Understand Virtual Particles in Quantum Gravity

[Hans de Vries]

Diagrammatica (a pun on the older CERN preprint 'Diagrammar'?) is hard to read.

I really love his popular (non-technical) book from 2003:

Facts and Mysteries in Elementary Particle Physics
https://www.amazon.com/dp/981238149X/?tag=pfamazon01-20

This is a "should have" book.


Regards, Hans.

 

[nrqed]

Do you claim it is always preserved? I do not see it; should we put down the equations, at the risk of killing the thread? It is a hell of dirac delta functions.

Maybe I am misinterpreting the issue (in which case I apologize!) but four-momentum is indeed always conserved at all vertices in Feynman diagrams. Always. This forces internal lines to be off-shell.

But I am probably missing the point of the discussion, in which case, again, I apologize!

Regards

Patrick

 

[nrqed]

IHans, about this explanation, could you be more concrete for instance in the case of electroweak interaction by exchange of a W or Z0 particle? My understanding is that two electrons, or two neutrinos, can interact by exchange of a Z0 even if their kinetic energy is a lot less than the mass of the Z0.

Yes they can. There is no limit on their kinetic energies in order to exchange a Z_0 (or anything else that they couple to). It only means that if the Z_0 exchanged will be very off-shell.

Patrick

 

[arivero]

A way to sneak in arbitrary large virtual masses into a vertex is to
have a very high mass virtual particle going in and out (the
virtual particle and its antiparticle) . They cancel each other out for
momentum conservation. Two of those become a loop then of course.

Yep, I see the mechanism. You start on shell having one E, p with E^2-p^2=m^2, then it divides to (E_a,p_a), (E_b,p_b) and because of the dirac deltas in the vertex you can tell that its sum equals the initial (E,p). Then your onshell tells that (E_a+E_b)^2-(p_a+p_b)^2=m^2. and as the quantities E_a^2-p_a^2 and E_b^2-p_b^2 do not need to coincide with the respective masses m_a, m_b it is said that these quantities, or the corresponding particles, are off-shell or "virtual".

My first little point is that the diagram is built with the dirac deltas and complete freedom to violate Energy-momentun conservation, then you integrate them out in your very first step in order to reduce the number of integrals you need to calculate. I could devise for instance a regularisation of these deltas, do the calculation (ahem) of the integrals, and finally to remove the regularisation so that energy-momentum preservation is imposed at the very end of the process.

The other point is if even after imposing the energy preservation we can claim that the off-shell particles are related to Heisenberg uncertainty. This is, if the probability of propagating a particle of mass m_a and offshell 4-momentum (E_a,p_a) is related via uncertainty to a "most probable" interval (\Delta t_a, \Delta x_a). I think it is so because for an on shell particle such interval is infinite (you can go as far as you wish for so much time as you wish).

 

[Hans de Vries]

Another possibility, within QFT, is along nrqed's remark that the Z or W is very
off-shell (E<<m). The difference should be due to the Weak interaction
then. This is where things become indeed more complicated to write down
mathematically.

For the Beta decay you were probably thinking about we get according
to "Diagrammatica, Apendix E" :

The vertex function for u --> W-, d Veltman gives:

ig\frac{1}{2\sqrt{2}} \gamma^\mu(1+\gamma^5)V_{ud}

So, there's the chiralty selection and the the Vud which denotes the factor
from the Cabibbo-Kobayashi-Maskawa matrix. The propagator for W is
given by:

\frac{\delta_{\mu\nu}}{p^2+M^2-i\epsilon}

So, well, this still hides the complexity. But the interaction term should
be the product of the u-quark spinor plane wave with the W-plane wave
where W- is:

W^-\ =\ \frac{1}{\sqrt{2}}\left(A^\mu_1-iA^\mu_2 \right)

(Weinberg 21.3.13, Volume 2) Where A is from the Yang Mills field. The
interaction can be found in the Yang Mills equivalent of the normal
covariant derivative:

\def\pds{\kern+0.1em /\kern-0.55em \partial}<br /> \def\lts#1{\kern+0.1em /\kern-0.65em #1}<br /> \lts{D}_\mu \ \equiv\ \pds_\mu - ie\kern+0.25em /\kern-0.75em A_\mu

Which in the Electroweak case becomes:

\def\pds{\kern+0.1em /\kern-0.55em \partial}<br /> \def\lts#1{\kern+0.1em /\kern-0.65em #1}<br /> \left( \pds_\mu - i\kern+0.25em /\kern-0.75em {\vec A}_\mu \cdot {\vec t}_L\right) u

Which can be read from the YM Lagrangian (Weinberg 21.3.11) where
t is the isospin. The real task is now to establish if this interaction can
be enough to mostly cancel the rest-mass energy...


