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It is common to read that the lifetime of a virtual particle is given by the uncertainty relation: $$\tau \sim \frac{\hbar}{E}$$ on the premise that the virtual particle 'borrows energy'. This statement is infact wrong (at least I think it is) since energy is conserved in Feynmann diagrams and thus no energy needs to be borrowed. Given this, how do we actually determine the lifetime of a virtual particle, and why is it not just the same as the real particle?

Quantum spaghettification
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  • Unlike real particles, virtual particles do not ''live'' in any sense, hence do not have a lifetime. See https://physics.stackexchange.com/a/377265/7924 and the link there. – Arnold Neumaier Dec 31 '17 at 12:58

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That relation you quote has noting to do with borrowing energy. It is just Heisenberg's uncertainty principle. However, in my humble opinion, it is best not to ascribe any "reality" to virtual particles. They are just pictorial representations of terms called propagators which appear when performing perturbation theory on a Quantum Field Theory. There is very little understanding to get along the route you took in your question.

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Virtual particles are defined only within the mathematical framework of Feynman diagram calculations for measurable quantities, as crossections and lifetimes.

feyndiag

This is the first order in an expansion to get the crossection of e-e- scattering.

The wavy line represents a mathematical term which is under an integral, it is called a photon because it has the quantum numbers of a photon but not the mass, the mass is off mass shell. Have a look at this lecture where the propagator representing mathematically the virtual line, has the mass of the named particle in the denominator, but the four vector describing within the integral the virtual particle is off mass shell.

It is all under an integration and not real. Real on mass shell particles are the incoming and out going dark lines. So there is no lifetime for individual virtual lines. Only for the total interaction a lifetime can be calculated.

anna v
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  • Hi Anna I asked this question a while back in which the answer suggests the propagator is replaced by the Breit-Wigner for an unstable particle. If these 'virtual particles' don't have lifetimes what is therefore the meaning of an 'unstable particle' in this context. – Quantum spaghettification Jun 02 '17 at 13:39
  • The real particles have lifetimes. Resonances have energy widths which can be translated to lifetimes using the time-energy uncertainty.( or the exact calculation, i.e. do all the integrations). The virtual lines in the feynman diagram vary over the integration limits , giving the breit wigner crossection widtha as the energy of the interaction is scanned. (see http://pdg.lbl.gov/2016/hadronic-xsections/rpp2016-sigma_R_ee_plots.pdf) – anna v Jun 02 '17 at 16:52
  • I am still really confused by this topic. I cannot see then why people keep saying that the lifetime of a virtual particle is $\Delta t=\hbar/\Delta E$ (where $\Delta E$ is the energy difference compared to the on-mass shell value). I have for example seen this as an explanation for Fermi's theory neutron decay (see here) please could you try and explain it? Thanks. – Quantum spaghettification Jun 06 '17 at 04:17
  • The fact that an uncertainty relation exists between energy and time does not mean that wherever in the mathematics of calculating crossections and widths there is an energy variable it has a meaning to substidute the HUP time for it. The meaning of the HUP exists only for the real incoming and outgoing . The mathematical constructs in the middel are just that, mathematical, not measurable. Have you read this https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/ ? – anna v Jun 06 '17 at 04:36
  • Could we not measure the distance between the decay vertex and the creation vertex to get the distance traveled by the virtual particle and use the momentum transfer to find a velocity - from which we could find a mean lifetime? – Quantum spaghettification Jun 06 '17 at 04:47
  • No, there is no "distance" measurable between the two vertices that make up the virtual line. It is all mathematics. – anna v Jun 06 '17 at 05:42
  • I can't see why people are use the relation $\Delta t=\hbar/\Delta E$ to explain why some reactions are point like then? Is this simply wrong or is there a reason behind it? As every, thanks. – Quantum spaghettification Jun 06 '17 at 09:14
  • I have not seen it used that way except in handwaving about Hawking radiation. Any connection with real measurable numbers comes with input and output real vertices. – anna v Jun 06 '17 at 09:18