18O | Nuclear Physics 101

on beta-delayed neutron-emission

August 13, 2023

GoLuckyRyan Basic 16N, 16O, 17N, 17O, 18O Leave a comment

We calculated the Q-value of the beta-delayed proton emission in this post. The decay channel is rare, because, when a $\left|nlj \right>$ neutron changes to a $\left|nlj \right>$ proton and emits an electron, the $\left|nlj \right>$ proton orbital must be empty, i.e. the nucleus is N > Z, and the proton emission threshold is larger than neutron threshold. After the beta decay, in most cases, N > Z for the daughter nucleus, so that to have a proton emission is rare.

The situation is different for the beta-delayed neutron emission. After the beta decay, the neutron shell could still have an excessive number of neutrons and undergoes neutron emission. The Q-value for beta-delayed neutron decay is

$\displaystyle Q_{\beta^{-}n} = m(Z,A) - m(Z+1,A-1) - m_e -m_n$

whenever $Q_{\beta^{-}n} > 0$ , the beta-delayed neutron emission is possible. Nuclei with just 2 or 3 neutrons away from the stability village have beta-delayed neutron emission. For example, any isotopes beyond these nuclei can undergo beta-delayed neutron emission ⁹Li, ¹²Be, ¹³B, ¹⁶C, ¹⁷N, ²²O, ²²F, ²⁶Ne, etc.. The 2-step process can be pictured like this:

The interesting thing about the beta-delayed neutron emission is the nature of the daughter nucleus right after the beta-decay, and the branching ratio for the gamma-decay and neutron emission. Normally, the daughter nucleus is bound and only goes to gamma decay to release excessive energy. When the daughter nucleus is unbound and neutron emission is possible, what is the nature of the nucleus? Since the neutron emission is not 100% but ~15%. This indicates the daughter nucleus is a compound nucleus, involving collective motion.

In particle configuration, beta decay is when a neutron converts to a proton in the same orbital, as if the neutron decay to proton + electron inside the orbital. For example, 16N beta decay, decays to the 6.14 MeV 3- state (68%) and to the ground state (26%) of 16O. The 3- state can be understood as a coupling of a proton p1/2 hole + a proton d5/2 particle, which can also couple to 8.87 MeV 2- state. The ground state is a sd-shell neutron to proton p1/2. The 7.11 MeV 1- is probably proton p1/2 + a proton s1/2 particle.

Taken from David E. Alburger, Phys. Rev. **111**, 1586 (1958)

17N can populate 17O (Sn = 4.14 MeV ) unbounded state between 4.5 to 6.0 MeV [H. Ohm, W. Rudolph, K.-L. Kratz, Nuclear Physics A 274, 45-52 (1976)]. The 0.0 (5/2+) MeV, 0.87 (1/2+) MeV and 3.06 (1/2-) MeV are also populated by the beta-decay. The ground state can be understood as a sd-shell neutron converting to proton p1/2. The 0.87 MeV state can be imagined as follows, 17N should have a fraction of the sd-shell neutron pair in (s1/2)^2. A s1/2 neutron in 17N converts to proton p1/2 and filled the p-shell, leaving a single s1/2 neutron on top of the 16O core. For the 3.06 MeV state, one of the p1/2 neutrons converts to a proton in the p1/2 orbital, leaving a p1/2 neutron hole in 17O. The spectroscopic factor for the 16O(d,p) reaction of the 3.06 MeV state is only 0.032 and the SF for the same state from 18O(p,d) reaction is 0.88 with L = 1. How to understand the branching ratio? How doe sit related to particle configuration?

Taken from A. R. Poletti and J. G. Pronko, Physical Review C 8, 1285 (1973)

To be precise, the 4.55 (3/2-), 5.08 (3/2+), 5.39 (3/2-), and 5.93 (1/2-) MeV are populated. These 4 states have been populated by the 16O(d,p) and 18O(p,d) reactions, here are the spectroscopic factors

Energy [MeV]	J-pi	L	SF (adding)	SF (removal)	branching ratio
4.55	3/2-	1	0.23	0.14	36.6 +- 2.6%
5.08	3/2+	2	1.25	0.13	0.6 +- 0.4 %
5.39	3/2-	??	??	??	55.5 +- 3.5%
5.93	1/2-	??	??	??	7.4 +- 0.5%

We can see that, there is 55% to populate 17O 5.39 MeV state, but this state is not known to be populated using a single neutron-adding/removal experiment. It is not clear whether the lack of information is due to experimental limitations or whether those states are not neutron single-particle states. If it is later, the situation becomes very interesting. What is the 5.39 MeV state of 17O?

