# Virtual particles and the Uncertainity Principle

## Main Question or Discussion Point

Recently, I was reading about Hawking Radiation in A Brief History of Time. It says that at no point can all the fields be zero and so there's nothing like empty space(quantum fluctuation etc.). Now, the reason mentioned was that virtual(force-carrier) particles cannot have both a precise rate of change and a precise position(Uncertainty Principle).

So, my question is : This video (https://www.youtube.com/watch?v=bKldI-XGHIw) says that virtual particles don't follow normal physical laws. So, how can we say that they obey the uncertainty principle?

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ChrisVer
Gold Member
By not obeying physical laws, I think he meant that they violate the $E^2=p^2+m^2$ relation and only...

Recently, I was reading about Hawking Radiation in A Brief History of Time. It says that at no point can all the fields be zero and so there's nothing like empty space(quantum fluctuation etc.). Now, the reason mentioned was that virtual(force-carrier) particles cannot have both a precise rate of change and a precise position(Uncertainty Principle).

So, my question is : This video (https://www.youtube.com/watch?v=bKldI-XGHIw) says that virtual particles don't follow normal physical laws. So, how can we say that they obey the uncertainty principle?
Virtual particles are a manifestation of the uncertainty principle. They "borrow" their energy for a brief period of time.

They don't necessairily follow the normal physical laws, because they're not stable excitations in the same way that their real counterparts are, which is what the normal physical laws typically describe.

@ChrisVer Yes. He said that. So, I am asking that if they don't follow this law how can we say that they follow the uncertainty principle?

@craigi Borrow energy from where?

jtbell
Mentor
Virtual particles are a manifestation of the uncertainty principle. They "borrow" their energy for a brief period of time.
They don't "borrow" energy. Energy and momentum are always conserved at every vertex of a Feynman diagram.

Virtual particles don't have to have the same mass as the corresponding "real" particles. That is, $E^2 - (pc)^2 \ne (m_{real}c^2)$. Particle physicists often use the term "off the mass shell" for them, whereas "real" particles are "on the mass shell" ($m = m_{real}$).

@craigi Borrow energy from where?
It's a common interpretation, that is used to intuitively describe the role of the HUP in virtual particle creation. The idea is that the energy is "borrowed" from the vaccum, for a short period of time.

For a more formal interpretation of the mathematics, I'd follow JTBell's description, above.

ChrisVer
Gold Member
It's not that HUP describes the virtual particles... It's just an interpretation of explaining how they exist... In fact a virtual particle is not measured- it just corresponds to the momentum transfer between two interacting particles... That certain particle does not obey the law I stated above... That law : $E^{2}-p^{2}=m^{2}$ in fact describes a hyperplane, the 4 variables (1 for E and 3 from p) must satisfy that law and then you are on the mass shell (That's the reason of the word shell)... The virtual particles don't verify that (they are not on the mass shell, but they lie out). The HUP just helps people understand how it happens- since virtual particles live for small enough times they can have great uncertainty of energies...so in fact your propagator cannot have a definite momentum (and so the 4momentum^2 equation above could not really make sense, since it wants a definite energy E and a definite momentum p to make sense). That's also why if you check the wikipedia for the virtual particles, it states that "particles which live long enough are considered real" because then you can have a really well determined energy and momentum.
By that I wanted to explain that it makes no sense to question me if the energy-momentum relation above doesn't work,why does HUP work... In fact I guess the question should be how can HUP and that relation coexist :P (hahaha)...

In order to see that the on-shell condition is not sattisfied, you can have a look at $e^{-}e^{+} \rightarrow e^{-}e^{+}$
given by the diagram you can see below:
There suppose the left, so you have the upper electron coming with $p_{1}$ momentum, and going out with [/itex]p_{2}[/itex] and the ones of the right one coming in with $p_{3}$ and going out with $p_{4}$ after "absorbing the virtual photon".
working out the kinematics, you can see that your photon is not null as you'd expect, but it can be either space or timelike ($q^{2}<0 , >0$)