For quantum numbers, chapter 1 on angular momentum and chapter 5 of Suhonen and chapters 5, 12 and 13 of Alex Brown. For a discussion of isospin, see for example Alex Brown's lecture notes chapter 12, 13 and 19.
| Quantum numbers | Energy [MeV] |
|---|---|
| \( 0s_{1/2}^{\pi} \) | -40.4602 |
| \( 0s_{1/2}^{\pi} \) | -40.4602 |
| \( 0s_{1/2}^{\nu} \) | -40.6426 |
| \( 0s_{1/2}^{\nu} \) | -40.6426 |
| \( 0p_{1/2}^{\pi} \) | -6.7133 |
| \( 0p_{1/2}^{\pi} \) | -6.7133 |
| \( 0p_{1/2}^{\nu} \) | -6.8403 |
| \( 0p_{1/2}^{\nu} \) | -6.8403 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\pi} \) | -11.5886 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
| \( 0p_{3/2}^{\nu} \) | -11.7201 |
We observe that with increasing value of \( j \) the degeneracy increases. For \( j=3/2 \) we end up diagonalizing the same matrix four times. With increasing value of \( j \), it is rather obvious that our insistence on using an uncoupled scheme (or just m-scheme) will lead to unnecessary labor from our side (or more precisely, for the computer). The obvious question we should pose ourselves then is whether we can use the underlying symmetries of the nuclear forces in order to reduce our efforts.
In order to understand the basics of the nucleon-nucleon interaction and the pertaining symmetries, we need to define the relevant quantum numbers and how we build up a single-particle state and a two-body state, and obviously our final holy grail, a many-boyd state.
We have that the operators for the orbital momentum are given by
$$
L_x=-i\hbar(y\frac{\partial }{\partial z}-z\frac{\partial }{\partial y})=
yp_z-zp_y,
$$
$$
L_y=-i\hbar(z\frac{\partial }{\partial x}-x\frac{\partial }{\partial z})= zp_x-xp_z,
$$
$$
L_z=-i\hbar(x\frac{\partial }{\partial y}-y\frac{\partial }{\partial x})=xp_y-yp_x.
$$
Since we have a spin orbit force which is strong, it is easy to show that the total angular momentum operator
$$
\hat{J}=\hat{L}+\hat{S}
$$
does not commute with \( \hat{L}_z \) and \( \hat{S}_z \). To see this, we calculate for example
$$
\begin{eqnarray}
[\hat{L}_z,\hat{J}^2]&=&[\hat{L}_z,(\hat{L}+\hat{S})^2] \\ \nonumber
&=&[\hat{L}_z,\hat{L}^2+\hat{S}^2+2\hat{L}\hat{S}]\\ \nonumber
&=& [\hat{L}_z,\hat{L}\hat{S}]=[\hat{L}_z,\hat{L}_x\hat{S}_x+\hat{L}_y\hat{S}_y+\hat{L}_z\hat{S}_z]\ne 0,
\end{eqnarray}
$$
since we have that \( [\hat{L}_z,\hat{L}_x]=i\hbar\hat{L}_y \) and \( [\hat{L}_z,\hat{L}_y]=i\hbar\hat{L}_x \).
We have also
$$
|\hat{J}|=\hbar\sqrt{J(J+1)},
$$
with the the following degeneracy
$$
M_J=-J, -J+1, \dots, J-1, J.
$$
With a given value of \( L \) and \( S \) we can then determine the possible values of
\( J \) by studying the \( z \) component of \( \hat{J} \).
It is given by
$$
\hat{J}_z=\hat{L}_z+\hat{S}_z.
$$
The operators \( \hat{L}_z \) and \( \hat{S}_z \) have the quantum numbers
\( L_z=M_L\hbar \) and \( S_z=M_S\hbar \), respectively, meaning that
$$
M_J\hbar=M_L\hbar +M_S\hbar,
$$
or
$$
M_J=M_L +M_S.
$$
Since the max value of \( M_L \) is \( L \) and for \( M_S \) is \( S \)
we obtain
$$
(M_J)_{\mathrm{maks}}=L+S.
$$
For nucleons we have that the maximum value of \( M_S=m_s=1/2 \), yielding
$$
(m_j)_{\mathrm{max}}=l+\frac{1}{2}.
$$
Using this and the fact that the maximum value of \( M_J=m_j \) is \( j \) we have
$$
j=l+\frac{1}{2}, l-\frac{1}{2}, l-\frac{3}{2}, l-\frac{5}{2}, \dots
$$
To decide where this series terminates, we use the vector inequality
$$
|\hat{L}+\hat{S}| \ge \left| |\hat{L}|-|\hat{S}|\right|.
$$
Using \( \hat{J}=\hat{L}+\hat{S} \) we get
$$
|\hat{J}| \ge |\hat{L}|-|\hat{S}|,
$$
or
$$
|\hat{J}|=\hbar\sqrt{J(J+1)}\ge |\hbar\sqrt{L(L+1)}-
\hbar\sqrt{S(S+1)}|.
$$
If we limit ourselves to nucleons only with \( s=1/2 \) we find that
$$
|\hat{J}|=\hbar\sqrt{j(j+1)}\ge |\hbar\sqrt{l(l+1)}-
\hbar\sqrt{\frac{1}{2}(\frac{1}{2}+1)}|.
$$
It is then easy to show that for nucleons there are only two possible values of
\( j \) which satisfy the inequality, namely
$$
j=l+\frac{1}{2}\hspace{0.1cm} \mathrm{or} \hspace{0.1cm}j=l-\frac{1}{2},
$$
and with \( l=0 \) we get
$$
j=\frac{1}{2}.
$$
Let us study some selected examples. We need also to keep in mind that parity is conserved. The strong and electromagnetic Hamiltonians conserve parity. Thus the eigenstates can be broken down into two classes of states labeled by their parity \( \pi= +1 \) or \( \pi=-1 \). The nuclear interactions do not mix states with different parity.
For nuclear structure the total parity originates from the intrinsic parity of the nucleon which is \( \pi_{\mathrm{intrinsic}}=+1 \) and the parities associated with the orbital angular momenta \( \pi_l=(-1)^l \) . The total parity is the product over all nucleons \( \pi = \prod_i \pi_{\mathrm{intrinsic}}(i)\pi_l(i) = \prod_i (-1)^{l_i} \)
The basis states we deal with are constructed so that they conserve parity and have thus a definite parity.
Note that we do have parity violating processes, more on this later although our focus will be mainly on non-parity viloating processes
Consider now the single-particle orbits of the \( 1s0d \) shell. For a \( 0d \) state we have the quantum numbers \( l=2 \), \( m_l=-2,-1,0,1,2 \), \( s+1/2 \), \( m_s=\pm 1/2 \), \( n=0 \) (the number of nodes of the wave function). This means that we have positive parity and
$$
j=\frac{3}{2}=l-s\hspace{1cm} m_j=-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}.
$$
and
$$
j=\frac{5}{2}=l+s\hspace{1cm} m_j=-\frac{5}{2},-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{5}{2}.
$$
Our single-particle wave functions, if we use the harmonic oscillator, do however not contain the quantum numbers \( j \) and \( m_j \). Normally what we have is an eigenfunction for the one-body problem defined as
$$
\phi_{nlm_lsm_s}(r,\theta,\phi)=R_{nl}(r)Y_{lm_l}(\theta,\phi)\xi_{sm_s},
$$
where we have used spherical coordinates (with a spherically symmetric potential) and the spherical harmonics
$$
Y_{lm_l}(\theta,\phi)=P(\theta)F(\phi)=\sqrt{\frac{(2l+1)(l-m_l)!}{4\pi (l+m_l)!}}
P_l^{m_l}(cos(\theta))\exp{(im_l\phi)},
$$
with \( P_l^{m_l} \) being the so-called associated Legendre polynomials.
Examples are
$$
Y_{00}=\sqrt{\frac{1}{4\pi}},
$$
for \( l=m_l=0 \),
$$
Y_{10}=\sqrt{\frac{3}{4\pi}}cos(\theta),
$$
for \( l=1 \) and \( m_l=0 \),
$$
Y_{1\pm 1}=\sqrt{\frac{3}{8\pi}}sin(\theta)exp(\pm i\phi),
$$
for \( l=1 \) and \( m_l=\pm 1 \),
$$
Y_{20}=\sqrt{\frac{5}{16\pi}}(3cos^2(\theta)-1)
$$
for \( l=2 \) and \( m_l=0 \) etc.
How can we get a function in terms of \( j \) and \( m_j \)? Define now
$$
\phi_{nlm_lsm_s}(r,\theta,\phi)=R_{nl}(r)Y_{lm_l}(\theta,\phi)\xi_{sm_s},
$$
and
$$
\psi_{njm_j;lm_lsm_s}(r,\theta,\phi),
$$
as the state with quantum numbers \( jm_j \).
Operating with
$$
\hat{j}^2=(\hat{l}+\hat{s})^2=\hat{l}^2+\hat{s}^2+2\hat{l}_z\hat{s}_z+\hat{l}_+\hat{s}_{-}+\hat{l}_{-}\hat{s}_{+},
$$
on the latter state we will obtain admixtures from possible \( \phi_{nlm_lsm_s}(r,\theta,\phi) \) states.
To see this, we consider the following example and fix
$$
j=\frac{3}{2}=l-s\hspace{1cm} m_j=\frac{3}{2}.
$$
and
$$
j=\frac{5}{2}=l+s\hspace{1cm} m_j=\frac{3}{2}.
$$
It means we can have, with \( l=2 \) and \( s=1/2 \) being fixed, in order to have \( m_j=3/2 \) either \( m_l=1 \) and \( m_s=1/2 \) or
\( m_l=2 \) and \( m_s=-1/2 \). The two states
$$
\psi_{n=0j=5/2m_j=3/2;l=2s=1/2}
$$
and
$$
\psi_{n=0j=3/2m_j=3/2;l=2s=1/2}
$$
will have admixtures from \( \phi_{n=0l=2m_l=2s=1/2m_s=-1/2} \) and \( \phi_{n=0l=2m_l=1s=1/2m_s=1/2} \).
How do we find these admixtures? Note that we don't specify the values of \( m_l \) and \( m_s \)
in the functions \( \psi \) since
\( \hat{j}^2 \) does not commute with \( \hat{L}_z \) and \( \hat{S}_z \).
We operate with
$$
\hat{j}^2=(\hat{l}+\hat{s})^2=\hat{l}^2+\hat{s}^2+2\hat{l}_z\hat{s}_z+\hat{l}_+\hat{s}_{-}+\hat{l}_{-}\hat{s}_{+}
$$
on the two \( jm_j \) states, that is
$$
\hat{j}^2\psi_{n=0j=5/2m_j=3/2;l=2s=1/2}= \alpha\hbar^2[l(l+1)+\frac{3}{4}+2m_lm_s]\phi_{n=0l=2m_l=2s=1/2m_s=-1/2}+
$$
$$
\beta\hbar^2\sqrt{l(l+1)-m_l(m_l-1)}\phi_{n=0l=2m_l=1s=1/2m_s=1/2},
$$
and
$$
\hat{j}^2\psi_{n=0j=3/2m_j=3/2;l=2s=1/2}= \alpha\hbar^2[l(l+1)+\frac{3}{4}+2m_lm_s]+ \phi_{n=0l=2m_l=1s=1/2m_s=1/2}+
$$
$$
\beta\hbar^2\sqrt{l(l+1)-m_l(m_l+1)}\phi_{n=0l=2m_l=2s=1/2m_s=-1/2}.
$$
This means that the eigenvectors \( \phi_{n=0l=2m_l=2s=1/2m_s=-1/2} \) etc are not eigenvectors of \( \hat{j}^2 \). The above problems gives a \( 2\times2 \) matrix that mixes the vectors \( \psi_{n=0j=5/2m_j3/2;l=2m_ls=1/2m_s} \) and \( \psi_{n=0j=3/2m_j3/2;l=2m_ls=1/2m_s} \) with the states \( \phi_{n=0l=2m_l=2s=1/2m_s=-1/2} \) and \( \phi_{n=0l=2m_l=1s=1/2m_s=1/2} \). The unknown coefficients \( \alpha \) and \( \beta \) are the eigenvectors of this matrix. That is, inserting all values \( m_l,l,m_s,s \) we obtain the matrix
$$
\left[ \begin{array} {cc} 19/4 & 2 \\ 2 & 31/4 \end{array} \right]$$
whose eigenvectors are the columns of
$$
\left[ \begin{array} {cc} 2/\sqrt{5} &1/\sqrt{5} \\ 1/\sqrt{5} & -2/\sqrt{5} \end{array}\right]$$
These numbers define the so-called Clebsch-Gordan coupling coefficients (the overlaps between the two basis sets). We can thus write
$$
\psi_{njm_j;ls}=\sum_{m_lm_s}\langle lm_lsm_s|jm_j\rangle\phi_{nlm_lsm_s},
$$
where the coefficients \( \langle lm_lsm_s|jm_j\rangle \) are the so-called Clebsch-Gordan coeffficients.
