On the microscopic origin of deformation + paper reading

February 4, 2023

GoLuckyRyan Basic , , , , , , , Leave a comment

Nuclear deformation is usually mimicked by macroscopic deformation potential, for example, a quadrupole deformation potential as in the Nilsson model. This provides a description without understanding the origin of the deformation potential. This macroscopic description is particularly unsatisfactory for the deformed light nuclei. Also, it does not answer the question that, are the whole nuclei deformed? or just the surface?

The paper Towards a unified microscopic description of nuclear deformation by P. Federman and S. Pittel [Physic Letters B 69, 385 (1977)] provides an interesting and reasonable understanding of the microscopic description.

The paper starts by comparing 20Ne and 20O. 20Ne is well deformed with \beta_2 = 0.7 and 20O is less deformed with \beta_2 = 0.27. Let’s assume the 16O core is inert in both nuclei. The difference between 20Ne and 20O is two protons in the sd-shell are replaced with two neutrons from 20Ne to 20O. The paper said, “The conclusion seems clear. Deformation in light nuclei is due to the T=0 neutron-proton interaction.“.

As we know from the two-nucleon system, there are 4 possible combinations with 2 nucleons. In those combinations, the T=0, J=1 pn pair is the only bound pair due to the tensor force. And the pair is not spatial isotropic. In a many-nucleons nucleus, a similar thing happens for NN pair. In 20O, and 20Ne, all valence nucleons are in d5/2-orbitals in the simplest picture. A T=0 pn pair, the spin of the proton and neutron must be aligned, which means the proton and neutron are orbiting in the same direction, and the total spin of the pn pair is J = 5, which is very spatially deformed. And in 20O, all valence nucleons are neutrons, only able to form T=1, J=0 nn pairs, in which the neutrons are orbiting oppositely and spatially spherical. The result is 20Ne is deformed and 20O is spherical, and the ultimate reason is the tensor force tends to make T=0 pn pair.

Wait… the total spin of 20Ne and 20O are both Zero. If it is the T=0 pn pair and the spin is not Zero, would the spin of 20Ne be non-zero? For even-even nuclei, protons and neutrons are paired up and formed J=0 pp and nn pair. Where are the T=0 pn pairs? The key is that, although the pp and nn are paired up, it does not exclude the T=0 pn pair.

For simplicity, let’s check the Slater determinate for 3 fermions. Suppose the 3 nucleon wave functions are p_a, p_b, n_a, where p, n stands for proton and neutron, \alpha, \beta are spin-up and spin-down. And all wavefunctions are in the same orbital.

\displaystyle \Psi = \frac{1}{3!} \begin{vmatrix} p_a(1) & p_b(1) & n_a(1) \\ p_a(2) & p_b(2) & n_a(2) \\ p_a(3) & p_b(3) & n_a(3) \end{vmatrix}

without loss of generality, we can collect terms of the 1-th particle.

\displaystyle 3 \Psi = p_a(1) \frac{p_b(2) n_a(3) - p_b(3) n_a(2) }{2} + \\ ~~~~~~~~p_b(1) \frac{p_a(2) n_a(3) - p_a(3) n_a(2) }{2} + \\~~~~~~~~ n_a(1) \frac{p_a(2) p_b(3) - p_a(3) p_b(2) }{2}

\displaystyle 3 \Psi = p_a(1) ( pn, T=1, J = 0 ) + \\~~~~~~~~~ p_b(1) (pn, T=0, J =1) +\\~~~~~~~~~ n_a(1) (pp, T=1, J= 0)

Now, imagine it is the triton wave function. We can see that the total wave function contains a pp J=0 pair, but there are also pn J=0 and J=1 pair. The detailed coupling of the total spin of the total wave function involves CG coefficient, and the pn T=0, J=1 pair should be coupled to a spin-down proton (j=1/2) and form J = 1/2.

We can imagine that in a wavefunction with 2 protons and 2 neutrons, there will be T=0 pn pair and T=1 pp/nn pair. While a wavefunction with 4 neutrons can never form pn pair.

