In mala's DataConverter class, in which the LDOS is processed from QE outputs, the default target_units=None in the add_snapshot function. This means that MALA performs no unit conversion by default on the QE outputs, which are in atomic units.
On the other hand, the default units in MALA are eV and Angstrom. For example, in the read_from_numpy_file function of of the LDOS class, the default units units="1/(eV*A^3)".
So if a user runs a QE calculation, processes the output from that using the DataConverter class, and then performs calculations on the LDOS using the LDOS class, there will be a units inconsistency and energies will be wrong.
Although it's possible to override the default units in either the DataConverter or LDOS classes, I strongly think that, if a user 'does nothing', i.e. uses everywhere defaults, then these should align to produce a consistent result. I would propose to account for this in the DataConverter class by making the default target_units="1/(Ry*Bohr^3)".
What do you think @RandomDefaultUser? If you agree I'll make a PR to address this issue.
In mala's
DataConverterclass, in which the LDOS is processed from QE outputs, the defaulttarget_units=Nonein theadd_snapshotfunction. This means that MALA performs no unit conversion by default on the QE outputs, which are in atomic units.On the other hand, the default units in MALA are eV and Angstrom. For example, in the
read_from_numpy_filefunction of of theLDOSclass, the default unitsunits="1/(eV*A^3)".So if a user runs a QE calculation, processes the output from that using the
DataConverterclass, and then performs calculations on the LDOS using theLDOSclass, there will be a units inconsistency and energies will be wrong.Although it's possible to override the default units in either the
DataConverterorLDOSclasses, I strongly think that, if a user 'does nothing', i.e. uses everywhere defaults, then these should align to produce a consistent result. I would propose to account for this in theDataConverterclass by making the defaulttarget_units="1/(Ry*Bohr^3)".What do you think @RandomDefaultUser? If you agree I'll make a PR to address this issue.