The Clebsch-Gordan coeffficients \( \langle lm_lsm_s|jm_j\rangle \) have some interesting properties for us, like the following orthogonality relations
$$
\sum_{m_1m_2}\langle j_1m_1j_2m_2|JM\rangle\langle j_1m_1j_2m_2|J'M'\rangle=\delta_{J,J'}\delta_{M,M'},
$$
$$
\sum_{JM}\langle j_1m_1j_2m_2|JM\rangle\langle j_1m_1'j_2m_2'|JM\rangle=\delta_{m_1,m_1'}\delta_{m_2,m_2'},
$$
$$
\langle j_1m_1j_2m_2|JM\rangle=(-1)^{j_1+j_2-J}\langle j_2m_2j_1m_1|JM\rangle,
$$
and many others. The latter will turn extremely useful when we are going to define two-body states and interactions in a coupled basis.
The orthogonality relation can be tested using the symbolic python package wigner. Let us test
$$
\sum_{m_1m_2}\langle j_1m_1j_2m_2|JM\rangle\langle j_1m_1j_2m_2|J'M'\rangle=\delta_{J,J'}\delta_{M,M'},
$$
The following program tests this relation for the case of \( j_1=3/2 \) and \( j_2=3/2 \) meaning that \( m_1 \) and \( m_2 \)
run from \( -3/2 \) to \( 3/2 \).
Summing up, for for the single-particle case, we have the following eigenfunctions
$$
\psi_{njm_j;ls}=\sum_{m_lm_s}\langle lm_lsm_s|jm_j\rangle\phi_{nlm_lsm_s},
$$
where the coefficients \( \langle lm_lsm_s|jm_j\rangle \) are the so-called Clebsch-Gordan coeffficients.
The relevant quantum numbers are \( n \) (related to the principal quantum number and the number of nodes of the wave function) and
$$
\hat{j}^2\psi_{njm_j;ls}=\hbar^2j(j+1)\psi_{njm_j;ls},
$$
$$
\hat{j}_z\psi_{njm_j;ls}=\hbar m_j\psi_{njm_j;ls},
$$
$$
\hat{l}^2\psi_{njm_j;ls}=\hbar^2l(l+1)\psi_{njm_j;ls},
$$
$$
\hat{s}^2\psi_{njm_j;ls}=\hbar^2s(s+1)\psi_{njm_j;ls},
$$
but \( s_z \) and \( l_z \) do not result in good quantum numbers in a basis where we
use the angular momentum \( j \).
For a two-body state where we couple two angular momenta \( j_1 \) and \( j_2 \) to a final angular momentum \( J \) with projection \( M_J \), we can define a similar transformation in terms of the Clebsch-Gordan coeffficients
$$
\psi_{(j_1j_2)JM_J}=\sum_{m_{j_1}m_{j_2}}\langle j_1m_{j_1}j_2m_{j_2}|JM_J\rangle\psi_{n_1j_1m_{j_1};l_1s_1}\psi_{n_2j_2m_{j_2};l_2s_2}.
$$
We will write these functions in a more compact form hereafter, namely,
$$
|(j_1j_2)JM_J\rangle=\psi_{(j_1j_2)JM_J},
$$
and
$$
|j_im_{j_i}\rangle=\psi_{n_ij_im_{j_i};l_is_i},
$$
where we have skipped the explicit reference to \( l \), \( s \) and \( n \). The spin of a nucleon is always \( 1/2 \) while the value of \( l \) can be deduced from the parity of the state.
It is thus normal to label a state with a given total angular momentum as
\( j^{\pi} \), where \( \pi=\pm 1 \).
Our two-body state can thus be written as
$$
|(j_1j_2)JM_J\rangle=\sum_{m_{j_1}m_{j_2}}\langle j_1m_{j_1}j_2m_{j_2}|JM_J\rangle|j_1m_{j_1}\rangle|j_2m_{j_2}\rangle.
$$
Due to the coupling order of the Clebsch-Gordan coefficient it reads as
\( j_1 \) coupled to \( j_2 \) to yield a final angular momentum \( J \). If we invert the order of coupling we would have
$$
|(j_2j_1)JM_J\rangle=\sum_{m_{j_1}m_{j_2}}\langle j_2m_{j_2}j_1m_{j_1}|JM_J\rangle|j_1m_{j_1}\rangle|j_2m_{j_2}\rangle,
$$
and due to the symmetry properties of the Clebsch-Gordan coefficient we have
$$
|(j_2j_1)JM_J\rangle=(-1)^{j_1+j_2-J}\sum_{m_{j_1}m_{j_2}}\langle j_1m_{j_1}j_2m_{j_2}|JM_J\rangle|j_1m_{j_1}\rangle|j_2m_{j_2}\rangle=(-1)^{j_1+j_2-J}|(j_1j_2)JM_J\rangle.
$$
We call the basis \( |(j_1j_2)JM_J\rangle \) for the coupled basis, or just \( j \)-coupled basis/scheme. The basis formed by the simple product of single-particle eigenstates
\( |j_1m_{j_1}\rangle|j_2m_{j_2}\rangle \) is called the uncoupled-basis, or just the \( m \)-scheme basis.
We have thus the coupled basis
$$
|(j_1j_2)JM_J\rangle=\sum_{m_{j_1}m_{j_2}}\langle j_1m_{j_1}j_2m_{j_2}|JM_J\rangle|j_1m_{j_1}\rangle|j_2m_{j_2}\rangle.
$$
and the uncoupled basis
$$
|j_1m_{j_1}\rangle|j_2m_{j_2}\rangle.
$$
The latter can easily be generalized to many single-particle states whereas the first
needs specific coupling coefficients and definitions of coupling orders.
The \( m \)-scheme basis is easy to implement numerically and is used in most standard shell-model codes.
Our coupled basis obeys also the following relations
$$
\hat{J}^2|(j_1j_2)JM_J\rangle=\hbar^2J(J+1)|(j_1j_2)JM_J\rangle
$$
$$
\hat{J}_z|(j_1j_2)JM_J\rangle=\hbar M_J|(j_1j_2)JM_J\rangle,
$$
The nuclear forces are almost charge independent. If we assume they are, we can introduce a new quantum number which is conserved. For nucleons only, that is a proton and neutron, we can limit ourselves to two possible values which allow us to distinguish between the two particles. If we assign an isospin value of \( \tau=1/2 \) for protons and neutrons (they belong to an isospin doublet, in the same way as we discussed the spin \( 1/2 \) multiplet), we can define the neutron to have isospin projection \( \tau_z=+1/2 \) and a proton to have \( \tau_z=-1/2 \). These assignements are the standard choices in low-energy nuclear physics.
This leads to the introduction of an additional quantum number called isospin. We can define a single-nucleon state function in terms of the quantum numbers \( n \), \( j \), \( m_j \), \( l \), \( s \), \( \tau \) and \( \tau_z \). Using our definitions in terms of an uncoupled basis, we had
$$
\psi_{njm_j;ls}=\sum_{m_lm_s}\langle lm_lsm_s|jm_j\rangle\phi_{nlm_lsm_s},
$$
which we can now extend to
$$
\psi_{njm_j;ls}\xi_{\tau\tau_z}=\sum_{m_lm_s}\langle lm_lsm_s|jm_j\rangle\phi_{nlm_lsm_s}\xi_{\tau\tau_z},
$$
with the isospin spinors defined as
$$
\xi_{\tau=1/2\tau_z=+1/2}=\left(\begin{array}{c} 1 \\ 0\end{array}\right),
$$
and
$$
\xi_{\tau=1/2\tau_z=-1/2}=\left(\begin{array}{c} 0 \\ 1\end{array}\right).
$$
We can then define the proton state function as
$$
\psi^p(\mathbf{r}) =\psi_{njm_j;ls}(\mathbf{r})\left(\begin{array}{c} 0 \\ 1\end{array}\right),
$$
and similarly for neutrons as
$$
\psi^n(\mathbf{r}) =\psi_{njm_j;ls}(\mathbf{r})\left(\begin{array}{c} 1 \\ 0\end{array}\right).
$$
We can in turn define the isospin Pauli matrices (in the same as we define the spin matrices) as
$$
\hat{\tau}_x =\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right),
$$
$$
\hat{\tau}_y =\left(\begin{array}{cc} 0 & -\imath \\ \imath & 0 \end{array}\right),
$$
and
$$
\hat{\tau}_z =\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right),
$$
and operating with \( \hat{\tau}_z \) on the proton state function we have
$$
\hat{\tau}_z\psi^p(\mathbf{r})=-\frac{1}{2}\psi^p(\mathbf{r}),
$$
and for neutrons we have
$$
\hat{\tau}\psi^n(\mathbf{r})=\frac{1}{2}\psi^n(\mathbf{r}).
$$
We can now define the so-called charge operator as
$$
\frac{\hat{Q}}{e} = \frac{1}{2}\left(1-\hat{\tau}_z\right)=\begin{Bmatrix} 0 & 0 \\ 0 & 1 \end{Bmatrix},
$$
which results in
$$
\frac{\hat{Q}}{e}\psi^p(\mathbf{r})=\psi^p(\mathbf{r}),
$$
and
$$
\frac{\hat{Q}}{e}\psi^n(\mathbf{r})=0,
$$
as it should be.
The total isospin is defined as
$$
\hat{T}=\sum_{i=1}^A\hat{\tau}_i,
$$
and its corresponding isospin projection as
$$
\hat{T}_z=\sum_{i=1}^A\hat{\tau}_{z_i},
$$
with eigenvalues \( T(T+1) \) for \( \hat{T} \) and \( 1/2(N-Z) \) for \( \hat{T}_z \), where \( N \) is the number of neutrons and \( Z \) the number of protons.
If charge is conserved, the Hamiltonian \( \hat{H} \) commutes with \( \hat{T}_z \) and all members of a given isospin multiplet (that is the same value of \( T \)) have the same energy and there is no \( T_z \) dependence and we say that \( \hat{H} \) is a scalar in isospin space.
We have till now seen the following definitions of a two-body matrix elements with quantum numbers \( p=j_pm_p \) etc we have a two-body state defined as
$$
|(pq)M\rangle = a^{\dagger}_pa^{\dagger}_q|\Phi_0\rangle,
$$
where \( |\Phi_0\rangle \) is a chosen reference state, say for example the Slater determinant which approximates
\( {}^{16}\mbox{O} \) with the \( 0s \) and the \( 0p \) shells being filled, and \( M=m_p+m_q \). Recall that we label single-particle states above the Fermi level as \( abcd\dots \) and states below the Fermi level for \( ijkl\dots \).
In case of two-particles in the single-particle states \( a \) and \( b \) outside \( {}^{16}\mbox{O} \) as a closed shell core, say \( {}^{18}\mbox{O} \),
we would write the representation of the Slater determinant as
$$
|^{18}\mathrm{O}\rangle =|(ab)M\rangle = a^{\dagger}_aa^{\dagger}_b|^{16}\mathrm{O}\rangle=|\Phi^{ab}\rangle.
$$
In case of two-particles removed from say \( {}^{16}\mbox{O} \), for example two neutrons in the single-particle states \( i \) and \( j \), we would write this as
$$
|^{14}\mathrm{O}\rangle =|(ij)M\rangle = a_ja_i|^{16}\mathrm{O}\rangle=|\Phi_{ij}\rangle.
$$
For a one-hole-one-particle state we have
$$
|^{16}\mathrm{O}\rangle_{1p1h} =|(ai)M\rangle = a_a^{\dagger}a_i|^{16}\mathrm{O}\rangle=|\Phi_{i}^a\rangle,
$$
and finally for a two-particle-two-hole state we
$$
|^{16}\mathrm{O}\rangle_{2p2h} =|(abij)M\rangle = a_a^{\dagger}a_b^{\dagger}a_ja_i|^{16}\mathrm{O}\rangle=|\Phi_{ij}^{ab}\rangle.
$$
Let us go back to the case of two-particles in the single-particle states \( a \) and \( b \) outside \( {}^{16}\mbox{O} \) as a closed shell core, say \( {}^{18}\mbox{O} \). The representation of the Slater determinant is
$$
|^{18}\mathrm{O}\rangle =|(ab)M\rangle = a^{\dagger}_aa^{\dagger}_b|^{16}\mathrm{O}\rangle=|\Phi^{ab}\rangle.
$$
The anti-symmetrized matrix element is detailed as
$$
\langle (ab) M | \hat{V} | (cd) M \rangle = \langle (j_am_aj_bm_b)M=m_a+m_b | \hat{V} | (j_cm_cj_dm_d)M=m_a+m_b \rangle,
$$
and note that anti-symmetrization means
$$
\langle (ab) M | \hat{V} | (cd) M \rangle =-\langle (ba) M | \hat{V} | (cd) M \rangle =\langle (ba) M | \hat{V} | (dc) M \rangle,
$$
$$
\langle (ab) M | \hat{V} | (cd) M \rangle =-\langle (ab) M | \hat{V} | (dc) M \rangle.
$$
This matrix element is given by
$$
\langle ^{16}\mathrm{O}|a_ba_a\frac{1}{4}\sum_{pqrs}\langle (pq) M | \hat{V} | (rs) M' \rangle a^{\dagger}_pa^{\dagger}_qa_sa_r a^{\dagger}_ca^{\dagger}_d|^{16}\mathrm{O}\rangle.
$$
We can compute this matrix element using Wick's theorem.
