Nuclear force between neutrons

[DrZoidberg]

Why is it that neutrons don't stick together? As far as I know the nuclear force is also acting between 2 neutrons and not just between neutrons and protons.
So why are there no ultra dense neutron objects consisting of a large number of neutrons, except of course for neutron stars where gravity pulls the particles together?

 

[Meir Achuz]

The neutron-neutron force is a bit two weak to form pure bound neutron states.

 

[tom.stoer]

Afaik it's due to combination the Pauli principle with the effective nucleon-nucleon interaction. Neutrons are fermions, that means their spins must be antiparallel. But the strong force is repulsive for antiparallel spins. By the same principle diprotons are forbidden.

 

[Meir Achuz]

Afaik it's due to combination the Pauli principle with the effective nucleon-nucleon interaction. Neutrons are fermions, that means their spins must be antiparallel. But the strong force is repulsive for antiparallel spins. By the same principle diprotons are forbidden.

No. The N-N force is not repulsive for anti-parallel spins.
It is still attractive, but a little bit weaker than for parallel spins.

 

[Dickfore]

In 3d, a sufficiently shallow potential well has no bound states.

 

[tom.stoer]

No. The N-N force is not repulsive for anti-parallel spins.
It is still attractive, but a little bit weaker than for parallel spins.

Then please explain the absence of a di-neutron and a di-proton state.

 

[ansgar]

di proton = electric repulsion

 

[tom.stoer]

di proton = electric repulsion

That doesn't matter; there is no di-neutron, either; electromagnetic forces are mostly irrelevant.

 

[jimgraber]

A quick, naive look at the chart of nuclides would seem to say that both the effective proton-proton force and the effective neutron-neutron force are repulsive but that the effective neutron-proton force is attractive. (regardless of spin?. or overwhelming spin?) what does chromodynamics say?
yours naively,
Jim Graber

 

[tom.stoer]

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

$$V_C(r) \sim \frac{e^{-m_\pi r}}{r}$$

$$V_S(r,S) \sim V_S(r) (\vec{s}_1\vec{s}_2)$$

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

For the spin-spin coupling one has

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

Therfeore

$$\langle \vec{s}_1 \vec{s}_2\rangle_{S=0} = -\frac{3}{4}$$

$$\langle \vec{s}_1 \vec{s}_2\rangle_{S=1} = +\frac{1}{4}$$

As the diproton or dineutron is symmetric in isospin it must by antisymmetric in spin which corresponds to S=0. So the term

$$V_S(r,S=0) \sim -\frac{3}{4} V_S(r)$$

has - minus sign whereas

$$V_S(r,S=1) \sim +\frac{1}{4} V_S(r)$$

has a + sign.

I do not know if for S=0 this results in a repulsive potential $$V_C(r) + V_S(r,S=0)$$. 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.

This goes into the direction of Dickfore and Meir Achuz.

 

[Meir Achuz]

Then please explain the absence of a di-neutron and a di-proton state.

Dickfore:
"Re: nuclear force between neutrons
In 3d, a sufficiently shallow potential well has no bound states."

 

[Orion1]

binding energy per nucleon...

Why is it that neutrons don't stick together?

Because of the Pauli exclusion principle.

Most odd-odd nuclei are unstable with respect to beta decay, because the decay products are even-even, and are therefore more strongly bound, due to nuclear pairing effects. Deuterium, however, benefits from having its proton and neutron coupled to a spin-1 state, which gives a stronger nuclear attraction; the corresponding spin-1 state does not exist in the two-neutron or two-proton system, due to the Pauli exclusion principle which would require one or the other identical particle with the same spin to have some other different quantum number, such as orbital angular momentum. But orbital angular momentum of either particle gives a lower binding energy for the system, primarily due to increasing distance of the particles in the steep gradient of the nuclear force. In both cases, this causes the diproton and dineutron nucleus to be unstable.

According to the semi-empirical mass formula (SEMF) or liquid drop model:

Diprotium binding energy per nucleon:
$$E_b = -26.5625 \; \text{MeV}$$

Dineutronium binding energy per nucleon:
$$E_b = -26.7005 \; \text{MeV}$$

Deuterium binding energy per nucleon:
$$E_b = -2.90318 \; \text{MeV}$$

According to quantum mechanics:

Deuterium binding energy per nucleon: spin 0 'singlet' - virtual
$$E_b = -30 \; \text{keV}$$

The negative signs indicates that the nuclei are repulsive and the model experiences spontaneous fission.

