2016-05-02 18:55:51 8 Comments

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?

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## 1 comments

## @AccidentalFourierTransform 2016-05-02 19:34:12

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.

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.

## @descheleschilder 2016-05-02 20:59:26

Maybe the Z decay suggests the Z is a combination of electrically charged particles, wich net charge 0.

## @AccidentalFourierTransform 2016-05-02 21:01:39

@descheleschilder in the Standard Model, the $Z$ bosons are

fundamental particles. This means that, as far as we know, the $Z$'s arenotcombinations of smaller particles.## @descheleschilder 2016-05-02 21:06:13

In QFT, E and p are independent variables. In my view this makes things pretty dynamic. And what about the uncertainty relations that go with t (E) and x(p)?

## @AccidentalFourierTransform 2016-05-02 21:08:23

@descheleschilder there are no uncertainty relations in QFT. See e.g., physics.stackexchange.com/questions/55860/… and physics.stackexchange.com/questions/191042/… I don't know what you mena by "

this makes things pretty dynamic". I said that thevacuumis not dynamic, not that everything is not dynamic. All the kets are dynamic except for $|\Omega\rangle$, because $E_\Omega=\boldsymbol p_\Omega=0$ by definition.## @descheleschilder 2016-05-02 21:09:53

@AFT Maybe the standard model is wrong and can be replaced by a preon theory like the rishon model of Haïm Harari.

## @AccidentalFourierTransform 2016-05-02 21:10:32

@descheleschilder well maybe, but that was not your original question. I answered your question in the framework of accepted theories.

## @descheleschilder 2016-05-03 00:11:39

I think you rely too much on mathematics. How can a photon bring into existence a pair of particles (like an electron and positron for example), all on mass shell (E and p have a well defined relation, unlike the virtual particles)? I can´t see how a creation operator (something unphysical) does the job. And how to explain the Casimir effect? By pretty dynamic I meant the vacuüm.

## @AccidentalFourierTransform 2016-05-03 18:24:55

@descheleschilder I changed the words "not dynamic" for "static". I hope it makes more sense to you now. But I truly don't know how to help you any further if you want me not to rely on mathematics. Mathematics is the language of physics: asking me not to rely on them is asking me to shut up!