2-state system (an invert problem)

Given 2 states with energy $E_1 < E_2$, and interaction energy between the two states $V$, the Hamiltonian of the system is

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

This system is already discussed in this post. The solution state in here again as:

The eigen-energies

$\epsilon_{\pm} = \bar{E} \pm \sqrt{dE^2 + V^2 }$

the eigen-states are

$\Phi_{+} = \alpha \phi_1 + \beta \phi_2$

$\Phi_{-} = -\beta \phi_1 + \alpha \phi_2$

where

$\displaystyle \alpha = \frac{dE + \sqrt{dE^2 + V^2 }}{\sqrt{(dE+\sqrt{dE^2+V^2})^2 + V^2}}, \beta = \frac{V}{\sqrt{(dE+\sqrt{dE^2+V^2})^2 + V^2}}$

The $\phi_{1,2}$ are the wavefunction of pure state of $E_{1,2}$.

This can be easily extended to 3-state system or $n$-state system. The number of coupling constants is $n(n-1)/2$.

Experimentally, we observed the mixed states and the spectroscopic factors originated from one of the pure state. For example, in a neutron transfer reaction, the transferred neutron may coupled with the 2+ state of the core. The neutron is sitting in a pure orbital ($\phi_1$), say 1d5/2 orbital, it couple with the 2+ state ($\phi_2$) with interaction energy $V$. The final states will have energy $\epsilon_{\pm}$ and we extract the spectroscopic factors ($\alpha^2, \beta^2$) for the 1d5/2 orbital $. To state more clear, in a transfer reaction, we may observed two excited states $A_+, A_-$ with transfer of $|0p_{1/2}\rangle$ neutron. This neutron assumed to couple with a core state $|C\rangle$ and a core excited state $|C^+\rangle$. And we supposed that the 2 observed states are: $|A_+ \rangle = \alpha |0p_{1/2}\rangle |C\rangle + \beta |0p_{1/2} \rangle |C^+\rangle = \alpha |\phi_1\rangle + \beta |\phi_2\rangle$ $|A_- \rangle = \beta |0p_{1/2}\rangle |C\rangle - \alpha |0p_{1/2} \rangle |C_+\rangle = \beta |\phi_1\rangle - \alpha |\phi_2\rangle$ We then want to find out the un-perturbed energy $E_{1,2}$ and the interaction $V$. And this is actually an easy problem by the diagonalization process. The Hamiltonian was diagonalized into $\displaystyle H = \begin{pmatrix} E_1 && V \\ V && E_2 \end{pmatrix} = P \cdot D \cdot P^{-1}$ where $\displaystyle P = \begin{pmatrix} \alpha && -\beta \\ \beta && \alpha \end{pmatrix} , det(P) = 1$ $\displaystyle D = \begin{pmatrix} \epsilon_+ && 0 \\ 0 && \epsilon_- \end{pmatrix}$ Thus, $E_1 = \alpha^2 \epsilon_+ \beta^2 \epsilon_-$ $E_2 = \beta^2 \epsilon_+ \alpha^2 \epsilon_-$ $\displaystyle V = \alpha \beta (\epsilon_+ - \epsilon_-) = \frac{1}{2}\sqrt{d\epsilon^2 - dE^2}$ Advertisements Spectroscopic factor & 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 $\left| \Phi_B\right>_j$ $S_{ij} = n_i \beta_{ij}^2$ 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 $\displaystyle \vec{v}_{\pm} = \frac{1}{\sqrt{(dE \pm \sqrt{dE^2+V^2})^2 +V^2}} ( dE \pm \sqrt{dE^2 +V^2}, V)$ 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. Nuclear correlation & Spectroscopic factor 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 $\left| 2n \right> = \alpha \left|\phi_1\right>\left|\phi_1\right>+\beta \left|\phi_2\right>\left|\phi_2\right>$ 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. Energies in a nucleus There are many kinds of energies, such as single particle energy, potential energy, kinetic energy, separation energy, and Fermi energy. How these energies are related? I summarized in the following picture that the occupation number as a function of kinetic energy. Since the nucleus is a highly interactive system, although the temperature of the ground state of a nucleus should be absolute Zero, but the Fermi surface is not sharp but diffusive. The Fermi energy of a nucleon , which is the maximum kinetic energy of a nucleon, is approximately ~35 MeV. The potential energy is approximately ~ 50 MeV per nucleon. There is an additional energy for proton due to Coulomb force, which is a Coulomb barrier. The separation energy is the different between the potential energy and Fermi energy. The single particle energy is the energy for each single particle orbit. The binding energy for a nucleon is the energy requires to set that nucleon to be free, i.e. the energy difference between the single particle energy and the potential energy. There is a minimum kinetic energy, using the 3D spherical well as an approximation. The n-th root of the spherical Bessel function can give the energy of n-orbit with angular momentum $l$, so that. $j_{l}(kR) = 0, k^2=\frac{2 m E}{\hbar^2}$ For $l=0$, $j_0(x) = \frac{sin(x)}{x}$, the 1-st root is $x = \pi/2$, then $k=\frac{\pi}{2R}$ $\frac{2 m E}{\hbar^2} = \frac{\pi^2}{4 R^2}$ use $R = 1.25 A^{1/3}$, $E = \frac{\pi^2 \hbar^2}{8 m 1.25^2 A^{2/3}}$ use $\hbar c = 200 MeV \cdot fm$, $mc^2 = 940 MeV$ $E = \frac{33.6}{A^{2/3}} MeV$ we can see, for $^{16}O$, the minimum KE, which is the 1s-orbit, is about 5 MeV. Shell model calculation and the USD, USDA, and USDB interaction 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.