According to mass defect:

Deuterium binding energy per nucleon: spin +1 'triplet'
$$E_b = 1.11226 \; \text{MeV}$$

In nuclear terms, the Deuterium nucleus is just barely stable and susceptible to fracture inside stellar cores. Although Diprotium and Dineutronium cannot form bound states due to the Pauli exclusion principle, they can experience a nuclear reaction via the weak nuclear force resulting in a bound state Deuterium.

Endothermic weak nuclear reaction:
$$^1_1p + ^1_1p \rightarrow ^2_1D + e^+ + \nu_e + 0.42 \; \text{MeV}$$

Exothermic weak nuclear reaction:
$$^1_0n + ^1_0n \rightarrow ^2_1D + e^- + \overline{\nu}_e + 3.00691 \; \text{MeV}$$


Reference:
http://en.wikipedia.org/wiki/Pauli_exclusion_principle"
http://en.wikipedia.org/wiki/Semi-empirical_mass_formula"
http://en.wikipedia.org/wiki/Deuterium#Quantum_properties"

 

[tom.stoer]

Because of the Pauli exclusion principle.

No quite; look at our discussion above and your own arguments below: it's a it's a bit more complicated than that. The Pauli exclusion principle only tells you that neither diproton nor dineutron can exist in a spin triplet state. Only via further calculations from a specific NN-potential it follows that the singlet state is not a bound state.

According to the semi-empirical mass formula (SEMF) or liquid drop model:

This is not a derivation, but only a "collection of facts". A derivation should start from an NN-potential.

According to quantum mechanics:

Deuterium binding energy per nucleon: spin 0 'singlet' - virtual
$$E_b = -30 \; \text{keV}$$

The negative signs indicates that the nuclei are repulsive and the model experiences spontaneous fission.

Which NN-potential is used?

 

[Vanadium 50]

I'm afraid Wikiscraping isn't sufficient here. This takes some actual understanding.

As pointed out, the n-n potential is attractive. It's just not attractive enough to bind.

More interesting is the case of the tetraneutron (4n). Calculations have oscillated back and forth over the years on whether it's expected to be bound or not. Experimental evidence for one is not very good - most searches are negative. It's absolutely certain that if it is bound, it's loosely bound (less than 0.7 MeV per nucleon, less than a tenth of typical nuclear binding).

I don't believe it exists. I find the lack of a signal in DCX reactions compelling evidence for its absence. Others have different opinions.

 

[tom.stoer]

I found this overview http://www.physics.umd.edu/courses/Phys741/xji/chapter6.pdf which presents a first introduction.

It is explained that at low energies S-wave scattering is dominant. A few remarks regarding scattering phases and scattering length follow. A virtual bound state for T=1 is mentioned, too. The conclusion is - again: "the potential is attractive but not attractive enough".

For further investigation I think the "Nijmegen group" and the work around the "CD Bonn potential" is interesting:

http://arxiv.org/abs/nucl-th/0410042v1
Partial-Wave Analyses of all Proton-Proton and Neutron-Proton Data Below 500 MeV
M.C.M. Rentmeester (1), R.G.E. Timmermans (2), J.J. de Swart (1) ((1) Radboud University Nijmegen, (2) KVI, Groningen)
(Submitted on 9 Oct 2004)
Abstract: In 1993 the Nijmegen group published the results of energy-dependent partial-wave analyses (PWAs) of the nucleon-nucleon (NN) scattering data for laboratory kinetic energies below Tlab=350 MeV (PWA93). In this talk some general aspects, but also the newest developments on the Nijmegen NN PWAs are reported. We have almost finished a new energy-dependent PWA and will discuss some typical aspects of this new PWA; where it differs from PWA93, but also what future developments might be, or should be.

