AV18 | Nuclear Physics 101

Deuteron Properties from AV18

May 9, 2023

GoLuckyRyan Basic AV18, deuteron Leave a comment

Haven’t posted anything in 2 months!! I have been focused to develop a few GUI DAQ programs using Qt6. And, I am buying a house! Anyway~

From the last post, we calculated the Deuteron wave function. The d-wave probability is $\int w^2(r) dr$ and it is 5.8 % (experimental is 6.8%). The binding energy is -2.224574 MeV while the experimental value is -2.224575 MeV, only 1 eV difference!

The full wave function is

$\displaystyle \Phi_M(r) = \frac{u(r)}{r} Y_{00} (\Omega) \chi_M + \frac{w(r)}{r} \sum_m C^{1M}_{2m1(M-m)} Y_{2m} (\Omega) \chi_{M-m}$

where $M= -1, 0, 1$ , The angular integrated square of wave function is

$\displaystyle \int \Phi_M(r) \Phi_M^*(r) d\Omega = \frac{1}{r^2} ( u^2(r) + w^2(r) )$

The matter radius or the root-mean-square radius is

$\displaystyle \left<r^2 \right> = \frac{1}{2} \sqrt{\int_{0}^{\infty} (u^2(r) + w^2(r) ) r^2 dr } = 1.967~ \textrm{fm}$

The factor of 2 is due to transforming from the relative motion to the center of mass motion and the masses of the proton and neutron are almost the same.

The dipole moment of deuteron has 2 contributions, one from the proton and neutron intrinsic spin, and the second part is from the orbital motion of the d-wave. The dipole moment is

$\displaystyle \vec{\mu} = \mu_N \left( \left(g_L \frac{\vec{L}}{2} + g_S \vec{S} \right) \cdot \vec{J} \right) \vec{J}$

The orbital part needs to be half in CM frame than the relative frame, or $g_l = 1/2$ .

The total spin $\vec{S}$ is

$\displaystyle \vec{\mu_S} = \mu_N \left( \frac{g_p \vec{s_p} + g_n \vec{s_n}}{S(S+1)} \cdot \vec{S} \right) \vec{S} = \mu_N g_S \vec{S}$

The $S(S+1)$ is the magnitude of the vector $\vec{S}$ . It is the same as projecting a classical vector $\vec{u}$ to the direction of $\vec{w}$ , we have to divide by $|w|^2$ as $\vec{w}$ is not a uni-vector.

$\displaystyle \vec{s_p} \cdot \vec{S} = \frac{1}{2} \left( S (S+1) + s_p(s_p+1) - s_n(s_n+1) \right) = 1, S = 1$

Thus,

$\displaystyle \vec{\mu_S} = \mu_N g_S \vec{S} = \mu_N \frac{g_p + g_n}{2} \vec{S}$

The projection to $\vec{J}$ is

$\displaystyle \frac{\vec{\mu}}{\mu_N} = \frac{1}{J(J+1)} \left( \frac{1}{2} \vec{L} \cdot \vec{J} + \frac{g_p + g_n}{2} \vec{S} \cdot \vec{J} \right) \vec{J}$

For $L = 0, S = 1, J = 1$ , we have

$\displaystyle \vec{\mu} = \mu_N \frac{g_p + g_n}{2} \vec{J}$

Thus, the magnetic moment is

$\displaystyle \mu = \left< \vec{\mu} \cdot \vec{J} \right> = \mu_N ( g_p + g_n)$

For $L = 2, S = 1, J = 1$ , we have

$\displaystyle \vec{\mu} = \mu_N \frac{1}{2}\left( \frac{3}{4} - \frac{g_p + g_n}{2} \right)\vec{J}$

Now, the magnetic moment is the sum of L=0 and L =2 state

$\mu = \mu_N (g_p + g_n) \int u^(r) dr + \mu_N \left( \frac{3}{4} - \frac{g_p + g_n}{2} \right) \int w^2(r) dr = 0.84699 ~\mu_N$

the experimental value is $0.857406~\mu_N$ .

The Quadrupole moment

$\displaystyle Q = e \sqrt{\frac{\pi}{5}} \left< \Phi_1 |r^2 Y_{20}(\Omega)| \Phi_1 \right>$

$\displaystyle \left< \Phi_1 |r^2 Y_{20}(\Omega)| \Phi_1 \right> \\= \int \left( \frac{ u^2(r)}{4\pi} + \frac{u(r) w(r)}{\sqrt{10 \pi}} Y_{20} + \frac{w^2(r)}{10} (Y^2_{20} + 3|Y_{21}|^2 + 6 |Y_{22}|^2) \right) r^2 Y_{20}dr d\Omega$

$\displaystyle \left< \Phi_1 |r^2 Y_{20}(\Omega)| \Phi_1 \right> = \int \left( \frac{u(r) w(r)}{\sqrt{10 \pi}} - \frac{w^2(r)}{20} \sqrt{\frac{5}{\pi}} \right) r^2 dr$

$\displaystyle Q = \frac{e}{20} \int \left( \sqrt{8} u(r) w(r) - w^2(r) \right) r^2 dr = 0.26966 ~ \textrm{e. fm}^2$

The experimental value is $0.2859$ e. fm².