In this simple demonstration, it is simply forming the wave function without considering the nuclear force. There are two questions: 1) In the Slater determinate, what is the percentage for T=0 and T=1 NN pairs? 2) with the nuclear force, what is that percentage?

The paper Probing Cold Dense Nuclear Matter by R. Subedi et al., Science 320, 1476 (2008), could give us some hints. The study found that in 12C, there are 18% pn pairs and only 2% of pp or nn pairs. There are more recent developments on the topic, for example, PRL 121, 242501 (2018), PLB 820, 10 (2021).

P. Federman and S. Pittel apply the same idea (deformation caused by T=0 pn pair from tensor force) on 100Zr, a very different and complex nucleus than 20Ne. 98Zr has Z = 40 and N = 58 and is spherical. The proton shell is semi-closed at 1p1/2 orbital. and the neutron shell is semi-closed at 2s1/2 ( on top of N=40, 0g9/2, 1d5/2, 2s1/2 ). But 100Zr is highly deformed with \beta_2 \approx 0.35 . The nuclear shape changed so much by just 2 neutron differences has drawn a lot of attention since its discovery in the 70s. The Interacting Boson Model and Shell model calculation has been tried to compute this sudden transition with quite good results, but there are still many discrepancies on the microscopic origin of the deformation. For example, is the deformation driven by protons or neutrons? and also what is the configuration.

As we mentioned before, the T=0 pair could be causing the deformation. In the case of 100Zr, it is the g-orbital pn pairs. Other proposed mechanisms are core polarization of 98Zr and the presence of a valence neutron in 0h11/2 orbital. In a recent experimental proposal, I wrote the following:

The above three mechanisms are intertwined. The interplay between these mechanisms is illustrated in Figure. 1. The neutrons in the 0g7/2 orbital lower the proton 0g9/2 binding energy while increasing the binding energy of the proton 1p1/2 and 0f5/2 with the action of the tensor force (attractive for J< − J> pair and repulsive for J< − J< or J> − J> pair), which reduces or even breaks the pf-g Z = 40 shell-gap, favoring the promotion of the protons to the 0g9/2 orbital from the pf-shell, and creates a core polarization and deformation. The core polarization of Z = 40 core promotes protons into the π0g9/2 orbital, enabling the coupling with the 0g7/2 neutron and forming T = 0 p-n pairs under the influence of tensor interaction. The g-orbital T = 0 p-n pair have their spins aligned and create a strong quadrupole deformation. Also, the presence of 0g9/2 protons lowers the effective single-particle energies (ESPEs) of the neutron 0g7/2 and 0h11/2 orbital via the so-called Type II shell evolution, which increases the occupancy for the neutron 0h11/2 orbital. The presence of 0h11/2 neutrons in turn provides a strong quadrupole deformation force. The deformation then enhances the occupation of valence orbitals and fragmentation in single-particle energies in return.

This is a bit complicated and hard to prove. But the tensor force plays an important role here. Without such, the chain reaction of the nucleon reconfiguration would not happen.

\beta_2 values around 100Zr.

We can see the A=100 nuclei, 100Mo has 2 protons at 0g9/2 and N=58, it is deformed with \beta_2 = 0.23. A deformation could promote neutrons to 0g7/2. 102Mo has 2 protons at 0g9/2 and 2 neutron at 0g7/2, so it is deformed with \beta_2 = 0.31 (Would 102Mo be more deformed?) 102Ru has 4 protons at 0g9/2, and N = 58. It is slightly deformed with \beta_2 = 0.17. 102Pd has 6 proton at 0g9/2 and N = 54 with \beta_2 = 0.14. In the opposite direction, 100Sr (Z=38, N=62) is very deformed with \beta_2 = 0.41.

Also, Z = 38 or Z = 42, all isotopes are deformed. Clearly, Z = 40 and N < 60 are NOT deformed and are the ANOMRALY. The shell Z = 40 closure clearly forbids proton-shell configuration mixing.

