TBMEs | Nuclear Physics 101

Experimental TBMEs

September 26, 2020

GoLuckyRyan Basic 18O, TBMEs Leave a comment

For a doubly magic + 2 nucleons system, the 2 nucleons are in $j_1$ and $j_2$ orbitals. If there is no residual interaction, all possible $J$ -states formed by $j_1$ -orbital and $j_2$ -orbital are degenerated. As the total Hamiltonian is a simple sum of “single-particle” Hamiltonian with the mean field, that nucleons “do” not interact with each others.

Lets denote the single-particle energies for the $j_1$ an $j_2$ orbitals are $\epsilon_1$ and $\epsilon_2$ , respectively. The interaction energy between the two valence nucleons is:

$\displaystyle E_{JM}(j_1j_2, j_1j_2) = \left<\Psi_{JM}(j_1J_2)|V_{12}| \Psi_{JM}(j_1J_2) \right>$

where $H$ is the nuclear Hamiltonian. And $\Psi_{JM}(j_1J_2)$ is the wave function of two nucleons that coupled to spin $J$ .

Sometimes, there could be 2 configurations couples to same $J$ . for example, (d5/2)(d5/2) and (d5/2)(s1/2) can both couple to spin 2, and 3. For example, see this post. In this case, the wave function is a linear combination of the two configurations. Thus, the Hamiltonian is a bit complicated,

$\displaystyle H = \begin{pmatrix} E_1 && V \\ V && E2 \end{pmatrix}$

$\displaystyle E_1 = \epsilon_1 + \epsilon_2 + E_{JM}(j_1j_2, j_1j_2)$
$\displaystyle E_2 = \epsilon_3 + \epsilon_4 + E_{JM}(j_3j_4, j_3j_4)$
$\displaystyle V =E_{JM}(j_1j_2, j_3j_4)$

The solution is in this post. And the observed energy levels are $e_\pm$ ,

$\displaystyle e_\pm = \bar{E} \pm \sqrt{ \Delta E^2 + V^2}$

and the states have spectroscopic factors,

$\displaystyle \Psi_+ = \alpha \Psi(j_1j_2) + \beta \Psi(j_3j_4)$
$\displaystyle \Psi_- = -\beta \Psi(j_1j_2) + \alpha \Psi(j_3j_4)$

The inverted formula are

$\displaystyle E_1 = \alpha^2 e_+ + \beta^2 e_-$

$\displaystyle E_2 = \beta^2 e_+ + \alpha^2 e_-$

$\displaystyle V = \alpha\beta(e_+ - e_-)$

Now, the theoretical background is laid down. Experimentally, lets take the 18O , and the 17O(d,p)18O reaction as an example.

The 17O is a single d5/2 neutron on top of 16O. Adding another neutron on 17O, the new neutron will couple to that d5/2 neutron, any state contains d5/2 will be populated. We restrict ourselves only to the d5/2 and s1/2 states. For the $J = 0$ state, we pick the ground state and the 5.34 state as the mixing between the (d5/2)(d5/2) and (s1/2)(s1/2).

$\displaystyle \Psi_0(j_1j_1) = [\phi_{5/2} \times \phi_{5/2}]_{J=0}$
$\displaystyle \Psi_0(j_2j_2) = [\phi_{1/2} \times \phi_{1/2}]_{J=0}$

The spectroscopic factor of the ground state is 1.22, and the SF of the 5.34 MeV state is 0.16. We have to normalize the SF.

$\displaystyle \Psi_(0.00) = \sqrt{\frac{1.22}{1.22+0.16}} \Psi_0(j_1j_1) + \sqrt{\frac{0.16}{1.22+0.16}} \Psi_0(j_2j_2) \\= 0.94 \Psi_0(j_1j_1) + 0.34 \Psi_0(j_2j_2)$

$\displaystyle \Psi_(5.34)= -0.34\Psi_0(j_1j_1) + 0.94 \Psi_0(j_2j_2)$

Next, we have to estimate the single-particle energy. The binding energy for the d5/2 neutron is $S_n(17O) = \epsilon_1= -4.14 MeV$ , And the binding energy for 2 neutrons and their interaction is $S_{2n}(18O) = -12.19 MeV$ . Thus, the 2-neutron interaction energy is $S_{2n}(18O) - 2 S_{n}(17O) = -3.90 MeV$ .