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.

Binding energy, Mass, Separation energy, and effective interaction energy

September 23, 2020

GoLuckyRyan Basic 18O, Binding Energy, separation energy Leave a comment

I hate to re-formulate the relations between them, so I write out in here.

The mass excess is the mass different between the atomic mass and the atomic number times the atomic mass unit $m_{\mu} = 931.49432 MeV/c^2$ .

$\displaystyle \Delta M(^{Z}A) = M(^{Z}A) + Z m_e- A m_{\mu}$

The binding energy is negative.

$\displaystyle M(^{Z}A) = (A-Z) m_n + Z m_p + BE(^{Z}A)$

$\displaystyle S_p(^{Z}A) =M(^{Z-1}A-1) + m_p - M(^{Z}A)$

$\displaystyle S_n(^{Z}A) =M(^{Z}A-1) + m_n - M(^{Z}A)$

Separation energy in term of Binding energy

$\displaystyle S_p(^{Z}A) =BE(^{Z-1}A-1) - BE(^{Z}A)$

$\displaystyle S_n(^{Z}A) =BE(^{Z}A-1) - BE(^{Z}A)$

Binding energy and Mass in term of separation energy

$\displaystyle BE(^{Z}A) = BE(^{Z-1}A) + S_n(^{Z}A+1) - S_p(^{Z}A+1)$

$\displaystyle BE(^{Z}A) = BE(^{Z+1}A) - S_n(^{Z+1}A+1) + S_p(^{Z+1}A+1)$

$\displaystyle M(^{Z}A) = M(^{Z-1}A) + S_n(^{Z}A+1) - S_p(^{Z}A+1) + m_p - m_n$

Introduce p-n separation energy that remove a proton and a neutron,

$\displaystyle S_{pn}(^{Z}A) = M(^{Z-1}A-2) + m_p + m_n - M(^{Z}A)$

Also, the 2p and 2n separation energies

$\displaystyle S_{2p}(^{Z}A) = M(^{Z-2}A-2) + 2m_p - M(^{Z}A)$

$\displaystyle S_{2n}(^{Z}A) = M(^{Z}A-2) + 2m_n - M(^{Z}A)$

in term of binding energy

$\displaystyle S_{pn}(^{Z}A) = BE(^{Z-1}A-2) - BE(^{Z}A)$

$\displaystyle S_{2p}(^{Z}A) = BE(^{Z-2}A-2) - BE(^{Z}A)$

$\displaystyle S_{2n}(^{Z}A) = BE(^{Z}A-2) - BE(^{Z}A)$

I am not sure how other people think, the separation energy is a more intuitive way to think of interaction energy than using binding energy.

For example, the energy of the d5/2 neutron in 17O is $S_n$ , the energy for the two d5/2 neutrons in 18O is $S_{2n}$ . See this post. Thus, the interaction energy of the two d5/2 neutrons in 18O is $S_{2n} - 2 S_n$ . This is the extra energy caused by the n-n interaction.

Following the same idea, the interaction energy of the pn pair, is $S_{pn} - S_p - S_n$ .

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.

Single-particle structure of 19F

January 28, 2019

GoLuckyRyan Basic 18O, 19F, 20Ne, Nilsson orbital, rotational band 2 Comments

About one year ago, I studied the nuclear structure of 19F. At that time, a lot of thing don’t understand and confused. But now, I have a better understanding.

19F is always fascinating because the so complicated energy levels for such a relatively simple system with only 1 proton, 2 neutrons on top of a 16O core, which usually treated as double magic rigid core.

It ground state is 1/2+, which is unusual.
It is also well-deformed with $\beta_2 \sim 0.4$ , deduced from rotational band.
It also has very low lying negative parity state of 0.110 MeV. This state is also the band head of K=1/2- rotational band.
The rotational band of K=1/2± does not following J(J+1) curve (see the last picture from this post), this indicate the rigid rotor assumption is not so good.