Deuteron Wave function using AV18

February 27, 2023

GoLuckyRyan Basic AV18, deuteron Leave a comment

It has been a long journey to this point. The basic facts for deuteron are discussed here, in which the tensor force plays an important role to bound the pn- pair and creating the d-wave. The coupled equation of the deuteron in the C.M. frame is derived here. The AV18 potential is used for the radial form of the central and tensor interaction. Now, the final step is solving it numerically.

The coupled equation is

$\displaystyle y''(r) = g(r) y(r), \\ g(r) = \frac{1}{\alpha} \begin{pmatrix} V_1(r) - E & \sqrt{8} V_T(r) \\ \sqrt{8} V_T(r) & \alpha \frac{6}{r^2} + V_2(r) - E \end{pmatrix}$

where $E < 0$ , $\alpha = \frac{\hbar^2}{2m_\mu} = 41.47~\text{MeV} , m_\mu = \frac{m_p m_n}{m_p + m_n}$ , and $\int u^2 + w^2 dr = 1$ . The boundary conditions are $u(0) = u(\infty) = w(0) = w(\infty) = 0$ .

The coupled equation can be solved using Numerov’s method. However, the boundary conditions alone are not sufficient to solve the equation. We need 4 more, which is the derivative of $u, w$ , the ratio of the two, and the eigen energy.

To solve it. we can use Numerov’s method with 2 sets of derivatives, and get 2 sets of solutions.

$u(\delta r) = u_a , w(\delta r) = w_a \rightarrow u_1(r) , w_1(r)$

$u(\delta r) = u_b , w(\delta r) = w_b \rightarrow u_2(r) , w_2(r)$

I used $\delta r = 0.01$ fm, $u_a = w_b = 1, u_b = w_a = 0.1$ . The actual solution would be

$u(r) = a u_1(r) + b u_2(r), \\w(r) = a w_1(r) + b w_2(r)$

Next, at $r \rightarrow \infty$ , the coupled equation becomes

$\displaystyle g(r \rightarrow \infty) = \frac{1}{\alpha} \begin{pmatrix} - E & 0 \\ 0 & \alpha \frac{6}{r^2} - E \end{pmatrix}$

and decoupled. The asymptotic solutions are

$\displaystyle f_s(r) = \exp( - \sqrt{|E|/\alpha} r), \\f_d(r) = \exp( - \sqrt{|E|/\alpha} r) \left(1 + \frac{3}{r \sqrt{|E|/\alpha}} + \frac{3}{r^2 |E|/\alpha} \right)$

Thus, we can relate

$\displaystyle u(r) = a u_1(r) + b u_2(r) \rightarrow \eta f_s(r), \\w(r) = a w_1(r) + b w_2(r) \rightarrow \zeta f_d(r)$

and also the derivative

$\displaystyle u'(r) = a u'_1(r) + b u'_2(r) \rightarrow \eta f'_s(r), \\w'(r) = a w'_1(r) + b w'_2(r) \rightarrow \zeta f'_d(r)$

Let $r = x$ be the maximum radial distance for the numerical solution. We have

$\displaystyle a u_1(x) + b u_2(x) = \eta f_s(x), \\a w_1(x) + b w_2(x) = \zeta f_d(x), \\ a u'_1(x) + b u'_2(x) = \eta f'_s(x), \\ a w'_1(x) + b w'_2(x) = \zeta f'_d(x)$

In matrix form

$\displaystyle A = \begin{pmatrix} u_1(x) & u_2(x) & f_s(x) & 0 \\ w_1(x) & w_2(x) & 0 & f_d(x) \\ u'_1(x) & u'_2(x) & f'_s(x) & 0 \\ w'_1(x) & w'_2(x) & 0 & f'_d(x) \end{pmatrix}, \vec{v} = (a, b, -\eta, -\zeta)$

such that

$\displaystyle A \vec{v} = 0$

For a non-zero solution for a null space equation, the determinant of the matrix $A$ must be zero. So, by solving the $u_i, w_i$ with different values of $E$ , the eigen energy can be found when $\det(A) = 0$ .

The eigen energy is found to be -2.224574206 MeV. (experimental value is -2.224575 MeV. )

Once we have the eigen energy. We can solve the null space equation $A\vec{v} = 0$ , in fact, we found that the 1st column vector and the 2nd column vector of the matrix $latex $ are proportional. Because of that, we can eliminate $a$ in the 4 coupled equation of $A\vec{v} = 0$ , and get

$\displaystyle b = \frac{f_s(x) u'_1(x) - f'_s(x) u_1(x)}{ u_2(x) u'_1(x) - u'_2(x) u_1(x)} \eta$

Treat $\eta$ as a free parameter. Eliminate $\eta$ , we have

$\displaystyle a = -b \frac{f_d(x) w'_2(x) - f'_d(x) w_2(x)}{ f'_d(x) w_1(x) - f_d(x) w'_1(x) }$

or, the A matrix is

$\displaystyle A(x) = \begin{pmatrix} 1.255 \times 10^8 & 9.693 \times 10^8 & 0.000964826 & 0 \\ -3.159 \times 10^9 & -2.4395 \times 10^{10} & 0 & 0.00144172 \\ 2.9065 \times 10^7 & 2.24486 \times 10^8 & -0.00022346 & 0 \\ -7.8296 \times 10^8& -6.04724\times 10^9 & 0 & -0.000351821 \end{pmatrix}$

We can see that the ratio between column 1 and column 2 is 0.129474.