Thus, we have

$e_1 = 0.00 - 12.19 = -12.19 MeV$
$e_2 = 5.34 - 12.19 = -6.85 MeV$
$\alpha = 0.94, \beta = 0.34$

And using the formula, we have,

$E_1 = -11.57 MeV = \epsilon_1 + \epsilon_1 + E_0(j_1j_1, j_1j_1)$
$E_2 = -7.47 MeV = \epsilon_2 + \epsilon_2 + E_0(j_2j_2, j_2j_2)$
$V = -1.71 MeV = E_0(j_1j_1, j_2j_2)$

Thus, the TBME of the d5/2-d5/2 neutrons coupled to J = 0 is -3.29 MeV. The TBME of the d5/2-s1/2 neutrons coupled to J = 0 is -1.71 MeV.

The $\epsilon_2$ is the single-particle energy of the s1/2 neutron. The 1/2+ state next to the 5/2+ ground state of 17O is 0.87 MeV. Thus, $\epsilon_2 = -3.27 MeV$ , then, $E_0(j_2j_2, j_2j_2) = -0.92 MeV$ .

For the 4+ state of 18O, it can only be coupled by (d5/2)(d5/2) neutron. The excited energy is 3.55 MeV, The TBME of J = 4 is -0.35 MeV.

For the 2+ state, we can repeat the same method. We take the 1.98 MeV and 3.92 MeV, with spectroscopic factors for the d5/2 neutron adding are 0.83 and 0.66 respectively. The 2+ state can be formed by (d5/2)(d5/2) and (d5/2)(s1/2) configuration.

$e_1 = 1.98 - 12.19 = -10.21 MeV$
$e_2 = 5.26 - 12.19 = -8.27 MeV$
$\alpha = 0.75, \beta = 0.67$

Thus,

$E_1 = -10.21 MeV = \epsilon_1 + \epsilon_1 + E_2(j_1j_1, j_1j_1)$
$E_2 = -8.27 MeV = \epsilon_1 + \epsilon_2 + E_2(j_1j_2, j_1j_2)$
$V = -0.96 MeV = E_2(j_1j_1, j_1j_2)$

Then, the $J=2$ (d5/2)(d5/2) TBME is -1.06 MeV, the $J = 2$ TBME for (d5/2)(s1/2) is -1.71 MeV, and the off-diagonal TBME is -0.96 MeV.

At last, the 3+ state can only be formed by (d5/2)(s1/2). The excited state in 18O is 5.375 MeV. $5.375 = \epsilon_1 + \epsilon_2 + E_3(j_1j_2, j_1j_2)$ , thus, the $J=0$ TBME for (d5/2)(s1/2) is 0.60 MeV.

In summary,

J	T		configuration	TBMEs [MeV]
0	1	diagonal	d5/2 – d5/2	-3.29
2	1	diagonal	d5/2 – d5/2	-1.06
4	1	diagonal	d5/2 – d5/2	-0.35
0	1	diagonal	s1/2 – s1/2	-0.92
0	1		(d5/2)(d5/2) – (s1/2)(s1/2)	-1.71
2	1	diagonal	(d5/2)(s1/2) – (d5/2)(s1/2)	-1.71
2	1		(d5/2)(d5/2) – (d5/2)(s1/2)	-0.96
3	1	diagonal	d5/2 – s1/2	0.60

The total Hamiltonian is

$\displaystyle H = \begin{pmatrix} H_{J=0} (2 \times 2) && 0 && 0 && 0 \\ 0 && H_{J=2} (2 \times 2) && 0 && 0 \\ 0 && 0 && H_{J=3} && 0 \\ 0 && 0 && 0 && H_{J=4} \end{pmatrix}$

$\displaystyle H_{J=0} = \begin{pmatrix} 2\epsilon_1-3.29 && -1.71 \\ -1.71 && 2 \epsilon_2-0.92 \end{pmatrix}$

$\displaystyle H_{J=2} = \begin{pmatrix} 2\epsilon_1-1.06 && -0.96 \\ -0.96 && \epsilon_1+\epsilon_2-1.71 \end{pmatrix}$

$\displaystyle H_{J=3} = \epsilon_1 + \epsilon_2 + 0.60$

$\displaystyle H_{J=4} = 2\epsilon_1-0.35$

We can compare these value with our previous estimation. For the (d5/2) (d5/2) configuration.