3 of the rotational bands of 19F, from M. Oyamada et al., PRC11 (1975) 1578

When looking the region around 19F. From 16O a double magic core, to 20Ne very deformed nucleus ( $\beta_2 \sim = 0.7$ ), and then go to 32Mg, the center of island of inversion. 16O, 18O, 18F, 20F are normal nuclei with normal shell order. 19F is a doorway nuclei for the complicated nuclear structure. That make the understanding the nuclear structure of 19F important.

it helps to understand how deformation happen in small system,
and how deformation can be described in particle-configuration.

The single-particle structure of the ground state of 19F has been studied using (p,2p) reaction. And found that the ground state to ground state transition only contain strength from 1s1/2 orbital. Thus, the wave function of 19F must bein the form

$\displaystyle |^{19}F\rangle = \alpha |\pi 1s_{1/2} \rangle |^{18}O\rangle + \beta |\pi 0d_{5/2}\rangle |^{18}O^* \rangle + ...$

The single-particle structures of the ground state and excited states are best studied using single-nucleon transfer. There are at least 4 directions, neutron-removal from 20F, neutron-adding from 18O, proton-removal from 20Ne, and proton-adding from 18O. Also, a proton/neutron-removal from 19F itself to study the ground state properties.

In above figure, we also plotted the spectroscopic factors from 2 reactions, which is taken from G. Th. Kaschl et al., NPA155(1970)417, and M. Yasue et al., PRC 46 (1992) 1242. These are the most significant studies. It was suprised that the negative parity state of 0.110 meV can be populated from the 18O(3He,d) reaction. These suggest the 18O is not a good core that it has about 10% strength of two-nucleon hole in its ground state. From the 20Ne(d, 3He) reaction, the 0.110 MeV state is highly populated. Given that the ground state of 20Ne is mainly proton 1s1/2 strength due to deformation. These result indicated that the negative parity state of 19F is due to a proton-hole. Thus we have a picture for the 0.110 MeV state of 19F:

The 18O(3He,d) reaction put a 0p1/2 proton in 18O 2-proton hole state to form the negative parity state. And 20Ne(d, 3He) reaction remove a 0p1/2 proton from 20Ne.

But this picture has a problem that, how come it is so easy to remove the 0p1/2 proton from 20Ne? the p-sd shell gap is known to be around 6 MeV! or, why the proton hole in 19F has such a small energy?

In current understanding, the 2-neutron are coupled to J=0 pair, and no contribution to low-lying states. But is it true?

I suspect, the underneath reason for the proton 1s1/2 orbital is lower is because the proton 0d5/2orbital was repealed by the 2-neutron in 0d5/2 orbital due to the tensor force. And somehow, the tensor force becomes smaller in 20F when the 0d5/2 orbital is half filled.

And because the proton is in 1s1/2 and the 0d5/2 is just 0.2 MeV away, a huge configuration mixing occurred. and then, a Nilsson orbit is formed with beta = 0.4. This is an example of NN-interaction driving deformation.

Rotational Band

January 18, 2019

GoLuckyRyan Basic 18O, 19F, Coriolis force, deformation, Nilsson orbital, rotation, Wigner-D 2 Comments

For deformed nuclei, it can be rotated in various angular momentum in Laboratory frame. Assume rigid body rotation, the energy is

$\displaystyle E_J = \frac{1}{2}I\omega^2 = \frac{1}{2I}J^2$

In QM, that becomes

$\displaystyle H = \sum_{i=1}^{3} \frac{\hbar^2}{2I_i} J_i^2$

For axial symmetry, $I_1 = I_2 = I$

$\displaystyle H = \frac{\hbar^2}{2I} (J^2 - J_3^2) + \frac{\hbar^2}{2I_3}J_3^2$

Remember, in deformed nuclei, the projection of $J$ along the symmetry axis in the body-frame is $K$ . The expected value of the Hamiltonian with state $|Nn_z m_l K \rangle$ in the body-frame is proportional to $J(J+1)$ for $J^2$ and $K$ for $J_3$ . i.e.