The null space is

$\displaystyle \vec{v} = (0.000918457, -0.000118917, 0.99969, 0.0248831)$

Thus, we have the deuteron wave function using AV18 potential in numerical form. Finally, normalized with a constraint

$\displaystyle \int u^2(r) + w^2(r) dr = 1$

We have the final wave function as

$\displaystyle \psi_M(r) = \frac{u(r)}{r} Y_{00}(\Omega) \chi_{1M}+ \frac{w(r)}{r} \sum_{m}Y_{2m}(\Omega) \chi_{1,M-m}, M = -1, 0, 1$

The radial functions are

The d-wave is 5.76%, and the experimental value is 6.8% from the numerical solution.

Place holder for Dipole moment, Quadrupole moment, and matter radius.

The dipole vector is

$\displaystyle \vec{\mu_d} = g_p \vec{\mu_p} + g_n \vec{\mu_n} + \frac{\vec{l}}{2}$

The quadrupole moment is

$\displaystyle Q_d = \frac{e}{5} \int\left( \frac{2 u w}{\sqrt{8}} + \frac{w^2}{4} \right) r^2 dr$

Deuteron pn interaction using the AV18 potential

February 12, 2023

GoLuckyRyan Basic AV18, NN-interaction Leave a comment

After getting the 18 components of the AV18 potential, the pn-interaction for deuteron can be formed.

The equation for the deuteron wavefunction, quoted in here, is

$\displaystyle \left( \alpha \frac{d^2}{dr^2} - V_1(r) + E \right) u(r)= \sqrt{8} V_T(r) w(r)$

$\displaystyle \left( \alpha \frac{d^2}{dr^2} - \alpha \frac{6}{r^2} - V_2(r) + E \right) w(r) = \sqrt{8}V_T(r) u(r)$

where $\alpha = \frac{\hbar^2}{2m_\mu} = 41.47~\text{MeV} , m_\mu = \frac{m_p m_n}{m_p + m_n}$ , and $V_2(r) = V_1(r) - 2 V_T - 3 V_{LS}$ .

Or , set $y(r) = ( u(r) , w(r) )$

$\displaystyle y''(r) = g(r) y(r), g(r) = \frac{1}{\alpha} \begin{pmatrix} V_1(r) - E & \sqrt{8} V_T(r) \\ \sqrt{8} V_T(r) & \alpha \frac{6}{r^2} + V_2(r) - E \end{pmatrix}$

The central term for $L=0, S=1, J=1, T=0, T_z = 0$

$\displaystyle V_1(r) = v_1 + (\tau_1 \cdot \tau_2) v_{\tau} + (\sigma_1 \cdot \sigma_2) v_\sigma + (\tau_1 \cdot \tau_2)(\sigma_1 \cdot \sigma_2) v_{\sigma \tau} \\ ~~~~~~~~ + T_{12} v_T + T_{12} (\sigma_1 \cdot \sigma_2) v_{\sigma T} + (\tau_{1z} + \tau_{2z})v_{\tau_{1z} + \tau_{2z}} \\ ~~~~~~~~ + V_{C1}(np) + (\sigma_1 \cdot \sigma_2) V_{MM}^{\sigma}(np)$

evaluate the $(\tau_1 \cdot \tau_2) = -3, (\sigma_1 \cdot \sigma_2) = 1, T_{12} = 0$ , the $T_{12}$ is zero as the pn par is isoscalar for deuteron.

$\displaystyle V_1(r) = v_1 -3 v_{\tau} + v_\sigma -3 v_{\sigma \tau} + V_{C1}(np) + (\sigma_1 \cdot \sigma_2) V_{MM}^{\sigma}(np)$

$\displaystyle V_{C1}(np) = \alpha \beta_n \frac{b^2}{384} \left( 15 (br)+ 15 (br)^2 + 6 (br)^3 + (br)^4 \right) \frac{e^{-br}}{r}, \\ ~~ b = 4.27~ \text{fm}^{-1}, \beta_n = 0.0189 ~\text{fm}^2, \alpha = \frac{e^2}{4\pi \epsilon_0 \hbar c}$

$\displaystyle V_{MM}^{\sigma}(np) = - \frac{\alpha}{4 m_p m_n} \mu_p \mu_n \frac{2}{3} F_{\delta}(r) \\ F_{\delta}(r) = \frac{b^3}{48} \left(3 + 3 (br) + (br)^2 \right) e^{-br}$

