Q: 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?

A: 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. 

