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According to the rules of qft there are virtual photons in the vacuüm. But how can this be if for the production of photons you need an electric charge?

Qmechanic
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According to the rules of qft there are virtual photons in the vacuüm.

No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way,

  • The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing is wrong: the vacuum doesn't change in time.

  • The vacuum at the point $\boldsymbol x$ is exactly the same vacuum at the point $\boldsymbol x+\Delta\boldsymbol x$ for any $\Delta\boldsymbol x$. This means that the picture of pairs that are created at some point, travel for some time, and annihilate at some other point is wrong: the vacuum is the same at every point in space.

If there were any photons in $|\Omega\rangle$, we would necessarily have $H|\Omega\rangle>0$, which is false, by definition. There are no photons, nor any other kind of particle, in the vacuum of any QFT.

Note that when non-experts allude to virtual particles, they imagine little blobs of whatever popping in and out of existence. There is absolutely nothing in the mathematics of QFT that suggest that this might be possible. When experts speak of virtual particles, they are not referring to particles at all. They use the word "particle" but they know what they are talking about. A virtual particle is just a contraction of fields that depend on variables over which you integrate, when using Dyson's perturbative series for the S-matrix. That's it. If you want to use the concept of "virtual particle" with any other meaning than that, well, you are using it wrong.

But how can this be if for the production of photons you need an electric charge?

This is not actually true, because for example a $Z$ boson can decay into three photons, $Z\to\gamma\gamma\gamma$, and $Z$ has no charge.

AccidentalFourierTransform
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