# Transition dipole moment: cause or result of interaction?

1. Oct 2, 2015

### maxverywell

According to the wikipedia article about the transition dipole moment:

When an atom or molecule interacts with an electromagnetic wave of frequency , it can undergo a transition from an initial to a final state of energy difference through the coupling of the electromagnetic field to the transition dipole moment.
https://en.wikipedia.org/wiki/Transition_dipole_moment

As I have understood it, the interaction of an atom with electric field causes a transition between its energy states, and this gives the atom a dipole moment ("transition dipole moment"). But it is this dipole moment that interacts which the electric field. This reasoning seems circular.

Moreover, before the transition is made, the dipole moment of the atom is zero. So how the interaction, which causes the transition, is possible in the first place? Or I have completely misunderstood this concept?

2. Oct 3, 2015

### fzero

You are using classical reasoning (but so is the wikipedia page). As you say the dipole moment is zero before and after the transition. There was an old picture of the interaction of light with atoms that suggested that the electric field of the incoming wave would induce a dipole moment that would let the atom interact via the classical coupling of the electromagnetic field with the dipole moment. I don't have a reference for this at hand, but it is probably mentioned in some classical EM texts. It is also essentially the picture that the wikipedia page is promoting with that figure of oscillating wavefunctions in the upper right hand corner of the page.

The classical picture is probably somewhat useful, but the proper quantum mechanical interaction is completely different because the electron and photon are treated as point particles. The interaction occurs instantaneously at a definite spacetime point and there is no time for a dipole moment to be induced. The photon does not actually interact with a dipole moment at all, but interacts directly with the electron via a coupling between the vector potential and the momentum operator $\mathbf{A}\cdot \hat{\mathbf{p}}$. I gave a reference and a brief explanation in this post last month that might be helpful to you.

3. Oct 3, 2015

### Jano L.

The transition dipole moment has two meanings: 1) off-diagonal element of electric moment matrix; 2) expected average of electric moment during the interaction.

In non-relativistic Schroedinger's equation describing interaction of an atom with EM field, the interaction is a continuous process that takes time. During this time the transition dipole moment in the sense 1) is constant, but the transition moment in the sense 2) changes in time; if the external EM field is periodic, the moment 2) oscillates in time.

EDIT
The Wikipedia text is misleading. EM field interacts with the electrons, not with transition dipole moment. Capability of the system to interact is partially and approximately quantified by 1). The presence of interaction at time $t$ in the model is thus due to non-vanishing 1) which is constant and does not vanish, even if 2) vanishes at this time.

Last edited by a moderator: May 7, 2017
4. Oct 3, 2015

### vanhees71

Also have a look here:

https://www.physicsforums.com/insights/sins-physics-didactics/

5. Oct 4, 2015

### DrDu

Thats not quite true. As $j=ev=e\dot{r}$, $E=-\frac{dA}{dt}$ and $d=er$, we can write $jA=dE+\frac{d}{dt}(dA)$, so both terms dE and jA are equivalent in a hamiltonian as the hamiltonians only differ by a total time derivative.
This also holds quantum mechanically as both hamiltonians are related by a unitary transformation. This argumentation can also be extended to higher multipole moments and leads to the consideration of the multipolar hamiltonian , see e.g. the book molecular quantum electrodynamics,
http://store.doverpublications.com/0486402142.html
What I find especially interesting is the fact that with this hamiltonian, the vector potential is eliminated from the hamiltonian (at the expense of the introduction of the polarization or dipole moment).

Last edited: Oct 4, 2015
6. Oct 4, 2015

### vanhees71

But the picture is correct within the semiclassical theory of radiation-atom interaction (semiclassical in the sense that you treat the em. field as a classical field and the electrons in the atom quantum mechanically). Then you can only have induced emission, i.e., an excited atomic state is strictly stable within this approximation. To describe also spontaneous emission, i.e., the fact that these excited states are not stable but can decay spontaneously under mission of a photon, you need to quantize also the radiation field, introducing the Bose-enhancement factors in the transition amplitudes. Spontaneous emission is an example for vacuum fluctuations of the electromagnetic field.

Spontaneous emission was discovered within old quantum theory by Einstein in 1917 when he derived Planck's radiation law in a stochastic/kinetic way and in the new quantum theory it was first described by Dirac in 1927, when he introduced annihilation and creation operators for photons.

7. Oct 4, 2015

### DrDu

Ups, I had copied something I didnt want to. See my corrected post.

8. Oct 4, 2015

### fzero

But the "transition dipole moment" is never an observable of any state involved in the transition, so it doesn't really have a direct classical interpretation. We can wave our hands around various classical pictures that have a bit in common with the result of the quantum calculation. That can be useful pedagogy, but they are not more valid than the quantum description. I was also referring to the proper calculation in QFT, so I would also argue that the nonrelativisitic quantum mechanics treatment can be misleading if you try to take it more seriously than the physics behind it.

9. Oct 4, 2015

### DrDu

The vector potential entering in the jA scheme isn't observable, while the dipole moment is a well defined observable. The calculation using multipole momenta, is also correct in QFT (see the book I cited). That it is not as useful in accelerator physics, doen't mean it is not correct.

10. Oct 5, 2015

### vanhees71

11. Oct 5, 2015

### fzero

To be more precise, the transition dipole moment can be defined for electronic transitions for hydrogenic atoms. But it is not an observable dipole moment of any state involved in the transition. The classical induced-dipole description isn't the one that emerges from QED. Molecules can have nontrivial dipole moments which should contribute directly in some way to certain transition dipole moments. I will try to track down a copy of that book at some point.

And yes, the transition dipole moment is technically an observable of scattering experiments. I was using a narrower point of view where the observable was computed in a single state that the system was in. At lowest order there is no intermediate state; the transition is direct from initial to final state.

12. Oct 5, 2015

### Jano L.

You seem to imply that the system changes state from one Hamiltonian eigenstate to another instantly, leaving no time for the expected average electric moment to have non-zero value. Is that what you are assuming?