Regards, Hans

 

[Careful]

The other point is if even after imposing the energy preservation we can claim that the off-shell particles are related to Heisenberg uncertainty. This is, if the probability of propagating a particle of mass m_a and offshell 4-momentum (E_a,p_a) is related via uncertainty to a "most probable" interval (\Delta t_a, \Delta x_a). I think it is so because for an on shell particle such interval is infinite (you can go as far as you wish for so much time as you wish).

I think you are asking here for something like a mean lifetime or path length of an off shell particle (in scattering processes). I don't know what it is worth but I found a springer paper on the web http://www.springerlink.com/content/m27154006404kl8j/
written by two russians in 2004. However, I would never speak about this in the context of Heisenberg uncertainty since there is no self adjoint operator for them (it is not an observable quantity). It is just so that in the path integral, these intermediate lines have a certain lifetime (with a certain weight factor) over which is integrated.

 

[arivero]

Which in the Electoweak case becomes:

\def\pds{\kern+0.1em /\kern-0.55em \partial}<br /> \def\lts#1{\kern+0.1em /\kern-0.65em #1}<br /> \left( \pds_\mu - i\kern+0.25em /\kern-0.75em {\vec A}_\mu \cdot {\vec t}_L\right) u

Which can be read from the YM Lagrangian (Weinberg 21.3.11) where
t is the isospin. The real task is now to establish if this interaction can
be enough to mostly cancel the rest-mass energy...

But we are speaking of the same thing! Problem is, you call A_\mu "interaction", I call it "Potential", or "potential energy". Your point is that even if a fermion is off-shell against its free equation, it is on shell for its interacting equation for some potential A_\mu. Now, this potential comes from nowhere, and it has a characteristic scale in units of Energy (it is a potential!). The inverse of this scale should be a characteristic time for off-shell propagation.

Now it is true that the general (sum of) non abelian gauge becomes more complicated than the naive electrostatic A_\mu=(V,0,0,0) but the mechanisms are the same, or should be.

 

[arivero]

I would never speak about this in the context of Heisenberg uncertainty since there is no self adjoint operator for them (it is not an observable quantity).

Well this is a typical objection, isn't it? Quantum mechanics (0+1 QFT) has no time operator, and in the same way 3+1 quantum field theory has not time nor position operator at all, I guess.

 

[Careful]

Well this is a typical objection, isn't it? Quantum mechanics (0+1 QFT) has no time operator, and in the same way 3+1 quantum field theory has not time nor position operator at all, I guess.

I don't see much problem with space operators, but for time yes. Btw, typical objections are usually there for a meaningful purpose !

 

[arivero]

I don't see much problem with space operators,:

Ah, but you should! In QFT space and time have the same issues. Consider 1+1 QFT. Or consider 3+1 as usual: do you see a space operator there?

 

[Careful]

Ah, but you should! In QFT space and time have the same issues. Consider 1+1 QFT. Or consider 3+1 as usual: do you see a space operator there?

I could think about constructing a suitable Hamiltonian theory (although, granted, it doesn't exist yet for interacting theories). In relativistic quantum mechanics, we dispose of the Newton Wigner operator (which becomes singular in the massless case). Actually, I think there is nothing wrong with introducing a classical tetrad, so that I can define eg. in an invariant way an electric field strength operator for the electromagnetic theory of which I can study localization properties. Anyway, I never gave this much thought so tell me if I miss something.


Careful

 

[CarlB]

"Diagrammatica" arrived in the mail and has not yet taught me anything I didn't already know. Kind of surprising, actually. I was hoping for a more revolutionary interpretation of Feynman diagrams.