http://arxiv.org/abs/nucl-th/0006014
The high-precision, charge-dependent Bonn nucleon-nucleon potential (CD-Bonn)
R. Machleidt (University of Idaho)
(Submitted on 9 Jun 2000)
Abstract: We present a charge-dependent nucleon-nucleon (NN) potential that fits the world proton-proton data below 350 MeV available in the year of 2000 with a chi^2 per datum of 1.01 for 2932 data and the corresponding neutron-proton data with chi^2/datum = 1.02 for 3058 data. This reproduction of the NN data is more accurate than by any phase-shift analysis and any other NN potential. The charge-dependence of the present potential (that has been dubbed `CD-Bonn') is based upon the predictions by the Bonn Full Model for charge-symmetry and charge-independence breaking in all partial waves with J <= 4. The potential is represented in terms of the covariant Feynman amplitudes for one-boson exchange which are nonlocal. Therefore, the off-shell behavior of the CD-Bonn potential differs in a characteristic and well-founded way from commonly used local potentials and leads to larger binding energies in nuclear few- and many-body systems, where underbinding is a persistent problem.

 

[Orion1]

tetraneutronium...


The model is based upon a 3 dimensional infinitely repulsive spherical finite potential well of nuclear radius $$r_0$$.

Nuclear potential well boundary conditions:
$$V(r) = \left\{ \begin{array}{ll} V_{0} &, \ 0 \le r \le r_{0} \\ 0 &, r_{0} < r \right. \end{array}$$

Matter wave wavenumber: (ref. 1)
$$k = \frac{\sqrt{2m E}}{\hbar}$$

Matter wavenumbers: (ref. 2)
$$K' = \frac{\sqrt{m(V_0 + E)}}{\hbar}$$ and $$k = \frac{\sqrt{mE}}{\hbar}$$

Deuterium spin 0 'singlet' virtual state energy:
$$\boxed{E_1 = -\frac{\pi \hbar^2 E_b}{3 \pi \hbar^2 - m_p \sigma_p E_b}}$$

$$E_b$$ - total binding energy
$$\sigma_p$$ - proton cross section

Deuterium binding energy per nucleon: spin 0 'singlet' virtual state
$$E_1 = -34.9685 \; \text{keV}$$

Deuterium binding energy per nucleon: spin +1 'triplet'
$$E_b = 1.11226 \; \text{MeV}$$

According to the semi-empirical mass formula (SEMF) or liquid drop model:

Tetraneutronium binding energy per nucleon:
$$E_b = -17.9688 \; \text{MeV}$$

Substituting in neutron mass and Helium binding energy:

Tetraneutronium spin 0 'singlet' virtual state energy:
$$\boxed{E_1 = -\frac{\pi \hbar^2 E_b}{3 \pi \hbar^2 - m_n \sigma_p E_b}}$$

Tetraneutronium binding energy per nucleon: spin 0 'singlet' virtual state
$$E_1 = -16.05825 \; \text{keV}$$

The negative signs indicates that the nuclei are repulsive and the model experiences spontaneous fission and is nearly as unstable as Diprotonium and Dineutronium and singlet Deuterium.

Tetraneutronium cannot form a stable bound state, however it can experience a nuclear reaction via the weak nuclear force resulting in a bound state Helium nucleus.

Exothermic weak nuclear reaction:
$$4 ^1_0n \rightarrow ^4_2He + 2 e^- + 2 \overline{\nu}_e + 29.8615 \; \text{MeV}$$

Reference:
http://en.wikipedia.org/wiki/Scattering_length"
http://mightylib.mit.edu/Course%20Materials/22.101/Fall%202004/Notes/Part3.pdf"
http://en.wikipedia.org/wiki/Semi-empirical_mass_formula"
http://en.wikipedia.org/wiki/Wavenumber"

 

[ansgar]

can you write down the NN potential or not dear Orion1?

 

[Orion1]

NN potentials...

can you write down the NN potential or not dear Orion1?

Negative, the reference does not state which NN potential is used in the differential scattering model and the NN potential for Quantum Chromodynamics (QCD) is unknown to me.

According to Wikipedia the most prevalent models are based on Yukawa, Paris, Argonne AV18, CD-Bonn and Nijmegen potentials.

My preference is the Yukawa potential:
$$V(r)= -g^2 \; \frac{e^{-m_{\pi} r}}{r}$$

$$V_0 = 18 \; \text{MeV}$$ - Deuterium spin = 0 'singlet'
$$V_0 = 36 \; \text{MeV}$$ - Deuterium spin = +1 'triplet'

Reference:
http://en.wikipedia.org/wiki/Nuclear_force#Nucleon-nucleon_potentials"
http://en.wikipedia.org/wiki/Yukawa_potential"

 

[Dickfore]

Actually, it was stated in the first statement of his previous post:

The model is based upon a 3 dimensional infinitely repulsive spherical finite potential well of nuclear radius $$r_0$$.

