Atom orbitals physics vs. chemistry

Question

Why draw chemists always $|2p_x\rangle$, $|2p_y\rangle$ and $|2p_z\rangle$ orbitals instead of $|2,1,m\rangle$? Since $|2p_x\rangle$ or $|2p_y\rangle$ are a superposition of energy eigenstates they shouldn‘t be stable or is this different in bound states?

Answer 1

If an operator has a degenerate eigenvalue $$ A|a_1\rangle = a|a_1\rangle \qquad A|a_2\rangle = a|a_2\rangle $$ Then for any linear combination of the degenerate eigenvectors we have \begin{align} A(\alpha |a_1\rangle + \beta|a_2\rangle) &= \alpha A|a_1\rangle + \beta A|a_2\rangle\\ &=\alpha a|a_1\rangle + \beta a|a_2\rangle\\ &=a(\alpha |a_1\rangle + \beta|a_2\rangle) \end{align} so the linear combination is also an eigenvector of $A$ with the same eigenvalue.

As the Hamiltonian for an isolated atom does not depend of $m$, the states $|2,1,m\rangle$ are degenerate and so any linear combination of them (such as $|2p_z\rangle$, etc.) will also be an eigenstates of the Hamiltonian. Note, however that, unlike the states $|2,1,m\rangle$, the states $|2p_z\rangle$, $|2p_x\rangle$ and $|2p_y\rangle$ are not eigenstates of the angular momentum operator as the different $m$ values denote different eigenvalues for $L_z$.