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Predicted stabilities of positronium and H atoms

  1. Aug 7, 2004 #1
    Hi, my studies of quantum mechanics aren't too deep (I'm still in highschool), but out of pure curiosity, I would like to see the mathematical derivations for the theoretical values for the decay rate /half-life of positronium (either for orthopositronium or parapositronium if they are different, whichever doesn't matter) as opposed to a regular hydrogen atom. I'd imagine this would involve Schrödinger's Equation and the wavefunction but not exactly sure. I'm only familiar with calculus (I-III) and some complex analysis but I can have some idea of what's going on. I tried googling for a while but that didn't work. Any links to any introductory quantum mechanics ebooks would also be extremely appreciated.
  2. jcsd
  3. Aug 8, 2004 #2


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    As it happens, I am reading the Martin & Shaw book called Particle Physics. Turn with me to page 101. :smile:

    "A good approximation to the spectrum of positronium can be obtained by firstly considering only the Coulomb interaction between the electron and positron. [Potential = - alpha/r, where alpha is the fine structure constant and r is the distance between electron and positron.] The energy levels then have the same kind of form as those for the hydrogen atom.

    E_n = -R/(2n^2).

    Here R is the Rydberg constant, n is the principal quantum number... m/2 is the reduced mass. The level spacings are thus half those in hydrogen... In each level, the orbital angular momentum L is restricted to L<= n-1 as in hydrogen, while the total spin is the sum of the electron and positron spins... so that S=0 or S=1. The resulting states corresponding to the n=1 and n=2 bands are easily deduced... together with the corresponding values of parity


    and C-parity


    ... The 3S_1 state with n=1 is called orthopositronium and the 1S_0 state with n=1 is called parapositronium... In the positronium spectrum, different states corresponding to the same values of n are not exactly degenerate, but are split owing to small spin-dependent interactions... In particular, ortho- and parapositronium are split by an amount which is measured to be 8.45x10^-4 eV, so that parapositronium is the ground state. This splitting is the sum of two contributions. The first of these arises because the magnetic moment of the positron... gives rise to a magnetic field which interacts with the magnetic moment of the electron... The second contribution to the energy shift has no analogue in hydrogen. In the latter, the e p interaction is due to one-photon exchange, which gives rise to the spin-spin and spin-orbit interactions as well as the Coulomb force. While the same diagram occurs for positronium, there is an additional 'annihilation diagram.' Because annihilation occurs at a point, the contribution from the latter will be proportional to |Psi_nlm(0)|^2 and will vanish except for S-waves; and since the photon has spin-1 and angular momentum is conserved, only positronium states with J=1 will be affected. It thus shifts the energies of the 3S_1 states only... The ground state is itself unstable, and can decay via electron-positron annihilation... The rate is to a good approximation proportional to the squared wave function at the origin, which vanishes for all except S-waves." The book does not explicity calculate the decay rate, but it does say that the ratio of the lifetimes of parapositronium and orthopositronium ought to be approximately the value of the dimensionless fine structure contant. Experimentally, the lifetime ratio is 0.88x10^-3.

    I have omitted some of the details in the book, but maybe what I included above will be of some help.

    {As you may have already realized, when I typed things like "3S_1," the 3 is a leading superscript and the 1 is a trailing subscript.}
    Last edited: Aug 8, 2004
  4. Aug 10, 2004 #3
    Unfortunately, the Schroedinger equation will not suffice here. What you need is a first order Feynman diagram, since particle/antiparticle annihilation must be treated by quantum field theory. Since you're in highschool, let me recommend the book "Introduction to Elementary Particles" by David Griffiths - some of the chapters there are easily accessible without advanced physics and the mathematical part is reasonably accessible. The lifetime of the spin singlet state (I don't know which is which, but this one is the shortest lived) is .125 ns
    For reasons that I don't want to get into, the spin singlet state must annihilate to an even number of photons (mostly two), and the triplet state to an odd number (mostly three, one is impossible). More photons [Feynman talk] mean more vertices mean more factors of the fine structure constant[/Feynman talk] mean a reduction in the rate by roughly 1/137 per photon - there are other factors contributing to the difference in rates, such as the difference in binding energy.
    As for the hydrogen atom, that one's stable - what decay did you have in mind?
  5. Aug 11, 2004 #4