This would mean:

$$V(r) = \left\{ \begin{array}{ll} V_{0} &, \ 0 \le r \le r_{0} \\ 0 &, r_{0} < r \right. \end{array}$$

EDIT:

A repulsive force is assumed here, hence no bound state is possible.

 

[tom.stoer]

My preference is the Yukawa potential

That does not really help. Yukawa is simple but far from realistic. In our context here it fails to explain the difference between S=0 and S=1 (Yukawa is explicitly spin independent).

Unfortunately you have to use the more complicated potentials like CD-Bonn etc.

 

[Orion1]

Quantum Chromodynamics...


What is the NN potential for Quantum Chromodynamics (QCD)?

Reference:
http://en.wikipedia.org/wiki/Quantum_chromodynamics"

 

[tom.stoer]

What is the NN potential for Quantum Chromodynamics (QCD)?

Reference:
http://en.wikipedia.org/wiki/Quantum_chromodynamics"

Writing that down is equivalent to booking a flight to Stockholm next fall.

Afaik there is no rigorous derivation of NN potentials from QCD. What people are doing is to look at the structure of the QCD operators and try to find a parameterization respecting their symmetry. The parameters are fitted based on experimately input. Phenomenologically successful potentials include many different spin, isospin and spin-orbit terms. In addition there are not only meson exchange terms but vector-meson terms as well.

All this can be motivated by QCD considerations, but a rigorous derivation is not available.

 

[tom.stoer]

One remark: a qm potential V(r) is strictly speaking the wrong picture in quantum field theory. In some cases it can be derived as a "static limit" from more complicated interactions, but this is already an approximation of q.m. of relativistic q.m.

 

[brother time]

I'm a new member here and thought I might ask a stupid question. So can Neutrons attract other neutrons? yes or no. And has the answer been proven or is it a theory that is likely or almost impossible to be wrong? Sorry if I can't understand completely what you are talking about (I am in grade 9).
cheers, BT

 

[brother time]

As an after thought, do neutrons attract and can they occupy the same space?
Sorry I didn't post this the first time,
cheers, BT

 

[tom.stoer]

OK, you are asking for an explanation of the formulas. I'll do my very best.

First of all you have to prepare an experiment for low energy neutron collisions; this is not possible in practice because there are no "neutrons beams" which you could use, but let's forget about that. If you bring two neutrons together you have to distinguish between parallel and anti-parallel spin orientation.

In case of parallel spins there is a "repulsive force" which is due to the Pauli exclusion principle. This force cannot be explained in terms of classical physics and has nothing to do with a potential or something like that.

In case of anti-parallel spins there seems to be an attractive force. But - and this is strange - the two neutrons do not stick together but start to depart again. If you now analyze these low-energy neutron collisions you can extract some information regarding the force between the neutrons from the scattering angles measured in a huge number of collisions. This analysis tells you that the force was attractive!

Now one could guess that there is a "repulsive core" of the potential which prevents the neutrons from sticking together. But the energy of the two neutrons is low, therefore they do not "penetrate enough" to feel the repulsive core.

If you start with quantum mechanics you learn how to calculate eigenfunctions in a square well potential in one dimension. This corresponds to something like a particle bound in the square well. What is strange is that in three dimensions there are no bound states in such a potential if the well is not deep enough. Something like that happens to the two neutrons. The force is attractive (in case of antiparallel spins) but not attractive enough to form a bound state.

 

[Orion1]

Schrodinger radial equation?...


According to Wikipedia, the quantum nuclear shell model is based upon the Woods–Saxon potential:
$$V(r) = - \frac{V_0}{1 + \exp(\frac{r-R}{a})}$$

Laplace operator:
$$\Delta f = \nabla^2 f = \nabla \cdot \nabla f$$

Hamiltonian:
$$\mathbf{H} = -\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{r},t)$$

A more realistic potential, such as Woods Saxon potential, would approach a constant at this limit. One main consequence is that the average radius of nucleons orbits would be larger in a realistic potential; This leads to a reduced term $l(l+1)\hbar^2 / 2 m r^2$ in the Laplace operator of the Hamiltonian.

