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The science behind the conversion of particles to other particles

  1. Nov 2, 2011 #1
    It seems kind of weird to me that particles convert between each other.

    As I understand it, if a positron and an electron "collide", two gamma ray photons are created, and the positron and electron are gone.

    1st - What is meant by a collision? The positron and electron are each described completely by a wavefunction -- what mathematical condition must these wavefunctions meet to constitute a "collision"?

    The classical view would be two marbles that have overlapping trajectories and roll into each other. The probability of a collision is partly a function of the marbles' cross-sectional area. Obviously, when the "marbles" are wavefunctions, this analogy doesn't translate well at all.

    2nd - Does the "conversion process" take place within Planck time? If so, then we can't know whether the conversion occupies finite time or is instantaneous. However, if the conversion has a duration longer than Planck time, we should be able to model this process, right?

    3rd - How is it that rest mass just disappears? As a thought experiment, if we had a highly sensitive "test particle" near the collision site that measured the gravitational force vector due to the two particles' masses, at the time of the particles' collision, the gravity probe would not register any instantaneous change in the magnitude of the force vector since total energy/mass is conserved, right? But d[force magnitude reading]/dt would be different now since the energy/mass of the photons now moves at c, correct?

    (If this 3rd point isn't clear, I can type up some equations trying to explain what I mean).
  2. jcsd
  3. Nov 2, 2011 #2
    1: My understanding is that by "collision" we mean that the particles' wavefunctions overlap.

    2: Don't think of annhilation processes as being like chemical reactions with intermediate steps and definate time-scales. What's going on here is that we have two or more quantum fields overlapping, except they aren't quite isomorphic, so their "addition" isn't linear. It's a special kind of wave interference.

    3: Rest mass doesn't disappear - it shows up as part of the energy of the gamma photon produced by the annihilation. The relativistic energy-momentum tensor is always conserved.

    Does this answer your questions?
  4. Nov 2, 2011 #3
    Hey thanks for replying! It doesn't quite answer everything though...

    The domain of a free electron or positron's wavefunction is infinite, so in a sense, they are always overlapping. Which means they would constantly be undergoing a collision, which does not happen.

    Is there any particular reason I shouldn't think of it like that? A collision is a definitive event. Before the collision we have a discrete positron and a discrete electron. After the collision we have two discrete photons. The two types of particles have very different properties and equations that describe them. The only question is whether the timescale of this collision event is below Planck scale. My guess is "most likely" but I would prefer a professional source or experiment to back this up if somebody knows of one -- it seems likely the LHC could determine the bounds on this timeframe.

    Yes... I know this. In fact, it is implicit in my question's formulation. But that doesn't answer the question. I may have to clarify this one with some math if nobody else sees what I'm asking.
    Last edited: Nov 2, 2011
  5. Nov 2, 2011 #4
    1. The wave-function measures the probability of a particle being in a particular position or momentum eigenstate. A collision occurs classically when position overlaps and momenta are different. So we're looking for the probability that the position overlaps and the momenta are different. When the probability of finding a particle in "colliding states" is greater then 0, then in a sense the collision is taking place in all of those states at once and you get a mixture of probability amplitudes for states that are daughter states of the collisions. It's like any other quantum system: if the wave-functions overlap then there's a probability that a collision will occur there. I guess I should have been more explicit.
  6. Nov 2, 2011 #5
    2. What we are talking about here are non-linear interactions between different quantum fields. It's helpful to think of these fields as being like quantum mechanical wavefunctions. Imagine you have two wave crests colliding. What do you get? You get the sum of the two waves, so when they move past each other you get what you started with, two waves. What I'm saying is that when these processes occur what's effectively happening is that instead of adding, the two waves might interact with each other by some other binary operation. So you come away with something other than two waves. Sometimes 1+1=0. How long does it take for this to occur? As long as it takes for the waves to pass through each other.

