I've read, from several sources, that in internal conversion -- an excited electron transferring its energy to another electron which is then emitted -- no intermediate gamma radiation is produced.

How can we know that? Given that the distance it would travel is on the Å scale, is there any way to detect (the absence of) such a photon?


3 Answers 3


Interesting question.

At least in some cases the expected rate for each process (with and without a intermediate photon) will be calculable (in some approximation as there are nuclear structure functions involved) and these calculations can be compared with experiment.

But you can see right away that if both processes are allowed the no-photon case must dominate, because the with-photon case has two additional QED verticies at leading order and the rate is therefore suppressed by $\alpha^2 \approx 1/(137)^2$.


All interactions are quantized, so a photon is transferred.

The difference is whether or not this is a virtual photon.

The way we know no gamma ray is released is because we can do out the maths for the interation, and there is no need for an on-shell photon for it to go ahead.


The reason they say this is because non-relativistic atomic systems are not relativistic QED, and it is no good to use QED methods which are best for electron-positron creation energies.

In the non-relativistic limit, the QED Lagrangian is approximated (in a way that breaks renormalization) by a nonrelativistic atom coupled to a relativistic field. You don't calculate field effects to higher than first order, or you get the renormalization trouble. In particular, you don't calculate the Coulomb force as due to intermediate particle exchange,

The field is always in Dirac gauge, where the two physical photon polarizations appear in addition to the instantaneous coulomb force. The photon creation events are first order perturbations, the coulomb force is a potential interaction.

The Feynman diagram for an internal conversion is then just potential scattering, it comes from a term in the action:

$$\int \psi^\dagger(x)\psi^\dagger(y) V(x-y) \psi(x)\psi(y)$$

Where $\psi$ is the nonrelativistic Schrodinger electron field, and

$$ V(x-y)={1\over |x-y|}$$

is the Coulomb force between electrons. The internal conversion just kicks one electron while dropping a second electron to the ground state, and this process is pure electrostatic, and does not require a photon (meaning a physical polarization photon).

In Feynman QED, the nonphysical polarization photons are responsible for the Coulomb force, and this is a nicer point of view for relativity. It is not the nicest point of view for this stuff, so people stick with the Dirac description.

  • $\begingroup$ Fine from the calculational part. But if one went into the quark level diagrams, is it not true that virtual photons and Ws are exchanged? Have a look at slide 7 in this presentation google.com/url?q=http://www.physics.hku.hk/~phys3321/… .The key being virtual, that no energy is taken away by an other on shell particle. $\endgroup$
    – anna v
    Commented Apr 13, 2012 at 8:45
  • $\begingroup$ @annav: I don't know what it means for something to be "really" this or "really" that if there are two ways to calculate that agree. In this case, it doesn't make any difference if you want to do it with virtual photons, and for non-relativistic physics, it is more convenient to use instantaneous Coulomb. I don't distinguish between two experimentally equivalent possibilities, and the Dirac method is why they say internal conversion involves no photon--- it is a pure V effect. $\endgroup$
    – Ron Maimon
    Commented Apr 13, 2012 at 16:27
  • $\begingroup$ Note that there's also the (gamma) photoelectric effect, where the nucleus emits a gamma and the gamma then kicks out a high energy photon. $\endgroup$ Commented Jun 25, 2014 at 3:11

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