all degenerate, so without an external reference (say a magnetic or electric field), the choice of coordinates is completely arbitrary.

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$\newcommand{\ket}[1]{\left|#1\right>}$

If you have two coordinate systems, $xyz$ and $x'y'z'$, you can construct the associated wave functions $\ket{p_x}$, $\ket{p_y}$, $\ket{p_z}$, and $\ket{p_{x'}}$, $\ket{p_{y'}}$, $\ket{p_{z'}}$. Then the latter set can be constructed as (generally complex) linear combinations of the former set.

So the arbitrary choice of coordinate system decides which orbitals you'll get to cover the full space of $\ell=1$ wave functions.

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