# Understanding Atomic Dipoles and Spontaneous Emission

1. Feb 18, 2013

### McLaren Rulez

Hi,

In quantum optics, the interaction between light and atoms is described by a Hamiltonian of the form d.E where d is the dipole moment of the atom. The picture given is basically that this is a vector and we take the the dot product with this and the electric field vector (whose direction comes from the polarization direction). I don't understand why the atom has this asymmetry.

1) If the atom is spherically symmetric, how do we get this dipole pointing in one specific direction? Can an experimentalist put an atom with its dipole pointing in a specific way?

2) If we look at spontaneous emission, the rate is given by the Einstein A coefficient which is reproduced using quantum optics. It is
$$\Gamma=\frac{\omega^{2}d^{2}}{3\pi\epsilon_{0} \hbar c^{3}}$$
Is this sponteanous emission spatially isotropic or is there more radiation in some directions compared to others?

I feel that I may have some misconceptions regarding the whole thing. Please do correct me if I do. Thank you :)

Last edited: Feb 18, 2013
2. Feb 19, 2013

### DrDu

1) The point is that d is an operator and, even if you start from a spherically symmetric electronic wavefunction, the operator induces a transition to a state which is not spherically symmetric. E.g. in a hydrogen atom from an s function to a p function.
2) The emission of a single atom will not be isotropic, however the formula for the Einstein A coefficient is averaged over many atoms and the radiation emitted spontaneously by a large ensemble of atoms is isotropic on the mean. This averaging leads to the factor 3 in the denominator.

3. Feb 19, 2013

### McLaren Rulez

Thanks DrDu. So the actual radiation pattern be the same as that due to the dipole antenna, I assume.

Also, is it possible to experimentally orient the atomic dipole in any direction we want? Or is it a random process?

Thank you

4. Feb 19, 2013

### DrDu

With an atom in a spherically symmetric ground state this is difficult.
You could use e.g. the Stark effect to split the final levels of the transition.
It is much easier with molecules whose molecular axes can be oriented, e.g. in a crystal or polymer matrix.

5. Feb 19, 2013

### Staff: Mentor

Why? If you take a dilute gas of atoms (where collisions are not important) and shine polarized light on it, you will get polarized atoms.

6. Feb 19, 2013

### DrDu

Admittedly true. I was more referring to spontaneous emission.

7. Feb 19, 2013

### McLaren Rulez

Thank you for the replies! I have a much better idea of the process now.