# What particle is created when a B- or E-field operator acts on the vacuum

I have some conceptual trouble understanding what the different field operators in QED do. According to Wikipedia, the field operators are given by

$$\mathbf A(\mathbf{r} )=\sum _{\mathbf {k} ,\mu }{\sqrt {\frac {\hbar }{2\omega V\epsilon _{0}}}}\left\{\mathbf {e} ^{(\mu )}a^{(\mu )}(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {e} }}^{(\mu )}{a^{\dagger }}^{(\mu )}(\mathbf {k} )e^{-i\mathbf {k} \cdot \mathbf {r} }\right\}\\\mathbf {E} (\mathbf {r} )=i\sum _{\mathbf {k} ,\mu }{\sqrt {\frac {\hbar \omega }{2V\epsilon _{0}}}}\left\{\mathbf {e} ^{(\mu )}a^{(\mu )}(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }-{\bar {\mathbf {e} }}^{(\mu )}{a^{\dagger }}^{(\mu )}(\mathbf {k} )e^{-i\mathbf {k} \cdot \mathbf {r} }\right\}\\\mathbf {B} (\mathbf {r} )=i\sum _{\mathbf {k} ,\mu }{\sqrt {\frac {\hbar }{2\omega V\epsilon _{0}}}}\left\{\left(\mathbf {k} \times \mathbf {e} ^{(\mu )}\right)a^{(\mu )}(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }-\left(\mathbf {k} \times {\bar {\mathbf {e} }}^{(\mu )}\right){a^{\dagger }}^{(\mu )}(\mathbf {k} )e^{-i\mathbf {k} \cdot \mathbf {r} }\right\}$$

Now as I understand this, each of these operators can act on the vacuum and create a particle.

My first question is: is this particle in each case a photon, which differs, however, by polarization?

Second, I wonder if a classical B-field is then "made up" by a coherent superpsition of (virtual) photons with the "B-field-type" polarization?

I am interested in these conceptual questions in the context of the conversion of an axion into a photon within an external B-field. This is described by the term

$$-g_{a\gamma\gamma}\mathbf E\cdot\mathbf B\,a$$.

So can I actually understand this equation as an axion interacting with one of the virtual "B-field-type" polarized photons present due to an external B-field to give then an "E-field-type" polarized photon? (By the clumsy expressions "B/"E-field-type" I mean photons that would be created by the respective operators given above)

In summary, I think my confusion comes from the fact that in QED, one usually deals with $$A_\mu$$, so in my mind, photons were associated with $$A_\mu$$, like in the Feynman rule saying that $$\overbracket{A_\mu|\mathrm p\rangle}$$ gives an external photon line. But in the axion coupling term, there are actually the E and B fields, and hence the question in the title: what do these field operators actually create?

Any clarification is very much appreciated, thank you very much in advance!

• Hi, actually I am interested in both. So on the one hand, what would be the particles created by the non interacting E/B-field itself, despite these not showing up in the QED Lagrangian (directly). On the other hand, I am also interested what is the difference between the term $-g_{a\gamma\gamma}\mathbf{E\cdot B} a$ and for example the usual term $-e\bar\psi\gamma_\mu A^\mu \psi$ which couples to $A^\mu$? So in very naive terms, what does it mean for a field to couple to the derivative of a field? Sorry for this being a little clouded... – moinde12 Feb 21 at 8:18