
#1
Jul2111, 02:05 AM

P: 520

Please teach me this:
I do not know whether interaction transfer boson particle(virtue particle) is mass onshell or mass offshell(meaning square(4p)=square(m) or= square(m)) or both case happening? Thank you very much in advance. 



#2
Jul2111, 06:03 AM

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hi ndung200790!
anything with creation and annihilation operators must be onshell so in the position representation, the virtual particles are onshell (and 3momentum is conserved) in the momentum representation, the virtual particles are offshell (and 4momentum is conserved) 



#3
Jul2111, 06:52 AM

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I don't understand this remark. Of course an creation operator wrt. the energymomentum eigenstates adds an onshell (asymptotically) free particle with the given momentum (and spin or helicity) to the state the creation operator adds on.
In terms of Feynman diagrams, which are nothing else than a special notation for the calculation of Smatrix elements in perturbation theory, these initial or finalstate asymptotically free particles are represented by the external legs. The "virtual particles" are represented by inner lines, connecting to vertices of the diagram. These lines stand for particle propagators. At each vertex, energymomentum conservation holds, and usually the four momenta of internal lines are offshell. It's in fact a problem in naive perturbation theory, when the kinematics of a process is such that an internal line's four momentum becomes onshell since there the propagator has a pole. The reason are usually infrared of collinear divergences when massless particles are involved. These divergences have to be remedied by an appropriate resummation of many diagrams (e.g., by the BlochNordsieck argument in QED). 



#4
Jul2111, 07:00 AM

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The virtue particle is mass onshell or offshell?
hi vanhees71!
apart from that, aren't you agreeing with me? 



#5
Jul2111, 07:49 AM

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I just said that I don't understand what you mean. What do you mean by "on shell" in the position representation?




#6
Jul2211, 04:23 AM

P: 520

Thanks all of you very much.Knowing that in momentum representation the virtue particle is mass offshell is important(it seem to me).




#7
Jul2311, 06:55 AM

P: 520

The position and 3momentum representation theory in normal Quantum Mechanics is already clear.But how about the position and momentum representation in Quantum Field Theory.Please give me a favour to explaint more detail.




#8
Jul2311, 09:21 PM

P: 520

At the moment, I think that the position and momentum representation relate with each other by Fourier transformation(Fourier integral transformation).Is that correct?




#9
Jul2411, 04:57 AM

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hi ndung200790!
the position and momentum representations are both Fourier integrals the position representation (of the propagator) is a Fourier integral over d^{3}p, see (6.2.1) at p274 the momentum representation is obtained by the trick of extending the physical variable p to a mathematical variable q, where q = p and q^{0} is mathematically convenient the momentum representation (of the propagator) is a Fourier integral over d^{4}q, see (6.2.18) at p277 



#10
Jul2511, 04:44 AM

P: 520

Considering propagator,we use Fourier integral over 4momentum.Please teach me why we consider the energy as mathematical convenient,because the energy is determined by 3momentum then it seems that 4momentum q is mass onshell but not offshell.




#11
Jul2511, 05:27 AM

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q^{0} is defined as p^{0} + s, where s is a new variable which can take any value (and where p^{0} is the energy determined by the 3momentum p) so q = (q^{0} , q) = (p^{0} + s , p), which is offshell see p276 of Weinberg "Quantum Theory of Fields", just below (6.2.15), viewable online (it says "Volume 2", bit it's actually Volume 1 ) at http://books.google.co.uk/books?id=3...ed=0CC4QuwUwAA (sorry, the references in my previous post were to the same book … i confused this with another thread, and thought you'd already mentioned you were reading Weinberg ) btw, the mathematical convenience is that integrating over d^{3}p is not Lorentz invariant (nothing that's 3D can be Lorentz invariant), so we invent a new 4D variable q that is Lorentz invariant, and we integrate over d^{4}q 



#12
Jul2511, 07:43 AM

P: 520

Then 4q vector may be spacelike vector,may be timelike vector?




#13
Jul2611, 02:22 AM

P: 520

In QFT book of Schroder&Peskin,chapter 12.4 ''Renormalization of Local Operators'' I do not understand why they can neglect the square of 4q vector comperision with square of massive boson,because I think that the square of 4q is arbitrary value(4q is mass offshell but arbitrary).Please be pleasure to teach me this.




#14
Jul2611, 04:38 PM

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sorry, i don't have acces to schroeder and peskin
can you find the equivalent place in weinberg's book ?(and yes, q can be spacelike null or timelike) 



#15
Jul2611, 05:00 PM

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