Perturbation treatment of hydrogen molecular ion

In summary, the book "molecular physics by wolfgang demtroder" uses some advanced quantum mechanics concepts, like perturbation theory. I found it difficult to understand because of the lack of explanations for concepts like derivatives and angles in spherical coordinates. I appreciate any help.
  • #1
patric44
308
40
Homework Statement
question about perturbation method treatment ?
Relevant Equations
in the pictures
hi guys
i am a the third year undergrad student and in this 2nd semester in my collage we should start taking quantum mechanics along with
molecular physics , our molecular physics professor choose a book that we are going to take which is " molecular physics by wolfgang Demtroder "
when i opened that book i found that it uses some advanced quantum mechanics like perturbation theory ... i found it really heavy for a student that just has some relativity intermediate quantum mechanics knowledge !
i started to get this perturbation theory thing ! but has some questions about this :
i get that he can represent the electronic distribution as a linear combination of functions with some coefficients that depend on the
position of the nuclei , how did he expand that integral and reached that partial derivative term its very unclear to me ?
quantum2.png
quantum3.png

i appreciate any help
thanks
 
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  • #2
Can you give us the explicit expression for the operator ##\hat H'##?
 
  • #3
Screenshot_2020-03-23-22-36-37.jpg
 
  • #4
This tells you that ##\hat{H'}## is the operator corresponding to the kinetic energy due to the motion of all of the nuclei. Suppose the ##k^{th}## nucleus has a mass ##M_k## and position ##\mathbf R_k##. Can you write down the kinetic energy operator ##\hat T_k## for this nucleus?
 
  • #5
TSny said:
This tells you that ##\hat{H'}## is the operator corresponding to the kinetic energy due to the motion of all of the nuclei. Suppose the ##k^{th}## nucleus has a mass ##M_k## and position ##\mathbf R_k##. Can you write down the kinetic energy operator ##\hat T_k## for this nucleus?
I guess it should be something like this :
Screenshot_2020-03-24-09-13-41.jpg

I'm not sure if the laplacian expression in spherical is right because the book didn't mention the angles in his treatment!
 
  • #6
Let's express the laplacians in cartesian coordinates and see if we can make sense of (2.10).

So, as you wrote, ##\hat{H'} = \sum_k \frac{1}{M_k} \nabla_k^2##.

If ##x_k##, ##y_k##, and ##z_k## denote the cartesian coordinates of the position of the ##k^{th}## nucleus, what does ##\hat{H'}## look like when expressed in terms of these coordinates?

Also, suppose you have two functions ##f(x)## and ##g(x)##. Expand the expression ##\frac{d^2}{dx^2}\left[f(x)g(x)\right]##.
 
  • #7
TSny said:
Let's express the laplacians in cartesian coordinates and see if we can make sense of (2.10).

So, as you wrote, ##\hat{H'} = \sum_k \frac{1}{M_k} \nabla_k^2##.

If ##x_k##, ##y_k##, and ##z_k## denote the cartesian coordinates of the position of the ##k^{th}## nucleus, what does ##\hat{H'}## look like when expressed in terms of these coordinates?

Also, suppose you have two functions ##f(x)## and ##g(x)##. Expand the expression ##\frac{d^2}{dx^2}\left[f(x)g(x)\right]##.
I guess something like this :
Screenshot_2020-03-24-17-23-58.jpg
 
  • #8
OK. Generalize this to get an expansion of

##\nabla_k^2 \left[f(\mathbf R_k) g(\mathbf R_k)\right]##
 
  • #9
TSny said:
OK. Generalize this to get an expansion of

##\nabla_k^2 \left[f(\mathbf R_k) g(\mathbf R_k)\right]##
Screenshot_2020-03-24-17-49-29.jpg

I guess some terms should vanish as they are not differentiable with respect to either x,y,z?
 
  • #10
patric44 said:
I guess some terms should vanish as they are not differentiable with respect to either x,y,z?
In the expression ##f(\mathbf R_k)##, ##\mathbf R_k## is the position vector of the ##k^{th}## molecule: ##\mathbf R_k = x_k \hat i +y_k \hat j+z_k \hat k##. The notation ##f(\mathbf R_k)## is just shorthand for ##f(x_k, y_k, z_k)##.

In your result for ##\nabla_k^2 \left[f(\mathbf R_k) g(\mathbf R_k)\right]##, your use of primes might be a little confusing. But I guess your subscripts, ##x##, ##y##, and ##z## indicate what derivative the prime denotes.

Can you tidy up your result for ##\nabla_k^2 (fg)## by expressing it in terms of ##\nabla_k^2 f##, ##\nabla_k^2 g##, ##\vec \nabla_k f## and ##\vec \nabla_k g##?
 