$\displaystyle E_J = \frac{\hbar^2}{2I} J(J+1) + E_K$

From body-frame to Lab-frame, we should apply the Wigner D-Matrix to the intrinsic wave function. ( I am not sure the following equation is correct, but the idea is rotating the body-frame wavefunction with Wigner D-Matrix to get the Lab-frame wave function. In Lab frame the total angular momentum must be a good Quantum number as rotational symmetry restored, so as $J_z = M$ . The problem of the following equation is that the J is not a good Q-number in Nilsson wavefunction )

$\displaystyle |JMK\rangle = \sum_{M} D_{MK}^{J} |Nn_zm_lK\rangle$

However, the Wigner D-Matrix does not conserve parity transform:

$\displaystyle D_{MK}^J \rightarrow (-1)^{J+K} D_{M-K}^{J}$

In order to restored the parity, we need to include $\pm K$ in the Lab-frame wave function.

$\displaystyle |JMK\rangle = \sum_{M} \left( D_{MK}^J \pm (-1)^{J+K} D_{M-K}^J \right) |Nn_zm_lK\rangle$

where + for positive parity, – for negative parity.

From the above equation, for $K^\pi = 0^+$ ( $0^-$ ), $J$ must be even (odd). For $K > 0$ , $J = K, K+1, K+2, ...$ .

rotaional Band of 205Fm.png

rotational band of 253No.png

rotational band of 19F.png

We can see for $K = 1/2$ , the $J = 5/2, 9/2, 11/2$ are lower to the main sequence. This was explained by adding an extra term in the rotation Hamiltonian that connect $\Delta K = 1$ .

$\displaystyle \langle JMK | H'(\Delta K = 1) |JMK \rangle$

$\displaystyle \rightarrow \langle D_{MK}^J | H' | D_{MK}^J \rangle+ \langle D_{M-K}^J |H'| D_{MK}^J \rangle + \langle D_{MK}^J | H' | D_{M-K}^J \rangle+ \langle D_{M-K}^J |H'| D_{M-K}^J \rangle$

The term with $\Delta K = 0$ vanished. And since $\latex K = 1$, the only non-zero case is $K = 1/2$ .

A possible form of the $H' (\Delta K = 1) = \frac{1}{2} \omega (J_+ + J_-)$ . These are the ladder operator to rise or lower the m-component by 1. In 19F case, we can think it is a single proton on top of 18O core. A rotation core affect the proton with an additional force, similar to Coriolis force on earth.

The nuclear structure of 19F

January 30, 2018

GoLuckyRyan Basic 18O, 19F, 20Ne, 3Hed, d3He, Nilsson orbital, p2p Leave a comment

Some facts about 19F:

Ground state spin-parity is 1/2+.
Has low-lying 1/2- state at 110 keV.
Magnetic dipole moment is 2.62885 μ0.
19F(p,2p) experiment reported only 2s-wave can fit the result. [M.D. High et al., PLB 41 (1972) 588]
19F(d, 3He) experiment reported the ground state is from 1s1/2 proton with spectroscopic factor of 0.38. [G. Th. Kaschl et al., NPA 155 (1970) 417]
18O(3He, d) experiment report the ground state is 1s1/2 proton with spectroscopic factor of 0.21. [C. Schmidt et al., NPA 155 (1970) 644]
There is a rotational band of 19F [C. F. Williamson et al., PRL 40 (1978) 1702]

My understanding [2018-01-30]

The 19F is deformed. The deformation is confirmed from rotation band.

The deformation distorted the spherical basis into deformed basis. In the simplest deformed basis, the cylindrical basis, the lowest s-d shell state is $|Nn_z m_l K\rangle =|220(1/2)\rangle$ , which is mixed with 1s1/2 (~33%) and 0d5/2 (~66%) orbits and the K, the intrinsic spin, is 1/2. Thus, the 19F ground state spin must be 1/2.

The 19F wave function can be written as (approximately)

$|^{19}F\rangle = \sqrt{0.2 \sim 0.4}|\pi 1s_{1/2} \times ^{18}O_{g.s}\rangle + \sqrt{0.8\sim0.6}|\pi 0d_{5/2} \times ^{18}O^*\rangle + ...$

Under proton transfer/pickup reactions, the selection of oxygen ground state force the transfer proton to be in 1s1/2 state. The founding of s-wave ground state spin of 1/2 of 19F agrees with this picture.

