Units of derivative coupling (non-adiabatic coupling) in output

Dear all,
In the calculation of the non-adiabatic couplings, the (closely related) derivative coupling are written in the output (see below), but the units are not specified, nor did I find it in the manual. Does anyone know what units are used here?
Thanks in advance,

Example output:
SF-CIS derivative coupling with ETF                   
  Atom         X              Y              Z        
   1      -0.141357      -0.001891      -0.285072     
   2      -0.132861      -0.003380       0.293653     
   3      -0.009087       0.000272      -0.003756     
   4       0.137841      -0.038263       0.764793     
   5       0.267727       0.082422      -0.850842     
   6       0.006263       0.001799      -0.019702     
   7      -0.549606      -0.088453       0.543794     
   8       0.277761       0.044026      -0.148716     
   9      -0.057628      -0.055786       0.311969     
  10       0.009594      -0.000041       0.002089     
  11       0.295452      -0.000028       0.068386     
  12      -0.000077      -0.002123      -0.002412     
  13       0.039318       0.015193      -0.362853     
  14      -0.005108       0.001291      -0.008390     
  15      -0.281295       0.004393      -0.112621     
  16       0.001311       0.002530      -0.001512     
  17       0.174245       0.050989      -0.316830     
  18       0.009030      -0.000077      -0.004075     
  19      -0.073574      -0.022727       0.234137     
  20       0.016164       0.004927      -0.051426     
  21       0.015890       0.004928      -0.050614     
 Derivative coupling time:  CPU 189.32 s  wall 39.17 s

One way to see it is to take the exact derivative coupling expression, $\mathbf{d}{IJ} = \left<\Psi{I}|\frac{\partial}{\partial\mathbf{R}}|\Psi_{J}\right>$, and rewrite it in finite difference form. For a single Cartesian coordinate $x$ of the vector, this would be

\frac{\left<\Psi_{I}(x)|\Psi_{J}(x + \Delta x)\right> - \left<\Psi_{I}(x)|\Psi_{J}(x - \Delta x)\right>}{2\Delta x} + O[(\Delta x)^2]

The overlap between states is unitless, meaning the numerator has no units, but the denominator has units of length. This means the derivative coupling vector has dimensions of inverse length. Since the output is in atomic units, the final units are in inverse bohr.

I adapted this idea from https://doi.org/10.1063/1.4820485, specifically equation 52b.

Thank you very much for your answer!
It looks indeed like the coupling has dimensions of inverse length. In the output however, some of the other length units are Angstroms instead of bohr, so do you know if this part is certainly atomic units (1/bohr) or could it be 1/Angstrom maybe?

I took a look at the code, and the derivative coupling is calculated as the non-adiabatic coupling between two states, scaled by the energy difference between the two states, just as in eqs. 9.5 and 9.6 (9.7.1 Nonadiabatic Couplings‣ 9.7 Nonadiabatic Couplings and Optimization of Minimum-Energy Crossing Points ‣ Chapter 9 Exploring Potential Energy Surfaces: Searches for Critical Points and Molecular Dynamics ‣ Q-Chem 5.3 User’s Manual). The NAC is printed as “CIS Force Matrix Element” (which is not the CIS gradient, but has the same units), so you can prove to yourself the above by taking this matrix and dividing by “Ej-Ei” above it. Since this energy difference is in hartree, and unit systems are never mixed internally (that would be a disaster), the units for length must be bohr, not angstrom. There are no conversion factors applied before the output is printed.

Another thing you can do, if you don’t trust me, is to see if you can reproduce some literature results. All of the derivative couplings in https://doi.org/10.1063/1.3665031 are given as inverse bohr, and they are the original authors of Q-Chem’s derivative coupling code, so your results should match the paper well if geometries are identical.

I have added a note to the manual so that in the future the units will be explicitly stipulated.

I’m definitely convinced now.
Thank you both very much for your help!