We have also defined matrix elements in the coupled basis, the so-called \( J \)-coupled scheme. In this case the two-body wave function for two neutrons outside \( {}^{16}\mbox{O} \) is written as
$$
|^{18}\mathrm{O}\rangle_J =|(ab)JM\rangle = \left\{a^{\dagger}_aa^{\dagger}_b\right\}^J_M|^{16}\mathrm{O}\rangle=N_{ab}\sum_{m_am_b}\langle j_am_aj_bm_b|JM\rangle|\Phi^{ab}\rangle,
$$
with
$$
|\Phi^{ab}\rangle=a^{\dagger}_aa^{\dagger}_b|^{16}\mathrm{O}\rangle.
$$
We have now an explicit coupling order, where the angular momentum \( j_a \) is coupled to the angular momentum \( j_b \) to yield a final two-body angular momentum \( J \).
The normalization factor is
$$
N_{ab}=\frac{\sqrt{1+\delta_{ab}\times (-1)^J}}{1+\delta_{ab}}.
$$
We note that, using the anti-commuting properties of the creation operators, we obtain
$$
N_{ab}\sum_{m_am_b}\langle j_am_aj_bm_b|JM\rangle \vert\Phi^{ab}\rangle=-N_{ab}\sum_{m_am_b}\langle j_am_aj_bm_b|JM\rangle\vert\Phi^{ba}\rangle.
$$
Furthermore, using the property of the Clebsch-Gordan coefficient
$$
\langle j_am_aj_bm_b|JM>=(-1)^{j_a+j_b-J}\langle j_bm_bj_am_a|JM\rangle,
$$
which can be used to show that
$$
|(j_bj_a)JM\rangle = \left\{a^{\dagger}_ba^{\dagger}_a\right\}^J_M|^{16}\mathrm{O}\rangle=N_{ab}\sum_{m_am_b}\langle j_bm_bj_am_a|JM\rangle|\Phi^{ba}\rangle,
$$
is equal to
$$
|(j_bj_a)JM\rangle=(-1)^{j_a+j_b-J+1}|(j_aj_b)JM\rangle.
$$
The implementation of the Pauli principle looks different in the \( J \)-scheme compared with the \( m \)-scheme. In the latter, no two fermions or more can have the same set of quantum numbers. In the \( J \)-scheme, when we write a state with the shorthand
$$
|^{18}\mathrm{O}\rangle_J =|(ab)JM\rangle,
$$
we do refer to the angular momenta only. This means that another way of writing the last state is
$$
|^{18}\mathrm{O}\rangle_J =|(j_aj_b)JM\rangle.
$$
We will use this notation throughout when we refer to a two-body state in \( J \)-scheme. The Kronecker \( \delta \) function in the normalization factor
refers thus to the values of \( j_a \) and \( j_b \). If two identical particles are in a state with the same \( j \)-value, then only even values of the total angular momentum apply. In the notation below, when we label a state as \( j_p \) it will actually represent all quantum numbers except \( m_p \).
The two-body matrix element is a scalar and since it obeys rotational symmetry, it is diagonal in \( J \), meaning that the corresponding matrix element in \( J \)-scheme is
$$
\langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle = N_{ab}N_{cd}\sum_{m_am_bm_cm_d}\langle j_am_aj_bm_b|JM\rangle
$$
$$
\times \langle j_cm_cj_dm_d|JM\rangle\langle (j_am_aj_bm_b)M | \hat{V} | (j_cm_cj_dm_d)M \rangle,
$$
and note that of the four \( m \)-values in the above sum, only three are independent due to the constraint \( m_a+m_b=M=m_c+m_d \).
Since
$$
|(j_bj_a)JM\rangle=(-1)^{j_a+j_b-J+1}|(j_aj_b)JM\rangle,
$$
the anti-symmetrized matrix elements need now to obey the following relations
$$
\langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle = (-1)^{j_a+j_b-J+1}\langle (j_bj_a) JM | \hat{V} | (j_cj_d) JM \rangle,
$$
$$
\langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle = (-1)^{j_c+j_d-J+1}\langle (j_aj_b) JM | \hat{V} | (j_dj_c) JM \rangle,
$$
$$
\langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle = (-1)^{j_a+j_b+j_c+j_d}\langle (j_bj_a) JM | \hat{V} | (j_dj_c) JM \rangle=\langle (j_bj_a) JM | \hat{V} | (j_dj_c) JM \rangle,
$$
where the last relations follows from the fact that \( J \) is an integer and \( 2J \) is always an even number.
Using the orthogonality properties of the Clebsch-Gordan coefficients,
$$
\sum_{m_am_b}\langle j_am_aj_bm_b|JM\rangle\langle j_am_aj_bm_b|J'M'\rangle=\delta_{JJ'}\delta_{MM'},
$$
and
$$
\sum_{JM}\langle j_am_aj_bm_b|JM\rangle\langle j_am_a'j_bm_b'|JM\rangle=\delta_{m_am_a'}\delta_{m_bm_b'},
$$
we can also express the two-body matrix element in \( m \)-scheme in terms of that in \( J \)-scheme, that is, if we multiply with
$$
\sum_{JMJ'M'}\langle j_am_a'j_bm_b'|JM\rangle\langle j_cm_c'j_dm_d'|J'M'\rangle
$$
from left in
$$
\langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle = N_{ab}N_{cd}\sum_{m_am_bm_cm_d}\langle j_am_aj_bm_b|JM\rangle\langle j_cm_cj_dm_d|JM\rangle
$$
$$
\times \langle (j_am_aj_bm_b)M| \hat{V} | (j_cm_cj_dm_d)M\rangle,
$$
we obtain
$$
\langle (j_am_aj_bm_b)M | \hat{V} | (j_cm_cj_dm_d)M\rangle=\frac{1}{N_{ab}N_{cd}}\sum_{JM}\langle j_am_aj_bm_b|JM\rangle\langle j_cm_cj_dm_d|JM\rangle
$$
$$
\times \langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle.
$$
We can now use the above relations to compute the Hartre-Fock energy in \( j \)-scheme. In \( m \)-scheme we defined the Hartree-Fock energy as
$$
\varepsilon_{pq}^{\mathrm{HF}}=\delta_{pq}\varepsilon_p+ \sum_{i\le F} \langle pi \vert \hat{V} \vert qi\rangle_{AS},
$$
where the single-particle states \( pqi \) point to the quantum numbers in \( m \)-scheme. For a state with for example \( j=5/2 \), this results in six identical values for the above potential.
Our Hartree-Fock basis is orthogonal by definition, meaning that we have
$$
\varepsilon_{p}^{\mathrm{HF}}=\varepsilon_p+ \sum_{i\le F} \langle pi \vert \hat{V} \vert pi\rangle_{AS},
$$
We have
$$
\varepsilon_{p}^{\mathrm{HF}}=\varepsilon_p+ \sum_{i\le F} \langle pi \vert \hat{V} \vert pi\rangle_{AS},
$$
where the single-particle states \( p=[n_p,j_p,m_p,t_{z_{p}}] \). Let us assume that \( p \) is a state above the Fermi level. The quantity \( \varepsilon_p \) could represent the harmonic oscillator single-particle energies.
Let \( p\rightarrow a \).
The energies, as we have seen, are independent of \( m_a \) and \( m_i \). We sum now over all \( m_a \) on both sides of the above equation and divide by \( 2j_a+1 \), recalling that \( \sum_{m_a}=2j_a+1 \). This results in
$$
\varepsilon_{a}^{\mathrm{HF}}=\varepsilon_a+ \frac{1}{2j_a+1}\sum_{i\le F}\sum_{m_a} \langle ai \vert \hat{V}\vert ai\rangle_{AS},
$$
$$
\varepsilon_{a}^{\mathrm{HF}}=\varepsilon_a+ \frac{1}{2j_a+1}\sum_{i\le F}\sum_{m_a} \langle ai \vert \hat{V} \vert ai\rangle_{AS},
$$
as
$$
\varepsilon_{a}^{\mathrm{HF}}=\varepsilon_a+ \frac{1}{2j_a+1}\sum_{n_i,j_i,t_{z_{i}}\le F}\sum_{m_im_a}
\langle (j_am_aj_im_i)M | \hat{V} | (j_am_aj_im_i)M\rangle_{AS},
$$
where we have suppressed the dependence on \( n_p \) and \( t_z \) in the matrix element.
Using the definition
$$
\langle (j_am_aj_bm_b)M \vert \hat{V} \vert (j_cm_cj_dm_d)M\rangle=\frac{1}{N_{ab}N_{cd}}\sum_{JM}\langle j_am_aj_bm_b|JM\rangle\langle j_cm_cj_dm_d|JM\rangle \langle (j_aj_b)J \vert \hat{V} \vert (j_cj_d)M\rangle_{AS},
$$
with the orthogonality properties of Glebsch-Gordan coefficients and that the \( j \)-coupled two-body matrix element is a scalar and independent of \( M \) we arrive at
$$
\varepsilon_{a}^{\mathrm{HF}}=\varepsilon_a+ \frac{1}{2j_a+1}\sum_{j_i\le F}\sum_J (2J+1)
\langle (j_aj_i)J \vert \hat{V} \vert (j_aj_i)M\rangle_{AS},
$$
In a similar way it is easy to show that the potential energy contribution to the ground state energy in \( m \)-scheme
$$
\frac{1}{2}\sum_{ij\le F}\langle (j_im_ij_jm_j)M | \hat{V} | (j_im_ij_jm_j)M\rangle_{AS},
$$
can be rewritten as
$$
\frac{1}{2}\sum_{j_i,j_j\le F}\sum_J(2J+1)
\langle (j_ij_j)J | \hat{V} | (j_ij_j)J\rangle_{AS},
$$
This reduces the number of floating point operations with an order of magnitude on average.
We are now going to define two-body and many-body states in an angular momentum coupled basis, the so-called \( j \)-scheme basis. In this connection
We will also study some specific examples, like the calculation of the tensor force.
We define an irreducible spherical tensor \( T^{\lambda}_{\mu} \) of rank \( \lambda \) as an operator with \( 2\lambda+1 \) components \( \mu \) that satisfies the commutation relations (\( \hbar=1 \))
$$
[J_{\pm}, T^{\lambda}_{\mu}]= \sqrt{(\lambda\mp \mu)(\lambda\pm \mu+1)}T^{\lambda}_{\mu\pm 1},
$$
and
$$
[J_{z}, T^{\lambda}_{\mu}]=\mu T^{\lambda}_{\mu}.
$$
Our angular momentum coupled two-body wave function obeys clearly this definition, namely
$$
|(ab)JM\rangle = \left\{a^{\dagger}_aa^{\dagger}_b\right\}^J_M|\Phi_0\rangle=N_{ab}\sum_{m_am_b}\langle j_am_aj_bm_b|JM\rangle|\Phi^{ab}\rangle,
$$
is a tensor of rank \( J \) with \( M \) components. Another well-known example is given by the spherical harmonics (see examples during today's lecture).
The product of two irreducible tensor operators
$$
T^{\lambda_3}_{\mu_3}=\sum_{\mu_1\mu_2}\langle \lambda_1\mu_1\lambda_2\mu_2|\lambda_3\mu_3\rangle T^{\lambda_1}_{\mu_1}T^{\lambda_2}_{\mu_2}
$$
is also a tensor operator of rank \( \lambda_3 \).
We wish to apply the above definitions to the computations of a matrix element
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle,
$$
where we have skipped a reference to specific single-particle states. This is the expectation value for two specific states, labelled by angular momenta \( J' \) and \( J \). These states form an orthonormal basis.
Using the properties of the Clebsch-Gordan coefficients we can write
$$
T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle=\sum_{J''M''}\langle \lambda \mu J'M'|J''M''\rangle|\Psi^{J''}_{M''}\rangle,
$$
and assuming that states with different \( J \) and \( M \) are orthonormal we arrive at
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle= \langle \lambda \mu J'M'|JM\rangle \langle \Phi^J_M|\Psi^{J}_{M}\rangle.
$$
We need to show that
$$
\langle \Phi^J_M|\Psi^{J}_{M}\rangle,
$$
is independent of \( M \).
