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Citing from this site:

Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable.

This made me think about two problems:

  1. A quantum state, as my professor once said, should not distinguish a specific phase from any other, i.e., what really matters is not ${| \psi \rangle }$ but rather ${| \psi \rangle }\langle \psi |$, and this last is invariant under the transformation ${| \psi \rangle } \mapsto e^{i\phi}{| \psi \rangle }$, which precisely corresponds to a particular phase choice.

  2. Let's assume that phase matters in describing orbitals. Then why are there just two possibilities for the overlapping: in-phase or out-of-phase? The phase is not discreet: $\phi \in [0, 2\pi)$, so I would expect a continuous.

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    $\begingroup$ The phase doesn't chage a state, but combination of two states clearly depends on phase: $|\psi_1 + \psi_2e^{i\theta}|^2=|\psi_1|^2 + |\psi_2|^2+2\Re (\psi_1^*\psi_2e^{i\theta})$. $\endgroup$
    – Roger V.
    Apr 15, 2021 at 11:54
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    $\begingroup$ A molecular orbital is typically constructed by a linear combination of atomic orbitals $\varphi^{\text{MO}}= \sum_\alpha e^{i\theta_\alpha}\phi_\alpha^{\text{AO}}$. And it is the relative phase of these atomic orbitals with respect to each other inside a linear combination that matters. $\endgroup$
    – Hans Wurst
    Apr 15, 2021 at 12:15

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First of all, quantum states are invariant under global phaseshifts $\psi \rightarrow e^{i\phi} \psi$. If you shift the phase between two contributions in a superposition, you get a different state.

You are right that you can build a continuous number of states $\psi_1 + e^{i\phi} \psi_2$, and they will all be different. However, there are only two choices of $\phi$ which diagonalize the hamiltonian in the subspace {$\psi_1,\psi_2$} (remember that the $\psi_{1,2}$ are the hydrogen groundstates at the first or second proton), and those are $\phi = \pi$ or $\phi=2 \pi$.

Just to clarify, the Hamiltoniam I'm talking about is

$$ H = \frac{p^2}{2m}+\frac{Ze}{|r-R_1|}+\frac{Ze}{|r-R_2|} = \begin{pmatrix} \epsilon & V\\ V & \epsilon \end{pmatrix}$$

The last equal sign is in the 2 state subspace.

Usually, if we want to describe actual systems in nature, it is helpful to know the eigenstates of the Hamiltonian (for several reasons). This is why people always talk about those two states and ignore the others.

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  • $\begingroup$ Ok, thanks. Lastly, how do you know there are just two possible phase choices? $\endgroup$
    – ric.san
    Apr 15, 2021 at 15:31
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    $\begingroup$ They correspond to the eigenvectors of the Hamiltonian. Basically the reason is that the diagonals of H are identical. Calculating H in this two state subspace would be a good exercise. $\endgroup$
    – curio
    Apr 15, 2021 at 15:37
  • $\begingroup$ Can you clarify the second equal sign, please? $\endgroup$
    – Mark_Bell
    Apr 15, 2021 at 15:46
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    $\begingroup$ If we restrict the Hilbertspace to the subspace spanned by $\psi_1, \psi_2$, the Hamiltonian can be written as $H= \sum_{ij} \langle \psi_i|H|\psi_j \rangle |\psi_i\rangle \langle \psi_j |$. This restriction is an approximation, but it is often justified because of energetic constraints in the physical system. $\endgroup$
    – curio
    Apr 15, 2021 at 15:59

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