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Why nucleons with parallel spins have greater nuclear force than the ones with anti-parallel spins? I just want a clear and easy explanation. Thank you!

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    $\begingroup$ There are experimental signatures that the nuclear force depends on the spin. For example if you consider deuteron- it is only found with $S=1$ (meaning that this configuration has a lower energy). $\endgroup$ – AMS Oct 23 '16 at 12:38
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There are no easy and simple answers when it comes to questions about the nuclear force. The force between two nucleons is a complicated residual interaction that leaks outside the color confinement walls of the QCD strong interaction. It is best visualized as due to exchanges of quark - antiquark pairs or mesons. There are multiple mesons involved in the nuclear force and the contribution of an individual meson depends strongly on the spin and parity of the meson under considerstion. The lightest meson is the pion, a pseudoscalar particle (spin 0 but odd parity). The force induced by the exchange of a pion is strongly spin dependent, in fact the pion exchange does not even contribute a force component for nuclei with spherical symmetry and spin 0. Since the deuteron lacks spherical symmetry, it does experience a binding force associated with pion exchange, and since the deuteron is a spin 1 particle, we infer that the pion contribution to the nucleon-nucleon force favors spin aligned over spin opposite orientation. The story does not end here.

There are mesons more massive than the pion that also contribute to nuclear binding. We know this to be the case because many of the most tightly bound nuclei are spin 0 and spherical. By symmetry considerations the pion exchange forces average to 0 for these nuclei, so the more massive omega, sigma, and rho mesons must provide the binding forces. The omega meson is an uncharged spin 1 odd parity (vector) meson. It is similar to a photon, except that it is massive (782 MeV). Like the coulomb interaction, the exchange of an omega meson contributes a central force as well as a spin-orbit force that is small in comparison to the central force. The sign of the central force from omega meson exchange is repulsive, however, so it cannot be solely responsible for the binding of these nuclei.

Another likely contribution comes from sigma meson exchange. The sigma meson is uncharged even parity and spin 0 (scalar). It's existence was inferred from its likely role in nuclear binding before it was established by experiments, but now it is listed in the particle data tables. Unlike the omega, the sigma meson is a broad resonance with an ill defined mass range (500-600 MeV). As a Yukawa - like interaction, it also yields a central force term and a small spin-orbit term. Unlike the omega central term, hovever, the sigma central term is attractive. This means that the weak central attraction in spherical nuclei must arrive from a cancellation between omega exchange and sigma exchange with the sigma exchange dominating.

There is an interesting difference between the spin orbit contributions from omega and sigma. Unlike the central components where the two exchanges are of opposite signs, the spin orbit contributions from sigma and omega exchanges are additative. This means that the importance of the spin-orbit force relative to the central force is enhanced in direct proportion to the degree of cancellation of the central force components. As a result of this complex interplay, the spin-orbit contribution in the nuclear force becomes a 10% effect unlike the case in atoms where it is a 1% effect.

The rho meson is another vector (spin 1) particle, but unlike the omega, the rho is a charge triplet (-1, 0, 1). It is said to have isospin 1. Because of its isospin, the rho contributions to both central and spin-orbit forces are proportional to the difference between the neutron and proton distributions in a nucleus, so it is less important than the omega. There are likely to be other even more massive mesons that play a small role in nuclear binding, but the spin dependence of these are likely to be similar to those already discussed.

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  • $\begingroup$ "As a result of this complex interplay, the spin-orbit contribution in the nuclear force becomes a 10% effect unlike the case in atoms where it is a 1% effect." I don't understand this part $\endgroup$ – Mockingbird Jul 17 '17 at 7:36
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    $\begingroup$ @Mockingbird In the binding of electrons to atoms, the spin-orbit splitting is of order $(v/c)^2$ or about 1% of the total binding energy for electrons in low Z atoms. This binding comes entirely from the attractive Coulomb interaction (photon exchange). In nuclei there is a force similar to the Coulomb force in atoms. It comes from the exchange of the $\omega$ meson and there is an accompanying spin-orbit splitting from this force (also about 1% or the central potential contribution because nucleons in a nucleus also move slowly like electrons in an atom). $\endgroup$ – Lewis Miller Jul 17 '17 at 13:09
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    $\begingroup$ ... The central potential contribution of the $\omega$ is repulsive, so something else must be responsible for the binding. That is the contribution of the $\sigma$ meson (a neutral scalar or spin 0 particle). This force also has central and spin-orbit components. The central component is attractive and cancels the repulsion of the $\omega$ force leaving an overall central attraction. The spin-orbit components of these two forces are additive due to the difference between scalar ($\sigma$) and vector ($\omega$) exchanges. It just works out that the overall spin-orbit splitting is 10%. $\endgroup$ – Lewis Miller Jul 17 '17 at 13:21
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Spin dependence of nuclear force is highly non trivial. As an example, exceptional stability of some particular nucleus with nucleon number (also known as magic number) 2, 8, 20, 28, 50, 82, and 126 can be explained by considering the spin-orbit interactions between nucleons.

