Where Can I Find Information on the Total Molecular Hamiltonian?

In summary: So the nuclear contribution to the Hamiltonian is still there.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,the B-O approx corresponds to neglecting terms
  • #1
Mayhem
354
253
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.
 
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
You have to add the nuclear part plus spin (if needed) and other corrections.
 
  • #3
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?

1654079423259.png
 
  • #4
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:
$$\hat{H} = -\frac{\hbar^2}{2m_e} \sum^{N_e}_{i=1} \nabla_i^2 -j_0 \sum^{N_e}_{i=1} \sum^{N_n}_{I=1} \frac{Z_I}{r_{Ii}}+\frac{1}{2}j_0 \sum^{N_e}_{i=j}\frac{1}{r_{ij}}$$
 
  • #5
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,
 
  • #6
  • #8
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.
 
  • Like
Likes Mayhem
  • #9
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.
 
  • #10
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.
 

FAQ: Where Can I Find Information on the Total Molecular Hamiltonian?

What is the total molecular Hamiltonian?

The total molecular Hamiltonian is a mathematical operator that describes the total energy of a molecule in terms of its constituent particles and their interactions. It takes into account the kinetic energy of the particles, as well as their potential energy due to their interactions with each other.

How is the total molecular Hamiltonian derived?

The total molecular Hamiltonian is derived from the Schrödinger equation, which is a fundamental equation in quantum mechanics. By solving the Schrödinger equation for a molecule, the total molecular Hamiltonian can be obtained as an operator acting on the wave function of the molecule.

What information can be obtained from the total molecular Hamiltonian?

The total molecular Hamiltonian provides important information about the energy levels and properties of a molecule. It can be used to calculate the molecular energy, as well as the vibrational and rotational energies of the molecule. It also provides insights into the molecular structure and dynamics.

How is the total molecular Hamiltonian used in molecular simulations?

In molecular simulations, the total molecular Hamiltonian is used to calculate the forces acting on the particles in a molecule. This information is then used to update the positions and velocities of the particles, allowing for the simulation of the molecule's behavior over time.

Are there any limitations to the total molecular Hamiltonian?

Yes, there are limitations to the total molecular Hamiltonian. It assumes that the particles in a molecule are point masses and neglects the effects of relativistic and quantum mechanical phenomena. Additionally, it can become computationally expensive for larger molecules, making approximations necessary in some cases.

Back
Top