Why does delocalization (only) occur in molecules represented by resonance hybrids? Before I start this question, I am aware tha

Electrons in all molecules are delocalized to some extent
Delocalization is a the phenomenon and resonance is an attempt to explain it.

When I asked this question from my professor, this is the explanation he gave me:

In VBT, it is difficult to approximate the molecular orbital with
localized bonding orbitals, so the electronic configuration is a
weighted average of the infinite possible electronic configurations
i.e Ψ= Aψ1 +Bψ2+Cψ3......
For most molecules, the most stable lewis structure has the majority
of the contribution and can sufficiently approximate the actual
electronic by itself. However, for some molecules, there are multiple
equally-stable lewis structures with an equal and significant
contribution and each of them has to be taken into account. For e.g,
for nitrate ion, Ψ=1/3 (ψ1 +ψ2 +ψ3)  where ψ1,2,3  represent the 3
equal lewis structures. As a result, the electrons MUST be delocalised
between the appropriate bonds

This explanation, in simple words, argues that  since the 3 lewis strucutes are identical/indistinguishable, the electron density must be equal and thus delocalised between the bonds by symmetry.
However, this doesn't really explain the physical reasoning behind delocalization. Why is (significant) delocalization more energetically stable for some molecules but not for others? And why dpes delocalization always occur in molecules with some kind of symmetry between the possible atoms for a double/triple bond?
A short answer is that we mostly use resonance structures to represent delocalization when the bond orders or formal charges predicted by VBT are not approximately integers, simply because non-integer bond orders and charges are more cumbersome to draw. A typical example is a carboxylate, which we could draw with 1.5 bonds between both C-O pairs and -0.5 charge on each O, but it is simpler to just draw a resonance structure with one C=O double bond and one C-O single bond and a full -1 charge on one oxygen.
As others have noted, electrons are still delocalized across the entire molecule in molecules for which we do not use resonance structures. The difference is that we can approximate the bond orders in those cases as integers, so only one structure is needed to represent that.