Electronic gradient of Schroedinger Equation

[ani4physics]

Hi all. I have a question that I am thinking about for a couple of days. Let's consider the time-independent Schroedinger equation for a molecule:

H0 [psi> = E0 [psi>

Now, we know that the unperturbed Hamiltonian consist of electronic kinetic energy operator, electron-electron repulsion operator, electron-nuclear attraction operator, and nuclear-nuclear repulsion operator (Within the Born-Oppenheimer approximation).

If we differentiate both sides of the equation with respect to the coordinate of electron i, then we we need to consider only the gradients of electronic kinetic energy operator, electron-electron repulsion operator, electron-nuclear attraction operator, and the wave function.

My question is: Is the gradient of the electronic KE operator with respect to coordinate of electron i = 0?

Please let me know. Thanks.

 

[DrDu]

Yes, it is, but note that you don't get rid of the KE operator as the KE still works on the derivative of the wavefunction.

 

[ani4physics]

Yes, it is, but note that you don't get rid of the KE operator as the KE still works on the derivative of the wavefunction.

Thanks a ton. Yes I understand that the KE operator still operates on the electronic gradient of the wavefunction. could you please give me a brief idea of how the gradient of KE operator with respect to electronic coordinate is zero. Thanks again.

 

[DrDu]

Well, don't misunderstand me. I just meant that the derivative and the kinetic energy operator commute. You may take this to mean that the derivative of the KE operator vanishes.