To show that
$$
\langle \Phi^J_M|\Psi^{J}_{M}\rangle,
$$
is independent of \( M \), we use the ladder operators for angular momentum.
We have that
$$
\langle \Phi^J_{M+1}|\Psi^{J}_{M+1}\rangle=\left((J-M)(J+M+1)\right)^{-1/2}\langle \hat{J}_{+}\Phi^J_{M}|\Psi^{J}_{M+1}\rangle,
$$
but this is also equal to
$$
\langle \Phi^J_{M+1}|\Psi^{J}_{M+1}\rangle=\left((J-M)(J+M+1)\right)^{-1/2}\langle \Phi^J_{M}|\hat{J}_{-}\Psi^{J}_{M+1}\rangle,
$$
meaning that
$$
\langle \Phi^J_{M+1}|\Psi^{J}_{M+1}\rangle=\langle \Phi^J_M|\Psi^{J}_{M}\rangle\equiv\langle \Phi^J_{M}||T^{\lambda}||\Phi^{J'}_{M'}\rangle.
$$
The double bars indicate that this expectation value is independent of the projection \( M \).
The Wigner-Eckart theorem for an expectation value can then be written as
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle\equiv\langle \lambda \mu J'M'|JM\rangle\langle \Phi^J||T^{\lambda}||\Phi^{J'}\rangle.
$$
The double bars indicate that this expectation value is independent of the projection \( M \).
We can manipulate the Clebsch-Gordan coefficients using the relations
$$
\langle \lambda \mu J'M'|JM\rangle= (-1)^{\lambda+J'-J}\langle J'M'\lambda \mu |JM\rangle
$$
and
$$
\langle J'M'\lambda \mu |JM\rangle =(-1)^{J'-M'}\frac{\sqrt{2J+1}}{\sqrt{2\lambda+1}}\langle J'M'J-M |\lambda-\mu\rangle,
$$
together with the so-called \( 3j \) symbols.
It is then normal to encounter the Wigner-Eckart theorem in the form
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle\equiv(-1)^{J-M}\left(\begin{array}{ccc} J & \lambda & J' \\ -M & \mu & M'\end{array}\right)\langle \Phi^J||T^{\lambda}||\Phi^{J'}\rangle,
$$
with the condition \( \mu+M'-M=0 \).
The \( 3j \) symbols obey the symmetry relation
$$
\left(\begin{array}{ccc} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3\end{array}\right)=(-1)^{p}\left(\begin{array}{ccc} j_a & j_b & j_c \\ m_a & m_b & m_c\end{array}\right),
$$
with \( (-1)^p=1 \) when the columns \( a,b, c \) are even permutations of the columns \( 1,2,3 \), \( p=j_1+j_2+j_3 \) when the columns \( a,b,c \) are odd permtations of the
columns \( 1,2,3 \) and \( p=j_1+j_2+j_3 \) when all the magnetic quantum numbers \( m_i \) change sign. Their orthogonality is given by
$$
\sum_{j_3m_3}(2j_3+1)\left(\begin{array}{ccc} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3\end{array}\right)\left(\begin{array}{ccc} j_1 & j_2 & j_3 \\ m_{1'} & m_{2'} & m_3\end{array}\right)=\delta_{m_1m_{1'}}\delta_{m_2m_{2'}},
$$
and
$$
\sum_{m_1m_2}\left(\begin{array}{ccc} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3\end{array}\right)\left(\begin{array}{ccc} j_1 & j_2 & j_{3'} \\ m_{1} & m_{2} & m_{3'}\end{array}\right)=\frac{1}{(2j_3+1)}\delta_{j_3j_{3'}}\delta_{m_3m_{3'}}.
$$
For later use, the following special cases for the Clebsch-Gordan and \( 3j \) symbols are rather useful \[ \langle JM J'M' |00\rangle =\frac{(-1)^{J-M}}{\sqrt{2J+1}}\delta_{JJ'}\delta_{MM'}. \] and \[ \left(\begin{array}{ccc} J & 1 & J \\ -M & 0 & M'\end{array}\right)=(-1)^{J-M}\frac{M}{\sqrt{(2J+1)(J+1)}}\delta_{MM'}. \]
Using \( 3j \) symbols we rewrote the Wigner-Eckart theorem as
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle\equiv(-1)^{J-M}\left(\begin{array}{ccc} J & \lambda & J' \\ -M & \mu & M'\end{array}\right)\langle \Phi^J||T^{\lambda}||\Phi^{J'}\rangle.
$$
Multiplying from the left with the same \( 3j \) symbol and summing over \( M,\mu,M' \) we obtain the equivalent relation
$$
\langle \Phi^J||T^{\lambda}||\Phi^{J'}\rangle\equiv\sum_{M,\mu,M'}(-1)^{J-M}\left(\begin{array}{ccc} J & \lambda & J' \\ -M & \mu & M'\end{array}\right)\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle,
$$
where we used the orthogonality properties of the \( 3j \) symbols from the previous page.
This relation can in turn be used to compute the expectation value of some simple reduced matrix elements like
$$
\langle \Phi^J||{\bf 1}||\Phi^{J'}\rangle=\sum_{M,M'}(-1)^{J-M}\left(\begin{array}{ccc} J & 0 & J' \\ -M & 0 & M'\end{array}\right)\langle \Phi^J_M|1|\Phi^{J'}_{M'}\rangle=\sqrt{2J+1}\delta_{JJ'}\delta_{MM'},
$$
where we used
$$
\langle JM J'M' |00\rangle =\frac{(-1)^{J-M}}{\sqrt{2J+1}}\delta_{JJ'}\delta_{MM'}.
$$
Similarly, using
$$
\left(\begin{array}{ccc} J & 1 & J \\ -M & 0 & M'\end{array}\right)=(-1)^{J-M}\frac{M}{\sqrt{(2J+1)(J+1)}}\delta_{MM'},
$$
we have that
$$
\langle \Phi^J||{\bf J}||\Phi^{J}\rangle=\sum_{M,M'}(-1)^{J-M}\left(\begin{array}{ccc} J & 1 & J' \\ -M & 0 & M'\end{array}\right)\langle \Phi^J_M|j_Z|\Phi^{J'}_{M'}\rangle=\sqrt{J(J+1)(2J+1)}
$$
With the Pauli spin matrices \( \sigma \) and a state with \( J=1/2 \), the reduced matrix element
$$
\langle \frac{1}{2}||{\bf \sigma}||\frac{1}{2}\rangle=\sqrt{6}.
$$
Before we proceed with further examples, we need some other properties of the Wigner-Eckart theorem plus some additional angular momenta relations.
The Wigner-Eckart theorem states that the expectation value for an irreducible spherical tensor can be written as
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle\equiv\langle \lambda \mu J'M'|JM\rangle\langle \Phi^J||T^{\lambda}||\Phi^{J'}\rangle.
$$
Since the Clebsch-Gordan coefficients themselves are easy to evaluate, the interesting quantity is the reduced matrix element. Note also that
the Clebsch-Gordan coefficients limit via the triangular relation among \( \lambda \), \( J \) and \( J' \) the possible non-zero values.
From the theorem we see also that
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle=\frac{\langle \lambda \mu J'M'|JM\rangle\langle }{\langle \lambda \mu_0 J'M'_0|JM_0\rangle\langle }\langle \Phi^J_{M_0}|T^{\lambda}_{\mu_0}|\Phi^{J'}_{M'_0}\rangle,
$$
meaning that if we know the matrix elements for say some \( \mu=\mu_0 \), \( M'=M'_0 \) and \( M=M_0 \) we can calculate all other.
If we look at the hermitian adjoint of the operator \( T^{\lambda}_{\mu} \), we see via the commutation relations that \( (T^{\lambda}_{\mu})^{\dagger} \) is not an irreducible tensor, that is
$$
[J_{\pm}, (T^{\lambda}_{\mu})^{\dagger}]= -\sqrt{(\lambda\pm \mu)(\lambda\mp \mu+1)}(T^{\lambda}_{\mu\mp 1})^{\dagger},
$$
and
$$
[J_{z}, (T^{\lambda}_{\mu})^{\dagger}]=-\mu (T^{\lambda}_{\mu})^{\dagger}.
$$
The hermitian adjoint \( (T^{\lambda}_{\mu})^{\dagger} \) is not an irreducible tensor. As an example, consider the spherical harmonics for
\( l=1 \) and \( m_l=\pm 1 \). These functions are
$$
Y^{l=1}_{m_l=1}(\theta,\phi)=-\sqrt{\frac{3}{8\pi}}\sin{(\theta)}\exp{\imath\phi},
$$
and
$$
Y^{l=1}_{m_l=-1}(\theta,\phi)=\sqrt{\frac{3}{8\pi}}\sin{(\theta)}\exp{-\imath\phi},
$$
It is easy to see that the Hermitian adjoint of these two functions
$$
\left[Y^{l=1}_{m_l=1}(\theta,\phi)\right]^{\dagger}=-\sqrt{\frac{3}{8\pi}}\sin{(\theta)}\exp{-\imath\phi},
$$
and
$$
\left[Y^{l=1}_{m_l=-1}(\theta,\phi)\right]^{\dagger}=\sqrt{\frac{3}{8\pi}}\sin{(\theta)}\exp{\imath\phi},
$$
do not behave as a spherical tensor. However, the modified quantity
$$
\tilde{T}^{\lambda}_{\mu}=(-1)^{\lambda+\mu}(T^{\lambda}_{-\mu})^{\dagger},
$$
does satisfy the above commutation relations.
With the modified quantity
$$
\tilde{T}^{\lambda}_{\mu}=(-1)^{\lambda+\mu}(T^{\lambda}_{-\mu})^{\dagger},
$$
we can then define the expectation value
$$
\langle \Phi^J_M|T^{\lambda}_{\mu}|\Phi^{J'}_{M'}\rangle^{\dagger} = \langle \lambda \mu J'M'|JM\rangle\langle \Phi^J||T^{\lambda}||\Phi^{J'}\rangle^*,
$$
since the Clebsch-Gordan coefficients are real. The rhs is equivalent with
$$
\langle \lambda \mu J'M'|JM\rangle\langle \Phi^J||T^{\lambda}||\Phi^{J'}\rangle^*=\langle \Phi^{J'}_{M'}|(T^{\lambda}_{\mu})^{\dagger}|\Phi^{J}_{M}\rangle,
$$
which is equal to
$$
\langle \Phi^{J'}_{M'}|(T^{\lambda}_{\mu})^{\dagger}|\Phi^{J}_{M}\rangle=(-1)^{-\lambda+\mu}\langle \lambda -\mu JM|J'M'\rangle\langle \Phi^{J'}||\tilde{T}^{\lambda}||\Phi^{J}\rangle.
$$
Let us now apply the theorem to some selected expectation values. In several of the expectation values we will meet when evaluating explicit matrix elements, we will have to deal with expectation values involving spherical harmonics. A general central interaction can be expanded in a complete set of functions like the Legendre polynomials, that is, we have an interaction, with \( r_{ij}=|{\bf r}_i-{\bf r}_j| \),
$$
v(r_{ij})=\sum_{\nu=0}^{\infty}v_{\nu}(r_{ij})P_{\nu}(\cos{(\theta_{ij})},
$$
with \( P_{\nu} \) being a Legendre polynomials
$$
P_{\nu}(\cos{(\theta_{ij})}=\sum_{\mu}\frac{4\pi}{2\mu+1}Y_{\mu}^{\nu *}(\Omega_{i})Y_{\mu}^{\nu}(\Omega_{j}).
$$
We will come back later to how we split the above into a contribution that involves only one of the coordinates.
This means that we will need matrix elements of the type
$$
\langle Y^{l'}||Y^{\lambda}|| Y^{l}\rangle.
$$
We can rewrite the Wigner-Eckart theorem as
$$
\langle Y^{l'}||Y^{\lambda}|| Y^{l}\rangle=\sum_{m\mu}\langle \lambda\mu lm|l'm'\rangle Y^{\lambda}_{\mu}Y^l_m,
$$
This equation is true for all values of \( \theta \) and \( \phi \). It must also hold for \( \theta=0 \).
