2 questions - virtual particles and quarks?

[rasp]

1) In non-mathmatical terms, please correct my description of vitual particles...

a) virtual particles are real but too small to be observed directly. i.e. their dimensions are within Plank's values and subject to Heisenberg's uncertainty principles.
b) virtual particles are real mathematically in that their presense satisfies the theory but like individual quarks they do not exist individually and independently and cannot be observed.
c) virtual particles are real and their role can be deduced by their effect on other elementary particles.

2) How do those in the know explain the concept that the quark is considered an elementary particle which doesn't exist independently? It seems to me frought with contradiction, analogous to a number system in which the number 1 is not defined.

 

[ansgar]

non of them are correct

virtual particles are only a calculation artefact due to perturbation theory. if we could calculate scattering amplitudes etc without perturbation theory, we would not need them.

physics is independent on which way we choose to calculate things

 

[chrispb]

Electric fields are made up of virtual photons.

 

[humanino]

Electric fields are made up of virtual photons.

Sure, that's why Faraday and Maxwell could not understand electric fields.

 

[ansgar]

Electric fields are made up of virtual photons.

are they? how do you know that? from perturbation theory?

 

[tom.stoer]

virtual particles are only a calculation artefact due to perturbation theory. if we could calculate scattering amplitudes etc without perturbation theory, we would not need them.

Exactly. This becomes clear if one quantizes a theory non-perturbatively (unfortunately this is not always possible). The formalism is completely different and does not use virtual particles (free propagators) at all.

Btw.: mathematically a virtual particle is not a single particle but "a rule how to integrate over an infinite number of particles".

 

[rasp]

Can you clear up my confusion about the single quark?

that the quark is considered an elementary particle which doesn't exist independently? It seems to me frought with contradiction, analogous to a number system in which the number 1 is defined but doesn't exist.

thanks for the answers on virtual particles

 

[humanino]

There are many difficulties associated with "Well identified non-flying objects" as Dokgarbagezer once called them.

Now the way you describe it makes it difficult to answer specifically. I could argue that nothing really "exists independently". Until an interaction has occurred, the evolution operator (or S matrix) reduces to identity operator in a mixed basis.

Can you exhibit a specific contradiction you encounter with the concept of confined particle ?

 

[guerom00]

Electric fields are made up of virtual photons.

Hmmm…

Virtual particles are associated with fluctuating quantities. The well known example are virtual photons from the EM vacuum : the vacuum state of the electromagnetic field contains, as its name state, zero photon. 0 is an eigenvalue of the number operator (a^+ a) i.e. (a^+ a) |0> = 0 |0>. Problem : the vacuum state |0> is _not_ an eigenstate of the operator (a^+ a)^2. So, to recap, the mean value of the number of photon in the vacuum state is 0 <N>=0 _but_ the mean value of the square of the number of photon is not zero <N^2>≠0. That means its variance is not zero therefore this quantity fluctuates.
Long story short : it means that in the vacuum state |0>, the number of photon (as well as the electric field, the magnetic field, etc…) fluctuate around its mean value which is zero.
What does that mean : You have an EM vacuum. It contains _a mean value_ of zero photon. It means that from time to time, it will contains 1 or 2 or 3, etc… photons. Those photons arising from the mere fluctuations of the number of _real_ photons (which is zero in the vacuum state) are those virtual photons.
For all intented purposes, virtual particles are merely particles which exists for a limited time before dissapearing.

 

[ansgar]

Hmmm…

Virtual particles are associated with fluctuating quantities. The well known example are virtual photons from the EM vacuum : the vacuum state of the electromagnetic field contains, as its name state, zero photon. 0 is an eigenvalue of the number operator (a^+ a) i.e. (a^+ a) |0> = 0 |0>. Problem : the vacuum state |0> is _not_ an eigenstate of the operator (a^+ a)^2. So, to recap, the mean value of the number of photon in the vacuum state is 0 <N>=0 _but_ the mean value of the square of the number of photon is not zero <N^2>≠0. That means its variance is not zero therefore this quantity fluctuates.
Long story short : it means that in the vacuum state |0>, the number of photon (as well as the electric field, the magnetic field, etc…) fluctuate around its mean value which is zero.
What does that mean : You have an EM vacuum. It contains _a mean value_ of zero photon. It means that from time to time, it will contains 1 or 2 or 3, etc… photons. Those photons arising from the mere fluctuations of the number of _real_ photons (which is zero in the vacuum state) are those virtual photons.
For all intented purposes, virtual particles are merely particles which exists for a limited time before dissapearing.

according to perturbation theory yes

 

[guerom00]

Virtual particles have nothing to do with perturbation theory. As I've said, they merely are the manifestation of fluctuating eigenvalues of the particle number operator.

