Dipole operator question

  • #1
McLaren Rulez
292
3
Hi,

In quantum optics, when we talk about atom field interaction with a classical field and quantized atom, we say that the Hamiltonian has an interaction part of the form [itex]\hat{d}.\vec{E}[/itex] where d is the dipole operator.

For a two level atom, the dipole operator has only off diagonal elements and these are of the form [itex]\langle g\mid\hat{d}\mid e\rangle[/itex]. Now, this expectation value still has a direction because we take its dot product with the polarization direction of the electric field. But, in any atom, both g and e are spherically symmetrical states so what "direction" is being talked about here?

Thank you :)
 

Answers and Replies

  • #2
36,103
13,028
If both g and e have a spherical symmetry (=> S-states), the dipole expectation value should be 0. You need an odd ##\Delta l## to get a non-zero expectation value - one symmetric and one antisymmetric wave function, I think.
 
  • #3
McLaren Rulez
292
3
Thank you mfb.

Pardon me if this is a silly question, I just revised the angular quantum number section of my QM text. I understand that if we are talking about states with nonzero l, then they are not spherically symmetric, correct? If so, what determines the asymmetry? For instance, how do which direction the p oribital's dumbell points along?

Also, when we talk about the emission of a photon, does this asymmetry also determine the direction of the photon? That is, there is a higher probability of emission in some directions compared to others?

Thank you.
 
Last edited:

Suggested for: Dipole operator question

Replies
2
Views
278
Replies
56
Views
1K
  • Last Post
Replies
21
Views
446
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
398
  • Last Post
Replies
3
Views
341
  • Last Post
Replies
19
Views
538
  • Last Post
Replies
5
Views
444
Replies
1
Views
430
Replies
7
Views
433
Top