In the above, we always use \beta_2 as a measure or indicator for deformation. But the \beta_2 of 16O is 0.35. Is 16O not spherical? That is exactly the reason why deformation is “hard” to understand for light nuclei. In my opinion, 16O is not spherical, as \beta_2 is pretty much a measure of the geometrical shape. However, 16O is shell-closure and has almost no configuration mixing (there are ~ 10% sd-shell components). And the shape deformation is caused by the T=0 pn pairs. However, where is the rotational band of 16O? Another thing is, in light nuclei, deformation, and configuration mixing can be separated. Configuration mixing will lead to deformation, but the reverse is not always true.

In fact, all light nuclei are more or less deformed!! Within the sd-shell, beta_2 > 0.1 and the most spherical nuclei is 40Ca. Also also, I think all even-odd nuclei are deformed as the unpaired nucleon has a deformed orbital.

The known smallest beta_2 nucleus is 206Pb with a value of 0.03.

In the case of 20Ne, the deformation could have different effects on different orbitals. i.e. the mean field for each nucleon could be different.

How to apply this idea? and how to predict the degree of deformation? Is it the only mechanism?

I should calculate the beta_2 for each orbital….

Electromagnetic origin of spin-spin interaction

November 29, 2020

GoLuckyRyan Basic Leave a comment

In this post, we calculated the spin-spin interaction. In this post, we are going to derive the form of the interaction. Why the interaction takes such a “complicated” form:

\displaystyle H_T = \frac{1}{r^3}\left( \vec{\sigma_1}\cdot \vec{\sigma_2} - 3 (\vec{\sigma_1} \cdot \hat{r}) (\vec{\sigma_2} \cdot \hat{r}) \right)

In general, the magnetic vector potential for a given current \vec{J} is

\displaystyle \vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int{\frac{\vec{J}(\vec{r'})}{|\vec{r}-\vec{r'}|}}d^3 x'

expand the 1 / |\vec{r} - \vec{r'}|

\displaystyle \frac{1}{|\vec{r}-\vec{r'}|} = \frac{1}{r} + \frac{\vec{r}\cdot \vec{r'}}{r^3} + ...

\displaystyle \vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \left( \frac{1}{r} \int{\vec{J}(\vec{r'})}d^3 x' + \frac{1}{r^3} \int{ (\vec{r} \cdot \vec{r'}) \vec{J}(\vec{r'})}d^3 x' + ... \right)

If the current \vec{J} is inside the integration volume, the first term will be zero. This assumption is true when the observation point \vec{r} is far far away from the current \vec{J} . And the 2nd term is the dipole term.

For a closed loop, and \vec{J} = I d\vec{r'} , we claim that ( I have no proof, but tested on few cases. ),

\int{ (\vec{r} \cdot \vec{J}) \vec{r'} }d^3x' = - \int{ (\vec{r} \cdot \vec{r'}) \vec{J} }d^3x'

Thus,

\displaystyle \int{ ( \vec{r} \cdot \vec{r'}) \vec{J} d^3 x' } = \frac{1}{2} \int{ ( \vec{r} \cdot \vec{r'}) \vec{J} - ( \vec{r} \cdot \vec{J}) \vec{r'} d^3 x' }

using vector identity,

\displaystyle \vec{A} \times \left( \vec{B} \times \vec{C} \right) = \left( \vec{A} \cdot \vec{C} \right) \vec{B} - \left( \vec{A} \cdot \vec{B} \right) \vec{C}

thus,

\displaystyle \int{ ( \vec{r} \cdot \vec{r'}) \vec{J} d^3 x' } \\ = \frac{1}{2} \int{ ( \vec{r} \cdot \vec{r'}) \vec{J} - ( \vec{r} \cdot \vec{J}) \vec{r'} d^3 x' } \\ = \frac{1}{2} \int{ \vec{r} \times \left( \vec{J} \times \vec{r'} \right) d^3 x' } \\ = \frac{1}{2} \int{ \left( \vec{r'} \times \vec{J} \right) \times \vec{r} d^3 x' }

Define the dipole moment as

\displaystyle \vec{\mu} = \frac{1}{2} \int{ \left( \vec{r'} \times \vec{J} \right) d^3 x' }

Then, the vector potential up to dipole moment is

\displaystyle \vec{A} = \frac{\mu_0}{4\pi} \vec{\mu} \times \frac{\vec{r}}{r^3} + ...

