What Determines the Direction of a Dipole Operator in Quantum Optics?

[McLaren Rulez]

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 \hat{d}.\vec{E} 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 \langle g\mid\hat{d}\mid e\rangle. 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 :)

 

[mfb]

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.

 

[McLaren Rulez]

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.