Using USDB interaction with pn formalism. The 18O, 19F ground state are

$|^{18}O\rangle = \sqrt{0.78} |(\nu0d_{5/2})^2 \times ^{16}O\rangle + \sqrt{0.17}|(\nu1s_{1/2})^2 \times ^{16}O\rangle + ...$

$|^{19}F\rangle = \\ \sqrt{0.22} |(\pi1s_{1/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.17}|(\pi1s_{1/2})(\nu1s_{1/2})^2 \times ^{16}O\rangle + \\ \sqrt{0.27}|(\pi0d_{5/2})(\nu0d_{5/2})^2 \times ^{16}O\rangle + ...$

The sum of $\pi1s_{1/2}$ is 0.39, and the ground state is 0.17. This close to the 18O(3He, d) experiment result, so that the interaction accurately reproduce the shell configuration.

The fact that the spectroscopic factor is much less then unity suggests the ground state configuration of 19F is not fit for single particle picture.

There are fill questions,

Why deform? due to the single 1d5/2 proton? Suppose adding a proton on 18O, the proton fill on 1d5/2 shell, and the d-shell creates a deformation on the sd shell, that shift the energy lower by mixing with s-shell?
in 19F(d, 3He) reaction, the sum of spectroscopic factor in sd-shell is just 1.54. This suggest large uncertainty. And the s-state SF is 0.38, almost a double for 18O(3He,d) reaction, How come?
in 19F(d,3He) reaction, the s:d ratio is 0.4:0.6, this is similar to prediction of Nilsson model, but difference from USDB calculation.
If the ground state has 1d5/2 proton, why the magnetic moment are so close to free proton? the l=2 should also contribute.
Is neutron shell also 1s1/2 ?
What is the $\beta_2$ ?
in 20Ne, will the proton also in 1s1/2 shell? 20Ne has $\beta_2 = 0.7$ very deformed.
Deformed DWBA?

The following is not organised thought.

According to the standard shell ordering, on top of 18O, an extra proton should fill up the 1d5/2 shell, and then the ground state spin of 19F should be 5/2. However, the ground state spin in 1/2. This is postulated to be due to deformation [mean field calculation, β2 = 0.275], 18O core excitation, or configuration mixing state [J.P. Elhot and A. M. Lane(1957)].

Under deformation, the conventional shell ordering is not suitable and may be an invalid picture to view the nucleus. So, talk about shell ordering is non-sense.

Since the 19F is 18O + 1s1/2 proton superposed with 1d5/2, there could be deformation. The spherical shape of 19F can be seen indirectly from the magnetic dipole momentum, the value is very close to that of a free proton of 2.78284734 μ0, only difference by 0.154 μ0, or 5.5%. How to solve this contradiction?

From the study of G. Th. Kaschl et al., the spectroscopic factor of the 19F(d,3He)18Og.s. channel is 0.38. The missing 1s1/2 strength most probably can be found in the higher excitation states. This indicates the ground state of 19F is a configuration mixing state. However, they also pointed out that caution is advisable with the absolute spectroscopic factor, this could be due to imperfect DWBA calculation.

The relative spectroscopic factors for the positive parity states, which normalised to the ground state, are agree with shell model prediction in sd-shell model space suggests that the core excitation should not play an important role.

From the USDB interaction, the shell ordering is normal, but the interaction result in a 1/2+ ground state. How?

What is the nature of the low lying 1/2- excited state in 19F?

20Ne(d,3He)19F reaction can populate this low lying state, suggests the p-shell proton pickup come from the nuclear surface.

( if (12C,13N) proton pickup reaction can populate this state, then, it can be confirmed that this is a surface p-shell proton, that it could be from 2p3/2. )

Spectroscopic factor & Occupation number

May 5, 2016

GoLuckyRyan Basic 18O, 2-state system, configuration mixing, Eigen System, monopole, nucleon-nucleon correlation, nucleon-nucleon interaction, occupation number, spectroscopic factor Leave a comment

[Update 2024-01-02] Please check the post: Sum rule of Spectroscopic factors. This is the correct way to relate the spectroscopic factor and the occupation number.