  • #11
TSny said:
In the expression ##f(\mathbf R_k)##, ##\mathbf R_k## is the position vector of the ##k^{th}## molecule: ##\mathbf R_k = x_k \hat i +y_k \hat j+z_k \hat k##. The notation ##f(\mathbf R_k)## is just shorthand for ##f(x_k, y_k, z_k)##.

In your result for ##\nabla_k^2 \left[f(\mathbf R_k) g(\mathbf R_k)\right]##, your use of primes might be a little confusing. But I guess your subscripts, ##x##, ##y##, and ##z## indicate what derivative the prime denotes.

Can you tidy up your result for ##\nabla_k^2 (fg)## by expressing it in terms of ##\nabla_k^2 f##, ##\nabla_k^2 g##, ##\vec \nabla_k f## and ##\vec \nabla_k g##?
Thank you so much
I guess the final expression should be as this :
Screenshot_2020-03-24-18-25-49.jpg

Thanks again that helped me alot, I thought that I would never get this scary looking equation
 
  • #12
I'm beginning to like perturbation theory :smile:
 
  • #13
You just about have it. You even included the summation over ##m## in the last term, which the text left out.

However, the argument of your summations in the last term is not written quite correctly. As written, you would have "cross terms" like ##\frac{\partial \phi_m}{\partial x_k} \frac{\partial \chi_m}{\partial y_k}## which have derivatives with respect to ##x_k## and ##y_k## in the same term. But your result in post #9 did not have such mixed terms in ##x_k## and ##y_k##. Try to express your last integral in terms of the gradients ##\vec \nabla_k \phi## and ##\vec \nabla_k \chi##.

We strongly encourage posters to type out mathematical expressions (using Latex if possible) rather than posting pictures of handwritten work. I should have mentioned this much earlier. Typing your expressions makes it much easier for helpers to quote specific parts of your work.
 
  • #14
TSny said:
You just about have it. You even included the summation over ##m## in the last term, which the text left out.

However, the argument of your summations in the last term is not written quite correctly. As written, you would have "cross terms" like ##\frac{\partial \phi_m}{\partial x_k} \frac{\partial \chi_m}{\partial y_k}## which have derivatives with respect to ##x_k## and ##y_k## in the same term. But your result in post #9 did not have such mixed terms in ##x_k## and ##y_k##. Try to express your last integral in terms of the gradients ##\vec \nabla_k \phi## and ##\vec \nabla_k \chi##.

We strongly encourage posters to type out mathematical expressions (using Latex if possible) rather than posting pictures of handwritten work. I should have mentioned this much earlier. Typing your expressions makes it much easier for helpers to quote specific parts of your work.
I almost had it, I think the confusing part that the book express all the derivative as d/dR, but I guess I has it right this time :
Screenshot_2020-03-24-19-19-04.jpg

I'm really sorry for not using Latex but I really don't know how to, I will learn it an I promise you that the next perturbation theory question will be in latex
 
  • #15
The next summation should be on "m", my bad
 
  • #16
Looks good!
 

FAQ: Perturbation treatment of hydrogen molecular ion

1. What is the hydrogen molecular ion?

The hydrogen molecular ion, also known as the dihydrogen cation, is a molecule composed of two hydrogen atoms with a net positive charge. It is the simplest molecular ion and plays an important role in understanding the behavior of more complex molecules.

2. What is perturbation treatment?

Perturbation treatment is a mathematical technique used to approximate the behavior of a system that is slightly different from a known, simpler system. In the case of the hydrogen molecular ion, perturbation treatment is used to calculate the energy levels and wavefunctions of the ion by treating the interaction between the two hydrogen atoms as a perturbation to the simpler hydrogen atom system.

3. Why is perturbation treatment used for the hydrogen molecular ion?

The hydrogen molecular ion is a challenging system to solve using traditional methods due to the strong interaction between the two hydrogen atoms. Perturbation treatment allows for a more accurate and efficient calculation of the ion's properties compared to other methods.

4. How does perturbation treatment work for the hydrogen molecular ion?

Perturbation treatment involves breaking down the system into two parts: the unperturbed system (in this case, the hydrogen atom) and the perturbation (the interaction between the two hydrogen atoms). The energy levels and wavefunctions of the perturbed system are then calculated by taking into account the effects of the perturbation on the unperturbed system.

5. What are the applications of perturbation treatment for the hydrogen molecular ion?

Perturbation treatment of the hydrogen molecular ion has applications in various fields such as quantum chemistry, atomic and molecular physics, and astrophysics. It can be used to understand the behavior of more complex molecules, as well as to study the properties of ions in interstellar space.

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