We have
$$
\langle Y^{l'}||Y^{\lambda}|| Y^{l}\rangle=\sum_{m\mu}\langle \lambda\mu lm|l'm'\rangle Y^{\lambda}_{\mu}Y^l_m,
$$
and for \( \theta=0 \), the spherical harmonic
$$
Y_m^l(\theta=0,\phi)=\sqrt{\frac{2l+1}{4\pi}}\delta_{m0},
$$
which results in
$$
\langle Y^{l'}||Y^{\lambda}|| Y^{l}\rangle=\left\{\frac{(2l+1)(2\lambda+1)}{4\pi(2l'+1)}\right\}^{1/2}\langle \lambda0 l0|l'0\rangle.
$$
Till now we have mainly been concerned with the coupling of two angular momenta \( j_{a} \) and \( j_{b} \) to a final angular momentum \( J \). If we wish to describe a three-body state with a final angular momentum \( J \), we need to couple three angular momenta, say the two momenta \( j_a,j_b \) to a third one \( j_c \). The coupling order is important and leads to a less trivial implementation of the Pauli principle. With three angular momenta there are obviously \( 3! \) ways by which we can combine the angular momenta. In \( m \)-scheme a three-body Slater determinant is represented as (say for the case of \( {}^{19}\mbox{O} \), three neutrons outside the core of \( {}^{16}\mbox{O} \)),
$$
|^{19}\mathrm{O}\rangle =|(abc)M\rangle = a^{\dagger}_aa^{\dagger}_ba^{\dagger}_c|^{16}\mathrm{O}\rangle=|\Phi^{abc}\rangle.
$$
The Pauli principle is automagically implemented via the anti-commutation relations.
$$
\vert([j_a\rightarrow j_b]J_{ab}\rightarrow j_c) J\rangle= \sum_{m_am_bm_c}\langle j_am_aj_bm_b|J_{ab}M_{ab}\rangle \langle J_{ab}M_{ab}j_cm_c|JM\rangle|j_am_a\rangle\otimes |j_bm_b\rangle \otimes |j_cm_c\rangle \ ,
\tag{1}
$$
that is, we couple first \( j_a \) to \( j_b \) to yield an intermediate angular momentum \( J_{ab} \), then to \( j_c \) yielding the final angular momentum \( J \).
That is, we can have
$$
\vert(j_a\rightarrow [j_b\rightarrow j_c]J_{bc}) J\rangle = \sum_{m_a'm_b'm_c'}\langle j_bm_b'j_cm_c'|J_{bc}M_{bc}\rangle \langle j_am_a'J_{bc}M_{bc}|J'M'\rangle|\Phi^{abc}\rangle .
$$
We will always assume that we work with orthornormal states, this means that when we compute the overlap betweem these two possible ways of coupling angular momenta, we get
$$
\begin{align}
\langle (j_a\rightarrow [j_b\rightarrow j_c]J_{bc}) J'M'| ([j_a\rightarrow j_b]J_{ab}\rightarrow j_c) JM\rangle = &
\delta_{JJ'}\delta_{MM'}\sum_{m_am_bm_c}\langle j_am_aj_bm_b|J_{ab}M_{ab}\rangle \langle J_{ab}M_{ab}j_cm_c|JM\rangle
\tag{2}\\
& \times \langle j_bm_bj_cm_c|J_{bc}M_{bc}\rangle \langle j_am_aJ_{bc}M_{bc}|JM\rangle .
\tag{3}
\end{align}
$$
$$
\begin{eqnarray}
\langle (j_a\rightarrow [j_b\rightarrow j_c]J_{bc}) J'M'| ([j_a\rightarrow j_b]J_{ab}\rightarrow j_c) JM\rangle & = & \delta_{JJ'}\delta_{MM'}\sum_{m_am_bm_c}\langle j_am_aj_bm_b|J_{ab}M_{ab}\rangle \langle J_{ab}M_{ab}j_cm_c|JM\rangle \\ \nonumber
& & \times \langle j_bm_bj_cm_c|J_{bc}M_{bc}\rangle \langle j_am_aJ_{bc}M_{bc}|JM\rangle \\ \nonumber
& =&(-1)^{j_a+j_b+j_c+J}\sqrt{(2J_{ab}+1)(2J_{bc}+1)}\left\{\begin{array}{ccc} j_a & j_b& J_{ab} \\ j_c & J & J_{bc} \end{array}\right\} ,
\end{eqnarray}
$$
where the symbol in curly brackets is the \( 6j \) symbol. A specific coupling order has to be respected in the symbol, that is, the so-called triangular relations between three angular momenta needs to be respected, that is
$$
\left\{\begin{array}{ccc} x & x& x \\ & & \end{array}\right\}\hspace{0.1cm}\left\{\begin{array}{ccc} & & x \\ x& x & \end{array}\right\}\hspace{0.1cm}\left\{\begin{array}{ccc} & x& \\ x & &x \end{array}\right\}\hspace{0.1cm}\left\{\begin{array}{ccc} x & & \\ & x &x \end{array}\right\}\hspace{0.1cm}
$$
The \( 6j \) symbol is invariant under the permutation of any two columns
$$
\begin{Bmatrix} j_1 & j_2 & j_3\\ j_4 & j_5 & j_6 \end{Bmatrix} = \begin{Bmatrix} j_2 & j_1 & j_3\\ j_5 & j_4 & j_6 \end{Bmatrix} = \begin{Bmatrix} j_1 & j_3 & j_2\\ j_4 & j_6 & j_5 \end{Bmatrix} = \begin{Bmatrix} j_3 & j_2 & j_1\\ j_6 & j_5 & j_4 \end{Bmatrix}.
$$
The \( 6j \) symbol is also invariant if upper and lower arguments are interchanged in any two columns
$$
\begin{Bmatrix} j_1 & j_2 & j_3\\ j_4 & j_5 & j_6 \end{Bmatrix} = \begin{Bmatrix} j_4 & j_5 & j_3\\ j_1 & j_2 & j_6 \end{Bmatrix} = \begin{Bmatrix} j_1 & j_5 & j_6\\ j_4 & j_2 & j_3 \end{Bmatrix} = \begin{Bmatrix} j_4 & j_2 & j_6\\ j_1 & j_5 & j_3 \end{Bmatrix}.
$$
The above properties of \( 6j \) symbols can again be tested using the symbolic python package wigner. Let us test the invariance
$$
\begin{Bmatrix} j_1 & j_2 & j_3\\ j_4 & j_5 & j_6 \end{Bmatrix} = \begin{Bmatrix} j_2 & j_1 & j_3\\ j_5 & j_4 & j_6 \end{Bmatrix}.
$$
The following program tests this relation for the case of \( j_1=3/2 \), \( j_2=3/2 \), \( j_3=3 \), \( j_4=1/2 \), \( j_5=1/2 \), \( j_6=1 \)
The \( 6j \) symbols satisfy this orthogonality relation
$$
\sum_{j_3} (2j_3+1) \begin{Bmatrix} j_1 & j_2 & j_3\\ j_4 & j_5 & j_6 \end{Bmatrix} \begin{Bmatrix} j_1 & j_2 & j_3\\ j_4 & j_5 & j_6' \end{Bmatrix} = \frac{\delta_{j_6^{}j_6'}}{2j_6+1} \{j_1,j_5,j_6\} \{j_4,j_2,j_6\}.
$$
The symbol \( \{j_1j_2j_3\} \) (called the triangular delta) is equal to one if the triad \( (j_1j_2j_3) \) satisfies the triangular conditions and zero otherwise.
A useful value is given when say one of the angular momenta are zero, say \( J_{bc}=0 \), then we have
$$
\left\{\begin{array}{ccc} j_a & j_b& J_{ab} \\ j_c & J & 0 \end{array}\right\}=\frac{(-1)^{j_a+j_b+J_{ab}}\delta_{Jj_a}\delta_{j_cj_b} }{\sqrt{(2j_{a}+1)(2j_{b}+1)}}
$$
$$
\vert (j_a\rightarrow [j_b\rightarrow j_c]J_{bc}) JM\rangle =\sum_{J_{ab}}(-1)^{j_a+j_b+j_c+J}\sqrt{(2J_{ab}+1)(2J_{bc}+1)}\left\{\begin{array}{ccc} j_a & j_b& J_{ab} \\ j_c & J & J_{bc} \end{array}\right\}| ([j_a\rightarrow j_b]J_{ab}\rightarrow j_c) JM\rangle.
$$
Can you find the inverse relation?
These relations can in turn be used to write out the fully anti-symmetrized three-body wave function in a \( J \)-scheme coupled basis.
If you opt then for a specific coupling order, say \( | ([j_a\rightarrow j_b]J_{ab}\rightarrow j_c) JM\rangle \), you need to express this representation in terms of the other coupling possibilities.
Note that the two-body intermediate state is assumed to be antisymmetric but not normalized, that is, the state which involves the quantum numbers \( j_a \) and \( j_b \). Assume that the intermediate two-body state is antisymmetric. With this coupling order, we can rewrite ( in a schematic way) the general three-particle Slater determinant as
$$
\Phi(a,b,c) = {\cal A} | ([j_a\rightarrow j_b]J_{ab}\rightarrow j_c) J\rangle,
$$
with an implicit sum over \( J_{ab} \). The antisymmetrization operator \( {\cal A} \) is used here to indicate that we need to antisymmetrize the state. Challenge: Use the definition of the \( 6j \) symbol and find an explicit
expression for the above three-body state using the coupling order \( | ([j_a\rightarrow j_b]J_{ab}\rightarrow j_c) J\rangle \).
We can also coupled together four angular momenta. Consider two four-body states, with single-particle angular momenta \( j_a \), \( j_b \), \( j_c \) and \( j_d \) we can have a state with final \( J \)
$$
|\Phi(a,b,c,d)\rangle_1 = | ([j_a\rightarrow j_b]J_{ab}\times [j_c\rightarrow j_d]J_{cd}) JM\rangle,
$$
where we read the coupling order as \( j_a \) couples with \( j_b \) to given and intermediate angular momentum \( J_{ab} \).
Moreover, \( j_c \) couples with \( j_d \) to given and intermediate angular momentum \( J_{cd} \). The two intermediate angular momenta \( J_{ab} \) and \( J_{cd} \)
are in turn coupled to a final \( J \). These operations involved three Clebsch-Gordan coefficients.
Alternatively, we could couple in the following order
$$
|\Phi(a,b,c,d)\rangle_2 = | ([j_a\rightarrow j_c]J_{ac}\times [j_b\rightarrow j_d]J_{bd}) JM\rangle,
$$
The overlap between these two states
$$
\langle([j_a\rightarrow j_c]J_{ac}\times [j_b\rightarrow j_d]J_{bd}) JM| ([j_a\rightarrow j_b]J_{ab}\times [j_c\rightarrow j_d]J_{cd}) JM\rangle,
$$
is equal to
$$
\begin{eqnarray}
\nonumber
& & \sum_{m_iM_{ij}}\langle j_am_aj_bm_b|J_{ab}M_{ab}\rangle \langle j_cm_cj_dm_d|J_{cd}M_{cd}\rangle \langle J_{ab}M_{ab}J_{cd}M_{cd}|JM\rangle \\
& & \times\langle j_am_aj_cm_c|J_{ac}M_{ac}\rangle \langle j_bm_bj_dm_d|J_{cd}M_{bd}\rangle \langle J_{ac}M_{ac}J_{bd}M_{bd}|JM\rangle \\ \nonumber
&= & \sqrt{(2J_{ab}+1)(2J_{cd}+1)(2J_{ac}+1)(2J_{bd}+1)}\left\{\begin{array}{ccc} j_a & j_b& J_{ab} \\ j_c & j_d& J_{cd} \\J_{ac} & J_{bd}& J\end{array}\right\}
, \nonumber
\end{eqnarray}
$$
with the symbol in curly brackets \( \{\} \) being the \( 9j \)-symbol. We see that a \( 6j \) symbol involves four Clebsch-Gordan coefficients, while the \( 9j \) symbol
involves six.