The tensor term is

$\displaystyle V_T(r) = v_t + (\tau_1 \cdot \tau_2) v_{t \tau} + T_{12} v_{t T} + V_{MM}^{t}(np)$

The $L\cdot S$ term is

$\displaystyle V_{LS}(r) = v_{LS} + (\tau_1 \cdot \tau_2) v_{LS \tau} + V_{MM}^{LS}(np)$

For the $V_2(r)$ , this is the equation for the $L=2$ channel, so, it has additional terms from $L^2$ and $(L\cdot S)^2$

$\displaystyle V_2(r) = V_1(r) - 2 v_T + (L\cdot S) v_{LS} + L^2 v_{L^2} + (L\cdot S)^2 v_{LS^2}, \\~~~~~~~~~~~~~ L=2, L(L+1) = 6, L \cdot S = -3$

The EM terms can be found here.

Here is the plot of the $V_1, \sqrt{8} V_T, V_2$ .

Now, we have everything for solving the coupled equation for $u(r), w(r)$ . However, it is not that trivial. A conventional approach would be searching the binding energy, so that $u(r), w(r)$ is almost flat at $r \rightarrow \infty$ . However, it is unlink solving the radial function of Woods-Saxon, in which the amplitude of the solution can be normalized afterward, as the eigen-equation does not care about the amplitude. Here, the initial condition is critical for solving the equations. Take the s-wave equation as an illustration, the rigth-hand side depends on $w(r)$ , and if the initial condition of $w(r)$ is too large, the $u''(r)$ will be too big and $u(r)$ would blow up more quickly. The same is true for the d-wave equation.

As a 2nd order differential equation, there are 2 initial (or boundary) conditions for fixing the solution. The $u(0)= w(0)= 0$ which is already fixed. Thus, we can have either $u(r_1) , w(r_1), r_1 \neq 0$ , or slope, or the 2nd derivative to be fixed. So, besides the binding energy, there are 2 more parameters needed to be searched.

I will work on it more in my free time….

AV18 (II)

July 3, 2022

GoLuckyRyan Basic AV18, NN-interaction Leave a comment

We read the paper of AV18 potential in this post. At the end, I lost grips on the paper at the end about the projection. After some study, in a simple term, the projection is an rearrangement of the potentials into the 18 terms of

$1, (\tau_1\cdot \tau_2), (\sigma_1\cdot \sigma_2), (\tau_1\cdot \tau_2)(\sigma_1\cdot \sigma_2) = 1, \tau, \sigma, \tau \sigma$
$S_{12}, S_{12}(\tau_1\cdot \tau_2) = t , t \tau$
$L\cdot S, (L\cdot S)(\tau_1\cdot \tau_2) = ls, ls \tau$
$L^2, L^2 (\tau_1\cdot \tau_2), L^2 (\sigma_1\cdot \sigma_2), L^2 (\tau_1\cdot \tau_2)(\sigma_1\cdot \sigma_2) = l2, l2\tau, l2 \sigma, l2 \sigma \tau$
$(L\cdot S)^2, (L\cdot S)^2(\tau_1\cdot \tau_2) = ls2, ls2 \tau$
$T_{12}, T_{12} (\sigma_1\cdot \sigma_2), T_{12} S_{12}, (\tau_{1z}+\tau_{2z}) = T, \sigma T, t T, \tau_z$

where $X_{12} = 3 x_{1z} x_{2z} - \vec{x}_1 \cdot \vec{x}_2, x_z = {-1, 1}$ (I know, $x_z$ should be -1/2 or 1/2, but for some reasons, it is -1, or 1 in here). The right hand side are the short form for the terms.

From the Table 2 in the paper, we can see that the central term of the mid-range potential has 3 parts for $T=1$ , which are $pp, np, nn$ , while it is 1 terms for $T=0$ . The central terms for $T=1$ rewrite as the sum of charge-independent (CI), charge-dependent (CD), and charge-asymmetric (CA) terms

$v_{S1,NN}^{c} = v_{S1}^{CI} + v_{S1}^{CD} T_{12} + v_{S1}^{CA} (\tau_{1z}+\tau_{2z})$

Write it into a matrix form,

$\displaystyle \begin{pmatrix} v_{S1,pp}^c \\ v_{S1, nn}^c \\ v_{S1, np}^c \end{pmatrix} = \begin{pmatrix} 1 & 2 & 2 \\ 1 & 2 & -2 \\ 1 & -4 & 0 \end{pmatrix} \begin{pmatrix} v_{S1}^{CI} \\ v_{S1}^{CD} \\ v_{S1}^{CA} \end{pmatrix}$

so that

$\displaystyle v_{S1}^{CI} = \frac{1}{3} \left( v_{S1,pp}^c + v_{S1,np}^c + v_{S1, nn}^c \right)$

$\displaystyle v_{S1}^{CD} = \frac{1}{12} \left( v_{S1,pp}^c - 2 v_{S1,np}^c + v_{S1, nn}^c \right)$

$\displaystyle v_{S1}^{CA} = \frac{1}{4} \left( v_{S1,pp}^c - v_{S1, nn}^c \right)$

A similar transform is also for $S = 1, T = 1$ tensor terms $v_{11,NN}^t$

and for the $T=0$ ,

$v_{S0}^{k} = v_{S0, NN}^{c} , k = CI, CD, CA$

Note that, the pion-exchange potential should also added to $v_{S1, NN}^c, v_{S1,NN}^t$ .

