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Why is positronium unstable?

  1. Nov 24, 2008 #1

    mrjeffy321

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    I know positronium is an unstable pairing of an electron and positron in a Hydrogen-like system with a very short lifetime on the order of around E-10 seconds (depending on its initial state).

    I would like to better-understand why positronium is so unstable, despite having bound energy states like those in stable atoms.

    I have heard that the short lifetime is due to the partial wave function overlap between the two particles (which form a particle-antiparticle pair). This, very brief, explanation makes some sense to me. If the wave function’s overlap, there is some probability of the two particles being in the same space at the same time. If this should ever occur (which it is bound to after a sufficient amount of time has passed) then the electron-positron pair will annihilate each other and destroy the positronium atom, in the process releasing a couple of photons in order to conserve energy and momentum.

    In order to get a better understanding (beyond what I just inferred) I asked my professor who is quite knowledgeable on the subject. The only kind of answer I could get was that ‘positronium is unstable because the electron and positron are annihilating each other’. This explanation, which feels somewhat unsatisfactorily basic, just leads me to believe that the two particles know the other is there and ‘wants’ to annihilate, so it does.

    Where can I find a more detailed description of the process?
     
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  3. Nov 25, 2008 #2

    malawi_glenn

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    I have also always thought that the wavefunctions are overlapping in the ground state thus allowing for annihilation.

    It should be to solve the same Schrödinger eq. as for the hydrogen atom.
     
  4. Nov 25, 2008 #3

    Vanadium 50

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    That's exactly it - there is a wavefunction overlap. (And not just for the ground state)
     
  5. Nov 25, 2008 #4

    malawi_glenn

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    Ah yes of course the other states ahve overlap, silly me :-)
     
  6. Nov 25, 2008 #5

    mrjeffy321

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    Solving the Schrödinger equation for positronium should be virtually identical to solving it for Hydrogen except that one needs to use the reduced mass of the electron (more-so) in order to correct for the altered energy levels.
    Like in Hydrogen there will be a small, but non-zero, probability of finding the electron in the “nucleus” (on in this case, the positron) when in one of the S orbitals (when l = 0 angular momentum). When this happens, the electron-positron pair annihilates and the positron atom decays into gamma rays.

    There are, however, bound states where (at least for Hydrogen, so I would assume also for positronium) there is zero probability of finding the electron in the “nucleus”. In these states then there would be no way for the wave functions to overlap and the electron-positron pair to annihilate. Would these correspond the metastable, longer-lived, states? From these states, the atom can relax by emitting a photon, and going to a lower energy level where there is an overlap in the wave functions and annihilation is possible.

    Would this be an accurate explanation?
     
  7. Nov 25, 2008 #6

    Vanadium 50

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    That's close. The one complication is that states may mix: for example 3D1 has no wavefunction overlap at the origin, but 3S1 does. Both have the same quantum numbers, so if there is any mixing at all, both states can annihilate. I don't know if they do mix - just that this would be the consequence if they did.
     
  8. Nov 25, 2008 #7
    Well, also those energy levels are excited. The positronium WANTS to minimize it's energy, so it WANTS to be in the 1S1 state. Just like hydrogen---you don't find random excited hydrogen atoms floating around...they all decay by emitting gamma rays.
     
  9. Nov 25, 2008 #8
    positronium lifetimes...


    para-positronium lifetime (S = 0):
    [tex]t_{0} = \frac{2 \hbar}{m_e c^2 \alpha^5}[/tex]

    [tex]\boxed{t_{0} = 1.244 \cdot 10^{-10} \; \text{s}}[/tex]
    [tex]t_{0} = 1.25 \cdot 10^{-10} \; \text{s}[/tex] - measured

    ortho-positronium lifetime (S = 1):
    [tex]t_{1} = \frac{9 \pi \hbar}{2 m_e c^2 \alpha^6 (\pi^2 - 9)}[/tex]

    [tex]\boxed{t_{1} = 1.386 \cdot 10^{-7} \; \text{s}}[/tex]
    [tex]t_{1} = 1.420 \cdot 10^{-7} \; \text{s}[/tex] - measured

    Reference:
    Positronium - Wikipedia
    Precision Study of Positronium: testing bound state QED theory
     
    Last edited: Nov 25, 2008
  10. Nov 26, 2008 #9

    Vanadium 50

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    Positronium doesn't want anything. It's an inanimate object.

    It also doesn't emit gamma rays in atomic transitions - they are mostly in the ultraviolet. Only when there is annihilation do you get gamma rays.

    There is no 1S1 state of positronium. Or anything else, for that matter. 1S1 means spin angular momentum is zero, orbital angular momentum is zero, and total angular momentum is 1. That's impossible. The ground state of positronium is 1S0.

    Finally, annihilation can in fact be comparable in rate to atomic transitions. The classic example is that the 3S1 state can make a magnetic dipole transition to 1S0, but in fact tends to annihilate instead.
     
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