Is there any case reported that seems to resemble the following: there is a particle and at some moment, the particle seems to break down into two or more particles that are all identical to the original particle?

  • additional information: I am not talking of antimatter/matter issues.
  • 2
    $\begingroup$ A related question physics.stackexchange.com/questions/13851 $\endgroup$ – Slaviks Mar 28 '12 at 10:58
  • $\begingroup$ Very loosely interpreted, isn't this how lasers work? One photon of a given frequency produces more photons of the same frequency. $\endgroup$ – Peter Shor Mar 28 '12 at 23:39
  • $\begingroup$ @PeterShor. No. The question is about decays, not interactions. Lazers are interacting photons and crystals/gas molecules and another story. $\endgroup$ – anna v Mar 29 '12 at 4:00
  • $\begingroup$ @annav is energy conserved in the case of lasers (interactions), then? As E=hf in quantum level, if a single photon can produce more photons of the same frequency, energy seems to be not conserved.... $\endgroup$ – user27515 Mar 29 '12 at 15:10
  • $\begingroup$ @user27515 In the laser phenomenon a single photon does not produce more photons of the same frequency in vacuum. The crystal/gas molecules are pumped up to a higher energy level by inputted energy. In lasing there is induced emission, i.e. a photon of the same frequency as the pumped up to a higher energy level molecules, induces a coherent relaxation to the ground state of the excited molecules and the laser beam is created.en.wikipedia.org/wiki/Laser . look at the "physics paragraph $\endgroup$ – anna v Mar 29 '12 at 16:00

In "normal" cases, no, this is not possible. You can easily understand why by considering this process in the center-of-mass frame (which is the rest frame of the original particle). In this frame, you would start with a single particle $X$ at rest, which has energy $m_Xc^2$, and wind up with 2 or more $X$, which will necessarily have an energy of at least $2m_Xc^2$. So energy conservation has to be violated by these sorts of reactions.

But consider a caveat: what do I mean by "normal"? Well, what we consider normal matter is made of massive particles. If you're looking at massless particles, on the other hand, the above argument doesn't apply because a massless particle doesn't have a rest frame. So you have to examine it from a lab frame (that is, any inertial frame). It should be easy to convince yourself that a massless particle of energy $E$ can decay into multiple instances of the same kind of particle with energies $\{E_1,\ldots,E_N\}$ such that $\sum_i E_i = E$ as long as all the products have momentum parallel to that of the original particle.

This type of process is called collinear branching and it actually does have to be taken into account when doing calculations in quantum field theory involving photons and gluons, the two known massless particles. Being "taken into account" is a little different from actually "occurring," though. The effect of this is not to say that photons and gluons actually split into multiples, but that the only meaningful things we can calculate are quantities which don't depend on whether the bosons are splitting or not. So the question of whether or not this decay into copies actually happens cannot be experimentally decided, according to QFT.

  • $\begingroup$ I don't understand this answer--- collinear branching can be a real process in virtual exchange. Why can't you say yes? $\endgroup$ – Ron Maimon Mar 29 '12 at 3:06
  • $\begingroup$ @RonMaimon Because of the word "virtual"? In virtual exchanges anything may go that conserves the quantum numbers but not energy/momentum. $\endgroup$ – anna v Mar 29 '12 at 4:03
  • $\begingroup$ @anna v: fair enough, ok, but 1 photon to 3 is only kinematically prohibited because of time dilation, not conservation of energy $\endgroup$ – Ron Maimon Mar 29 '12 at 4:05
  • $\begingroup$ @RonMaimon take a real photon: suppose it does split into two photons ( I think three because of spin polarisation, may be wrong) as it is running along. These products, if they have an angle between them will have an invariant mass, and one can go to the center of mass of that invariant mass. There, the original photon will have an unbalanced momentum, since it cannot be at rest. It is only collinearly, 0 angles, that this could happen to a real photon but I have never heard of an experiment ( change of frequency of light in monochromatic? ). There must be something else inhibiting this. $\endgroup$ – anna v Mar 29 '12 at 4:13
  • $\begingroup$ @anna v: THe time dilation forbids this--- nothing else (and yes, the third photon is to have opposite helicity to conserve angular momentum in the "decay"). It just is not allowed. I remember t'Hooft considered the idea that one photon are 3 collinear ones might be gravitationally identical in one of the holography papers in the 1980s, but this idea is not consistent with naive field theory amplitudes for 3 collinear photons vs. one photon (3 photons are not as easy for an atom to absorb if it is resonant with the 1). I don't think they are identical, it's just a weird kinematic thing. $\endgroup$ – Ron Maimon Mar 29 '12 at 4:48

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