    Why aren't there intermediate states? We talk about intermediate Feynman diagrams which we can resolve into individual processes but ultimately all these intermediate processes are is terms in a perturbation series - mere polynomial approximations to some even more nonlinear function. The higher up in the series you get, the more nonlinear the interaction is. x^4+x^4 does not equal (2x)^4, but it is a better approximation to (2x)^4 then x^6+x^6.
  7. Nov 2, 2011 #6
    3. Sorry, I hadn't read through all of your question. I don't know. It's an interesting question. I think you'd need a quantum theory of gravity to predict what would happen.
  8. Nov 2, 2011 #7
    The wave function measures nothing. A wave function completely specifies the state of a quantum-mechanical system. This function has the property that

    [tex]\Psi^{*}(\mathbf{r}_1, \mathbf{r}_2, t)\Psi(\mathbf{r}_1, \mathbf{r}_2, t)dx_1 dy_1 dz_1 dx_2 dy_2 dz_2[/tex]

    is the probability that particle 1 lies in the volume element [itex]dx_1\ dy_1\ dz_1[/itex] located at [itex]\mathbf{r}_1[/itex] and particle 2 lies in the volume element [itex]dx_2\ dy_2\ dz_2[/itex] located at [itex]\mathbf{r}_2[/itex] at time t. Do you agree?

    The wave function describing a positron-electron system is:

    [tex]\Psi(\mathbf{r}_p, \mathbf{r}_e, t)[/tex]

    The probability that both the positron and electron are measured to lie within region [itex]\mathbf{R}[/itex] is:

    [tex]\int_{\mathbf{R}} d^3 \mathbf{r}_p \int_{\mathbf{R}} d^3 \mathbf{r}_e \Psi^{*}(\mathbf{r}_p, \mathbf{r}_e, t) \Psi(\mathbf{r}_p, \mathbf{r}_e, t)[/tex]

    Denoting [itex]l[/itex] as the minimum width of [itex]\mathbf{R}[/itex], then obviously:

    [tex]\lim_{l \to \infty} P(\text{collision}) = 0[/tex]

    where collision is defined as the event that causes [itex]\Psi(\mathbf{r}_p, \mathbf{r}_e, t)[/itex] to no longer exist — [itex]\Psi(\mathbf{r}_p, \mathbf{r}_e, t) = 0\quad \forall\ t > t_{\text{collision}}[/itex].

    Hence, [itex]P(\text{collision})[/itex] increases as [itex]\mathbf{R}[/itex] becomes smaller. In fact, there should exist an [itex]f(\mathbf{r}_p, \mathbf{r}_e, t)[/itex] such that

    [tex]P(\text{collision}) = f(\mathbf{r}_p, \mathbf{r}_e, t)[/tex]

    So, now my original question #1 can be stated more explicitly: What is [itex]f(\mathbf{r}_p, \mathbf{r}_e, t)[/itex]?
    Last edited: Nov 2, 2011
  9. Nov 4, 2011 #8
    ...anybody else have any insight into this?
  10. Nov 4, 2011 #9


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    What you write is true in nonrelativistic quantum mechanics, but things are not so simple in QFT. In some sense, it is still morally true, but read on:

    The question about localizing overlapping wavefunctions to within some volume is extremely thorny. In fact, in QFT interactions are assumed to take place *Formally* at a point (where say you have a vertex where an electron and positron annihilate and destruction operators are inserted at spacetime point x, and a corresponding photon is created). If you try to make this more precise by localizing the wavefunctions you will get badly divergent answers. The reason you get badly divergent answers, is b/c in QFT particles will undergo vacuum fluctuations and infinitely pair produce. So you are no longer asking a question about just a positron and electron, but instead have to deal with the wavefunctions of infinitely many positrons, electrons and photons all contributing to the combined wavefunction in some incredibly complicated way. In practise and even in principle, we can't and don't ever calculate this exactly. So there really isn't an answer to your question.

    So since we can't actually do that, or to make a sharp time dependant model of individual wavefunctions, instead we only answer questions about the positron and electron in the infinite future and infinite past, and infer that they interacted in some way by the structure of the amplitudes (or in short, by how they differ from the case where they don't interact at all, and continue as free particle plane wave states forever). In practise, this means measuring cross sections and things like that in a particle accelerator.
    Last edited: Nov 4, 2011
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