 

[ansgar]

Virtual particles have nothing to do with perturbation theory. As I've said, they merely are the manifestation of fluctuating eigenvalues of the particle number operator.

show please

 

[guerom00]

Show what ?
You are thinking of the virtual particles used in the Feynman diagrams I guess… Those are different things; those are indeed purely mathematical artifacts.

 

[ansgar]

Show what ?
You are thinking of the virtual particles used in the Feynman diagrams I guess… Those are different things; those are indeed purely mathematical artifacts.

can you show this fluctuations of the eigenvalues for the number operator?

 

[humanino]

can you show this fluctuations of the eigenvalues for the number operator?

Non-perturbatively of course...

 

[guerom00]

It does not commute with the Hamiltonian I guess.
This is a very general idea : each time you have an eigenvalue of whatever operator, you have to ask yourself whether this eigenvalue will fluctuate or not. One way is to see whether it commutes with the Hamiltonian or calculate its variance.

 

[ansgar]

It does not commute with the Hamiltonian I guess.
This is a very general idea : each time you have an eigenvalue of whatever operator, you have to ask yourself whether this eigenvalue will fluctuate or not. One way is to see whether it commutes with the Hamiltonian or calculate its variance.

you GUESS??

I can show you that the particle number operator indeed commutes with the hamiltonian, that is a basic exersisce in canonical quantization of fields...

 

[humanino]

Maybe I misunderstand, it seems to me I only know how to define the number operator perturbatively. Say on the lattice, I have a field configuration, if I want to define an occupation number, I need to decompose the field into appropriate modes, and I will claim I know the occupation of each mode once my decomposition (in some sense) approximates the configuration well enough. But the number operator is not necessary a priori on the lattice, neither is it defined non-perturbatively. For instance, I can get arbitrary high mode if I request arbitrary high precision.

 

[the_house]

the vacuum state of the electromagnetic field contains, as its name state, zero photon. 0 is an eigenvalue of the number operator (a^+ a) i.e. (a^+ a) |0> = 0 |0>.

If you define the vacuum like that, then the vacuum expectation value of both N and N^2 are trivially 0, are they not?

 

[chrispb]

In a free field theory, the true ground state of the theory, |Omega>, corresponds with |0>, the eigenstate of the number operator. Therefore, if there are 0 photons at one particular time, then later, there still won't be any photons. Obviously, this isn't true if you couple the photon to a source or something of the sort.

However, in an interacting field theory, like QED, |0> is no longer equal to |Omega>. In a weakly coupled/perturbative theory, |Omega> is some linear combination of eigenstates of the number operator. (This is still true in strongly coupled theories, but things get messier.) To put this a different way, we can decompose |Omega> in terms of any complete basis; why not choose the number basis? Well, we don't really learn too much from it, except for the fact that there simultaneously are and are not photons in the ground state, in the good old quantum sort of way :)

I know that electric fields are made up of virtual photons because I took a QFT course :P

 

[humanino]

I know that electric fields are made up of virtual photons because I took a QFT course :P

We can describe a classical field configuration using the perturbative technics of QFT, that's just not very efficient, and it is not fair to say that it reveals something fundamental about Nature. You are free to use GR to compute the trajectory of a non-relativistic mass, but remember : physics is all about the validity of clever approximations.

 

[the_house]

In a free field theory, the true ground state of the theory, |Omega>, corresponds with |0>, the eigenstate of the number operator. [...]However, in an interacting field theory, like QED, |0> is no longer equal to |Omega>.

That's correct, but I'm still a little hazy on how this shows that virtual particles are a real, nonperturbative phenomenon, rather than an artifact of a perturbative calculation.