# Phase of photon from excited atom

1. Mar 16, 2013

### johne1618

Imagine that an excited hydrogen atom in the symmetric 2s state emits a photon.

We use a birefringent crystal to split the photon into vertical and horizontal components.

Is the phase relation between these components random or is it the same for every photon emitted by a similarly excited hydrogen atom?

Last edited: Mar 16, 2013
2. Mar 17, 2013

### Jano L.

I do not know what to say about photons, but if you consider harmonic plane EM wave instead, the answer should be simple. Assume that the Maxwell equations for dielectric are valid and that the wave goes left to right. The EM field has to be continuous across the boundary, so both components of the field are in phase with the field outside. This means that the two waves have to have the same phase just right of the boundary surface.

As the waves will move into the crystal medium with different velocities and different wavelengths, they can't have the same phase everywhere. However, their phase difference at any point will be constant, and equality will happen at special equidistant points.

3. Mar 17, 2013

### johne1618

I don't think I have expressed my question very well.

I am simply asking what is the polarization state of a photon emitted by an individual excited hydrogen atom transitioning from say the symmetric 2s state to the ground state.

Is the photon polarization state well-defined (i.e. the same for all similarly prepared excited atoms) or is it random?

If it is well-defined is it linear, circular or elliptical?

Last edited: Mar 17, 2013
4. Mar 18, 2013

### DrDu

The example is not a good one as the transition from 2s to 1s is forbidden as long as either spin-orbit coupling or multiphoton transitions aren't taken into account.
For e.g. a 2p_z to 1s transition the polarisation will be well defined but angle dependent given in terms of vector spherical harmonics:
http://en.wikipedia.org/wiki/Vector_spherical_harmonics

5. Mar 18, 2013

### Jano L.

The polarization of the spontaneous emission usually depends on two important things:

1) the direction of oscillations of the charges (matrix element of dipole moment operator between two eigenfunctions considered), given e.g. by some vector $\mathbf s$, or some other quantity (multipole elements...)
2) the direction of the line joining the atom and the observer, given e.g. by the vector $\mathbf n$.

The polarization is then in the direction of the vector $\mathbf n\times (\mathbf n \times \mathbf s )$.

In your case, the matrix element of the dipole moment is zero. In similar cases there can be still some radiation, and it still will be perpendicular to $\mathbf n$, but it will be related to those other matrix elements of the pair of two eigenfunctions considered or some else property of the atom. But the case you consider, I think that DrDu is right, if you have both functions radially symmetric and do not consider other mechanisms, no radiation is predicted by the formalism.

6. Mar 18, 2013

### johne1618

Thanks for the replies!