To see the role of spin dependence, the deuteron nucleus is a good laboratory. The total angular momentum for the deuteron (or in general for a nucleus) is usually denoted by $I$, is given by
$$\hat{\vec{I}} = \hat{\vec{L}}+\hat{\vec{S_{p}}}+\hat{\vec{S_{n}}} \tag{1}$$

For the bound deuteron state $L = 0$ and $I=\hat{\vec{S_{p}}}+\hat{\vec{S_{n}}}=\hat{\vec{S}}$. A priori we can have $\hat{\vec{S}}=0$ or $1$ (following the rules for addition of angular momentum, with $\hat{\vec{S_{p,n}}}=1/2$).

The simplest form that a spin-dependent scalar potential could assume is $$V_{\text{spin}} = \frac{V(r)}{\hbar^{2}}\left(\hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}\right)\tag{2}$$

There we assumed that, how $V_{\text{spin}}$ has spatial dependence is encoded in $V(r)$. Now, in terms of Casimir's $V_{\text{spin}}$ can we written as, $$V_{\text{spin}}=\frac{V(r)}{2\hbar^{2}}\left(\hat{\vec{S^{2}}}-\hat{\vec{S_{p}^{2}}}-\hat{\vec{S_{n}^{2}}}\right)\tag{3}$$ For generic eigen functions $|S,S_{p},S_{n},S_{z}\rangle$, the expentation value of $\hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}$ is given by $$\langle \hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}\rangle=\frac{\hbar^{2}}{2}\left(S(S+1)-S_{p}(S_{p}+1))-S_{n}(S_{n}+1)\right) $$ Now, since $S_{p,n}=1/2$, we obtain $$\langle \hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}\rangle = \frac{\hbar^{2}}{2}\left(S(S+1)-\frac{3}{2}\right)=\begin{cases} +\frac{\hbar^{2}}{4}\quad\text{triplet state},\quad|S=1,\frac{1}{2},\frac{1}{2}\rangle \\ -\frac{3\hbar^{2}}{4}\quad\text{singlet state},\quad|S=0,\frac{1}{2},\frac{1}{2}\rangle \end{cases}$$

Now if $V(r)$ is attractive potential (i.e. $V(r)<0$) then

  • strength of $V_{\text{spin}}$ increases for $S=1$,
  • while it is reduced for $S=0$ configuration.
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Because of nature. Together with their positions, their isospins etc., the spins of the nucleons are degrees of freedom you might want to consider. Since the aim is to reproduce experimental results, as @AMS said, using a potential which has the right spin dependence you are able to get some reasonable values for the various observables.

For instance, in the specific case of the deuteron, in a simplified version, you know that the total spin is $S=1$ and the total isospin is $T=0$.

You can describe the nucleon-nucleon interaction by a potential of the form $$ V_{NN}=V_C(r)+V_\sigma(r)\vec{\sigma}_1\cdot\vec{\sigma}_2+V_\tau(r)\vec{\tau}_1\cdot\vec{\tau}_2+V_{\sigma\tau}(r)\vec{\sigma}_1\cdot\vec{\sigma}_2\vec{\tau}_1\cdot\vec{\tau}_2. $$ Where the first term is the Colombian part, the second depends on the spin, the third on the isospin and the last one on both.

Since $S=1$ and $T=0$ you get that $$ \vec{\sigma}_1\cdot\vec{\sigma}_2|\psi_d\rangle=|\psi_d\rangle,\;\;\;\;\vec{\tau}_1\cdot\vec{\tau}_2|\psi_d\rangle=-3|\psi_d\rangle\;\;\;\;\text{and}\;\;\;\;\vec{\sigma}_1\cdot\vec{\sigma}_2\vec{\tau}_1\cdot\vec{\tau}_2|\psi_d\rangle=-3|\psi_d\rangle, $$ where $|\psi_d\rangle$ is the wavefunction of the deuteron. Now that you have the potential for this simplified model of the deuteron (with only S wave), you can solve the Schrödinger equation and get a value for the binding energy. In this way you can test your phenomenological potential and verify that a spin dependence is needed.

Maybe the answer is the the first three words, but I hope that looking at an example might be more clarifying.

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protected by Qmechanic Oct 25 '16 at 12:26

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