I state the previous estimation in here, with the single-particle energy already subtracted.

$\displaystyle H'_{J=0} = \begin{pmatrix} -3.171 && -1.831 \\ -1.831 && -1.057 \end{pmatrix}$

$\displaystyle H'_{J=2} = \begin{pmatrix} -0.725 && -0.959 \\ -0.959 && -1.27 \end{pmatrix}$

$\displaystyle H'_{J=3} =0$

$\displaystyle H'_{J=4} = -0.302$

The 4+ state interaction action energy is -0.35 MeV. In the previous estimation, it is $-2/7 V_0 \bar(R) = -0.302$ . which is very good agreement.

The 3+ state is actually repulsive. In the previous estimation, we ignored the mutual interaction, because the (d5/2) and (s1/2) states are eigen state and there is no mutual interaction. Possible issue mixing with d3/2 or the interaction is not a delta function.

The $J=0$ TBME for the (d5/2)(d5/2) state is -3.29 MeV and in previous estimation, it is $-3V_0 \bar{R} = -3.17 MeV$ , good agreement. The $J=0$ TBME for the (d5/2)(d5/2) – (s1/2)(s1/2) is -1.71 MeV, and the $J=0$ TBME for (s1/2)(s1/2) is -0.92 MeV, they agree with the previous estimation of -1.83 MeV and -1.057 MeV, respectively.

For the $J = 2$ TBMEs, the off-diagonal term is -0.92, which is agreed with the $-12\sqrt{7}/35 V_0 \bar{R} = -0.96 MeV$ . However, the diagonal terms are -1.06 and -0.71, which is a bit different from $-24/35 V_0 \bar{R} = -0.74$ and $-6/5 V_0 \bar{R} = 1.27$ . Nut we have to notice that the 2+ state is offset in previous calculation.

Shell model calculation on 18O

May 15, 2020

GoLuckyRyan Basic 18O, 19F, 20Ne, Clebsch-Gordon, Shell Model, shell model calculation, TBMEs Leave a comment

This post is copy from the book Theory Of The Nuclear Shell Model by R. D. Lawson, chapter 1.2.1

The model space is only the 0d5/2 and 1s1/2, and the number of valence nucleon is 2. The angular coupling of the 2 neutrons in these 2 orbitals are

$\displaystyle (0d5/2)^2 = 0, 2, 4$
$\displaystyle (0d5/2)(1s1/2) = 2,3$
$\displaystyle (1s1/2)^2 = 0$

Note that for identical particle, the allowed J coupled in same orbital must be even due to anti-symmetry of Fermion system.

The spin 3, and 4 can only be formed by (0d5/2)(1s1/2) and (0d5/2)^2 respectively.

Since the Hamiltonian commute with total spin, i.e., the matrix is block diagonal in J that the cross J matrix element is zero,

$\displaystyle H = h_1 + h_2 + V$

$\displaystyle \left< J | V|J' \right> = V_{JJ'} \delta_{JJ'}$

or to say, there is no mixture between difference spin. The Hamiltonian in matrix form is like,

$\displaystyle H = \begin{pmatrix} M_{J=0} (2 \times 2) & 0 & 0 & 0 \\ 0 & M_{J=2} (2 \times 2) & 0 & 0 \\ 0 & 0 & M_{J=3} (1 \times 1) & 0 \\ 0 & 0 & 0 & M_{J=4} (1 \times 1) \end{pmatrix}$

The metrix element of J=3 and J=4 is a 1 × 1 matrix or a scalar.

$\displaystyle M_{J=3,4} = \left<J| ( h_1 + h_2 + V) | J \right> = \epsilon_1 + \epsilon_2 + \left< j_1 j_2 | V | j_3 J_4 \right>$

where $\epsilon_i$ is the single particle energy.

