# Distance between energy states?

1. Sep 14, 2015

I'm doing some personal research on how matter interacts with radiation. Specifically, I am looking through the treatment of Bransden and Joachain. I've taken two semesters of quantum in the past (a while ago), but now I'm coming across something that I've either never seen or never stopped to question. In discussing Einstein Coefficients, I keep seeing this |rba|2 term, where b and a both refer to energy states, with Eb > Ea. Now, I'm having some trouble visualizing any significance to this term. It is defined in the usual way, as the sum of |xi|2, but I just have no idea what this means, as energy levels don't ever really have definite locations, as far as I understand. Could anyone help shed some intuition on what this "distance" means?

Also, sorry in advance if this belongs in the homework/coursework forum, I wasn't quite sure.

2. Sep 14, 2015

### Jilang

Do you have a page number as there are nearly 700!

3. Sep 14, 2015

### fzero

That term is the matrix element for the dipole operator, which is proportional to the coordinate vector $\mathbf{r}$. It is not exactly a physical distance, but it is analogous to the classical idea that a dipole moment can be formed if a charge is displaced with respect to a neutral configuration according to $\mathbf{P} = \sum_i q_i (\mathbf{r}_i - \mathbf{r}_e)$.

You can find a derivation of that matrix element at the chapter http://quantummechanics.ucsd.edu/ph130a/130_notes/node417.html. Briefly, the interaction of the atom with an photon is determined in terms of the electromagnetic field of the photon through an interaction Hamiltonian that is proportional to $\mathbf{A} \cdot \mathbf{p}$ where $\mathbf{A}$ is the EM vector potential and $\mathbf{p}$ is the momentum operator. For a photon of specific wavevector $\mathbf{k}$, the vector potential can be expanded as a plane wave $a e^{i\mathbf{k}\cdot \mathbf{r}}$, but the dipole approximation assumes that $\mathbf{k}\cdot \mathbf{r} \ll 1$ so that we can treat $e^{i\mathbf{k}\cdot \mathbf{r} }\sim 1$. Then the interaction operator is essentially the momentum operator, but we can write
$$-\frac{i}{m} \mathbf{p} = [H_0, \mathbf{r}],$$
where $H_0$ is the unperturbed Hamiltonian (see the notes for a bit more justification of this). In an expectation value then
$$\langle b| \mathbf{p} |a\rangle \sim \langle b| [H_0, \mathbf{r}]|a\rangle = \langle b| (E_b - E_a) \mathbf{r}|a\rangle ,$$
so the nontrivial data are the matrix elements $\mathbf{r}_{ab} = \langle b| \mathbf{r}|a\rangle$.

4. Sep 14, 2015

Jilang, yes, this stuff starts on page 166.

Fzero, thank you for the extensive response! Is there a tie to the existence of an actual dipole anywhere, or is the dipole analogy just because of the dot product between the bra and ket vectors, forming a similar shape?

5. Sep 14, 2015

### fzero

In classical EM scattering, the plane wave $e^{i \mathbf{k}\cdot \mathbf{r}}$ is also expanded into spherical harmonics and this type of expansion is known as a multipole expansion. The lowest order term in the expansion for the vector potential is called the dipole term, so the same terminology is used here in the quantum case. This dipole nature is required because of the vector-nature of the potential.

The initial and final states need not have electric dipole moments of their own. The transition matrix element is really a hybrid object that measures how the initial and final states are related.

6. Sep 14, 2015

Ok, so the reference to a dipole is only mentioned to denote the level of approximation. And the term I originally brought up, |rab|2, is simply the result of applying this dipole-approximation operator to the original state. Is that right?

7. Sep 15, 2015

### fzero

Almost, The matrix element $\langle b|\mathbf{r} |a\rangle$ also depends on the final state $|b\rangle$. The way to think about it is the following. You would recall that the quantity $|\langle \phi | \psi \rangle|^2$ can be thought of as the probability that a measurement finds the system in the state $|\phi\rangle$ given that it was in the state $|\psi\rangle$ at some point before the measurement. Here we can think of applying the dipole operator to the initial state to give some intermediate state $|\psi \rangle = \mathbf{r} |a\rangle$. Then $|\langle b| \mathbf{r} | a\rangle|^2= |\langle b| \psi \rangle|^2$ is the probability to measure the final state given that intermediate state. Typically the initial and final states of interest are eigenstates of the unperturbed Hamiltonian, while the intermediate state can be expressed as a linear combination of essentially all energy eigenstates with the allowed angular momentum quantum numbers.

8. Sep 17, 2015