The magnetic field is

\displaystyle \vec{B} = \nabla \times \vec{A} = \frac{\mu_0}{4\pi} \nabla \times \left( \vec{\mu} \times \frac{\vec{r}}{r^3} +... \right)

using another vector identity,

\displaystyle \nabla \times (\vec{a} \times \vec{b} ) = \vec{a} (\nabla \cdot \vec{b} ) - \vec{b} (\nabla \cdot \vec{a} ) + (\vec{b} \cdot \nabla) \vec{a} - (\vec{a} \cdot \nabla) \vec{b}

\displaystyle \vec{B} =\frac{\mu_0}{4\pi} \left( \vec{\mu} (\nabla \cdot \frac{\vec{r}}{r^3} ) - \frac{\vec{r}}{r^3} (\nabla \cdot \vec{\mu} ) + (\frac{\vec{r}}{r^3} \cdot \nabla) \vec{\mu} - (\vec{\mu} \cdot \nabla) \frac{\vec{r}}{r^3} \right)

Since the dipole moment is a constant,

\displaystyle \vec{B} =\frac{\mu_0}{4\pi} \left( \vec{\mu} (\nabla \cdot \frac{\vec{r}}{r^3} ) - (\vec{\mu} \cdot \nabla) \frac{\vec{r}}{r^3} \right)

The first term is the divergence of inverse field \hat{r}/r^2 , which is 4 \pi \delta(r) . The second term is a directional derivative,

\displaystyle (\vec{\mu} \cdot \nabla) \frac{\hat{r}}{r^2} = \mu_x \frac{\partial}{\partial x} \frac{\hat{r}}{r^2} + \mu_y \frac{\partial}{\partial y} \frac{\hat{r}}{r^2}+ \mu_z \frac{\partial}{\partial z} \frac{\hat{r}}{r^2}

\displaystyle \frac{\partial}{\partial x} \frac{\hat{r}}{r^2} = \frac{\partial}{\partial x} \left( \frac{x}{(x^2+y^2+z^2)^{3/2}}, \frac{y}{(x^2+y^2+z^2)^{3/2}}, \frac{z}{(x^2+y^2+z^2)^{3/2}} \right) \\ = \left( \frac{r^2 - 3 x^2}{r^5}, \frac{-3xy}{r^5}, \frac{-3xz}{r^5} \right)

Thus, for the x-component

\displaystyle \frac{r^2 - 3 x^2}{r^5} \mu_x + \frac{-3xy}{r^5} \mu_y +\frac{-3xz}{r^5} \mu_z = \frac{1}{r^5} \left( r^2 \mu_x - 3x (\vec{r} \cdot \vec{\mu} ) \right)

Therefore, the direction derivative is

\displaystyle (\vec{\mu} \cdot \nabla) \frac{\hat{r}}{r^2} =\frac{1}{r^5} \left( r^2 \vec{\mu} - 3 \vec{r} (\vec{r} \cdot \vec{\mu} ) \right) = \frac{1}{r^3} \left(\vec{\mu} - 3 \hat{r} ( \hat{r} \cdot \vec{\mu} ) \right)

We have the magnetic field generated by a dipole is

\displaystyle \vec{B} =\frac{\mu_0}{4\pi} \left( 4\pi \delta(r) \vec{\mu}+ \frac{3 \hat{r} ( \hat{r} \cdot \vec{\mu} ) - \vec{\mu}}{r^3} \right)

If we set \vec{\mu} = \hat{z}, the field look likes

\displaystyle \vec{B} = \frac{1}{r^3}(3 \cos\theta \sin\theta, 3\cos^2\theta -1 )

Plotting r^3 \vec{B}.

