Change of Coordinates (Two-Body Problem)

Click For Summary
SUMMARY

The discussion focuses on the change of coordinates in the two-body problem as presented in Griffiths' Quantum Mechanics, specifically Problem 5.1. The key equations derived include the expressions for the positions of the two particles, \(\mathbf{r_1}\) and \(\mathbf{r_2}\), in terms of the center of mass \(\mathbf{R}\) and the relative position \(\mathbf{r}\). The reduced mass \(\mu\) is defined as \(\mu \equiv \frac{m_1m_2}{m_1 + m_2}\). The challenge arises in correctly expressing the gradients \(\nabla_1\) and \(\nabla_2\) in terms of \(\nabla_R\) and \(\nabla_r\).

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the two-body problem.
  • Familiarity with the concept of reduced mass in classical mechanics.
  • Knowledge of vector calculus, specifically gradient operations.
  • Experience with coordinate transformations in physics.
NEXT STEPS
  • Study the derivation of the Schrödinger equation for two-body systems in quantum mechanics.
  • Learn about the implications of reduced mass in various physical systems.
  • Explore vector calculus techniques, focusing on gradient transformations.
  • Investigate coordinate transformations in classical mechanics and their applications in quantum mechanics.
USEFUL FOR

Students of quantum mechanics, physicists working on two-body problems, and anyone interested in advanced topics in vector calculus and coordinate transformations.

cepheid
Staff Emeritus
Science Advisor
Gold Member
Messages
5,197
Reaction score
38
Context: reducing a two-body problem to a one body problem (each body is a quantum particle, although this is not relevant to the parts of the problem I am stuck on

Homework Statement



Griffiths QM, Problem 5.1

Typically the interaction potential depends only on the vector [itex]\mathbf{r} = \mathbf{r_1} - \mathbf{r_2}[/itex] between the two particles. In that case, the Schrödinger equation separates if we change variables from [itex]\mathbf{r_1}, \mathbf{r_2}[/itex] to [itex]\mathbf{r}[/itex] and [itex]\mathbf{R} \equiv (m_1\mathbf{r_1} + m_2\mathbf{r_2})/m_1+m_2[/itex] (the centre of mass).

(a) Show that [itex]\mathbf{r_1} = \mathbf{R} + (\mu/m_1)\mathbf{r}[/itex], [itex]\mathbf{r_2} = \mathbf{R} - (\mu/m_2)\mathbf{r}[/itex], and [itex]\nabla_1 = (\mu/m_2)\nabla_R + \nabla_r[/itex], [itex]\nabla_2 = (\mu/m_1)\nabla_R - \nabla_r[/itex], where

[tex]\mu \equiv \frac{m_1m_2}{m_1 + m_2}[/tex]

is the reduced mass of the system.

Homework Equations



Already given

The Attempt at a Solution



(a)

(i) [tex]\mathbf{r_1} = \mathbf{r} + \mathbf{r_2} = \mathbf{r} + \left(\frac{m_1 + m_2}{m_2}\mathbf{R} - \frac{m_1}{m_2}\mathbf{r_1} <br /> \right)[/tex]

[tex]\mathbf{r_1}\left(1 + \frac{m_1}{m_2}\right) = \mathbf{r_1}\left(\frac{m_1 + m_2}{m_2}\right) = \left(\frac{m_1 + m_2}{m_2}\right)\mathbf{R} + \mathbf{r}[/tex]

[tex]\longRightarrow \mathbf{r_1} = \mathbf{R} + \frac{m_2}{m_1 + m_2}\mathbf{r} = \mathbf{R} + \frac{\mu}{m_1}\mathbf{r}[/tex]

(ii) [tex]\mathbf{r_2} = \mathbf{r_1} - \mathbf{r} = \mathbf{R} + \frac{\mu}{m_1}\mathbf{r} - \mathbf{r}[/tex]

[tex]= \mathbf{R} + \left(\frac{\mu}{m_1} - 1 \right) \mathbf{r} = \mathbf{R} + \left(\frac{m_2}{m_1 + m_2} - \frac{m_1 + m_2}{m_1 + m_2} \right)\mathbf{r}[/tex]

[tex]= \mathbf{R} - \left(\frac{m_1}{m_1 + m_2}\right)\mathbf{r} = \mathbf{R} - \frac{\mu}{m_2}\mathbf{r}[/tex]

(iii) Now here's where I got stuck. I thought to start in cartesian coordinates:

[tex]\nabla_1 \equiv \mathbf{\hat{x}} \frac{\partial}{\partial x_1} + \mathbf{\hat{y}} \frac{\partial}{\partial y_1} + \mathbf{\hat{z}} \frac{\partial}{\partial z_1}[/tex]

and then based on the relationships between [itex]x_1[/itex], [itex]x_R[/itex], and [itex]x_r[/itex] obtained in the first part, I could find the answer. But if I do that, I just get:

[tex]\nabla_1 = \nabla_R + \frac{\mu}{m_1}\nabla_r[/tex]

which is clearly not the right answer. Any tips for this section?
 
Last edited:
Physics news on Phys.org
It works fine over here. For instance,

[tex]\frac{\partial}{\partial x_1}=\frac{\partial R_x}{\partial x_1}\frac{\partial}{\partial R_x}+\frac{\partial r_x}{\partial x_1}\frac{\partial}{\partial r_x}=\frac{\mu}{m_2}\frac{\partial}{\partial R_x}+\frac{\partial}{\partial r_x}[/tex]
 
You're right. It turns out I was just being dumb. Thanks.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K