Then, the projections to the 18 terms are ( I still don’t know why)

$\displaystyle v_1 = \frac{1}{16} \left( 9 v_{11}^{CI} + 3 v_{01}^{CI} + 3 v_{10}^{CI} + v_{00}^{CI} \right)$
$\displaystyle v_\tau= \frac{1}{16} \left( 3 v_{11}^{CI} + v_{01}^{CI} - 3 v_{10}^{CI} - v_{00}^{CI} \right)$
$\displaystyle v_{\sigma}= \frac{1}{16} \left( 3 v_{11}^{CI} - 3 v_{01}^{CI} + v_{10}^{CI} - v_{00}^{CI} \right)$
$\displaystyle v_{\sigma \tau}= \frac{1}{16} \left( v_{11}^{CI} - v_{01}^{CI} - v_{10}^{CI} + v_{00}^{CI} \right)$
$\displaystyle v_{t} = \frac{1}{4} \left ( 3 v_{11}^{t} + v_{10,NN}^{t} \right) , ~~ v_{11}^{t} = \frac{1}{3} \left(v_{11,pp}^t + v_{11,np}^t + v_{11, nn}^t \right)$
$\displaystyle v_{t\tau} = \frac{1}{4} \left ( v_{11}^{t} - v_{10,NN}^{t} \right)$
$\displaystyle v_{LS} = \frac{1}{4} \left ( 3 v_{11,NN}^{LS} + v_{10,NN}^{LS} \right)$
$\displaystyle v_{LS\tau} = \frac{1}{4} \left ( v_{11,NN}^{LS} - v_{10,NN}^{LS} \right)$
$\displaystyle v_{L^2} = \frac{1}{16} \left( 9 v_{11,NN}^{L^2} + 3 v_{01,NN}^{L^2} + 3 v_{10}^{L^2} + v_{00}^{L^2} \right)$
$\displaystyle v_{L^2\tau}= \frac{1}{16} \left( 3 v_{11}^{L^2} + v_{01}^{L^2} - 3 v_{10}^{L^2} - v_{00}^{L^2} \right)$
$\displaystyle v_{L^2\sigma}= \frac{1}{16} \left( 3 v_{11}^{L^2} - 3 v_{01}^{L^2} + v_{10}^{L^2} - v_{00}^{L^2} \right)$
$\displaystyle v_{L^2\sigma \tau}= \frac{1}{16} \left( v_{11}^{L^2} - v_{01}^{L^2} - v_{10}^{L^2} + v_{00}^{L^2} \right)$
$\displaystyle v_{LS^2} = \frac{1}{4} \left ( 3 v_{11,NN}^{LS^2} + v_{10,NN}^{LS^2} \right)$
$\displaystyle v_{LS^2\tau} = \frac{1}{4} \left ( v_{11,NN}^{LS^2} - v_{10,NN}^{LS^2} \right)$
$\displaystyle v_{T} = \frac{1}{4} \left ( 3 v_{11}^{CD} + v_{01}^{CD} \right)$
$\displaystyle v_{\sigma T} = \frac{1}{4} \left ( v_{11}^{CD} - v_{01}^{CD} \right)$
$\displaystyle v_{t T} = \frac{1}{12} \left ( v_{11_pp}^{t} - 2v_{11,np}^{t} + v_{11,nn}^{t}\right)$
$\displaystyle v_{\tau_{1z} + \tau_{2z}} = \frac{1}{4} \left ( v_{01,pp}^{c} - v_{01,nn}^{c}\right)$

Here are the plot for each terms

The central term $v_1(r)$ is very large when $r < 0.4~$ fm. In fact $v_1(0) = 2031 MeV$ The $v_1(r)$ becomes repulsive for $r < 0.68$ fm. Its minimum is ~ -59 MeV at 0.85 fm. The next biggest component is $v_{LS}$ , then $v_{L^2}$ . Here is a plot for the first few biggest terms.

With the 18 terms of the AV18 NN-potential, we can form the central term, the tensor term, the LS terms, the L² term, the (LS)² term.