The Hamiltonian of a dipole \vec{\mu_1} under a magnetic field is

\displaystyle H_B = - \vec{\mu_1} \cdot \vec{B}

the minus sign indicates the dipole has to anti-parallel to the field for minimum energy. Now, the magnetic field is generated by another dipole \vec{\mu_2} in “spin-spin” interaction, therefore, the Hamiltonian is

\displaystyle H_B = \frac{\mu_0}{4\pi} \left( \frac{ \vec{\mu_1} \cdot \vec{\mu_2} - 3 (\hat{r} \cdot \vec{\mu_1}) ( \hat{r} \cdot \vec{\mu_2}) }{r^3} - 4\pi \delta(r) \vec{\mu_1} \cdot \vec{\mu_2} \right)

The first term is the spin-spin interaction when r \neq 0 , and the second term is the contact term when the two dipole collided. However, the contact term is not “correct” because the volume integration of the magnetic field is 2/3 \mu_0 \vec{\mu} instead of \mu_0 \vec{\mu} . This is due to the dipole momentum at r=0 . (I integrated the magnetic field of single coil, the dipole moment is \pi a^2 , where a is the coil radius and current of 1 is used. The numerical integration give 1.06 \times 2/3\pi a^2 , 6% difference and far from \pi a^2. ) Thus, the total Hamiltonian is

\displaystyle H_B = \frac{\mu_0}{4\pi} \left( \frac{ \vec{\mu_1} \cdot \vec{\mu_2} - 3 (\hat{r} \cdot \vec{\mu_1}) ( \hat{r} \cdot \vec{\mu_2}) }{r^3} - \frac{8\pi}{3} \delta(r) \vec{\mu_1} \cdot \vec{\mu_2} \right)

Spin-spin (Tensor) interaction

October 14, 2020

GoLuckyRyan Basic , , , , 9 Comments

The spin-spin interaction, also called as tensor interaction, as the strength depends on the relative direction of the spins. The mathematical form of the interaction is

\displaystyle V_T(r) = \frac{1}{r^3} \frac{3(\vec{\sigma_1}\cdot\vec{r_1})(\vec{\sigma_2}\cdot\vec{r_2})-r^2(\vec{\sigma_1}\cdot\vec{\sigma_2})}{r^2}

For spin-1/2, we have.

\displaystyle \vec{\sigma} = \left(\sigma_x, \sigma_y, \sigma_z \right) = \left( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right)

The \vec{\sigma} is also called Pauli matrices.

The term \vec{\sigma}\cdot \vec{r} , we can expand,

\displaystyle \vec{\sigma}\cdot \vec{r} = \sigma_x x + \sigma_y y + \sigma_z z = \begin{pmatrix} z & x + i y \\ x-iy & -z \end{pmatrix}

If we express in spherical coordinate,

\displaystyle \vec{\sigma}\cdot \vec{r} =r \begin{pmatrix} \cos(\theta) & \sin(\theta)e^{i\phi} \\ \sin(\theta)e^{-i\phi} & - \cos(\theta) \end{pmatrix}

The eigen values are 1 and -1, with eigen vectors are

\displaystyle v_{-1} = (\sin(\theta), -e^{-i\phi}\cos(\theta) )

\displaystyle v_{1} = (\cos(\theta), e^{-i\phi}\sin(\theta) )

with

\displaystyle v_i \cdot v_j^* = \delta_{ij}

The v_{1} is a “condensed” vector of the \vec{r'} = (\cos(\theta), \sin(\theta) \sin(\phi), \sin(\theta) \cos(\phi)) = (z, y, x) . Since only the the z-component is known, the x,y-components condensed as a phase \phi.