Suppose the residual interaction is an attractive delta interaction

$\displaystyle V = - 4\pi V_0 \delta(r_i - r_j )$

Be fore we evaluate the general matrix element,

$\displaystyle \left< j_1 j_2 | V | j_3 j_4 \right>$

We have to for the wave function $\left|j_1 j_2 \right>$ ,

$\displaystyle \left| j_1 j_2 \right> = \\ \frac{1}{\sqrt{2(1+\delta_{j_1 j_2})}} \\ \sum_{m_1 m_2} C_{j_1 m_1 j_2 m_2}^{JM} \left( \phi_{j_1m_1}(1) \phi_{j_2m_2}(2) + (-1)^T \phi_{j_1m_1}(2) \phi_{j_2m_2}(1) \right)$

where $T$ is the isospin, and the single particle wave function is

$\displaystyle \phi_{jm} = R_l(r) \sum_{\kappa \mu} C_{l \kappa s \mu}^{jm} Y_{l \kappa} ( \hat{r} ) \chi_{\mu}$

Since the residual interaction is a delta function, the integral is evaluated at $r_1 = r_2 = r$ , thus the radial function and spherical harmonic can be pulled out in the 2-particle wave function $\left|j_1 j_2 \right>$ at $r_1 = r_2 = r$ is

$\displaystyle \left| j_1 j_2 \right> = \\ \frac{1}{\sqrt{2(1+\delta_{j_1 j_2})}} R_{l_1}(r) R_{l_2}(r) \\ \sum_{m_1 m_2 \kappa_1 \mu_1 \kappa_2 \mu_2} C_{j_1 m_1 j_2 m_2}^{JM} C_{l_1 \kappa_1 s \mu_1}^{j_1 m_1} C_{l_2 \kappa_2 s \mu_2}^{j_2 m_2} Y_{l_1 \kappa_1} ( \hat{r} ) Y_{l_2 \kappa_2} ( \hat{r} ) \\ \left( \chi_{\mu_1}(1) \chi_{\mu_2}(2) + (-1)^T \chi_{\mu_1}(2) \chi_{\mu_2}(1) \right)$

Using the product of spherical Harmonic,

$\displaystyle Y_{l_1 \kappa_1}(\hat{r})Y_{l_2 \kappa_2}(\hat{r}) = \sum_{LM} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2L+1)}} C_{l_1 0 l_2 0 }^{L0} C_{l_1 \kappa_1 l_2 \kappa_2}^{LM} Y_{LM}(\hat{r})$

using the property of Clebsch-Gordon coefficient for spin half system

$\displaystyle \chi_{SM_S} = \sum_{\mu_1 \mu_2} \chi_{\mu_1}(1) \chi_{\mu_2}(2)$

where

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

which is equal to $T = 1$

For $T = 0$

$\displaystyle \chi_{1,1} = \chi_{1/2}(1) \chi_{1/2}(2)$

With some complicated calculation, the J-J coupling scheme go to L-S coupling scheme that

$\displaystyle \left| j_1 j_2 \right> =\sum_{L S M_L M_S} \alpha_{LS}(j_1 j_2 JT ) C_{LM_L S M_S}^{LS} Y_{LM_L}(\hat{r}) \chi_{S M_S} R_{l_1}(r) R_{l_2}(r)$

with

$\displaystyle \alpha_{LS}(j_1j_2 JT) = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2L+1)}} \\ \frac{1-(-1)^{S+T}}{\sqrt{2(1+\delta_{j_1 j_2} \delta_{l_1 l_2})}} C_{l_1 0 l_2 0}^{L 0} \gamma_{LS}^{J}(j_1 l_1;j_2 l_2)$

$\displaystyle \gamma_{LS}^{J}(j_1 l_1;j_2 l_2) = \sqrt{(2j_1+1)(2j_2+1)(2L+1)(2S+1)} \\ \begin{Bmatrix} l_1 & s & j_1 \\ l_2 & s & j_2 \\ L & S & J \end{Bmatrix}$

Return to the matrix element

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right>$

Since the matrix element should not depends on $M$ , thus, we sum on M and divide by $(2J+1)$ ,