The central term is

$\displaystyle V_c(r) = v_1 + (\tau_1\cdot \tau_2) v_\tau + (\sigma_1 \cdot \sigma_2) v_\sigma + (\tau_1\cdot \tau_2)(\sigma_1 \cdot \sigma_2) v_{\sigma \tau} \\ ~~~~~~~ + T_{12} v_T + T_{12}(\sigma_1 \cdot \sigma_2) v_{\sigma T} + (\tau_{1z} + \tau_{2z}) v_{(\tau_{1z} + \tau_{2z})} + v_{EM}^c(NN)$

$\displaystyle v_{EM}^c(pp) = V_{C1}(pp) + V_{DF} + V_{VP} + V_{C2} + (\sigma_1 \cdot \sigma_2)V_{MM}^{\sigma}(pp)$

$\displaystyle v_{EM}^c(np) = V_{C1}(np)+(\sigma_1 \cdot \sigma_2) V_{MM}^{\sigma}(np)$

$\displaystyle v_{EM}^c(nn) =(\sigma_1 \cdot \sigma_2) V_{MM}^{\sigma}(nn)$

The spin-tensor term

$\displaystyle V_t(r) = v_t + (\tau_1\cdot \tau_2) v_{t\tau} + T_{12} v_{tT} + v_{EM}^t(NN)$

The LS term

$\displaystyle V_{LS}(r) = v_{LS} + (\tau_1\cdot \tau_2) v_{LS\tau} + v_{EM}^{LS}(NN)$

The (LS)² term

$\displaystyle V_{LS^2}(r) = v_{LS^2} + (\tau_1\cdot \tau_2) v_{LS^2 \tau}$

The L² term

$\displaystyle V_{L^2}(r) = v_{L^2} + (\tau_1\cdot \tau_2) v_{L^2 \tau} + (\sigma_1 \cdot \sigma_2) v_{L^2\sigma} + (\tau_1\cdot \tau_2)(\sigma_1 \cdot \sigma_2) v_{L^2\sigma \tau}$

The total potential is

$\displaystyle V(r) = V_c(r)+ S_{12} V_t(r) \\ ~~~~~~~ + (L \cdot S) V_{LS}(r) + (L \cdot S)^2 V_{LS^2}(r) + L(L+1) V_{L^2}(r)$

where

$\displaystyle T_{12}(NN) = 3 \tau_{1z} \tau_{2z} - \tau_1 \cdot \tau_2 , ~~\tau_1 \cdot \tau_2 = 4 T -3, ~~ \tau_z = {-1, 1}$

$T_{12}(pp) = T_{12}(nn) = 2, T_{12}(np) = -4$

$\displaystyle L\cdot S = \frac{1}{2} \left( J(J+1) - L(L+1) - S(S+1) \right)$

And the spin-tensor can be found in here.

Argonne V18 potential for NN data

February 7, 2022

GoLuckyRyan papers reading AV18, NN-interaction 5 Comments

The Argonne V18 (or AV18) potential plays a crucial part in DWBA calculation for transfer reaction. It is the potential to describe the interaction between a NN pair, for instance, a deuteron.

This post is a summary of the reference Accurate nucleon-nucleon potential with charge independent breaking by Robert. B. Wiringa (Argonne since 1981), Vincent. G. J. Stoks (University of Adelaide, retired?), and Rocco. Schiavilla (Old Dominion University).

Some keys points

Before AV18, NN potential descripts either T=0 or T=1, but not both. This is because the stronge interaction is charge-independent breaking (so is charge dependent?).
The AV18 is an improvement based on AV14.
The AV18 is non-relativistic, good from 0 MeV to 350 MeV.
The AV18 fits the Nijmegen NN scattering database, the reduced chi-square is 1.09.
The AV18 includes
- stronge interaction that depends on S, T, and Tz (base on AV14)
- charge-independent part (14 operators)
- charge-independent breaking part
  - 3 operators on charge dependent
  - 1 operator on chage-asymmetric
- complete electromagnetci potential (this term gives 18 keV repulsion in the case of deuteron, and deuteron mass has 1 keV accuracy.)
  - Coulomb
  - Darwin-Foldy
  - Vacuum polarization
  - magnetic moment with finite size effects
The accurate NN potential is also crucial and fundamental to a reasonable descrption of NNN potential. For example, a consistant description for 3He and 3H system.
Use local operator structure of AV14 and Urbaba V14 (no momentum dependent?).
No one-boson-excahnge.

The electron magnetic terms are:

$\displaystyle v_{EM}(NN) = V_{C1}(NN) + (V_{C2} + V_{DF}+V_{VP})\delta_{Np} + V_{MM}(NN)$

The Coulomb interaction:

$\displaystyle V_{C1}(pp) = \alpha' \frac{F_C(r)}{r}, ~ \alpha' = \frac{2k\alpha}{m_p v_{lab}}, ~ \alpha = \frac{e^2}{4\pi \epsilon_0 \hbar c}$

$\displaystyle V_{C2} = -\frac{\alpha}{2 m_p^2}\left[ \left( \nabla^2 + k^2 \right) \frac{F_C(r)}{r} + \frac{F_C(r)}{r}\left( \nabla^2 + k^2 \right) \right] \approx - \frac{\alpha \alpha'}{m_p} \left( \frac{F_C(r)}{r} \right)^2$

The Darwin-Foldy term

$\displaystyle V_{DF} = - \frac{\alpha}{4m_p^2} F_{\delta}(r)$

The vaccume polarization

$\displaystyle V_{VP} = \frac{2\alpha \alpha'}{3\pi} \frac{F_C(r)}{r} \int_1^{\infty} e^{-2m_e r x}\left(1+\frac{1}{2x^2} \right) \frac{\sqrt{x^2-1}}{x^2} dx$