The term (\vec{\sigma_1}\cdot\vec{r_1})(\vec{\sigma_2}\cdot\vec{r_2}) is a tricky one. It involves 2 spins in two spaces. The product has 9 terms, \sigma_1(i) \sigma_2(j) r_i r_j . The \sigma_1 and \sigma_2 act on particle-1 and particle-2 separately. In fact, when we talk about a 2-spin system, we can form a direct-product space, \sigma_1(i) \sigma_2(j) = \sigma_1(i) \otimes \sigma_2(j) , in matrix from, for example

\sigma_1(x) \otimes \sigma_2(y) = \begin{pmatrix} 0 & \sigma_y \\ \sigma_y & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & i \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{pmatrix}

The states in the direct-product space are

\displaystyle \left|\uparrow \uparrow \right> = (1, 0, 0, 0)

\displaystyle \left|\uparrow \downarrow \right> = (0, 1, 0, 0)

\displaystyle \left|\downarrow \uparrow \right> = (0, 0, 1, 0)

\displaystyle \left|\downarrow \downarrow \right> = (0, 0, 0, 1)

So, the eigen states for total spin 1 are (1, 0, 0, 0), \frac{1}{\sqrt{2}}(0, 1, 1, 0), (0,0,0,1) , and for total spin 0 are \frac{1}{\sqrt{2}} (0, 1, -1, 0) . We set

\displaystyle \chi = \left( (1, 0, 0, 0), \frac{1}{\sqrt{2}}(0, 1, 1, 0), (0,0,0,1) , \frac{1}{\sqrt{2}} (0, 1, -1, 0) \right)

In similar fashion, the term \vec{\sigma_1}\cdot \vec{\sigma_2} = \sigma_1(x)\otimes\sigma_2(x) + \sigma_1(y)\otimes\sigma_2(y)+\sigma_1(z)\otimes\sigma_2(z)

\displaystyle \vec{\sigma_1}\cdot \vec{\sigma_2} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}

Combine everything, we have

\displaystyle \frac{3(\vec{\sigma_1}\cdot\vec{r_1})(\vec{\sigma_2}\cdot\vec{r_2})-r^2(\vec{\sigma_1}\cdot\vec{\sigma_2})}{r^2} =\\ \frac{1}{r^2}\begin{pmatrix} 3z^2-r^2 & 3z(x+iy) & 3z(x+iy) & 3(r^2 - 2y^2-z^2+2ixy) \\ ... & r^2 - 3z^2 & r^2-3z^2 & -3z(x+iy) \\ ... & ... & r^2-3z^2 & -3z(x+iy) \\ ... & ... & ... & 3z^2 - r^2 \end{pmatrix}

Next, it is easy to calculate the matrix elements of the spin-spin interaction with the states \chi.

\displaystyle \left<\chi| V_T | \chi \right> = \frac{1}{r^2}\begin{pmatrix} 3z^2-r^2 & 3\sqrt{2} z(x+iy) & 3(x+iy)^2 & 0 \\ ... & 2(r^2 - 3z^2) & -3\sqrt{2}z (x+iy) & 0 \\ ... & ... & 3z^2-r^2 & 0 \\ ... & ... & ... & 0 \end{pmatrix}

We can see that, all matrix elements with spin-0 are 0. It is what we expect, as the spin-0 result no tensor force, as no spin in the system. For the diagonal elements:

\displaystyle \left< \chi_{Sm_S}| V_T | \chi_{Sm_S} \right> = \begin{cases} 2\left(1- 3 \frac{z^2}{r^2} \right), & S=1, m_S= 0 \\ 3 \frac{z^2}{r^2} -1, & S=1, m_S=\pm 1\\ 0, & S=0 \end{cases}

We can also calculate the matrix element of the \vec{\sigma_1}\cdot \vec{\sigma_2},

\left< \chi | \vec{\sigma_1}\cdot \vec{\sigma_2} |\chi \right> = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}

we can also calculate in other way. First, we convert to the spinor that we familiar,

s_i = \frac{1}{2} \sigma_i

where the eigen value are 1/2, which is the correct eigen value for spin-1/2 system. And set

S = s_1 + s_2

so that the eigen values are 1 and 0.

