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?

  • 1
    $\begingroup$ 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

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.
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.