A \( 9j \) symbol is invariant under reflection in either diagonal
$$
\begin{Bmatrix} j_1 & j_2 & j_3\\ j_4 & j_5 & j_6\\ j_7 & j_8 & j_9 \end{Bmatrix} = \begin{Bmatrix} j_1 & j_4 & j_7\\ j_2 & j_5 & j_8\\ j_3 & j_6 & j_9 \end{Bmatrix} = \begin{Bmatrix} j_9 & j_6 & j_3\\ j_8 & j_5 & j_2\\ j_7 & j_4 & j_1 \end{Bmatrix}.
$$
The permutation of any two rows or any two columns yields a phase factor \( (-1)^S \), where
$$
S=\sum_{i=1}^9 j_i.
$$
As an example we have
$$
\begin{Bmatrix} j_1 & j_2 & j_3\\ j_4 & j_5 & j_6\\ j_7 & j_8 & j_9 \end{Bmatrix} = (-1)^S \begin{Bmatrix} j_4 & j_5 & j_6\\ j_1 & j_2 & j_3\\ j_7 & j_8 & j_9 \end{Bmatrix} = (-1)^S \begin{Bmatrix} j_2 & j_1 & j_3\\ j_5 & j_4 & j_6\\ j_8 & j_7 & j_9 \end{Bmatrix}.
$$
A useful case is when say \( J=0 \) in
$$
\left\{\begin{array}{ccc} j_a & j_b& J_{ab} \\ j_c & j_d & J_{cd} \\ J_{ac} & J_{bd}& 0\end{array}\right\}=\frac{\delta_{J_{ab}J_{cd}} \delta_{J_{ac}J_{bd}}}{\sqrt{(2J_{ab}+1)(2J_{ac}+1)}} (-1)^{j_b+J_{ab}+j_c+J_{ac}} \begin{Bmatrix} j_a & j_b & J_{ab}\\ j_d & j_c & J_{ac} \end{Bmatrix}.
$$
The tensor operator in the nucleon-nucleon potential is given by
$$
\begin{array}{ll}
&\\
\langle lSJ\vert S_{12}\vert l'S'J\rangle =&
(-)^{S+J}\sqrt{30(2l+1)(2l'+1)(2S+1)(2S'+1)}\\
&\times\left\{\begin{array}{ccc}J&S'&l'\\2&l&S\end{array}\right\}
\left(\begin{array}{ccc}l'&2&l\\0&0&0\end{array}\right)
\left\{\begin{array}{ccc}s_{1}&s_{2}&S\\s_{3}&s_{4}&S'\\
1&1&2\end{array}
\right\}\\
&\times\langle s_{1}\vert\vert \sigma_{1}\vert\vert s_{3}\rangle
\langle s_{2}\vert\vert \sigma_{2}\vert \vert s_{4}\rangle,
\end{array}
$$
and it is zero for the \( ^1S_0 \) wave.
How do we get here?
To derive the expectation value of the nuclear tensor force, we recall that the product of two irreducible tensor operators is
$$
W^{r}_{m_r}=\sum_{m_pm_q}\langle pm_pqm_q|rm_r\rangle T^{p}_{m_p}U^{q}_{m_q},
$$
and using the orthogonality properties of the Clebsch-Gordan coefficients we can rewrite the above as
$$
T^{p}_{m_p}U^{q}_{m_q}=\sum_{m_pm_q}\langle pm_pqm_q|rm_r\rangle W^{r}_{m_r}.
$$
Assume now that the operators \( T \) and \( U \) act on different parts of say a wave function. The operator \( T \) could act on the spatial part only while the operator \( U \) acts only on the spin part. This means also that these operators commute.
The reduced matrix element of this operator is thus, using the Wigner-Eckart theorem,
$$
\langle (j_aj_b)J||W^{r}||(j_cj_d)J'\rangle\equiv\sum_{M,m_r,M'}(-1)^{J-M}\left(\begin{array}{ccc} J & r & J' \\ -M & m_r & M'\end{array}\right)
$$
$$
\times\langle (j_aj_bJM|\left[ T^{p}_{m_p}U^{q}_{m_q} \right]^{r}_{m_r}|(j_cj_d)J'M'\rangle.
$$
Starting with
$$
\langle (j_aj_b)J||W^{r}||(j_cj_d)J'\rangle\equiv\sum_{M,m_r,M'}(-1)^{J-M}\left(\begin{array}{ccc} J & r & J' \\ -M & m_r & M'\end{array}\right)
$$
$$
\times\langle (j_aj_bJM|\left[ T^{p}_{m_p}U^{q}_{m_q} \right]^{r}_{m_r}|(j_cj_d)J'M'\rangle,
$$
we assume now that \( T \) acts only on \( j_a \) and \( j_c \) and that \( U \) acts only on \( j_b \) and \( j_d \).
The matrix element \( \langle (j_aj_bJM|\left[ T^{p}_{m_p}U^{q}_{m_q} \right]^{r}_{m_r}|(j_cj_d)J'M'\rangle \) can be written out,
when we insert a complete set of states \( |j_im_ij_jm_j\rangle\langle j_im_ij_jm_j| \) between \( T \) and \( U \) as
$$
\langle (j_aj_bJM|\left[ T^{p}_{m_p}U^{q}_{m_q} \right]^{r}_{m_r}|(j_cj_d)J'M'\rangle=\sum_{m_i}\langle pm_pqm_q|rm_r\rangle\langle j_am_aj_bm_b|JM\rangle\langle j_cm_cj_dm_d|J'M'\rangle
$$
$$
\times \langle (j_am_aj_bm_b|\left[ T^{p}_{m_p}\right]^{r}_{m_r}|(j_cm_cj_bm_b)\rangle\langle (j_cm_cj_bm_b|\left[ U^{q}_{m_q}\right]^{r}_{m_r}|(j_cm_cj_dm_d)\rangle.
$$
The complete set of states that was inserted between \( T \) and \( U \) reduces to \( |j_cm_cj_bm_b\rangle\langle j_cm_cj_bm_b| \)
due to orthogonality of the states.
Combining the last two equations from the previous slide and and applying the Wigner-Eckart theorem, we arrive at (rearranging phase factors)
$$
\langle (j_aj_b)J||W^{r}||(j_cj_d)J'\rangle=\sqrt{(2J+1)(2r+1)(2J'+1)}\sum_{m_iM,M'}\left(\begin{array}{ccc} J & r & J' \\ -M & m_r & M'\end{array}\right)
$$
$$
\times\left(\begin{array}{ccc} j_a &j_b & J \\ m_a &m_b &-M \end{array}\right)
\left(\begin{array}{ccc} j_c &j_d &J' \\ -m_c &-m_d &M' \end{array}\right)
\left(\begin{array}{ccc} p & q & r \\ -m_p&-m_q &m_r \end{array}\right)
$$
$$
\times\left(\begin{array}{ccc} j_a &j_c &p \\ m_a &-m_c &-m_p \end{array}\right)\left(\begin{array}{ccc} j_b &j_d &q \\ m_b &-m_d &-m_q \end{array}\right)\langle j_a||T^p||j_c\rangle \times \langle j_b||U^q||j_d\rangle
$$
which can be rewritten in terms of a \( 9j \) symbol as
$$
\langle (j_aj_b)J||W^{r}||(j_cj_d)J'\rangle=\sqrt{(2J+1)(2r+1)(2J'+1)}\langle j_a||T^p||j_c\rangle \langle j_b||U^q||j_d\rangle\left\{\begin{array}{ccc} j_a & j_b& J \\ j_c & j_d & J' \\ p & q& r\end{array}\right\}.
$$
From this expression we can in turn compute for example the spin-spin operator of the tensor force.
In case \( r=0 \), that is we two tensor operators coupled to a scalar, we can use (with \( p=q \))
$$
\left\{\begin{array}{ccc} j_a & j_b& J \\ j_c & j_d & J' \\ p &p & 0\end{array}\right\}=\frac{\delta_{JJ'} \delta_{pq}}{\sqrt{(2J+1)(2J+1)}} (-1)^{j_b+j_c+2J} \begin{Bmatrix} j_a & j_b & J\\ j_d & j_c & p \end{Bmatrix},
$$
and obtain
$$
\langle (j_aj_b)J||W^{0}||(j_cj_d)J'\rangle=(-1)^{j_b+j_c+2J}\langle j_a||T^p||j_c\rangle\langle j_b||U^p||j_d\rangle \begin{Bmatrix} j_a & j_b & J\\ j_d & j_c & p \end{Bmatrix}.
$$
Another very useful expression is the case where the operators act in just one space. We state here without showing that the reduced matrix element
$$
\langle j_a||W^{r}||j_b\rangle=\langle j_a||\left[T^p\times T^q\right]^{r}||j_b\rangle= (-1)^{j_a+j_b+r}\sqrt{2r+1} \sum_{j_c}\begin{Bmatrix} j_b & j_a & r\\ p & q & j_c \end{Bmatrix}
$$
$$
\times \langle j_a||T^p||j_c\rangle \langle j_c||T^q||j_b\rangle.
$$
The tensor operator in the nucleon-nucleon potential can be written as
$$
V=\frac{3}{r^{2}}\left[ \left[ {\bf \sigma}_1 \otimes {\bf \sigma}_2\right]^
{(2)} \otimes\left[{\bf r} \otimes {\bf r} \right]^{(2)}\right]^{(0)}_0
$$
Since the irreducible tensor
\( \left[{\bf r} \otimes {\bf r} \right]^{(2)} \)
operates only on the angular quantum numbers and
\( \left[{\bf \sigma}_1 \otimes {\bf \sigma}_2\right]^{(2)} \)
operates only on
the spin states we can write the matrix element
$$
\begin{eqnarray*}
\langle lSJ\vert V\vert lSJ\rangle & = &
\langle lSJ \vert\left[ \left[{\bf \sigma}_1 \otimes {\bf \sigma}_2\right]^{(2)} \otimes
\left[{\bf r} \otimes {\bf r} \right]^{(2)}\right]^{(0)}_0\vert l'S'J\rangle \\
& = &
(-1)^{J+l+S}
\left\{\begin{array}{ccc} l&S&J \\ l'&S'&2\end{array}\right\}
\langle l \vert\vert\left[{\bf r} \otimes {\bf r} \right]^{(2)} \vert \vert l'\rangle\\
& &
\times \langle S\vert\vert\left[{\bf \sigma}_1 \otimes {\bf \sigma}_2\right]^{(2)} \vert\vert S'\rangle
\end{eqnarray*}
$$
We need that the coordinate vector \( {\bf r} \) can be written in terms of spherical components as
$$
{\bf r}_\alpha = r\sqrt{\frac{4\pi}{3}} Y_{1\alpha}
$$
Using this expression we get
$$
\begin{eqnarray*}
\left[{\bf r} \otimes {\bf r} \right]^{(2)}_\mu &=& \frac{4\pi}{3}r^2
\sum_{\alpha ,\beta}\langle 1\alpha 1\beta\vert 2\mu \rangle Y_{1\alpha} Y_{1\beta}
\end{eqnarray*}
$$
The product of two spherical harmonics can be written as
$$
Y_{l_1m_1} Y_{l_2m_2}=\sum_{lm}\sqrt{\frac{(2l_1+1)(2l_2+1)(2l+1)}{4\pi}}
\left(\begin{array}{ccc} l_1&l_2&l \\ m_1&m_2&m\end{array}\right)
$$
$$
\times \left(\begin{array}{ccc} l_1&l_2&l \\ 0 &0 &0\end{array}\right)
Y_{l-m}(-1)^m.
$$
Using this relation we get
$$
\begin{eqnarray*}
\left[{\bf r} \otimes {\bf r} \right]^{(2)}_\mu &=&
\sqrt{4\pi}r^2
\sum_{lm}
\sum_{\alpha ,\beta} \langle 1\alpha 1\beta\vert 2\mu \rangle \\
&&\times \langle 1\alpha 1\beta\vert l-m \rangle
\frac{(-1)^{1-1-m}}{\sqrt{2l+1}}
\left(\begin{array}{ccc} 1&1&l \\ 0 &0 &0\end{array}\right)Y_{l-m}(-1)^m\\
&=& \sqrt{4\pi}r^2
\left(\begin{array}{ccc} 1&1&2 \\ 0 &0 &0\end{array}\right)
Y_{2-\mu}\\
&=& \sqrt{4\pi}r^2 \sqrt{\frac{2}{15}}Y_{2-\mu}
\end{eqnarray*}
$$
We can then use this relation to rewrite the reduced matrix element containing the position vector as
$$
\begin{eqnarray*}
\langle l \vert\vert\left[{\bf r} \otimes {\bf r} \right]^{(2)} \vert \vert l'\rangle
& = &
\sqrt{4\pi}\sqrt{ \frac{2}{15}}r^2 \langle l \vert\vert Y_2 \vert \vert l'\rangle \\
& = &\sqrt{4\pi}\sqrt{ \frac{2}{15}} r^2 (-1)^l
\sqrt{\frac{(2l+1)5(2l'+1)}{4\pi}}
\left(\begin{array}{ccc} l&2&l' \\ 0&0&0\end{array}\right)
\end{eqnarray*}
$$
Using the reduced matrix element of the spin operators defined as
$$
\begin{eqnarray*}
\langle S\vert \vert\left[{\bf \sigma}_1 \otimes {\bf \sigma}_2\right]^{(2)} \vert \vert S' \rangle
& = &
\sqrt{(2S+1)(2S'+1)5}
\left\{\begin{array}{ccc} s_1&s_2&S \\s_3&s_4&S' \\ 1&1&2\end{array}\right\}\\
&\times&
\langle s_1 \vert \vert {\bf \sigma}_1 \vert \vert s_3\rangle
\langle s_2 \vert\vert {\bf \sigma}_2 \vert \vert s_4\rangle
\end{eqnarray*}
$$
and inserting these expressions for the two reduced matrix elements we get
$$
\begin{array}{ll}
&\\
\langle lSJ\vert V\vert l'S'J\rangle =&(-1)^{S+J}\sqrt{30(2l+1)(2l'+1)(2S+1)(2S'+1)}\\
&\times\left\{\begin{array}{ccc}l&S &J \\l'&S&2\end{array}\right\}
\left(\begin{array}{ccc}l&2&l'\\0&0&0\end{array}\right)
\left\{\begin{array}{ccc}s_{1}&s_{2}&S\\s_{3}&s_{4}&S'\\
1&1&2\end{array}
\right\}\\
&\times\langle s_{1}\vert\vert \sigma_{1}\vert\vert s_{3}\rangle
\langle s_{2}\vert\vert \sigma_{2}\vert \vert s_{4}\rangle.