\vec{\sigma_1}\cdot \vec{\sigma_2} = 4 s_1 \cdot s_2 = 2\left( S^2 - s_1^2 - s_2^2 \right)

Thus, we get

\displaystyle \left< \chi| \vec{\sigma_1}\cdot \vec{\sigma_2} | \chi\right> = \begin{cases} 1, & S=1\\ -3, & S=0 \end{cases} = 4 S - 3

This agrees with the matrix method.

The diagonal result can be written in spherical coordinate,

\displaystyle \left< \chi_{Sm_S}| V_T | \chi_{Sm_S} \right> = \begin{cases} 2\left(1- 3 \cos^2(\theta) \right), & S=1, m_S= 0 \\ 3 \cos^2(\theta) -1, & S=1, m_S=\pm 1\\ 0, & S=0 \end{cases}

We can see the angular dependent of the interaction. Let plot the 3 \cos^2(\theta) -1 ) for m_S = \pm 1 in below

At \theta = 0, \pi the force is strongest. This can picture as two magnets, aligned to the z-axis (m_S=\pm1). When the are aligned tail to head, or \theta = 0, \pi, the attraction is highest. At \theta = \pi/2 , when their head-to-head, tail-to-tail, they are repulsive.

Consider this distortion effect in deuteron. The proton and neutron couples to spin-1 and both are in the s-orbitals. Due to the tensor force, this s-orbital is no longer an eigenstate and it squeezes the orbital along the z-axis due to attraction and expands the x-y plan due to the repulsion. The total wave function would be oblate. Thus, the total wave function has to be mixed with the next orbital with the same parity, which is the d-orbital. And this is the case for m_s = \pm1.

For the case of m_s = 0 , the situation reversed, the tensor force expands the wavefunction along the z-axis and squeezes it on the x-y plane, making the total wavefunction be prolate. And since the magnitude is 2 times stronger for the m_s = 0 state than that of the m_s = \pm1 state. The deformation would be stronger.

[update 20230718]

In the case the total spin S= 1, m_S=0, the situation is like this:

This geometrical illustration explained why the tensor force change sign.

However, the x-y component of the total spin vector is unknown, so, the total spin can be rotated 90 degrees and look like this:

In this case, all tensor force is repulsive. Thus, the 2-spin system is not an axial deformed but also deformed on the x-y plane. However, after taking an average of all possible rotations, the strength on the x-y plane should be zero, and the deformation is axial.

We can see the tensor force, no matter how the spin is orientated, whenever tip to toe, it is attractive. whenever in parallel, repulsive. The m_S or the z-axis is artificially imposed so that we can do the calculation and make the tensor force operational. Or to say, the tensor force is rotationally symmetric.

The reason why the strength for the m_s = 0 case is twice that of the m_s = 1 is due to the internal degree of freedom on the x-y plane. When we sum up the strength, for m_s = 1, \theta = 0, \phi = 0, only 1 configuration is possible and has a strength of 2, but for m_s = 0, there are 4 configurations that have a strength of -1 each and sum up to -4. Similar for the \theta = \pi/2 as illustrated below.

But the problem is, the spin can rotate in any direction on the x-y plane, not just 4. Should it be 2\pi more instead of 2? Any better explanation?

In this post, the radial component of the tensor force AV18 p-n potential is negative, which explained why most nuclei tend to be prolated (longer in the z-axis) for axial deform.

Since different m_s states should degenerate without any external magnetic field. So that the different m_s states should be equally occupied. And the strength of the m_s = 0 state is twice that of the m_s = \pm 1 states, thus, the deformation effect along the z-axis of the tensor force should be canceled, but not on the x-y plane. As we discussed, for the m_s = 0 state, the x-y plane has no force on average due to rotation, but that is not true for the m_s = \pm1 states, the x-y plane is always attractive ( given the radial tensor force is negative), thus, the x-y plane is contracted, and make the whole nuclei prolate.

