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Another doubt regarding wavefunction.

  1. May 6, 2007 #1
    Will the wave function of a particle be the same in the two cases:

    1) When the particle is isolated i.e. as a free or independent particle.

    2) When the particle is bonded with other similar particles to constitute a composite particle.

    For example, consider an isolated quark and a quark which is bonded with other quarks by strong force to form a proton. Will the wave function of the two quarks be similar? Or will the wave function of the quark in the latter case be the same as the wave function of the proton sice that quark is always a part of that proton?

    Will the wave function of any particle be changed in accordance to the composite particle when it is bonded with other similar particles rather than when it is in isolation or independent?

    Kindly clarify.
  2. jcsd
  3. May 6, 2007 #2
    I am not sure there is such thing as isolated quarks, so I have nothing to say about your specific example. I may recommend you an old book "Quantum mechanics" by H.J.Lipkin. Composite particles are discussed there at a reasonably democratic level. Fascinating reading!
    Lipkin considers a deuteron. As you may know, this neucleus of deuterium, an isotope of hydrogen, is a bound state of a neutron and a proton. The proton in the deuteron, for example, is in the potential of nuclear forces of the neutron, so its wave function is different from that of an isolated proton. Lipkin considers the following question: How description of protons and neutrons as particles is compatible with the description of deuterons as composite particles. For example, neutrons and protons are fermions. Deutrons apparently should be bosons, as they have an integer spin. However, you cannot stuff too many fermions in a limited volume unless they have high energy, according to the Pauli principle. Can you do that with deutrons? If yes, then you'll have a lot of fermions (protons and neutrons) stuffed tight as well. Lipkin derives commutation relations for creation/annihilation operators of deuteron from those of protons and neutrons. It turns out that these relations for deutrons pretty much coincide with those of bosons unless deutron density is high. So deutrons behave as bosons in most cases, but not quite. Of course, it is difficult to imagine high-density deuteron matter, but the same calculations may be applied to Cooper pairs in the the BCS theory of superconductivity, and there the fact that Cooper pairs are not bosons is crucial.
  4. May 6, 2007 #3


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    Firstly don't confuse the "wave function" of a particle with the particles "state". I say this because your question is in part incomplete.

    One describes the range of particle behaviors by resolving its spectrum of wave functions. In this sense you may utilize the same wave-function bases for both bound and free particles. And then until and unless you assume that some specific observables have been measured or are known to be constrained then the full span of possible wave functions is possible.

    Then there is the question of time-dependent wave functions and time-independent wave functions. Once the Hamiltonian (essentially but not quite the Energy operator) of the particle is established you may then resolve the wave functions in an eigen-basis of the Hamiltonian which means you are resolving the particle in terms of its available energies.

    You will have resolved the particle's description as a superposition of stationary modes. In this sense then the difference between free and bound cases is vitally important because this is dictated by the Hamiltonian.

    Remember that we do not "measure the wave function of a particle". We rather utilize wave functions to express what we have measured or will measure in observing the particle.

    To answer you question about single vs. composite particles it doesn't matter per se whether they are bounded or not. A composite of particles is a larger system and thus has a distinct space of wave functions. There are physical conditions which dictate how distinct they are, namely the statistics of the particles, e.g. Fermions vs Bosons. You must be careful when you work within this composite description because distinguishing which particle of the composite to which you refer in some assertions presupposes you have selected some observable which separates it from the others (i.e. the particle which has spin up, or the particle which is moving to the left, or the red quark, etc).

    When you so stipulate you must then be sure not to ask questions about the indicated particle which rely on observables which fail to commute with the implicit observable you use to index the particle. This is how people get confused in trying to understand say EPR experiments. Such questions are ill formed and trying to answer them is like asking who shaves the Barber.

    Finally I caution you again to not take the wave-function too seriously as a physical object. It is a mathematical construct which we use to model the behavior of the observables associated with a particle. It is like your credit record. It isn't physically out there somewhere. It is rather represents knowledge of past and future behavior under a specific subset of possible circumstances.

    James Baugh
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