# Why exactly is it useful to construct the real-valued p-orbitals?

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?

• 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. – Pieter Mar 7 '18 at 9:37

• 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.
• 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. – Emilio Pisanty Mar 8 '18 at 12:36