Started from independent particle model, the Hamiltonian of a nucleus with mass number $A$ is

$H_A = \sum\limits_{i}^{A} h_i + \sum\limits_{i, j>i} V_{ij}$

we can rewrite the Hamiltonian by isolating a nucleon

$H_A = H_B + h_1 + V_{B1}$

than, we can use the basis of $H_B$ and $h_1$ to construct the wavefunction of the nucleus A as

$\left|\Phi_A\right>_{J} = \sum\limits_{i,j} \beta_{ij} \left[ \left|\phi_i\right>\left|\Phi_{B}\right>_{j} \right]_{J}$

where the square bracket is anti-symmetric angular coupling between single particle wavefunction $\left|\phi_i\right>$ and wavefunction $\left|\Phi_B\right>_{j}$ . The $\beta_{ij}$ is the spectroscopic amplitude.

The square of the spectroscopic amplitude times number of particle $n_i$ at state $\left|\phi_i\right>$ is the spectroscopic factor of the nucleon at state $\left|\phi_i\right>$ and nucleus B at state $\left| \Phi_B\right>_j$

$S_{ij} = n_i \beta_{ij}^2$ [comment 2024-01-02] This is an ad hox construct.

The occupation number is the sum of the spectroscopic factors of the nucleus B

$\sum\limits_{j} S_{ij} = n_i$

After the definition, we can see, when the nucleon-core interaction $V_{B1}$ is neglected, the spectroscopic factor is 1 and the occupation number is also 1.

Since the $\beta$ is coefficient for changing basis from $\left|\Phi_A\right>_J$ into basis of $\left[ \left|\phi_i\right>\left|\Phi_{B}\right>_{j} \right]_{J}$ Thus, the matrix $\beta$ is unitary

$\beta\cdot \beta^\dagger = 1$

thus, each column of row vector of $\beta$ is normalized.

The properties of $\beta$ can be found by solving the eigen system of the $H_A$ from the core $H_C = H_B + h_1$ . The core hamiltonian is diagonal. The nucleon-core interaction introduce diagonal terms and off-diagonal terms. When only diagonal terms or monopole term exist, only the eigen energy changes but the eignestate unchanges. Therefore, the configuration mixing is due to the off-diagonal terms.

However, when there are off-diagonal terms, the change of diagonal terms will changes the mixing.

In degenerate 2 states system, the Hamiltonian be

$H= \begin{pmatrix} E_1 & V \\ V & E_2 \end{pmatrix}$

The eigen energy are $\bar{E} \pm \sqrt{ dE^2 + V^2}$ , where $\bar{E} = (E_1+E_2)/2$ and $dE = (E_1 - E_2)/2$ , The eigen vector are

$\displaystyle \vec{v}_{\pm} = \frac{1}{\sqrt{(dE \pm \sqrt{dE^2+V^2})^2 +V^2}} ( dE \pm \sqrt{dE^2 +V^2}, V)$

2-state mixing.PNG

Since the eigenvector must be orthogonal. If we set one eigenvector be

$\displaystyle \vec{v}_+ = { \alpha, \beta}, \alpha^2 + \beta^2 = 1$

The other eigen vector must be

$\displaystyle \vec{v}_- = { -\beta, \alpha}$

One can check that

$\displaystyle \frac{V}{\sqrt{(dE + \sqrt{dE^2+V^2})^2 +V^2}} = \frac{ dE - \sqrt{dE^2 +V^2} }{\sqrt{(dE - \sqrt{dE^2+V^2})^2 +V^2}}$

When the states are degenerated, $E_1 = E_2 = E$ , the eigen energy is $E \pm V$ , and the eigen state is $\frac{1}{\sqrt{2}} (1,1)$

As we can see, the eigen state only depends on the difference of the energy level, thus, we can always subtract the core energy and only focus on a single shell. For example, when we consider 18O, we can subtract the 16O binding energy.

In above figure, we fixed the $E_1$ and $E_2$ . We can see the spectroscopic factor decreases for the lower energy state (red line), and the state mixing increase.

We now fixed the $V$ and $E2$ . When $E_1 < E_2$ , the particle stays more on the $E_1$ state, which is lower energy. The below plot is the eigen energy. Even though the mixing is mixed more on the excited state, but the eigen energy did not cross.

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.

Nuclear Physics 101

on beta-delayed neutron-emission

Experimental TBMEs

Binding energy, Mass, Separation energy, and effective interaction energy

Shell model calculation on 18O

Single-particle structure of 19F

Rotational Band

The nuclear structure of 19F

Spectroscopic factor & Occupation number

Correlation energy

Nuclear correlation & Spectroscopic factor

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