Quantum Hamiltonian?:
$$\mathbf{H} = - \frac{l(l+1) \hbar^2}{2 m r^2} \nabla^2 + V(\mathbf{r},t)$$

Nuclear potential well boundary conditions:
$$V(r) = \left\{ \begin{array}{ll} V_{0} &, \ 0 \le r \le r_{0} \\ 0 &, r_{0} < r \right. \end{array}$$

Model the nucleus with a potential which is zero inside the nuclear radius and infinite outside that radius?:
$$U(r) = \left\{ \begin{array}{ll} 0 &, \ 0 \le r \le r_{0} \\ \infty &, r_{0} < r \right. \end{array}$$

The nuclear Schrodinger radial equation for a infinite-walled spherical potential is?:
$$- \frac{\hbar^2}{2m} \left( \frac{d^2R}{dr^2} + \frac{2}{r} \frac{dR}{dr} \right) + \left[ V(r) + \frac{l(l+1) \hbar^2}{2 m r^2} \right]R = ER$$

Integration via substitution?

Nuclear infinite-walled spherical Schrodinger radial equation?:
$$\boxed{- \frac{\hbar^2}{2m} \left( \frac{d^2R}{dr^2} + \frac{2}{r} \frac{dR}{dr} \right) + \left[ - \frac{V_0}{1 + \exp(\frac{r-R}{a})} + \frac{l(l+1) \hbar^2}{2 m r^2} \right]R = ER}$$

Reference:
http://en.wikipedia.org/wiki/Nuclear_shell_model#Deforming_the_potential"
http://en.wikipedia.org/wiki/Woods_Saxon_potential"
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/shell.html#c1"
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/sphwel.html#c1"
http://en.wikipedia.org/wiki/Laplace_operator"
http://en.wikipedia.org/wiki/Hamiltonian_%28quantum_mechanics%29"

 

[tom.stoer]

Orion,

again this is a potential (the Woods–Saxon potential) that does not help in our context!

If you read the Wikipedia article carefully, you will find the statement (in the first sentence!)

"The Woods–Saxon potential is a mean field potential for the nucleons (protons and neutrons) inside the atomic nucleus, which is used to approximately describe the forces applied on each nucleon, in the shell model for the structure of the nucleus." That means that is definately not valid for nuclei with a small number of nucleons! "Mean field" means that it is an effective potential "generated by all nucleons 1...N-1 and felt by nucleon N". Think about a pool table where you have thousands of balls. If one ball moves between these balls it feels something like an effective force, a "mean force" created by "averaging" over all other balls. But we are not interested in this averaging procedure but in the exact force between two balls. So the Woods–Saxon potential does certainly not apply for any two-nucleon system!

In addition this potential is neither spin nor iso-spin dependent. But we are discussing here the following states
- diproton with two protons, antiparallel spins = total spin S=1 (triplet)
- dineutron with two neutrons, antiparallel spins = total spin S=1 (triplet)
- para-deuteron with one proton and one neutron, antiparallel spins = total spin S=1 (triplet)
- ortho-deuteron with one proton and one neutron, parallel spins = total spin S=0 (singlet)

Potentials taking into account neither spin nor isospin are not able to distinguish between these states!

 

[brother time]

thanks Tom and Orion. both replies helped.
cheers, BT

 

[Orion1]

quantized residual nuclear force...


Given that the nuclear scattering experimentally derived nuclear potential well depths are:
$$V^{(0)}_0 = 14.3 \; \text{MeV}$$
$$V^{(1)}_0 = 38.5 \; \text{MeV}$$

Quantized nuclear potential well depth spin decay energy: (s = 0 -> s = 1)
$$E_{\gamma} = \hbar \omega = (V^{(1)}_0 - V^{(0)}_0) = (38.5 - 14.3) \; \text{MeV} = 24.2 \; \text{MeV}$$

Is this nuclear reaction possible via a quantized residual strong nuclear force spin decay?
$$^2_1D^0 \rightarrow ^2_1D^1 + \gamma( 24.2 \; \text{MeV})$$

ortho-deuteron -> para-deuteron + gamma photon

Reference:
http://www.phys.washington.edu/users/savage/Class_560/lec560_3/node1.html#figwell"