The magnetic moment term

$\displaystyle V_{MM} = - \frac{\alpha}{2m_p^2} \mu_p^2 \left( \frac{2}{3} F_{\delta}(r) \vec{\sigma}_i \cdot \vec{\sigma}_j + \frac{F_t(r)}{r^3} S_{ij} \right) - \frac{\alpha}{2m_p^2}(4\mu_p-1) \frac{F_{ls}(r)}{r^3} \vec{L} \cdot \vec{S}$

where $S_{ij} = 3 \sigma_{zi}\sigma_{zj} - (\vec{\sigma}_i \cdot \vec{\sigma}_j)$ , see spin-spin interaction.

The $F_i (r)$ are shart-range functions that takes into account for the finite size of the nucleon charge distribution. They are dervied by assuming the nuclear form factors are well represented by a dipole form:

$\displaystyle \left(1 + \frac{q^2}{b^2} \right)^{-2}, ~ b = 4.27~\textrm{fm}^{-1}$

$\displaystyle F_C(r) = 1 - \left( 1 + \frac{11}{16}x + \frac{3}{16} x^2 + \frac{1}{48} x^3\right) e^{-x} , ~ x = br$

$\displaystyle F_{\delta}(r) = - \nabla^2\left( \frac{F_C(r)}{r} \right) = -\frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{d}{dr}\left(\frac{F_C(r)}{r}\right)\right) \\~~~~~~~~= b^3\left(\frac{1}{16} + \frac{1}{16}x + \frac{1}{48}x^2 \right) e^{-x}$

$\displaystyle F_t(r) = \frac{r^3}{3}\left[\frac{d^2}{dr^2}\left( \frac{F_C(r)}{r} \right) - \frac{1}{r} \frac{d}{dr}\left( \frac{F_C(r)}{r} \right) \right]\\~~~~~~~~= 1 - \left( 1 + x + \frac{1}{2}x^2 + \frac{1}{6} x^3 + \frac{1}{24} x^4 + \frac{1}{144}x^5\right) e^{-x}$

$\displaystyle F_{ls}(r) = r^2 \frac{d}{dr}\left( \frac{F_C(r)}{r} \right) = 1 - \left( 1 + x + \frac{1}{2}x^2 + \frac{7}{48} x^3 + \frac{1}{48} x^4 \right) e^{-x}$

Note that the relationships between $F_t, F_{ls}$ to $F_C$ are somewhat different that the paper. I confirmed the relationship using Mathematica.

Here is the plot for the short-range functions

For the neutron-proton, the Coulomb terms are

$\displaystyle V_{C1}(np) = \alpha \beta_n \frac{F_{np}(r)}{r}, \beta_n = 0.0189~\textrm{fm}^2$

$\displaystyle F_{np}(r) = b^2 (15x + 15x^2 + 6 x^3 + x^4) \frac{e^{-x}}{384}$

And the magnetic moment for np-pair is

$\displaystyle V_{MM}(np) = - \frac{\alpha}{4m_pm_n} \mu_p \mu_n \left( \frac{2}{3} F_{\delta}(r) \vec{\sigma}_i \cdot \vec{\sigma}_j + \frac{F_t(r)}{r^3} S_{ij} \right) \\~~~~~~~~~~~~~~~~ - \frac{\alpha}{2m_n m_r}\mu_n \frac{F_{ls}(r)}{r^3} \left( \vec{L} \cdot \vec{S} + \vec{L} \cdot \vec{A} \right), \vec{A} = \frac{1}{2}\left(\vec{\sigma}_i - \vec{\sigma}_j\right)$

The $m_r = \frac{m_n m_p}{m_n + m_p}$ is the nucleon reduced mass. And the $\vec{A}$ is charge-asymmetric force, which mixes spin-singlet and spin-triplet state.

For the neutron-neutron pair, only have the magnetic moment terms, which is only the spin-spin interaction.

$\displaystyle V_{MM}(nn) = - \frac{\alpha}{4m_n^2} \mu_n^2 \left( \frac{2}{3} F_{\delta}(r) \vec{\sigma}_i \cdot \vec{\sigma}_j + \frac{F_t(r)}{r^3} S_{ij} \right)$

The charge-dependent one-pion-exchange potential is

$\displaystyle v^{\pi}(pp) = f_{pp}^2 v_\pi(m_{\pi^0})$

$\displaystyle v^{\pi}(np) = f_{pp} f_{nn} v_\pi(m_{\pi^0}) + (-1)^{T+1} 2 f_c^2 v_\pi(m_{\pi^{\pm}})$

$\displaystyle v^{\pi}(nn) = f_{nn}^2 v_\pi(m_{\pi^0})$

where the coupling constant $f_{pp} = - f_{nn} = f_c = f , f^2 = 0.075$ .

