How to simplify the diatomic molecule Hamiltonian using an expansion?

Click For Summary

Homework Help Overview

The discussion revolves around simplifying the Hamiltonian for a diatomic molecule, specifically focusing on the potential energy terms and their expansion in terms of a small parameter related to the displacement from equilibrium. The subject area includes quantum mechanics and molecular physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss expanding the Hamiltonian and potential energy terms to second order in a small displacement variable, questioning how to correctly apply the expansion and what terms can be neglected. There are attempts to clarify the order of approximations needed for the parameters involved.

Discussion Status

Participants are actively engaging with the problem, offering insights into the necessary expansions and the implications of neglecting certain terms. There is a recognition of the need for agreement on the form of the Hamiltonian and the assumptions made regarding the small parameters.

Contextual Notes

There is mention of a typo affecting the clarity of the discussion, specifically regarding the order of approximation for the small parameter. The participants are focused on ensuring that their approaches align with the assumptions made in the original problem statement.

MathematicalPhysicist
Science Advisor
Gold Member
Messages
4,662
Reaction score
372

Homework Statement


I have the diatomic molecule hamiltonian given by:
$$-\hbar^2/(2\mu)d^2/dr^2+\hbar^2\ell(\ell+1)/(2\mu r^2)+(1/4)K(r-d_0)^2$$

Now it's written in my solutions that if we put:
$$K\equiv 2\mu \omega_0^2, \hbar^2\ell(\ell+1)/(2\mu d_0^4)\equiv \gamma_{\ell} K, r-d_0\equiv \rho$$

Expand to second order in ##\rho## and drop terms in ##\gamma_{\ell}^2## since ##\gamma_{\ell}\ll 1##, to get:
$$-\hbar^2/(2\mu)d^2/dr^2+(1/2)\mu \omega_0^2[(1+12\gamma_{\ell})(\rho - 4\gamma_{\ell}d_0)^2+4\gamma_{\ell}d_0^2]$$

How to get the last expression explicitly?

Homework Equations

The Attempt at a Solution


I thought of expanding ##1/(\rho+d_0)^2 \approx 1/(d_0^2)[1-2\rho/d_0+3\rho^2/d_0^2]##
But I don't see how did they get this expression for the Hamiltonian?
edit: I have edited and corrected the typo.
 
Last edited:
Physics news on Phys.org
MathematicalPhysicist said:
Expand to second order in ##\rho## and drop terms in ##\gamma_{\ell}$ since $\gamma_{\ell}\ll 1##, to get:
Due to the typo, we can't tell what order of approximation in ##\gamma_{\ell}## is to be made.

I thought of expanding ##1/(\rho+d_0)^2 \approx 1/(d_0^2)[1-2\rho/d_0+3\rho^2/d_0^2]##
Sounds good.
But I don't see how did they get this expression for the Hamiltonian?
The only thing that is required is for their way of writing H to agree with your way of writing H to the assumed orders of approximation in ##\rho## and ##\gamma_{\ell}##. They are probably choosing to write H in a particular way in order to simplify further analysis.
 
TSny said:
Due to the typo, we can't tell what order of approximation in ##\gamma_{\ell}## is to be made.

Sounds good.
The only thing that is required is for their way of writing H to agree with your way of writing H to the assumed orders of approximation in ##\rho## and ##\gamma_{\ell}##. They are probably choosing to write H in a particular way in order to simplify further analysis.
I corrected the typo, we should drop terms of order ##\gamma_{\ell}^2## since ##\gamma_{\ell}\ll 1##.
 
MathematicalPhysicist said:
I corrected the typo, we should drop terms of order ##\gamma_{\ell}^2## since ##\gamma_{\ell}\ll 1##.
OK.

You are on the right track to use your expansion ##1/(\rho+d_0)^2 \approx 1/(d_0^2)[1-2\rho/d_0+3\rho^2/d_0^2]## in the expression

##V(r) \doteq \hbar^2\ell(\ell+1)/(2\mu r^2)+(1/4)K(r-d_0)^2##

Then you can manipulate your expression for ##V(r)## into their corresponding expression. (Keep in mind that any terms of order ##\gamma_{\ell}^2## may be neglected.) The motivation for doing all this is that, in their expression, the potential energy part of ##H## is just that of a 1D simple harmonic oscillator.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 0 ·
Replies
0
Views
1K