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Complex potential?

  1. Jul 12, 2006 #1
    Yet another question from question from Griffiths. This time I was able to do the problem, but I’m not able to understand what it means.

    Chapter 1, problem 1.15

    He talks about unstable particles for which the probability of finding it somewhere is not one, but an exponentially decreasing function of time.

    [tex]\int_{-\infty}^{\infty}|\Psi(x,t)|^2dx = e^\frac{-t}{\tau}[/tex] ,

    To arrive at this result (in a “crude” way), we assume that the potential energy fucntion has an imaginary part, [itex]\Gamma[/itex].

    [tex]V = V_0 - \imath\Gamma[/tex],

    where [itex]V_0[/itex] is the true potential energy and [itex]\Gamma[/itex] a positive real constant.

    What I don’t understand is the nature of this potential. What’s a complex potential, and does gamma have a physical meaning?
    Last edited: Jul 12, 2006
  2. jcsd
  3. Jul 12, 2006 #2


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    have you related Gamma to the lifetime tau?
    Gamma is bascially the decay rate of the particle. The complex part of the potential is related to the fact that the probability of observing the particle "decays" with time. Basically, that the particle has a probability of "disappearing". Of course, whatthis means is that a complex potential describes an *unstable* particle. At some point it will decay into *other particles* but since you have only an equation describing this particle, you must take this effect into account by introducing a complex part in the potential (a full treatment would have to include the other particles as well as a mean for the number of particles to change and would thus require a full-fledged quantum field theory).
  4. Jul 12, 2006 #3
    Thanks for the reply. :smile:

    Yes, [tex]\tau = \frac{\hbar}{2\Gamma}[/tex]. Gamma, of course, has units of energy.

    That part was clear, i.e. to describe a unstable particle you need a complex potential.

    (A naïve question)So will this QFT treatment bring back the real (both physically and mathematically) potential?
    Last edited: Jul 12, 2006
  5. Jul 12, 2006 #4
    Is that a generalization? Complex potentials have mathematical (functional) significance as entities in gaussian space, but can their real and imaginary parts be necessarily ascribed a physical signifiance?

    If this has something to do with renormalization (as an analogous mathematical step in procedure or even spirit) then I would be interested to know what you mean by "bring back".
  6. Jul 12, 2006 #5
    That's what I want to know. But my earlier statement was within the limits of the problem.

    I did say it was a naive question. To make it clear, will the advanced treatment make the complex potential a sort of "trick" to describe an unstable particle. Sorry, I'm unable to explain it any better, at the moment. :redface:
  7. Jul 12, 2006 #6
    I am not exactly sure what this 'advanced treatment' actually does...perhaps we need to be directed to a paper/book where it is dealt with rigorously.

    Given that the probability integral has an exponential time-dependance, it should be possible to get back a suitable wavefunction without resorting to the trick. If this is possible, it would be interesting to know whether such a wavefunction is unique (not in terms of complex amplitude for normalization) and if not, what class of wavefunctions satisfy such a relation.

    My guess is that we need to look at the wavemechanics of radioactive decay....
  8. Jul 13, 2006 #7
    One more question...

    When we say, [tex]\tau = \frac{\hbar}{2\Gamma}[/tex], we don't actually take into consideration the nature of the particle. As per this formula, every particle in the given potential will have the same lifetime. Is that correct?
  9. Jul 13, 2006 #8
    What if the "nature" is absorbed into [itex]\Gamma[/itex]?

    EDIT: That's a wild guess.
    Last edited: Jul 13, 2006
  10. Jul 13, 2006 #9


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    To answer one of your previous questions: yes, in a complete (QFT) treatment, the equivalent of th epotentials would be real and the decay of a certain particle would lead to the corresponding creation of other particles with the correct energies, quantum numbers, etc. That's the power of QFT.

    For this question: I don't think there is a way in this formalism to describe two particles (even if they have the same decay rate). This is a trick which, (unless I am mistaken and someone else will correct me) will work only to describe one unstable particle. If you had two particles and an imaginary term in the potential, the combined wavefunction would be decaying and there is no way to untangle the decay of one particle vs the other. As far as I can see anyway.
  11. Jul 15, 2006 #10
    Messy, isn't it? And that kids, is why we use QFT :biggrin:
  12. Jul 15, 2006 #11
    Thanks for the clarification, nrqed.

    More reasons to get to QFT ASAP. :biggrin:
  13. Jul 15, 2006 #12
    Neutrino, there's a post on QFT prereqs somewhere on the High Energy forum...check it out first.
  14. Jul 15, 2006 #13
    Yes, I've been following that for sometime now. But there is still a long way before I get to that point.
  15. Jul 15, 2006 #14
    Off Topic: Nice blog there mate. How did you type those equations? I want to put up equations on my blog too (but I am too lazy). Oh and I'm putting a link to your blog on my links page (http://spinor.sitesled.com/ [Broken]).
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