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    For the record, the short-lived singlet state (electron and positron spins opposed) of positronium, which decays to a pair of photons for reasons to do with the conservation rules that electromagnetism respects, is parapositronium.
  6. Aug 11, 2004 #5
    Also for the record, it's quite easy to google up some useful info: http://positron.physik.uni-halle.de/panet/applications/intro/positronium.shtml [Broken]
    Last edited by a moderator: May 1, 2017
  7. Aug 11, 2004 #6


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    Thanks for that link. I am puzzled by this sentence at the website: "The electron density in matter must be sufficiently low in order to allow the formation of positronium." Can anyone elaborate on what sort of experimental conditions the author is talking about? (Maybe it has to do with what comes later on the page, about semiconductor surfaces and so on.)
  8. Aug 12, 2004 #7

    The decay rate of para-positronium:
    [tex]\lambda_p = \frac{2 M_e c^2 \alpha^5}{\hbar}[/tex]

    The half-life of para-positronium:
    [tex]T_{1/2} = \frac{\ln (2)}{\lambda} = \frac{\hbar \ln (2)}{2 M_e c^2 \alpha^5}[/tex]

    [tex]T_{1/2} = \frac{\hbar \ln (2)}{2 M_e c^2 \alpha^5}[/tex]

    The decay rate of ortho-positronium:
    [tex]\lambda_o = \frac{2 M_e c^2 \alpha^6 ( \pi^2 - 9)}{9 \pi \hbar}[/tex]

    The half-life of ortho-positronium:
    [tex]T_{1/2} = \frac{\ln (2)}{\lambda} = \frac{9 \pi \hbar \ln (2)}{2 M_e c^2 \alpha^6 ( \pi^2 - 9)}[/tex]

    [tex]T_{1/2} = \frac{9 \pi \hbar \ln (2)}{2 M_e c^2 \alpha^6 ( \pi^2 - 9)}[/tex]

    Classical Hydrogen decay rate:
    [tex]\frac{dr}{dt} = \frac{1}{3c^3} \left( \frac{2 K_e q^2}{M_e r} \right)^2[/tex]

    The para-positronium decay rate was first developed by Pirenne and Wheeler.

    The ortho-positronium decay rate was first developed by Ore-Powell.

    The para+ortho-positronium half-life equasions were developed by Orion1.

    Based upon the Orion1 equasion, what is the decay rate for Classical Hydrogen?

    Based upon the Orion1 equasions, what is the half-life of para and ortho positronium?

    http://www.streed.com/physics/positronium/ [Broken]
    [ref] J. A. Wheeler, Ann. N. Y. Acad. Sci. 48 , 219 (1946); J. Pirenne, Arch. Sci. Phys. Nat.
    28 , 233 (1946); 29 , 121, 207 & 265 (1947).
    [ref] A. Ore and J. L. Powell, Phys. Rev. 75 , 1696 (1949).
    Last edited by a moderator: May 1, 2017
  9. Aug 12, 2004 #8
    I believe it means that if there's a lot of electrons around, the positron will annihilate one of them before slowing down enough to allow the formation of positronium. He states somewhere that in matter, the decay rate of the triplet state (orthopositronium) is higher by a couple orders of magnitude as the positron annihilates with another neighboring electron with a more favorable spin state to a final state of two photons.
  10. Aug 12, 2004 #9


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    Thank you for elaborating. My thought was that in any solid or liquid, there would be lots of electrons, so I didn't know if the implication was that positronium would be easier to make in a plasma or something 'thin' like that.
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