if an atom in its ground state is coupled to an electromagnetic field it can absorb a photon if the EM field contains one with the right frequency. These transitions depends on $⟨f|H_i|i⟩$ (from Fermi's golden rule) with $|i⟩$ the initial state, $|f⟩$ the final one and $H_i=d\cdot E$ the interaction between the field and the atom (with $d$ the dipolar momentum and $E$ the electric field).

If we use $_F$ for the field and $_A$ for the atom we have:


If $|⟨f|H_i|i⟩|^2=0$ the transition is not possible, for example when $_A⟨f|d|i⟩_A=0$. But is it also possible that $_F⟨f|E|i⟩_F=0$ ?

If we look at transition with one photon (with the right impulsion p) we have $_F⟨0|E|1,p⟩_F\neq0$.

Is $_F⟨0|E|2,p'⟩_F$ (with p'=p/2) also different from zero? I have tried to make these calculations for a Klein-Gordon field : I find $_F⟨0|\phi(x)|1,p⟩_F=e^{ipx}\neq0$, but I'm not sure whether $_F⟨0|\phi(x)|2,p⟩_F=0$ or if I've made a mistake.

I think it's strange since I only heard that two photon linear absorption was forbidden by consideration on the atom states, but I know that non-linear two photons absorption is possible and it seems to make sense with $⟨0|\phi(x)|2,p⟩=0$ and $⟨0|\phi^n(x)|2,p⟩\neq0$

So here is my question, am I right? What prevents linear two photons transition?

EDIT : Thanks to Emilio Pisanty for making everything more clear, I'm not very good at notations.


I'm not sure where you think two-photons transitions come in.

The matrix element simplifies as $$\langle f|H_i|i\rangle=\langle f|\mathbf{d}\cdot\mathbf{E}|i\rangle=\mathbf{E}\cdot\langle f|\mathbf{d}|i\rangle$$ because $\mathbf{E}$ is a constant vector and is therefore not an atomic operator. The matrix element $\langle f|E|i\rangle=E\langle f|i\rangle$ is irrelevant in this problem.

A two-photon transition arises in second-order perturbation theory, where you will have factors of the form $\langle f|H_i|k\rangle\langle k|H_i|i\rangle$, for some intermediate (possibly virtual) state $|k\rangle$. These allow transitions when $\langle f|H_i|i\rangle=0$ but are far less likely, as they scale with $E^2$ instead of $E$. In general, two-photon transitions - also called quadrupole transitions - will have different selection rules to dipole (single-photon) transitions such as changes in $l$ of 0 or 2, and with the two together you get a broader range of allowed final states.

Other than that, I have no idea what your question actually is - please clarify it!

If you want to quantize the field , then you need to split your initial and final states into whatever atomic states ($|i\rangle_\text{A}$ and $|f\rangle_\text{A}$) and field states ($|i\rangle_\text{F}$ and $|f\rangle_\text{F}$) you're considering. The matrix element will then split as $$\langle f|H_i|i\rangle={}_\text{F}\langle f|\mathbf{E}|i\rangle_\text{F}\cdot{}_\text{A}\langle f|\mathbf{d}|i\rangle_\text{A}.$$ Spontaneous emission is possible since ${}_\text{F}\langle f|\mathbf{E}|i\rangle_\text{F}$ can be nonzero for $|f\rangle_\text{F}$ having one photon, with a vacuum $|i\rangle_\text{F}$.

The important thing is that $E$ is sandwiched in between field states, while $d$ is sandwiched between atomic states. Keep close tabs on which Hilbert spaces each operator acts on!

  • $\begingroup$ Is E a constant if the electromagnetic field is also quantized ? $\endgroup$ – agemO Mar 18 '13 at 19:01
  • $\begingroup$ For example if we look at spontaneous emission in a field with no photon we cannot say E=0 -> spontaneous emission impossible, E has to be an operator. $\endgroup$ – agemO Mar 18 '13 at 19:17
  • $\begingroup$ Thanks this is exactly what I mean you are much more clear than I was : $⟨f|Hi|i⟩=F⟨f|E|i⟩F⋅A⟨f|d|i⟩A$ : often we only use $A⟨f|d|i⟩A=0$ to say that a transition is not possible, but is it possible to have forbidden transitions due to $F⟨f|E|i⟩F=0$ ? That is exactly my point $\endgroup$ – agemO Mar 19 '13 at 0:20
  • $\begingroup$ Given $|i\rangle_\text{F}$, there are always final states inaccessible through single-photon radiation for which ${}_\text{F}\langle f|E|i\rangle_\text{F}=0$. However, that only restricts the final state, but any initial state can always receive extra photons so some transition will always be allowed. $\endgroup$ – Emilio Pisanty Mar 19 '13 at 1:29

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