The virtue particle is mass on-shell or off-shell?


by ndung200790
Tags: virtual particles
ndung200790
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#1
Jul21-11, 02:05 AM
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Please teach me this:
I do not know whether interaction transfer boson particle(virtue particle) is mass on-shell or mass off-shell(meaning square(4-p)=square(m) or= -square(m)) or both case happening?
Thank you very much in advance.
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tiny-tim
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Jul21-11, 06:03 AM
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hi ndung200790!

anything with creation and annihilation operators must be on-shell

so in the position representation, the virtual particles are on-shell (and 3-momentum is conserved)

in the momentum representation, the virtual particles are off-shell (and 4-momentum is conserved)
vanhees71
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Jul21-11, 06:52 AM
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I don't understand this remark. Of course an creation operator wrt. the energy-momentum 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 S-matrix elements in perturbation theory, these initial- or final-state 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, energy-momentum conservation holds, and usually the four momenta of internal lines are off-shell. 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 on-shell 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 Bloch-Nordsieck argument in QED).

tiny-tim
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Jul21-11, 07:00 AM
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The virtue particle is mass on-shell or off-shell?


hi vanhees71!
Quote Quote by vanhees71 View Post
At each vertex, energy-momentum conservation holds
but not in the position representation (coordinate-space representation), only in the momentum representation?
apart from that, aren't you agreeing with me?
vanhees71
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Jul21-11, 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?
ndung200790
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#6
Jul22-11, 04:23 AM
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Thanks all of you very much.Knowing that in momentum representation the virtue particle is mass off-shell is important(it seem to me).
ndung200790
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#7
Jul23-11, 06:55 AM
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The position and 3-momentum 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.
ndung200790
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Jul23-11, 09:21 PM
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At the moment, I think that the position and momentum representation relate with each other by Fourier transformation(Fourier integral transformation).Is that correct?
tiny-tim
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Jul24-11, 04:57 AM
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hi ndung200790!
Quote Quote by ndung200790 View Post
At the moment, I think that the position and momentum representation relate with each other by Fourier transformation(Fourier integral transformation).Is that correct?
no

the position and momentum representations are both Fourier integrals

the position representation (of the propagator) is a Fourier integral over d3p, 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 q0 is mathematically convenient

the momentum representation (of the propagator) is a Fourier integral over d4q, see (6.2.18) at p277
ndung200790
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#10
Jul25-11, 04:44 AM
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Considering propagator,we use Fourier integral over 4-momentum.Please teach me why we consider the energy as mathematical convenient,because the energy is determined by 3-momentum then it seems that 4-momentum q is mass on-shell but not off-shell.
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Jul25-11, 05:27 AM
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Quote Quote by ndung200790 View Post
Considering propagator,we use Fourier integral over 4-momentum.Please teach me why we consider the energy as mathematical convenient,because the energy is determined by 3-momentum …
no, the "energy" q0 is not determined by the 3-momentum q (= p)

q0 is defined as p0 + s, where s is a new variable which can take any value

(and where p0 is the energy determined by the 3-momentum p)

so q = (q0 , q) = (p0 + s , p), which is off-shell

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 d3p 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 d4q
ndung200790
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Jul25-11, 07:43 AM
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Then 4-q vector may be space-like vector,may be time-like vector?
ndung200790
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Jul26-11, 02:22 AM
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In QFT book of Schroder&Peskin,chapter 12.4 ''Renormalization of Local Operators'' I do not understand why they can neglect the square of 4-q vector comperision with square of massive boson,because I think that the square of 4-q is arbitrary value(4-q is mass off-shell but arbitrary).Please be pleasure to teach me this.
tiny-tim
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#14
Jul26-11, 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 space-like null or time-like)
samalkhaiat
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Jul26-11, 05:00 PM
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Quote Quote by tiny-tim View Post
hi ndung200790!

anything with creation and annihilation operators must be on-shell

so in the position representation, the virtual particles are on-shell (and 3-momentum is conserved)
How did reach this conclusion? why should on-shell/off-shell "status" of the fields depends on the representations?
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Jul26-11, 05:13 PM
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Quote Quote by samalkhaiat View Post
How did reach this conclusion?
creation and annihilation operators are only defined for on-shell particles
why should on-shell/off-shell "status" of the fields depends on the representations?
"off-shell" isn't a property of the field, it's a property of the variable, q


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