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What does this mean? spin +1 ( triplet ) and is thus a boson.

  1. Nov 14, 2012 #1
    What does this mean? " spin +1 ("triplet") and is thus a boson."


    Can anyone explain the best they can what the following means? Its in response to a question I asked about a certain substance.

    " spin +1 ("triplet") and is thus a boson."
  2. jcsd
  3. Nov 14, 2012 #2
    Re: What does this mean? " spin +1 ("triplet") and is thus a boson."

    It would be easier if you gave some sort of context, but I assume that you are asking about some two particle system, perhaps a atom like hydrogen, or perhaps something else. However it comes about, the composite particle you asked about has two particles with 1/2 spin. For such a particle/atom/molecule there are three states that can have a total spin of 1. Those are the states where the spin quantum number, s is 1 and m={-1,0,1}. There is another state where s=0 and m=0 which is called a singlet. The composite particle/atom/molecule would be considered a Boson since any particle with total spin equal to an integer follows Bose-Einstein statistics. If the individual particles with 1/2 spin were considered in an uncoupled state, they would follow Fermi-Dirac statistics and be called Fermions.

    There is a very good chance that none of that cleared up your confusion since your question could cover a number of situations. You will probably get a better answer if you are more specific.
  4. Nov 14, 2012 #3
    Re: What does this mean? " spin +1 ("triplet") and is thus a boson."

    Hey thanks for that!

    More specific? hmmm.

    How about lets take Deuterium as an example. This would be considered a Boson because why?
  5. Nov 14, 2012 #4
    Re: What does this mean? " spin +1 ("triplet") and is thus a boson."

    If only the nucleus is considered it should be clear since there is one proton and one neutron. If the whole atom is being considered, I don't think that it is a Boson, but I am not an expert. Sometimes funny things happen in composite particles that I don't quite understand.
  6. Nov 14, 2012 #5
    Re: What does this mean? " spin +1 ("triplet") and is thus a boson."

    Ahh yes ok.

    Here is the exact quote I am talking about:

    "The deuteron has spin +1 ("triplet") and is thus a boson."

    The deuteron is just the nucleus of deuterium. Contains one proton and one neutron.

    So, its called a boson just because the nucleus has a even mass number and has a spin +1?

    What does the "spin + 1" and "triplet" mean in this regards?
  7. Nov 15, 2012 #6
    Re: What does this mean? " spin +1 ("triplet") and is thus a boson."

    It is a boson because the spin is an integer and not a half integer.

    http://en.wikipedia.org/wiki/Spin_(physics [Broken])

    Spin is an intrinsic angular momentum that particles have. In the case of protons and neutrons, the spin is 1/2. The composite spin is the additive value of the spins of the individual particles. A simple way to think about it is that the spins pointing in the same direction will add to +1 and the spins pointed in the opposite direction will cancel. This is not really true because of a symmetry requirement. Whenever you have two spin 1/2 particles in a state where the total spin is 1, there are actually three configurations that can give this result. I'm not really giving much detail here, but I bet if you look up identical particles or symmetric spin states you'll find something. If you still have questions, I'll answer/wait until you take a quantum class:)
    Last edited by a moderator: May 6, 2017
  8. Nov 16, 2012 #7
    Re: What does this mean? " spin +1 ("triplet") and is thus a boson."

    The "funny things" don't occur if the system is in a ground state. The deuterium nucleus can be considered a boson when it is in the ground state. When the deuterium nucleus in in the ground state, the spins of the proton and the neutron point in opposite directions. The total spin of the nucleus is then zero. Furthermore, the two particles are entangled. The nucleus can then be considered a boson.

    If the deuterium nucleus is in an excited state, then it may not act like a boson. The excited state occurs if the orbit of either the proton or the neutron or both has extra energy. Deuterium nucleii in excited states may not act like bosons or fermions, since excited states are distinguishable. So the statement that you read should have been qualified by adding, "in the ground state".

    A "composite particle" can't act precisely like a fundamental particle unless all the components are entangled in a certain way. What I mean by "acting like a fundamental particle" is that the de Broglie relations apply.

    The de Broglie relation show how the wave properties of a system relate to the particle properties of a system. They really describe rules of amplitude quantization, but this was more completely discussed in some previous posts. The two basic "de Broglie" relations are:


    where E is the energy of the particle, p is the momentum of the particle, f is the frequency of the wave, λ is the wavelength of the wave, and h is Planck's constant.

    There are several important caveats to their use. For purposes of this discussion, the most important caveat is this. The relations were hypothesized for fundamental particles, not composite particles. One can extrapolate the de Broglie relations to systems of entangled particles. A group of unentangled particles may not have a corporate wavelength or a corporate frequency at all.

    The rules concerning spin in atoms and nuclei are usually specific to the ground states of these systems. You can rely on an atom a boson if the components have an integral number of quantized spins and if the atom is in a ground state. You can rely on an atom being a fermion if the components have an odd number of half states and if the atom is in a ground state.

    All this has nothing to do with the problem of isotopic effect in biology. The isotopic effect in chemistry has to do with the masses of the nuclei being different, not the entanglement of the nuclei.

    Here is a link describing some aspects of the isotopic effect. We note that the isotopic effect is not specific to biochemistry. Furthermore, it has nothing to do with bosons, fermions, entangled particles, ore wave-particle duality.

    “An isotopic substitution can greatly modify the reaction rate when the isotopic replacement is in a chemical bond that is broken or formed in the rate limiting step. In such a case, the change is termed a primary isotope effect. When the substitution is not involved in the bond that is breaking or forming, a smaller rate change, termed a secondary isotope effect is observed. Thus, the magnitude of the kinetic isotope effect can be used to elucidate the reaction mechanism. If other steps are partially rate-determining, the effect of isotopic substitution will be masked. This masking of the intrinsic isotope effect has been referred to as 'kinetic complexity'.
    Isotopic rate changes are most pronounced when the relative mass change is greatest, since the effect is related to vibrational frequencies of the affected bonds. For instance, changing a hydrogen atom to deuterium represents a 100% increase in mass, whereas in replacing carbon-12 with carbon-13, the mass increases by only 8%. The rate of a reaction involving a C-H bond is typically 6 to 10 times faster than the corresponding C-D bond, whereas a 12C reaction is only ~1.04 times faster than the corresponding 13C reaction (even though, in both cases, the isotope is one atomic mass unit heavier).
    Isotopic substitution can modify the rate of reaction in a variety of ways. In many cases, the rate difference can be rationalized by noting that the mass of an atom affects the vibrational frequency of the chemical bond that it forms, even if the electron configuration is nearly identical. Heavier atoms will (classically) lead to lower vibration frequencies, or, viewed quantum mechanically, will have lower zero-point energy. With a lower zero-point energy, more energy must be supplied to break the bond, resulting in a higher activation energy for bond cleavage, which in turn lowers the measured rate (see, for example, the Arrhenius equation).”

    I thought that I would clarify the entanglement issue for you before the PF mentor locks the topic. Again.
    Last edited: Nov 16, 2012
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