Schrodinger equation molecules

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So in SI units all I have to do with the kinetic energy term, bracketed T_n, is replace it with this:

[tex]- \sum^M_{A=1} ({\frac{h^2 \cdot m_e}{2 \cdot m_A} \cdot \nabla_A^2})[/tex]

Where m_e is the mass of an electron, m_A is the mass of nucleus A and h (sorry I couldn't find h bar but that's what I meant) is the reduced Planck constant.

And put that into the second Hamiltonian and I would have created a complete SI unit version of the first?
 
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Big-Daddy you seem to have some very big misunderstandings. What book have you been studying quantum from?
 
I don't study quantum mechanics on its own yet. I have just read the chapters in Atkins' Physical Chemistry.

Is my above expression correct?
 
Big-Daddy said:
I don't study quantum mechanics on its own yet. I have just read the chapters in Atkins' Physical Chemistry.

Is my above expression correct?
If it is supposed to be kinetic energy, what are the units on the expression you put forth?

In any event, physical chemistry texts are really horrid on the whole at introducing quantum mechanics. If you want a book that is better, but still chemistry oriented, I'd recommend getting McQuarrie's Quantum Chemistry.
 
Sorry, I noticed my answer isn't dimensionally sound. How about this one:

[tex]- \sum^M_{A=1} ({\frac{h^2}{2 \cdot m_A} \cdot \nabla_A^2})[/tex]

I will look into certain books which show some development of more quantum ideas.
 
Big-Daddy said:
Sorry, I noticed my answer isn't dimensionally sound. How about this one:

[tex]- \sum^M_{A=1} ({\frac{h^2}{2 \cdot m_A} \cdot \nabla_A^2})[/tex]

I will look into certain books which show some development of more quantum ideas.
Close, it should be hbar but otherwise, so long as M is the number of particles in the system, that is fine.
 
Thanks. And is energy (when we solve for it) a function of the position, or a numerical value for the system as a whole?
 
Big-Daddy said:
Thanks. And is energy (when we solve for it) a function of the position, or a numerical value for the system as a whole?
It's the latter.