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = \frac{1}{2J+1} \sum_M \left< j_1 j_2 J M | V | j_3 j_4 J M \right>$

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = \frac{-4\pi V_0 \bar{R}}{2J+1} \sum_{LS} \alpha_{LS}(j_1j_2JT) \alpha_{LS}(j_3j_4JT) \\ \sum_{M M_L M_S} (C_{LM_SSM_S}^{JM})^2 \int Y_{LM}^* Y_{LM} d\hat{r}$

with

$\displaystyle \bar{R} = \int R_{j_1} R_{j_2} R_{j_3} R_{j_4} dr$

( i give up, just copy the result ), for $T = 1$ ,

$\displaystyle \left< j_1 j_2 J M | V | j_3 j_4 J M \right> = (-1)^{j_1+j_3+l_2+l_4 + n_1+n_2+n_3+n_4}\\ (1+(-1)^{l_1+l_2+l_3+l_4}) (1 + (-1)^{l_3+l_4+J}) \\ \frac{V_0 \bar{R}}{4)2J+1)} \sqrt{\frac{(2j_1+1)(2j_2+1)(2j_3+1)(2j_4+1)}{(1+\delta_{j_3j_4})(1+\delta_{j_1j_2})}} \\ C_{j_1(1/2)j_2(-1/2)}^{J0} C_{j_3(1/2)j_4(-1/2)}^{J0}$

The block matrix are

$\displaystyle M_{J=0} = \begin{pmatrix} 2 \epsilon_d - 3 V_0 \bar{R} & -\sqrt{3} V_0 \bar{R} \\ -\sqrt{3} V_0 \bar{R} & 2 \epsilon_s - V_0 \bar{R} \end{pmatrix}$

$\displaystyle M_{J=2} = \begin{pmatrix} 2 \epsilon_d - \frac{24}{35} V_0 \bar{R} & -\frac{12\sqrt{7}}{35} V_0 \bar{R} \\ -\frac{12\sqrt{7}}{35} V_0 \bar{R} & \epsilon_d + \epsilon_s - \frac{6}{5}V_0 \bar{R} \end{pmatrix}$

$\displaystyle M_{J=3} = \epsilon_d + \epsilon_s$

$\displaystyle M_{J=4} = 2 \epsilon_d - \frac{2}{7} V_0 \bar{R}$

To solve the eigen systems, it is better to find the $\epsilon_d, \epsilon_s, V_0 \bar{R}$ from experimental data. The single particle energy of the d and s-orbtial can be found from 17O, We set the reference energy to the binding energy of 16O,

$\epsilon_d = BE(17O) - BE(16O) = -4.143 \textrm{MeV}$

$\epsilon_s = -4.143 + 0.871 = -3.272 \textrm{MeV}$

the ground state of 18O is

$E_0 = BE(18O) - BE(16O) = -12.189 \textrm{MeV}$

Solving the $M_{J=0}$ , the eigen value are

$\displaystyle \epsilon_d + \epsilon_s - 2 (V_0 \bar{R}) \pm \sqrt{ (\epsilon_s - \epsilon_d + V_0 \bar{R})^2 + 3 (V_0 \bar{R})^2 }$

Thus,

$V_0 \bar{R} = 1.057 \textrm{MeV}$

The solution for all status are

$E(j=0) = -12.189 ( 0 ) , 0.929 (0d_{5/2})^2 + 0.371 (1s_{1/2})^2$

$E(j=2) = -9.820 ( 2.368 ) , 0.764 (0d_{5/2})^2 + 0.645 (0d_{5/2})(1s_{1/2})$

$E(j=4) = -8.588 ( 3.600) , (0d_{5/2})^2$

$E(j=2) = -7.874 ( 4.313) , 0.645 (0d_{5/2})^2 -0.764 (0d_{5/2})(1s_{1/2})$

$E(j=3) = -7.415 (4.773) , (0d_{5/2})(1s_{1/2})$

$E(j=0) = -6.870 (5.317 ) , 0.371 (0d_{5/2})^2 - 0.929 (1s_{1/2})^2$

Annotation 2020-05-14 235102

The 2nd 0+ state is missing in above calculation. This is due to core-excitation that 2 p-shell proton promotes to d-shell.

