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Detection of a non-interacting particle

  1. Nov 19, 2009 #1
    In another thread, which I do not want to derail, I claimed that a free, non-interacting particle is not physical and that it cannot be detected or observed. It was pointed out to me that apparently I'm the only one on here who doubts about the possibility of detecting a non-interacting particle's energy, mass or spin.

    Now, I stand on my ground and insist on the non detectability of free particles. What I'd like someone to explain me is how to detect such a prticle.

    Please, before answering consider that you cannot see see such a particle (it does not interact with light), cannot bend its trajectory (no interaction with the EM field in general), you cannot have a bound state, you cannot pile it up and make a ball out of such matter, and so on and so on.

    Am I really the only one who knows that you cannot detect a non interacting particle?
     
  2. jcsd
  3. Nov 19, 2009 #2
    I think you have to precise and distinguish a never-interacting particle (i.e., worse than a neutrino) and a free (possibly charged) particle described with a free solution (plane wave). What do you exactly imply?
     
  4. Nov 19, 2009 #3
    A free particle, as an object described by a plane wave, i.e. a solution of the free field equations (pick your favourite field, scalar, spinor or vector) is what you mention as being "worse than a neutrino". The two concepts coincide.
     
  5. Nov 19, 2009 #4

    DarMM

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    A plane wave is just a form the wave function of the particle can take and doesn't really have anything to do with being a free particle in the QFT sense. I can make a highly interacting particle such as a proton have a roughly plane-wave wavefunction by sending it off on its own in a vacuum, where it will be in free motion.
     
  6. Nov 19, 2009 #5

    Fredrik

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    A non-interacting particle would be undetectable by any reasonable definition of "non-interacting".

    This started with a discussion about the approach taken by Weinberg in his QFT book, when I said that the particles that you get almost immediately are non-interacting, and Bob said it doesn't mean that they're undetectable. The next thing Weinberg does in the book is to add an interaction term to the Hamiltonian so that the irreducible representations are no longer independent of each other. This makes the particles "interacting", but it appears to be the same Hilbert space as before, so maybe Bob just means that the interacting particles in this theory are the same particles as the ones in the free-particle theory. It's just the Hamiltonian that's different, not the particles.
     
  7. Nov 19, 2009 #6
    No. An electron decoupled from the quantized EMF is simply "incapable" of emitting photons if accelerated. It does not mean the electron is not capable of interaction with a detector (atoms). That is why I insist in distinguishing a never-interacting particle (non-observable whatever properties you assign to it) and a free particle - solution of decoupled QED equations. The latter is observable, of course. The non-relativistic (and relativistic) calculations of particle penetration in materials in QM use the usual Coulomb potential of interaction with atoms of medium. Similarly, free photons are also observable. They are just external (known) fields in the material charge equations of motion, they can excite and ionise atoms, although they cannot create an electron-positron pair.
     
    Last edited: Nov 19, 2009
  8. Nov 19, 2009 #7
    A free particle in a QFT sense is also observable - it is the zeroth approximation for calculations, it makes sense as an asymptotic state at t = ± ∞.

    Free particle wave functions are involved in a free particle field in QFT: the corresponding c/a operators create/annihilate exactly these states.

    Instead of considering a "highly interacting proton" (what does that mean? A strongly interacting proton or what?), let us consider a free atom. Its total wave function is a product of a plane wave describing the center of mass free motion and an atomic wave function depending on relative (internal) variables. Let us note that the corresponding equations are decoupled, yet both describe an interacting system – an atom. (What I propose in QED is the same thing.)

    Now, we can construct two non-overlapping wave-packets corresponding to two distant atoms in the initial state and make them scatter off each other. The Coulomb atom-atomic potential contribution is zero in the initial state because of non-overlapping total wave functions. With time the atoms approach each other and scatter, the final state being a superposition of excites states of our again distant and free atoms. This is an exemplary model of interacting systems free in the In- and Out-states and free of conceptual and mathematical difficulties. Everything here is observable and meaningful. A rigorous physical QFT can be constructed in a similar way. What we have seen up to now are just sorry attempts to patch a failed self-action ansatz. In particular, they say that free particles are not observable. What physics becomes in hands of some?!
     
    Last edited: Nov 19, 2009
  9. Nov 19, 2009 #8
    Fermi postulated the neutrino existance long before it was detected, because of the missing energy and electron energy spectrum in beta decay. So we could postulate another undetectable particle in a similar way. However, if the particle was (were) truly non-interacting, no interaction could create it either.
    Bob S
     
  10. Nov 19, 2009 #9
    Bravo, Bob_S!!! You hit the point! Nice argument!

    No need to assign to a particle some properties (spin, momentum, mass, etc.) if it declared to be non-observable (never-interacting). The zeroth-order approximations in QFT possess certain properties just because they are observable.
     
    Last edited: Nov 19, 2009
  11. Nov 19, 2009 #10
    Also, if the particle has some energy, it will have gravitational interactions.
     
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