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𝑷𝒆𝒓𝒊𝒐𝒅𝒊𝒄 𝑫𝑭𝑻 𝒂𝒏𝒅 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝑫𝑭𝑻

 





𝑷𝒆𝒓𝒊𝒐𝒅𝒊𝒄 𝑫𝑭𝑻 𝒂𝒏𝒅 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝑫𝑭𝑻 are built on the same theory. They make completely different assumptions about your system. Choosing the wrong one is not a minor approximation. It is the wrong calculation.

Here is the distinction that matters:

Molecular DFT treats a finite system. Your molecule sits in a box surrounded by vacuum. Basis functions are Gaussian functions centred on atoms. The calculation ends at the boundary of the molecule. This is correct for isolated molecules, gas-phase reactions, excitation spectra, NMR chemical shifts, and anything where the system has a natural physical boundary.

Periodic DFT treats an infinite system. You define a unit cell, apply periodic boundary conditions, and the code tiles that cell infinitely in all directions. The crystal extends forever, which is physically correct for a solid, a surface, or any extended material. The basis is typically plane waves or a mixed scheme. This is correct for bulk crystals, surfaces, 2D materials, and any system where the long-range periodic potential is part of the physics.

The mistake worth naming clearly:
Computing a fragment cut from a crystal in molecular DFT and treating the result as representative of the solid. When a molecule or cluster is placed inside a crystal, three things happen that a finite calculation cannot capture.
The periodic potential of the crystal shifts orbital energies. Hybridisation between the fragment and its neighbours modifies the band edges and their character. Long-range electrostatics from the extended structure shift absolute energy levels in ways that are entirely absent when the same fragment sits in vacuum.

The result is that band gaps, charge transfer character, optical absorption onsets, and effective masses computed on a fragment are not the same as those of the crystal. Sometimes the difference is quantitative. Sometimes it is qualitative.

The choice of periodic versus molecular is not a software preference. It is a statement about what physical system you are computing. Get that wrong and the rest of the calculation, however carefully run, is answering the wrong question.

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