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?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Create virtual charges which then create virtual photons? PS - All this virtual particles business is quite bogus. These are introduced to have interpretations of individual Feynman diagrams which are actually not physical at all. $\endgroup$– PraharCommented May 2, 2016 at 19:04
-
3$\begingroup$ The definitive answer to this has yet to be written. It is something I am thinking about, but I have yet to find a way to do it justice. For now just note that virtual particles of any sort do not exist - they are a mathematical technique used in doing QFT calculations. The vacuum is not fluctuating in the sense that it is not time dependent. $\endgroup$– John RennieCommented May 2, 2016 at 19:12
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
10
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.
answered May 2, 2016 at 19:34
AccidentalFourierTransformAccidentalFourierTransform
55.3k20 gold badges140 silver badges266 bronze badges
-
$\begingroup$ Maybe the Z decay suggests the Z is a combination of electrically charged particles, wich net charge 0. $\endgroup$ Commented May 2, 2016 at 20:59
-
$\begingroup$ @descheleschilder in the Standard Model, the $Z$ bosons are fundamental particles. This means that, as far as we know, the $Z$'s are not combinations of smaller particles. $\endgroup$ Commented May 2, 2016 at 21:01
-
$\begingroup$ 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)? $\endgroup$ Commented May 2, 2016 at 21:06
-
$\begingroup$ @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 the vacuum is 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. $\endgroup$ Commented May 2, 2016 at 21:08
-
3$\begingroup$ @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! $\endgroup$ Commented May 3, 2016 at 18:24