$\displaystyle v_\pi(m) = \left(\frac{m}{m_s} \right)^2 \frac{1}{3} mc^2\left[ Y_{\mu}(r) \vec{\sigma}_i \cdot \vec{\sigma}_j + T_{\mu}(r) S_{ij} \right] , m_s = m_{\pi^{\pm}} = 139.57~\textrm{MeV}/c^2$

$\displaystyle Y_\mu(r) = \frac{e^{\mu r}}{\mu r} \left( 1- e^{-cr^2} \right), c = 2.1~\textrm{fm}^{-2}$ ,

In here, $\mu = \frac{mc}{\hbar}$ , and this $c$ should be the speed of light in vaccum.

$\displaystyle T_\mu(r) = \left(1 + \frac{3}{\mu r} + \frac{3}{\mu^2r^2} \right) \frac{e^{-\mu r}}{\mu r} \left( 1- e^{-cr^2} \right)^2$

The remaining intermediate- and short-range phenomenological potential has central, L-square, tensor, spin-orbital, and square of spin-orbital terms:

$\displaystyle v^R_{ST}(NN) = v^c_{ST,NN}(r) + v^{l2}_{ST,NN}(r) L^2 + v^t_{ST, NN}(r) S_{12} \\~~~~~~~~~~~~~+ v^{ls}_{ST, NN}(r) \vec{L}\cdot\vec{S} + v^{ls2}_{ST, NN}(r) \left( \vec{L}\cdot\vec{S} \right)^2$

where

$v^i_{ST,NN}(r) = I^i_{ST,NN} T_\mu^2(r) + \left[ P^i_{ST,NN} + \mu r Q^i_{ST,NN} + (\mu r)^2 R^i_{ST,NN} \right] \frac{1}{1+ e^\frac{r-r_0}{a}}$

in here, the $\mu = \frac{1}{3}(m_{\pi^0} + 2m_{\pi^{\pm}}) \frac{c}{\hbar}$ , which is different from above, while $T_\mu(r)$ is the same as above. The $I^i_{ST,NN}, P^i_{ST,NN}, Q^i_{ST,NN}, R^i_{ST,NN}$ are parameters to be fitted.

The AV18 required that

$\displaystyle v^t_{ST,NN}(r=0) = 0 \rightarrow P^t_{ST,NN} = 0$

$\displaystyle \frac{\partial v^{i\neq t}_{ST,NN}}{\partial r} |_{r=0} = 0 \rightarrow Q^{i\neq t}_{ST,NN} \approx 0.55 P^{i\neq t}_{ST,NN}$

The overall potential is the sum of the electromagnetic term, the one-pion-exchange, and the effective intermediate- and short-range terms

$\displaystyle v(NN) = v^{EM}(NN) + v^{\pi}(NN) + v^{R}_{ST}(NN)$

The middle part of the paper discusses on the fitting and compares calculation using AV18 and the experimental results.

Next, the potential project (or mapping, transform?) into 18 terms

$\displaystyle v_{ij} = \sum_{p=1,18} v_p(r_ij) O^P_{ij}$

where 14 terms are charge-independent:

$\displaystyle O^{P=1,14}_{ij} = 1,~\tau_i \cdot \tau_j, ~ \sigma_i \cdot \sigma_j, ~ (\tau_i \cdot \tau_j)(\sigma_i \cdot \sigma_j), \\~~~~~~~~~~~~~~ S_{ij}, ~ S_{ij} (\tau_i \cdot \tau_j), \\~~~~~~~~~~~~~~ L\cdot S, ~L\cdot S (\tau_i \cdot \tau_j), \\~~~~~~~~~~~~~~ L^2 ,~L^2(\tau_i \cdot \tau_j), L^2(\sigma_i \cdot \sigma_j), L^2 (\tau_i \cdot \tau_j)(\sigma_i \cdot \sigma_j), \\~~~~~~~~~~~~~~ (L\cdot S)^2, (L\cdot S)^2(\tau_i \cdot \tau_j)$

And 4 charge-independent breaking terms:

$\displaystyle O^{P=15,18}_{ij} = T_{ij}, T_{ij} (\sigma_i \cdot \sigma_j), S_{ij} T_{ij}, (\tau_{zi} + \tau_{zj}), ~~~ T_{ij} = 3 \tau_{zi}\tau_{zj} - (\tau_i \cdot \tau_j)$

The $T_{ij}$ is the isospin-isospin operator analoy to the spin-spin operator $S_{ij}$ .

( I am not so sure how to “project”…. )

Since the major improvement of AV18 is the charge-independent breaking term, Here is the idea why it is the case:

The above plot, it shows the radial potential due to the 4 charge-independent breaking terms, and the central Coulomb potential for pp pair. The iso-tensor and spin-iso-tensor terms are ~4 times bigger.

The last part of the paper discusses the application of AV18 on deuteron.

[20230212] A follow-up post on AV18 is here.

Nuclear Physics 101

Deuteron Properties from AV18

Deuteron Wave function using AV18

Deuteron pn interaction using the AV18 potential

AV18 (II)

Argonne V18 potential for NN data

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