PRL paper

May 26, 2020

GoLuckyRyan Basic , , , , Leave a comment

My PhD work is finally published today!

https://doi.org/10.1103/PhysRevLett.124.212502

https://phys.org/news/2020-05-proton-world-difference.html

There are few messages in this paper:

  1. Since the 24O is a doubly magic, by adding an extra proton, a conventional thinking would be no change of the magicity. i.e. when a proton is being knockout from 25F, the SF of 24O ground state would be large and not fragmented.
  2. But experiment on 25F(p,2p) tell the otherwise, the strength of 24O is small comparing to 24O excited state. That means the 25F composites a proton + (24O ground state and excited states ).
  3. Thus, there is large configuration mixing in the neutron sd-shell of 25F. This could means the end of N=16 magicity
  4. Using (p,2p) reaction to study neutron shell is a new method. This methodology is working because of the proton shell is a magic + 1 structure. Thus, the same method could be used to study 41Sc, 49Sc, etc.

Meson Theory on Strong Nuclear Force

January 20, 2016

GoLuckyRyan Basic, nuclear, theory , , , , Leave a comment

The theory is very difficult, I only state the result.

There are 4 main mesons for strong nuclear force.

type mass [MeV] range symmetry force type
 \pi  135  long  pseudo scalar  -\sigma_{1}\cdot\sigma_{2},-S_{12}
 \sigma  400 ~ 2000  medium  scalar -1, -L\cdot S
 \rho  775  short  vector  -2\sigma_1\cdot\sigma_2, +S_{12}
 \omega  783  short vector  +1, -3L\cdot S

Lorentz Force and Stress tensor

March 7, 2011

GoLuckyRyan others, relativity, theory, Working on , , , 1 Comment

F_{ij} =\begin {pmatrix} 0 & D_1 & D_2 & D_ 3\\ -D_1 & 0& H_3 & -H_2 \\ -D_2 & - H_3 & 0 & H_1 \\ -D_3 & H_2 & -H_1 & 0 \end {pmatrix}

G_{ij} =\begin {pmatrix} 0 & B_1 & B_2 & B_ 3\\ -B_1 & 0& -E_3 & E_2 \\ -B_2 & E_3 & 0 & -E_1 \\ -B_3 & -E_2 & E_1 & 0 \end {pmatrix}

The field equation are:

\partial /\partial x_i F_{i j } = - J_i

\partial /\partial x_i G_{i j } =0

That is the result from last time.

the conservation of charge is:

\partial/\partial x_i J_i = 0

thus the 4-Laplacian of the F-field is :

\partial^2/\partial x_i^2 F_{ij} = 0

The physical meaning of the simplification of the field equation by the field tensor is, a gradient in the tensor field is equation to the minus of 4-current, or zero. recall that the gradient in 3-D vector space, the conservation of charge density is :

\nabla \cdot \vec{J} = - \partial \rho/\partial t

we have the same form in the 4-D tensor space. the creation of field is conservation of the 4-charge displacement, if we integrate the 4-current. i dun know what physical meaning of the G-field. personally, i believe that the F-field and G-field can be related by some transform.

we have another interesting things. we can write the Lorentz force into the field tensor:

\vec{f}_j = \frac{d\vec{P}_j}{d\tau}=q \frac{d\vec{X}_j}{d\tau}F_{ij}

the reason why we can write this, i don’t know. any physical meaning? i don’t know. may be we can think in this way, the force depends on the motion of the 4-vector and the field and the charge. thus, it is natural to multiple them together to get the force. But why not the G field? never the less, the field tensor reduce the number of Field qualities into 2.

the Lorents Force can be more simple

\vec{f}_j = J_i F_{ij}

that the force is created by the field and the current.

The Electromagnetic stress tensor can also be related with the 4-force by :

\vec{f}_j = - \partial/\partial x_i T_{ij}

Thus, combined with the Field tensor:

\partial/\partial x_iT_{ij}+ J_iF_{ij} = 0