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I am quite comfortable with the complex wavefunctions for the three p-orbitals ($ p_{0},p_{1},p_{-1}$) and the construction of the real-valued versions ($ p_{z},p_{x},p_{y}$) from the $ \pm m $ superpositions (there are plenty of questions here relating to that). My question is not regarding the construction of these orbitals, but their usage. I have heard it said that the real-valued wavefunctions are more useful in situations where there is a preferred coordinate system such as with molecular bonding, but I am not sure why. What exactly is the benefit of choosing to work with the so-called $ p_{x} $ and $ p_{y} $ orbitals rather than the complex $ p_{1} $ and $ p_{-1} $ versions?

Furthermore, is it true to say that - given that the $ p_{x,y} $ orbitals are superpositions of the $ p_{1,-1} $ orbitals - the two different cases cannot exist simultaneously? In other words, an electron cannot be simultaneously in a $ p_{x} $ state and a $ p_{1} $ state?

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    $\begingroup$ Angular momentum is only a good quantum numer in spherical or cylindrical symmetry. Where there is a preferred coordinate system such as with molecular bonding, the spherical harmonics are not eigenstates. $\endgroup$ – Pieter Mar 7 '18 at 9:37
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Because the real-valued orbitals are eigenstates in those situations and the complex ones are not.

More specifically, the complex-valued orbitals are only eigenstates if the system has perfect rotational symmetry. Once you introduce a preferred direction in the hamiltonian, however (say, you perturb the system with an electric field), you break the degeneracy, and the eigenstates of the now-nondegenerate levels will be the real-valued orbitals.


As to your final question, it's actually ill-posed, and it depends sensitively on what you mean by "be in a state".

  • If you mean that the system's quantum state is the given one, then obviously not - the wavefunctions are different.
  • If you mean whether you can form a superposition of the two states, then yes, this is perfectly possible (but in this instance not normally helpful).
  • If you mean that measuring in the given basis will give a nonzero result, then yes: the $p_x$ orbital has nonzero support on $p_+$, and vice versa.
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  • $\begingroup$ Ah right, I see, that makes sense. Now I'm curious as to the underlying reason why the eigenstates happen to be real in the non-rotationally symmetric case? Clearly in the general case, eigenstates need not be entirely real. What is it about such a Hamiltonian that gives rise to real eigenstates? I sense I need to spend some time playing around with different potentials in the Schrodinger equation... $\endgroup$ – JeneralJames Mar 8 '18 at 11:32
  • $\begingroup$ Two facts - that the position-representation Schrödinger equation is a real-valued PDE, and that (once you introduce a perturbing real-valued potential) the eigenspaces are non-degenerate. On the other hand, if your break the degeneracy using a magnetic field (through a $\hat{\mathbf p}\cdot \mathbf A$ term in the hamiltonian, which is not real-valued in the position representation) instead of a scalar potential, then the eigenstates are the $p_\pm$ orbitals. $\endgroup$ – Emilio Pisanty Mar 8 '18 at 12:36

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