Where Can I Find Information on the Total Molecular Hamiltonian?

Mayhem
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Homework Statement
Describe the total molecular Hamiltonian
Relevant Equations
$$\hat{H} = -\frac{\hbar^2}{2m_e} \sum^{N_e}_{i=1} \nabla_i^2 -j_0 \sum^{N_e}_{i=1} \sum^{N_n}_{I=1} \frac{Z_I}{r_{Ii}}+\frac{1}{2}j_0 \sum^{N_e}_{i=j}\frac{1}{r_{ij}}$$
Hello.

As an assignment, I have to explain the total molecular Hamiltonian. Problem is, I can't find it anywhere in my book (Atkins, Physical Chemistry: Quanta, Matter, and Change, 2nd Edition), even when I access the index for "Hamiltonian -> polyatomic molecules". They do give the electronic Hamiltonian, which is $$\hat{H} = -\frac{\hbar^2}{2m_e} \sum^{N_e}_{i=1} \nabla_i^2 -j_0 \sum^{N_e}_{i=1} \sum^{N_n}_{I=1} \frac{Z_I}{r_{Ii}}+\frac{1}{2}j_0 \sum^{N_e}_{i=j}\frac{1}{r_{ij}}$$

Any pointers? If anyone has the book, and I overlooked something, a page reference would be nice, because I just can't find it.
 
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You have to add the nuclear part plus spin (if needed) and other corrections.
 
dextercioby said:
You have to add the nuclear part plus spin (if needed) and other corrections.
My book gives this definition of the hamiltonian in the context of nuclear spin. Does this look familiar?

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Mayhem said:
My book gives this definition of the hamiltonian in the context of nuclear spin. Does this look familiar?

View attachment 302221
This is not what you are after.

Can you identify the terms in the Hamiltonian you posted?
Mayhem said:
$$\hat{H} = -\frac{\hbar^2}{2m_e} \sum^{N_e}_{i=1} \nabla_i^2 -j_0 \sum^{N_e}_{i=1} \sum^{N_n}_{I=1} \frac{Z_I}{r_{Ii}}+\frac{1}{2}j_0 \sum^{N_e}_{i=j}\frac{1}{r_{ij}}$$
 
DrClaude said:
This is not what you are after.

Can you identify the terms in the Hamiltonian you posted?
Yes. From terms left to right on the RHS:

1) describes the kinetic energy of the electrons. To elaborate, this is similar for the kinetic energy of one particle, except here we sum over all electrons (Ne), which each have a corresponding gradient dot product, which gives us the direction the particle is moving in.

2) Attraction to nuclei. Again summing over all electrons in the system, but also particles in the nucleus.

3) Mutual repulsion,
 
Mayhem said:
Yes. From terms left to right on the RHS:

1) describes the kinetic energy of the electrons. To elaborate, this is similar for the kinetic energy of one particle, except here we sum over all electrons (Ne), which each have a corresponding gradient dot product, which gives us the direction the particle is moving in.

2) Attraction to nuclei. Again summing over all electrons in the system, but also particles in the nucleus.

3) Mutual repulsion,
Correct. So what is missing is the equivalent of 1 and 3 for the nuclei.
 
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DrClaude said:
Correct. So what is missing is the equivalent of 1 and 3 for the nuclei.
Using the Born-Oppenheimer approximation, wouldn't term 1 just equal 0? i.e. it has no kinetic energy relative to the rest of the system.
 
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Mayhem said:
Using the Born-Oppenheimer approximation, wouldn't term 1 just equal 0? i.e. it has no kinetic energy relative to the rest of the system.
First, you are looking for the full Hamiltonian, which is independent of of any approximation used to get manageable solutions to the time-independent Schrödinger equation.

Second, no, the kinetic energy term for the nuclei does not go away with the Born-Oppenheimer approximation. This term is where molecular vibrations and rotations will come from.

Writing the full wave function ##\Psi(\{\mathbf{R}\},\{\mathbf{r}\})## (function of the collection of the position of the nuclei and the electrons) as a function of a nuclear and an electronic wave function,
$$
\Psi(\{\mathbf{R}\},\{\mathbf{r}\}) = \psi_\mathrm{N} (\{\mathbf{R}\}) \psi_\mathrm{e} (\{\mathbf{r}\};\{\mathbf{R}\}),
$$
the B-O approx corresponds to neglecting terms in the Hamiltonian where ##\psi_\mathrm{e}## is differentiated with respect to ##\mathbf{R}## and vice versa. But derivatives of ##\psi_\mathrm{N}## with respect to ##\mathbf{R}## remain.
 
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