Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reason a Dineutron is not bound?

  1. Aug 24, 2010 #1
    Hello, I am learning very, very basic quantum from the internet, and I have a question about the reason why dineutrons cannot exist. I know that the standard answer is that they aren't bound, but I don't understand why they are not, whereas a proton-electron system is.

    Here is the context in which this question occurred:
    http://www.vias.org/physics/bk6_04_04.html

    Is the strong nuclear attraction insufficient to provide bounds for the wavefunction of the neutrons? Why is this the case, when electrostatic interactions are enough for other wave-particle bounds?

    Can two similar particles with mass just not bound one another? Why is it that adding a proton makes a bound state possible?

    As I'm sure everyone can tell by now, I am a bit confused, so please go easy on me :] Thank you very much for your help.
     
  2. jcsd
  3. Aug 24, 2010 #2
    I can't answer...but some clues here
    http://en.wikipedia.org/wiki/Dineutron

    I was just trying to refresh myself on simple boundary conditions and found an excellent analogy between a vibrating string and an electron bound in a box....
    back in a minute..yea it's exactly like yours.....
     
    Last edited: Aug 24, 2010
  4. Aug 24, 2010 #3
    The three part chart in your post is just like a vibrating string length L between rigid supports. Two waves moving in opposite directions (same frequency, speed and amplitude) along such a string are of the form Ysin(kx + wt) + Ysin(kx -wt) which combines via a trig identity to the form 2YSinkxCostwt.....where w = 2(pi)f and k = 2(pi)/(lamdda)b

    By analogy this IS in fact the Heisenberg matter wave for a confined particle with different symbols!!!!!....and the probability is the square of this function... all you have to do is realize the function must have nodes (zeros) at the boundaries.....L = n(lamda)/2 and you are on your way....

    It's a simple and insightful introduction from classic to quantum....I left out a number of simple steps....(once you see them)

    I found the description in my old undergraduate physics text, Halliday and Resnick, the very last chapter...."Waves and Particles"....
    Maybe someone will recognize this intoductory approach and post an available online source....I did a quick Wikipedia look but could not find this one....
     
    Last edited: Aug 24, 2010
  5. Aug 24, 2010 #4
    For me it is a mystery
    Diproton is almost stable
    Dineutron should be MORE stable than diproton (because neutrons dont repel)
     
  6. Aug 24, 2010 #5

    phyzguy

    User Avatar
    Science Advisor

    Note that the mystery doesn't end with the dineutron. Take a look at the chart of the nuclides and explain to me why some nuclei are stable and others are not. For example, why do all light nuclei have stable isotopes except Technetium, which has none? And why do the unstable nuclei have the half-lives that they do? And are any of them really stable, or do they just have half-lives too long to measure? There is no good theory today - the strong interaction is just not well enough understood.
     
  7. Aug 24, 2010 #6

    tom.stoer

    User Avatar
    Science Advisor

    First one should neglect the electromagnetic force; then - regarding the strong force only - the diproton and the dineutron are two degenrate states of isospin =+1 or -1 respectively. Both diproton and dineutron do not form a bound state!

    The deuteron which has isospin = 0 (one proton and one neutron form a state I=0 as the two isospins are antiparallel) is a bound state. How can this be?

    The reason is that describing the strong force via a scalar potential is a very crude approximation. Some nucleon potentials (like Woods-Saxon) that one single nucleon n=1 "feels" inside a nucleus are mean field approximations only. Such a potential is generated by averaging over all other nucleons N=2, ..., N forming the nucleus. These potentials are not sufficient to describe the nucleon-nucleon force.

    Instead one has to use two-nucleon potentials which have a rather complex structure: they have several spin- and isospin-dependent parts. One isospin-neutral part is always attractive, but there is an spin-and isospin-dependent part which is attractive for the deuteron but repulsive for diproton and dineutron. The reason is that for the diproton and the dineutron (with parallel isospin) the two spins always have to be antiparallel in order to fulfill the Pauli principle; (for the deuteron both parallel and antiparallel spins would be possible in principle since the two isospins are different and therefore the Pauli principle is respected by both spin states).

    So the nucleon-nucleon force is still attractive for antiparalell spins, but too weak to allow for a bound state.

    Here are some notes a posted several month ago:

    Looking a phenomenologically successful nucleon-nucleon potentials derived from one-pion exchange one finds

    [tex]V_C(r) \sim \frac{e^{-m_\pi r}}{r}[/tex]

    [tex]V_S(r,S) \sim V_S(r) (\vec{s}_1\vec{s}_2)[/tex]

    plus further terms, e.g. tensor and LS.

    For the spin-spin coupling one has

    [tex]\langle \vec{s}_1 \vec{s}_2\rangle = \frac{1}{2}\left[S(S+1) - s_1(s_1+1) -s_2(s_2+1)\right][/tex]

    Therefore

    [tex]\langle \vec{s}_1 \vec{s}_2\rangle_{S=0} = -\frac{3}{4}[/tex]

    [tex]\langle \vec{s}_1 \vec{s}_2\rangle_{S=1} = +\frac{1}{4}[/tex]

    As both diproton and dineutron are symmetric in isospin they must by antisymmetric in spin which corresponds to S=0. So the term

    [tex]V_S(r,S=0) \sim -\frac{3}{4} V_S(r) [/tex]

    has a minus sign whereas

    [tex]V_S(r,S=1) \sim +\frac{1}{4} V_S(r) [/tex]

    has a plus sign.

    I do not know if for S=0 this results in a repulsive potential [tex]V_C(r) + V_S(r,S=0)[/tex]. I have found my old notes indicating that at very low energies the scattering phases for NN scattering indicate an attractive potential for both S=0 and S=1, but that attraction is not strong enough to form a bound state.

    The reason why an attractive potential does not allow for bound state is well known from ordinary quantum mechanics. In one dimension an attractive potential always allowes for at least one bound state; in three dimension this is no longer true, an attractive but very flat potential has scattering solutions only.
     
    Last edited: Aug 24, 2010
  8. Sep 2, 2012 #7
    From 1965BA1A.pdf Pg3 (178)
    Here it is important to emphasize that
    such states do not have so definite a physical meaning
    as the quasistationary states, but are essentially a
    mathematical concept. This is most simply seen from
    the classic example of two neutrons in a JS state.
    There is no bound state of two neutrons. If, however,
    the interaction between neutrons were a trifle stronger,
    a bound state would appear. This closeness to the
    possibility of having a bound state leads to a number
    of characteristic features in the interaction of two
    neutrons at small energies (for example, to an increase
    of the cross section for scattering of neutrons
    by neutrons). It is this sort of situation that is described
    by the term "virtual state."
     
  9. Sep 2, 2012 #8

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    This thread is two years old.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook