# I Frustrated spontaneous emission

1. Jul 3, 2017

### Trollfaz

I have heard of frustrated spontaneous emission that somehow says that an atom that normally emit light will cease to do so when its surroundings is incapable of absorbing light. How is this possible, and is this experimentally proven?

2. Jul 3, 2017

### vanhees71

I'm not sure I understand your question, because it's not a point whether radiation can be absorbed or not but whether it can be emitted. There are many ways a decay can be surpressed.

I refer also to the usual electromagnetic spontaneous transitions in atoms as a "decay" of a state due to interaction with the quantized radiation field; note that spontaneous emission is the very point, where you really need photons, i.e., the quantized em. field, while for nearly everything else you can come very far with the semiclassical approximation, where the em. field is taken as a classical field.

Most common are the dipole selection rules, i.e., for electric dipole radiation the angular-momentum quantum number must stay the same or change by 1 (transitions $l=0 \rightarrow l=0$ are always forbidden) and the magnetic quantum number must stay the same or change by 1. All other transitions are "forbidden" (or maybe allowed but suppressed due to higher-order multipole transitions):

https://en.wikipedia.org/wiki/Selection_rule#Angular_momentum

Another possibility is that you put the atom in a cavity. Then an decay, "allowed" in free space, can become forbidden or at least suppressed, if the corresponding frequency of the photon is not close enough to a resonance frequency of the em. field in this cavity.

3. Jul 4, 2017

### Trollfaz

Does placing an atom in a mirror cavity prevents it from emitting light?

4. Jul 4, 2017

5. Jul 5, 2017

### vanhees71

The Wheeler-Feynman absorber theory has never been extended to a consistent quantum theory. In standard QED, putting an atom in a cavity, suppresses radiation whose wave length doesn's "fit" into the cavity.

6. Jul 5, 2017

### Jilang

It's very interesting. How does it know in advance that it won't fit?

7. Jul 8, 2017

### vanhees71

Nothing needs to know that it doesn't fit. In the cavity there's simply no state that describes a photon at that frequency.

8. Jul 8, 2017

### Jilang

Yes, the boundary conditions would dictate that. How would that be different for an infinite cavity?

9. Jul 9, 2017

### vanhees71

The "infinite cavity" has its own problems, as you know when studying QFT in the high-energy-particle context. One way to get well defined observables (S-matrix elements) is indeed to first use a finite "quantization volume". In this context it's wise to use periodic boundary conditions, because this admits the definition of a well-defined momentum operator. So take a cube with length $L$ as the quantization volume. We consider free photons and use the formalism starting from the fully gauge fixed description, i.e., we describe the field by a four-potential $A^{\mu}$ subject to the radiation-gauge condition (only possible for free fields)
$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
The remaining two field components fulfill the wave equation
$$\Box \vec{A}=0.$$
Now we look for plain-wave solutions, leading to
$$\vec{A}_{\vec{k}}=A \vec{\epsilon}_{\vec{k},\lambda} \exp(-\mathrm{i} k \cdot x)|_{k^0=\omega_{\vec{k}}=|\vec{k}|}+\text{c.c.}.$$
$\lambda$ labels the two unit vectors (polarization vectors of the wave) perpendicular to $\vec{k}$ since the gauge condition imposes that the waves are transvers:
$$\vec{k} \cdot \vec{\epsilon}_{\vec{k},\lambda}=0.$$
The boundary conditions impose
$$\vec{k} \in \frac{2 \pi}{L} \mathbb{Z}^3,$$
i.e., we have a discrete set of "allowed" momenta. The grid becomes the finer the larger the size of the quantization volume gets, and in the limit of $L \rightarrow \infty$.

A general field is given by the Fourier series
$$\vec{A}(t,\vec{x})=\sum_{\lambda=1}^2 \sum_{\vec{k} \ in 2 \pi/L \mathbb{Z}^3} [A_{\lambda}(\vec{k}) \epsilon_{\vec{k} \lambda} \exp(-\mathrm{i} k \cdot x)|_{k^0=|\vec{k}|}+\text{c.c.}]$$
The quantization is then straight forward, using the usual Lagrange-Hamilton procedure.

In the infinite-volume limit the sum goes over to an integral over $\vec{k} \in \mathbb{R}^3$.