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It seems that all wave functions studied in physical chemistry are orthogonal (e.g. particle in a box, hydrogen atomic orbitals). Does this come about because we purposefully make them orthogonal, or are they derived that way naturally? Can there be useful wave functions that are not orthogonal?

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    $\begingroup$ They are eigenfunctions of a hermitian operator. $\endgroup$ – user26143 Nov 22 '13 at 9:10
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In general, orthogonal wavefunctions are much easier to treat. In some cases they appear naturally, but usually, the orthogonality is imposed as a constrain while constructing the wavefunction.

For example, if you construct electronic wavefunction in the atomic orbital basis, you try to construct the orthogonal basis. This guarantees that the AOs are linearly independent. (Implication, not equivalence). Would you fail to fulfill this, the solution might still be possible, but much more difficult.

If you manage to solve the eigenvalue - eigenvector problem, the solutions are by definition orthogonal to each other. This is the case for the examples you provided.

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  • $\begingroup$ When AOs are linearly independent, does this mean that we are ignoring any interaction between the AOs? How does this relate to Huckel MO theory when neighboring AOs' interactions are taken into account? $\endgroup$ – halcyon Nov 23 '13 at 19:11
  • $\begingroup$ Sorry, I messed some things together. The atomic orbitals on one given atom are orthogonal (have zero overlap). Orbitals on two different atoms usually do have non-zero overlap (can be zero due to symmetry). $\endgroup$ – ssavec Nov 24 '13 at 19:28
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    $\begingroup$ Eigenvectors are not orthogonal "by definition". It's just that two eigenvectors of a Hermitian operator $A$, to different eigenvalues, are necessarily orthogonal (else e.g. decomposing in the middle of $A^2$ would lead to contradiction). They're however typically normalised by definition, thus orthonormal. $\endgroup$ – leftaroundabout Apr 29 '14 at 20:16
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    $\begingroup$ One should be more clear about if the topic is orthonormal BASISes or WAVEFUNCTIONs. The answer seems to talk about the first, the question about the second. $\endgroup$ – Greg Oct 17 '14 at 3:38
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I will try to add another point to the discussion which is statistics. Quantum mechanics is a non deterministic theory which can only make assumptions about expectation values and probabilities of obtaining a special value. In general your wave functions or vectors correspond to states while the projection of one wave function onto another (aka the scalar product of both functions where the explicit version depends on the norm of the vector space) gives you probability amplitudes and its square a probability. Now probabilities are between 0 and 1 - the same goes for its square at the extreme points 0 and 1. Orthogonality just means in that sense that for a chosen eigenbasis your probability of switching the basis vector by measurement is zero. The normalization occurs as the probability of staying in that eigenstate upon measuring the same operator again must be one (as the probability of reaching another basis vector is zero) but it cannot be greater than one - this is why you always use properly normalized wave functions. To be precise you normalize every time that you are working with a mapping between scalar products, projections and states - you could also see parallels between the time correlation function and a scalar product in statistical mechanics.

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