\end{array}
$$
Normally, we start we a nucleon-nucleon interaction fitted to reproduce scattering data. It is common then to represent this interaction in terms relative momenta \( k \), the center-of-mass momentum \( K \) and various partial wave quantum numbers like the spin \( S \), the total relative angular momentum \( {\cal J} \), isospin \( T \) and relative orbital momentum \( l \) and finally the corresponding center-of-mass \( L \). We can then write the free interaction matrix \( V \) as
$$
\langle kKlL{\cal J}ST\vert\hat{V}\vert k'Kl'L{\cal J}S'T\rangle.
$$
Transformations from the relative and center-of-mass motion
system to the lab system will be discussed
below.
To obtain a \( V \)-matrix in a h.o. basis, we need the transformation
$$
\langle nNlL{\cal J}ST\vert\hat{V}\vert n'N'l'L'{\cal J}S'T\rangle,
$$
with \( n \) and \( N \) the principal quantum numbers of the relative and
center-of-mass motion, respectively.
$$
\vert nlNL{\cal J}ST\rangle= \int k^{2}K^{2}dkdKR_{nl}(\sqrt{2}\alpha k)
R_{NL}(\sqrt{1/2}\alpha K)
\vert klKL{\cal J}ST\rangle.
$$
The parameter \( \alpha \) is the chosen oscillator length.
The most commonly employed sp basis is the harmonic oscillator, which in turn means that a two-particle wave function with total angular momentum \( J \) and isospin \( T \) can be expressed as
$$
\begin{array}{ll}
\vert (n_{a}l_{a}j_{a})(n_{b}l_{b}j_{b})JT\rangle =&
{\displaystyle
\frac{1}{\sqrt{(1+\delta_{12})}}
\sum_{\lambda S{\cal J}}\sum_{nNlL}}
F\times \langle ab|\lambda SJ \rangle\\
&\times (-1)^{\lambda +{\cal J}-L-S}\hat{\lambda}
\left\{\begin{array}{ccc}L&l&\lambda\\S&J&{\cal J}
\end{array}\right\}\\
&\times \left\langle nlNL| n_al_an_bl_b\right\rangle
\vert nlNL{\cal J}ST\rangle ,\end{array}
\tag{4}
$$
where the term
\( \left\langle nlNL| n_al_an_bl_b\right\rangle \)
is the so-called Moshinsky-Talmi transformation coefficient (see chapter 18 of Alex Brown's notes).
The term \( \langle ab|LSJ \rangle \) is a shorthand for the \( LS-jj \) transformation coefficient,
$$
\langle ab|\lambda SJ \rangle = \hat{j_{a}}\hat{j_{b}}
\hat{\lambda}\hat{S}
\left\{
\begin{array}{ccc}
l_{a}&s_a&j_{a}\\
l_{b}&s_b&j_{b}\\
\lambda &S &J
\end{array}
\right\}.
$$
Here
we use \( \hat{x} = \sqrt{2x +1} \).
The factor \( F \) is defined as \( F=\frac{1-(-1)^{l+S+T}}{\sqrt{2}} \) if
\( s_a = s_b \) and we .
The \( \hat{V} \)-matrix in terms of harmonic oscillator wave functions reads
$$
\langle (ab)JT|\hat{V}|(cd)JT\rangle=
{\displaystyle \sum_{\lambda \lambda ' SS' {\cal J}}\sum_{nln'l'NN'L}
\frac{\left(1-(-1)^{l+S+T}\right)}{\sqrt{(1+\delta_{ab})
(1+\delta_{cd})}}}
$$
$$
\times\langle ab|\lambda SJ\rangle \langle cd|\lambda 'S'J\rangle
\left\langle nlNL| n_{a}l_{a}n_{b}l_{b}\lambda\right\rangle
\left\langle n'l'NL| n_{c}l_{c}n_{d}l_{d}\lambda ' \right\rangle
$$
$$
\times \hat{{\cal J}}(-1)^{\lambda + \lambda ' +l +l'}
\left\{\begin{array}{ccc}L&l&\lambda\\S&J&{\cal J}
\end{array}\right\}
\left\{\begin{array}{ccc}L&l'&\lambda '\\S&J&{\cal J}
\end{array}\right\}
$$
$$
\times\langle nNlL{\cal J}ST\vert\hat{V}\vert n'N'l'L'{\cal J}S'T\rangle.
$$
The label \( a \) represents here all the single particle quantum numbers
\( n_{a}l_{a}j_{a} \).


| Project Status (12/10/2017 - 17:32:49) | |||
| Project File: | system.xmp | Implementation State: | Programming File Generated |
| Module Name: | system |
|
No Errors |
| Product Version: | EDK 14.6 |
|
267 Warnings (6 new) |
| XPS Reports | [-] | ||||
| Report Name | Generated | Errors | Warnings | Infos | |
| Platgen Log File | Sun Dec 10 16:39:39 2017 | 0 | 40 Warnings (1 new) | 180 Infos (0 new) | |
| Libgen Log File | |||||
| Simgen Log File | Sun Dec 10 16:03:48 2017 | 0 | 35 Warnings (35 new) | 233 Infos (233 new) | |
| BitInit Log File | Sun Dec 10 17:32:49 2017 | 0 | 4 Warnings (0 new) | 54 Infos (0 new) | |
| System Log File | Sun Dec 10 17:32:49 2017 |
| XPS Synthesis Summary (estimated values) | [-] | |||||
| Report | Generated | Flip Flops Used | LUTs Used | BRAMS Used | Errors | |
| system | Sun Dec 10 16:43:01 2017 | 17283 | 14958 | 49 | 0 | |
| system_clock_generator_0_wrapper | Sun Dec 10 16:39:34 2017 | 2 | 0 | |||
| system_push_button_1_wrapper | Sat Dec 9 03:27:04 2017 | 76 | 44 | 0 | ||
| system_ddr2_sdram_wrapper | Sat Dec 9 02:36:48 2017 | 10421 | 8127 | 33 | 0 | |
| system_mb_plb_wrapper | Sat Dec 9 02:34:01 2017 | 162 | 501 | 0 | ||
| system_cam_iic_1_wrapper | Sat Dec 2 22:41:14 2017 | 402 | 438 | 0 | ||
| system_cam_iic_0_wrapper | Sat Dec 2 22:40:48 2017 | 402 | 438 | 0 | ||
| system_cam_ctrl_1_wrapper | Sat Dec 2 22:40:21 2017 | 780 | 656 | 0 | ||
| system_cam_ctrl_0_wrapper | Sat Dec 2 22:39:55 2017 | 780 | 656 | 0 | ||
| system_hdmi_0_wrapper | Sat Dec 2 22:39:30 2017 | 752 | 857 | 0 | ||
| system_dvma_0_wrapper | Sat Dec 2 22:39:01 2017 | 1302 | 1366 | 0 | ||
| system_proc_sys_reset_0_wrapper | Sat Dec 2 22:38:14 2017 | 69 | 53 | 0 | ||
| system_mdm_0_wrapper | Sat Dec 2 22:38:07 2017 | 126 | 123 | 0 | ||
| system_rs232_uart_0_wrapper | Sat Dec 2 22:35:26 2017 | 145 | 131 | 0 | ||
| system_lmb_bram_wrapper | Sat Dec 2 22:35:07 2017 | 16 | 0 | |||
| system_ilmb_cntlr_wrapper | Sat Dec 2 22:34:55 2017 | 2 | 6 | 0 | ||
| system_dlmb_cntlr_wrapper | Sat Dec 2 22:34:49 2017 | 2 | 6 | 0 | ||
| system_dlmb_wrapper | Sat Dec 2 22:34:43 2017 | 1 | 1 | 0 | ||
| system_ilmb_wrapper | Sat Dec 2 22:34:37 2017 | 1 | 1 | 0 | ||
| system_microblaze_0_wrapper | Sat Dec 2 22:34:11 2017 | 1860 | 1552 | 0 |
| Device Utilization Summary (actual values) | [-] | ||||
| Slice Logic Utilization | Used | Available | Utilization | Note(s) | |
| Number of Slice Registers | 14,736 | 28,800 | 51% | ||
| Number used as Flip Flops | 14,716 | ||||
| Number used as Latch-thrus | 20 | ||||
| Number of Slice LUTs | 13,404 | 28,800 | 46% | ||
| Number used as logic | 12,312 | 28,800 | 42% | ||
| Number using O6 output only | 11,436 | ||||
| Number using O5 output only | 431 | ||||
| Number using O5 and O6 | 445 | ||||
| Number used as Memory | 1,015 | 7,680 | 13% | ||
| Number used as Dual Port RAM | 536 | ||||
| Number using O6 output only | 24 | ||||
| Number using O5 output only | 96 | ||||
| Number using O5 and O6 | 416 | ||||
| Number used as Shift Register | 479 | ||||
| Number using O6 output only | 478 | ||||
| Number using O5 output only | 1 | ||||
| Number used as exclusive route-thru | 77 | ||||
| Number of route-thrus | 533 | ||||
| Number using O6 output only | 493 | ||||
| Number using O5 output only | 32 | ||||
| Number using O5 and O6 | 8 | ||||
| Number of occupied Slices | 6,279 | 7,200 | 87% | ||
| Number of LUT Flip Flop pairs used | 19,079 | ||||
| Number with an unused Flip Flop | 4,343 | 19,079 | 22% | ||
| Number with an unused LUT | 5,675 | 19,079 | 29% | ||
| Number of fully used LUT-FF pairs | 9,061 | 19,079 | 47% | ||
| Number of unique control sets | 1,674 | ||||
| Number of slice register sites lost to control set restrictions |
2,989 | 28,800 | 10% | ||
| Number of bonded IOBs | 173 | 480 | 36% | ||
| Number of LOCed IOBs | 173 | 173 | 100% | ||
| IOB Flip Flops | 310 | ||||
| Number of BlockRAM/FIFO | 59 | 60 | 98% | ||
| Number using BlockRAM only | 59 | ||||
| Number of 36k BlockRAM used | 59 | ||||
| Total Memory used (KB) | 2,124 | 2,160 | 98% | ||
| Number of BUFG/BUFGCTRLs | 9 | 32 | 28% | ||
| Number used as BUFGs | 9 | ||||
| Number of IDELAYCTRLs | 3 | 16 | 18% | ||
| Number of BSCANs | 1 | 4 | 25% | ||
| Number of BUFIOs | 8 | 56 | 14% | ||
| Number of DSP48Es | 3 | 48 | 6% | ||
| Number of PLL_ADVs | 2 | 6 | 33% | ||
| Average Fanout of Non-Clock Nets | 3.70 |
| Performance Summary | [-] | |||
| Final Timing Score: | 0 (Setup: 0, Hold: 0, Component Switching Limit: 0) | Pinout Data: | Pinout Report | |
| Routing Results: | All Signals Completely Routed | Clock Data: | Clock Report | |
| Timing Constraints: | All Constraints Met |
| Detailed Reports | [-] | |||||
| Report Name | Status | Generated | Errors | Warnings | Infos | |
| Translation Report | Current | Sun Dec 10 16:50:56 2017 | 0 | 147 Warnings (0 new) | 6 Infos (0 new) | |
| Map Report | Current | Sun Dec 10 16:54:30 2017 | 0 | 111 Warnings (3 new) | 1249 Infos (0 new) | |
| Place and Route Report | Current | Sun Dec 10 16:56:36 2017 | 0 | 4 Warnings (0 new) | 4 Infos (0 new) | |
| Post-PAR Static Timing Report | Current | Sun Dec 10 16:57:08 2017 | 0 | 0 | 5 Infos (0 new) | |
| Bitgen Report | Current | Sun Dec 10 16:58:27 2017 | 0 | 5 Warnings (3 new) | 0 |
| Secondary Reports | [-] | ||
| Report Name | Status | Generated | |
| WebTalk Report | Current | Sun Dec 10 16:58:30 2017 | |
| WebTalk Log File | Current | Sun Dec 10 16:58:33 2017 |


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Iterate over the own and inherited enumerable properties of an object, and return an object with properties that evaluate to true from the callback. Exit early by returning
false
.
npm i for-own --save
npm test
var forOwn = require('for-own');
var obj = {a: 'foo', b: 'bar', c: 'baz'};
var values = [];
var keys = [];
forOwn(obj, function (value, key, o) {
keys.push(key);
values.push(value);
});
console.log(keys);
//=> ['a', 'b', 'c'];
console.log(values);
//=> ['foo', 'bar', 'baz'];
Jon Schlinkert
Copyright (c) 2014 Jon Schlinkert, contributors. Released under the MIT license
This file was generated by verb-cli on September 20, 2014.