In the sd- shell, there are 2 protons and 2 neutrons coupled to the lowest state. which is the same s-d shell configuration as 20Ne. The energy is

$E_{sd} = B(20Ne) - B(16O) = -33.027 \textrm{MeV}$

In the p-shell, the configuration is same as 14C, the energy is

$E_{p} = B(14C) - B(16O) = 22.335 \textrm{MeV}$

Thus, the energy for the 2-particle 2-hole of 18O is

$E(0^+_2) = -10.692 + 8 E' \textrm{MeV}$ ,

where $E'$ is the p-sd interaction, there are 4 particle in sd-shell and 2 hole in p-shell, thus, total of 8 particle-hole interaction.

The particle-hole can be estimate using 19F 1/2- state, This state is known to be a promotion of a p-shell proton into sd-shell.

In the sd-shell of 19F, the configuration is same as 20Ne. In the p-shell of 19F , the configuration is same as 15N, the energy is

$E_{p} = B(15N) - B(16O) = 12.128 \textrm{MeV}$

Thus, the energy for the 1/2- state of 19F relative to 16O is

$E_{1/2} = -20.899 + 4 E' \textrm{MeV}$

And this energy is also equal to

$E_{1/2} = -20.899 + 4 E' \textrm{MeV} = BE(19F) - BE(16O) + 0.110 = -20.072 \textrm{MeV}$

Thus,

$E' = 0.20675 \textrm{MeV}$

Therefore, the 2nd 0+ energy of 18O is

$E(0^+_2) = -9.038' \textrm{MeV} = 3.151 \textrm{MeV}$

Compare the experimental value of 3.63 MeV, this is a fair estimation.

It is interesting that, we did not really calculate the radial integral, and the angular part is calculated solely base on the algebra of J-coupling and the properties of delta interaction.

And since the single particle energies and residual interaction $V_0 \bar{R}$ are extracted from experiment. Thus, we can think that the basis is the “realistic” orbital of d and s -shell.

The spectroscopic strength and the wave function of the 18O ground state is $0.929 (0d_{5/2})^2 + 0.371 (1s_{1/2})^2$ . In here the basis wavefunctions $0d_{5/2}, 1s_{1/2}$ are the “realistic” or “natural” basis.

Correlation energy

April 26, 2016

GoLuckyRyan Basic 18O, correlation, neutron, TBMEs Leave a comment

The correlation is from the off-diagonal terms of the residual interaction, which is the TBME of the shell model. Consider following Hamiltonian of these nuclei

$H(Z,N-1) = H(Z,N-2) + h_n + R_n$

$H(Z,N) = H(Z,N-2) + h_{n_1} + h_{n_2} + R_{n_1} + R_{n_2} + R_{nn}$

the correlation energy between the two neutrons is the term $R_{nn}$ . Be aware that this is a residual interaction, not the nucleon-nucleon interaction $V_{ij}$ .

$V_{ij}= \begin{pmatrix} V_{11} & V_{12} & V_{1C} \\ V_{21} & V_{22} & V_{2C} \\ V_{C1} & V_{C2} & V_{CC} \end{pmatrix}$

Thus,

$R_{n} = \begin{pmatrix} V_{11} - U_1 & V_{1C} \\ V_{C1} & V_{CC}-U_C \end{pmatrix}$ ,

where $U_i$ is mean field.

The separation energies is proportional to the terms

$S_n(Z,N-1) \sim h_n + V_n$

$S_{2n}(Z,N)\sim h_{n_1} + h_{n_2} + V_{n_1} + V_{n_2} + V_{nn}$

Thus, the neutron-neutron correlation energy is

$\Delta_{pn}(N,Z) = 2*S_n(Z,N-1) - S_{2n}(N,Z)$

For 18O, $S_n(^{17}O) = 4.1431$ MeV, and $S_{2n}(^{18}O) = 12.1885$ MeV, thus, $\Delta_{2n}(^{18}O) = 3.9023$ MeV.

In the shell model calculation, the single particle energy of the 1d5/2 and 2s1/2 neutron are -4.143 MeV and -3.27 MeV respectively. The residual interaction is

$V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

The eigenenergy of the ground state of 18O is -10.539 on top of 16O.

