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?

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    $\begingroup$ To be blunt: many elementary chemistry texts wish to minimize the use of complex numbers, and they get away by taking real linear combinations. The convention in physics is to stick to spherical harmonics, which are complex functions. $\endgroup$ Feb 5, 2022 at 15:31

1 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 $L_z$.


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