BLACK SCORPION, THE (director: Edward Ludwig; screenwriters: Robert Blees/David Duncan/Paul Yawitz; cinematographer: Lionel Lindon; editor: Richard L. van Enger; cast: Richard Denning (Hank Scott), Mara Corday (Teresa Alverez), Carlos Rivas (Artur Ramos), Mario Navarro (Juanito), Carlos Múzquiz (Dr. Velazco), Pascual García Peña (Jose de la Cruz), Arturo Martínez (Major Cosio), Pedro Galván (Father Delgado); Runtime: 88; 1957)
Reviewed by Dennis Schwartz
This is a staid, low-budget B- sci-fi'er about giant scorpions that come out of a volcano in a remote part of Mexico and attack humans and cattle in their lust for blood. It is a poorly done remake of "Them!", worth seeing only for the efforts put forth by the great animator Willis O' Brien, the animator of "King Kong," who did this film when he was in his seventies.
When a volcano strikes a remote Mexican area, whose power is estimated to be as great as the eruption from Vesuvius, two geologist are sent to do a preliminary investigation, the American, Dr. Hank Scott (Richard Denning) and his Mexican friend, Dr. Artur Ramos (Carlos Rivas). They discover strange happenings, such as a dead policeman who emptied his gun before dying, crushed cars, and huge footprints. They go to the nearest village and meet Father Delgado. He tells them that there are strange things also happening in the village, that the natives superstitiously believe that the volcano was caused by a "demon bull," a symbol for evil.
In the field investigating the effects of the volcano, the two scientists discover a pretty lady ranch owner, Teresa Alverez (Mara Corday), who has fallen off her horse. Hank instantly falls in love with her and the two become very close in the next few days, as the scientists are guests in her luxurious ranch. The scientists collect samples and send them to the leading scientist in Mexico City, Dr. Velazco (Carlos Múzquiz). Meanwhile, Artur finds a fossil of a scorpion in an obsidian stone, and when he breaks the rock open, a small live scorpion comes out, which is of a species that has been extinct for thousands of years. At about the same time, a telephone repair man spots a giant scorpion attacking him and tells this to Theresa, before he gets crushed to death.
>From Dr. Velazco, we learn that giant scorpions have emerged from the volcano and have come back to plague the Earth. He goes on to say, they only come out at night and are slow-moving. When it is discovered where the scorpions nest, the two scientists go down the deep cave by cable car and take photographs of it, and then have the army dynamite the cave and seal it.
But the scientists soon learn the danger is not over, one giant scorpion remains on the loose and has gone on a blood lust, since they have to have blood every 3-4 days. A plan is coordinated between the scientists and the army, as they aim to hit it with a projectile and electrocute it with six hundred thousand volts, and from its throat obtain its poisonous venom so that they can have the poison available to inject any other scorpions that might pop up.
The star of the film, Richard Denning, made a career of appearing in B- sci-fi films, with his best known role being in "The Creature From The Black Lagoon."
REVIEWED ON 7/22/2000 GRADE: C-
Dennis Schwartz: "Ozus' World Movie Reviews"
http://www.sover.net/~ozus
<EMAIL_ADDRESS>© ALL RIGHTS RESERVED DENNIS SCHWARTZ
The review above was posted to the
rec.arts.movies.reviews newsgroup (de.rec.film.kritiken for German reviews).
The Internet Movie Database accepts no responsibility for the contents of the
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Please direct comments/criticisms of the review to relevant newsgroups.
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Related links: index of all rec.arts.movies.reviews reviews


Checkbox tests
Basic
should be unchecked
should be checked
Serialization
should be unchecked
should be checked
Bindings
should be checked
should be checked
Scroll
.
scroll check
.


IncludeOS allows you to run your application in the cloud without an operating system. IncludeOS adds operating system functionality to your application allowing you to create performant, secure and resource efficient virtual machines.
IncludeOS applications boot in tens of milliseconds and require only a few megabytes of disk and memory.
To run a service with IncludeOS on Linux or macOS you do not need to install IncludeOS, however you need to install a few dependencies depending on the service you will run. You can start by trying out our simplest hello_world service. For this service you will need the following dependencies.
With the above dependencies you should be able to build an application within minutes.
$ conan config install https://github.com/includeos/conan_config.git
$ git clone https://github.com/includeos/hello_world.git
$ mkdir build_hello_world
$ cd build_hello_world
$ conan install ../hello_world -pr
$ source ./activate.sh
$ cmake ../hello_world
$ cmake --build .
$ boot hello
The hello world booted service should look like this:
================================================================================
IncludeOS 0.14.1-1093 (x86_64 / 64-bit)
+--> Running [ Hello world - OS included ]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hello world
[ main ] returned with status 0
For detailed instructions see the GitHub README. Once installed we suggest looking at and booting a few of the demo-examples to familarize yourself with the system.
We strive to make it easy to create fast and useful services. The below code will set up a simple TCP echo service and happily talk to anyone connecting.
#include <os>
#include <iostream>
#include <net/interfaces>
void Service::start()
{
// Get the IP stack thats already been automatically configured
auto& inet = net::Interfaces::get(0);
// Setup a TCP echo server on port 7 (echo port)
auto& server = inet.tcp().listen(7);
server.on_connect([] (auto conn) {
// Log incomming connections on the console:
std::cout << "Connection " << conn->to_string() << " established\n";
// When data is received, echo back
conn->on_read(1024, [conn] (auto buf) {
conn->write(buf);
});
});
}
The network configuration of the virtual machine can reside in a JSON file, named config.json, placed in the same folder. It should look something like this, depending on your need:
{
"net" : [
{
"iface": 0,
"config": "dhcp-with-fallback",
"address": "<IP_ADDRESS>",
"netmask": "<IP_ADDRESS>",
"gateway": "<IP_ADDRESS>"
}
]
}
For security related inquires please send email to<EMAIL_ADDRESS>You can use our PGP key to encrypt the email.


Education
Tecnológico de Monterrey, Campus Monterrey. Bachelor’s Degree in Computer Science (August 2017 - Expected December 2021).
Skills
Languages
- Spanish (Native)
- English, B2 (BULATS, 2017)
- French, A2 ( DELF, 2017)
Projects
Bot made for replying tweets of people when it's mentioned
Game made in java to make people conscious about the reality of the pandemic
Model using Matlab, R and Python to predict the final position of a driver
Extracurricular Experience
Awards
2018 & 2019. Outstanding teacher assistant (social service: Niños Regios).


{# From bootstrap_admin 0.4.0 #} {% load i18n %} {% if app_list %} {% for app in app_list %}
{% trans "You don't have permission to edit anything." %}


March 10, 2019 | 19:56
Migrate to Hugo from Jekyll
Move static content to static Jekyll has a rule that any directory not starting with _ will be copied as-is to the _site output. Hugo keeps all static content under static. You should therefore move it all there. With Jekyll, something that looked like
▾ <root>/ ▾ images/ logo.png should become
▾ <root>/ ▾ static/ ▾ images/ logo.png Additionally, you’ll want any files that should reside at the root (such as CNAME) to be moved to static.
Read more


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data class Subcommand : ParameterHelp
name
- The name for this command
help
- The command's description
Subcommand(name: String, help: String) |
val help: String
The command's description |
|
val name: String
The name for this command |


Chapter 7 Final Words
To be continued.
Bai, Yin, Bin Xu, Yuanchao Ma, Guodong Sun, and Yu Zhao. 2012. “Will You Have a Good Sleep Tonight?: Sleep Quality Prediction with Mobile Phone.” In Proceedings of the 7th International Conference on Body Area Networks, 124–30. ICST (Institute for Computer Sciences, Social-Informatics; Telecommunications Engineering).
Brick, Cameron A., Darbi L. Seely, and Tonya M. Palermo. 2010. “Association Between Sleep Hygiene and Sleep Quality in Medical Students.” Behavioral Sleep Medicine 8 (2): 113–21. doi:10.1080/15402001003622925.
Buysse, Daniel J, Charles F Reynolds, Timothy H Monk, Susan R Berman, and David J Kupfer. 1989. “The Pittsburgh Sleep Quality Index: A New Instrument for Psychiatric Practice and Research.” Psychiatry Research 28 (2). Elsevier: 193–213.
Irish, Leah A, Christopher E Kline, Heather E Gunn, Daniel J Buysse, and Martica H Hall. 2015. “The role of sleep hygiene in promoting public health: A review of empirical evidence.” Sleep Medicine Reviews 22. Elsevier Ltd: 23–36. doi:10.1016/j.smrv.2014.10.001.
Kelleher, John D., Brian Mac Namee, and Aoife D’Arcy. 2015. Fundamentals of Machine Learning for Predictive Data Analytics: algorithms, worked examples, and case studies. 1. London: The MIT Press.
Kowarik, Alexander, and Matthias Templ. 2016. “Imputation with the R Package VIM.” Journal of Statistical Software 74 (7): 1–16. doi:10.18637/jss.v074.i07.
Liu, Huan, and Hiroshi Motoda. 1998. Feature selection for knowledge discovery and data mining. doi:10.1007/978-1-4615-5689-3.
Mastin, David F, Jeff Bryson, and Robert Corwyn. 2006. “Assessment of Sleep Hygiene Using the Sleep Hygiene Index.” Journal of Behavioral Medicine 29 (3). Springer: 223–27.
Posner, Donn, and Philip R. Gehrman. 2011. “Chapter 3 - Sleep Hygiene.” In Behavioral Treatments for Sleep Disorders, edited by Michael Perlis, Mark Aloia, and Brett Kuhn, 31–43. Practical Resources for the Mental Health Professional. San Diego: Academic Press. doi:http://dx.doi.org/10.1016/B978-0-12-381522-4.00003-1.
Tang, Jiliang, Salem Alelyani, and Huan Liu. 2014. “Feature Selection for Classification: A Review.” Data Classification: Algorithms and Applications, 37–64. doi:<IP_ADDRESS>9.5195.
van Buuren, Stef, and Karin Groothuis-Oudshoorn. 2011. “mice: Multivariate Imputation by Chained Equations in R.” Journal of Statistical Software 45 (3): 1–67. http://www.jstatsoft.org/v45/i03/.
Wentz, Dave, and Myron Wentz. 2011. The Healthy Home: Simple Truths to Protect Your Family from Hidden Household Dangers. Vanguard.


public class DestinyDto
extends java.lang.Object
implements java.io.Serializable
| Modifier and Type | Field and Description |
|---|---|
Star |
bodyMaster
The body master.
|
int |
bornDay
The born day.
|
int |
bornMonth
The born month.
|
Data.GroundTime |
bornTimeGround
The born time ground.
|
int |
bornYear
The born year.
|
java.util.List<Cell> |
cells
The cells.
|
Config.ConfigType |
configType
The config type.
|
Ground |
destinyGround
The destiny ground.
|
Star |
destinyMaster
The destiny master.
|
Data.FiveElement |
fiveElement
The five element.
|
int |
fiveElementNumber
The five element number.
|
boolean |
isDoubleMonth
The is double month.
|
java.util.Map<MajorStar,Data.StarEnergyLevel> |
majorStarEnergy
The major star energy.
|
boolean |
pathClockwise
The path clockwise.
|
Data.Sex |
sex
The sex.
|
Data.ShadowLight |
shadowLight
The shadow light.
|
Ground |
sonDou
The son dou.
|
java.util.Map<java.lang.String,Star> |
starReactionMap
The star reaction map.
|
Ground |
yearGround
The year ground.
|
Sky |
yearSky
The year sky.
|
| Constructor and Description |
|---|
DestinyDto(Destiny destiny)
Instantiates a new destiny dto.
|
public final Config.ConfigType configType
public final Data.Sex sex
public final Sky yearSky
public final Ground yearGround
public final int bornYear
public final int bornMonth
public final int bornDay
public final Data.GroundTime bornTimeGround
public final boolean isDoubleMonth
public final Data.ShadowLight shadowLight
public final Data.FiveElement fiveElement
public final int fiveElementNumber
public final java.util.List<Cell> cells
public final Star destinyMaster
public final Star bodyMaster
public final Ground sonDou
public final java.util.Map<java.lang.String,Star> starReactionMap
public final java.util.Map<MajorStar,Data.StarEnergyLevel> majorStarEnergy
public final boolean pathClockwise
public final Ground destinyGround
public DestinyDto(Destiny destiny)
destiny
- the destiny