The non-correlated binding energy of 18O is 2*-4.143 = -8.286 MeV.

Therefore, the theoretical correlated energy is 2.258 MeV.

Nuclear correlation & Spectroscopic factor

April 25, 2016

GoLuckyRyan Basic 18O, configuration, configuration mixing, core polarization, correlation, long-ranged correlation, neutron, residual interaction, shell model calculation, short-ranged correlation, spectroscopic factor, TBMEs Leave a comment

In the fundamental, correlation between two objects is

$P(x,y) \neq P(x) P(y)$

where $P$ is some kind of function. To apply this concept on nuclear physics, lets take a sample from 18O. 18O can be treated as 16O + n + n. In the independent particle model (IPM), the wave function can be expressed as

$\left|^{18}O\right>= \left|^{16}O\right>\left|\phi_a\right>\left|\phi_b\right>= \left|^{16}O\right>\left|2n\right>$

where the wave function of the two neutrons is expressed as a direct product of two IPM eigen wave functions, that they are un-correlated. Note that the anti-symmetry should be taken in to account, but neglected for simplicity.

We knew that IPM is not complete, the residual interaction has to be accounted. According to B.A. Brown, Lecture Notes in Nuclear Structure Physics [2011], Chapter 22, the 1s1/2 state have to be considered. Since the ground state spins of 18O and 16O are 0, thus, the wavefunction of the two neutrons has to be spin 0, so that only both are in 1d5/2 or 2s1/2 orbit. Thus, the two neutrons wave function is

when either $\alpha$ or $\beta$ not equal 0, thus, the two neutrons are correlation. In fact, the $\alpha = 0.87$ and $\beta = 0.49$ .

The spectroscopic factor of the sd-shell neutron is the coefficient of $\alpha$ times a isospin-coupling factor.

From the above example, the correlation is caused by the off-diagonal part of the residual interaction. To be more specific, lets take 18O as an example. The total Hamiltonian is

$H_{18} = H_{16} + h_1 + h_2 + V$

where $h_1 = h_2 = h$ is the mean field or single particle Hamiltonian

$h\left|\phi_i\right>= \epsilon_i\left|\phi_i\right>$

since $H_{16}$ is diagonal and not excited (if it excited, then it is called core polarization in shell model calculation, because the model space did not included 16O.), i.e. $H_{16} = \epsilon_{16} I$ , we can neglect it in the diagonalization of the $h_1+h_2 + V$ and add back at the end. In the 1d5/2 and 2s1/2 model space, in order to form spin 0, there is only 2 basis,

$\left|\psi_1\right> = \left|\phi_1\right>\left|\phi_1\right>$ and

$\left|\psi_2\right> = \left|\phi_2\right>\left|\phi_2\right>$

express the Hamiltonian in these basis,

$V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

Because of the diagonalization, the two states $\left|\psi_1\right>$ and $\left|\psi_2\right>$ are mixed, than the two neutrons are correlated. This is called configuration mixing.

According to B.A. Brown,

the configuration mixing on the above is long-ranged correlation (LRC). It is near the Fermi surface and the energy is up to 10 MeV.

The short-ranged correlation (SRC) is caused by the nuclear hard core that scattered a nucleon to highly single particle orbit up to 100 MeV.

The LRC is included in the two-body-matrix element. The SRC is included implicitly through re-normalization of the model space.

There is a correlation due to tensor force. Since the tensor force is also short-ranged, sometimes it is not clear what SRC is referring from the context. And the tensor force is responsible for the isoscalar pairing.

The discrepancy of the experimental spectroscopic factor and the shall model calculation is mainly caused by the SRC.

Shell model calculation and the USD, USDA, and USDB interaction

February 28, 2016

GoLuckyRyan Basic, theory Binding Energy, Carbon, configuration, diagonalization, ESPS, Gamow-Teller, interaction, model space, N=16, neutron, new Magic number, OXBASH, oxygen, renormalization, separation energy, Shell Model, single particle, single particle energy, spectroscopic factor, spin-orbital, TBMEs, USD Leave a comment

Form the mean field calculation, the single particle energies are obtained. However, the residual interaction is still there that the actual state could be affected. Because the residual interaction produces the off-diagonal terms in the total Hamiltonian, and that mixed the single particle state.

The Shell Model calculation can calculate the nuclear structure from another approach. It started from a assumed nuclear Hamiltonian, with a basis of wavefunctions. The Hamiltonian is diagonalized with the basis, then the eigenstates are the solution of the wavefunctions and the nuclear structure, both ground state and excited states. The basis is usually the spherical harmonic with some radial function. Or it could be, in principle, can take from the result of mean field calculation. Thus, the Shell Model calculation attacks the problem directly with only assumption of the nuclear interaction.

However, the dimension of the basis of the shell model calculation could be very huge. In principle, it should be infinitely because of the completeness of vector space. Fro practical purpose, the dimension or the number of the basis has to be reduced, usually take a major shell. for example the p-shell, s-d shell, p-f shell. However, even thought the model space is limited, the number of basis is also huge. “for $^{28}$ Si the 12-particle state with M=0 for the sum of the $j_z$ quantum numbers and $T_z=0$ for the sum of the %Latex t_z$ quantum numbers has dimension 93,710 in the m-scheme” [B. A. Brown and B. H. Wildenthal, Ann. Rev. Nucl. Part. Sci. 38 (1998) 29-66]. Beside the huge dimensions and the difficult for diagonalizing the Hamiltonian, the truncation of the model space also affect the interaction.

We can imagine that the effective interaction is different from the actual nuclear interaction, because some energy levels cannot be reached, for example, the short range hard core could produce very high energy excitation. Therefore, the results of the calculation in the truncated model space must be “re-normalized”.

There are 4 problems in the shell model calculation:

the model space
the effective interaction
the diagonalization
the renormalization of the result

The shell model can also calculate the excited state with $1\hbar \omega$ (1 major shell). This requires combination of the interactions between 2 major shell.

For usage, say in the code OXBASH, user major concern is the choice of the interaction and model space. The shell model are able to calculate

The binding energy
The excitation energies
- The nucleons separation energies
The configuration of each state
The magnetic dipole matrix elements
The Gamow-Teller (GT) transition
The spectroscopic factor
…… and more.

The W interaction (or the USD) for the s-d shell was introduced by B.H. Wildenthal around 1990s. It is an parametric effective interaction deduced from fitting experimental energy levels for some s-d shell nuclei. Before it, there are some theoretical interactions that require “no parameter”, for example the G-matrix interaction is the in-medium nucleon-nucleon interaction.

The problem for the USD interaction is the interpretation, because it is a black-box that it can reproduce most of the experimental result better than theoretical interactions, but no one know why and how. One possible way is translate the two-body matrix elements (TBME) back to the central, spin-orbit, tensor force. It found that the central and spin-orbit force are similar with the theoretical interactions, but the tensor force could be different. Also, there could be three-body force that implicitly included in the USD interaction.

In 2006, B.A. Brown and W.A. Richter improved the USD interaction with the new data from the past 20 years [B.A. Brown, PRC 74, 034315(2006)]. The new USD interaction is called USDA and USDB. The difference between USDA and USDB is the fitting (something like that, I am not so sure), but basically, USDA and USDB only different by very little. Since the USDB has better fitting, we will focus on the USDB interaction.

The single particle energy for the USDB is

$1d_{3/2} = 2.117$
$2s_{1/2} = -3.2079$
$1d_{5/2} = -3.9257$

in comparison, the single particle energies of the neutron of 17O of $2s_{1/2} = -3.27$ and $1d_{5/2} = -4.14$ .

Can to USD interaction predicts the new magic number N=16?

Yes, in a report by O. Sorlin and M.-G. Porquet (Nuclear magic numbers: new features far from stability) They shows the effective single particle energy of oxygen and carbon using the monopole matrix elements of the USDB interaction. The new magic number N=16 can be observed.

Nuclear Physics 101

Experimental TBMEs

Shell model calculation on 18O

Correlation energy

Nuclear correlation & Spectroscopic factor

Shell model calculation and the USD, USDA, and USDB interaction

Recent Posts

Calendar

Archives

Keywords

# of Visitors

Login