Morten Hjorth-Jensen, National Superconducting Cyclotron Laboratory and Department of Physics and Astronomy, Michigan State University, East Lansing, MI 48824, USA
Date: Jun 27, 2017
Copyright 2013-2017, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license
Outline.
Discussion of single-particle and two-particle quantum numbers, uncoupled and coupled schemes
Discussion of angular momentum recouplings and the Wigner-Eckart theorem
Applications to specific operators like the nuclear two-body tensor force
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.
When solving the Hartree-Fock project using a nucleon-nucleon interaction in an uncoupled basis (m-scheme), we found a high level of degeneracy. One sees clear from the table here that we have a degeneracy in the angular momentum $j$, resulting in $2j+1$ states with the same energy. This reflects the rotational symmetry and spin symmetry of the nuclear forces.
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.
For the single-particle states, due to the fact that we have the spin-orbit force, the quantum numbers for the projection of orbital momentum $l$, that is $m_l$, and for spin $s$, that is $m_s$, are no longer so-called good quantum numbers. The total angular momentum $j$ and its projection $m_j$ are then so-called good quantum numbers.
This means that the operator $\hat{J}^2$ does not commute with $\hat{L}_z$ or $\hat{S}_z$.
We also start normally with single-particle state functions defined using say the harmonic oscillator. For these functions, we have no explicit dependence on $j$. How can we introduce single-particle wave functions which have $j$ and its projection $m_j$ as quantum numbers?
We have that the operators for the orbital momentum are given by
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does not commute with $\hat{L}_z$ and $\hat{S}_z$. To see this, we calculate for example
with the the following degeneracy
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
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
or
Since the max value of $M_L$ is $L$ and for $M_S$ is $S$ we obtain
Using this and the fact that the maximum value of $M_J=m_j$ is $j$ we have
To decide where this series terminates, we use the vector inequality
or
It is then easy to show that for nucleons there are only two possible values of $j$ which satisfy the inequality, namely
and with $l=0$ we get
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
and
where we have used spherical coordinates (with a spherically symmetric potential) and the spherical harmonics
for $l=m_l=0$,
for $l=1$ and $m_l=0$,
for $l=1$ and $m_l=\pm 1$,
and
as the state with quantum numbers $jm_j$. Operating with
and
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
and
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
on the two $jm_j$ states, that is
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and
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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
whose eigenvectors are the columns of
These numbers define the so-called Clebsch-Gordan coupling coefficients (the overlaps between the two basis sets). We can thus write
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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 $3/2$.
In [1]:
from sympy import S
from sympy.physics.wigner import clebsch_gordan
# Twice the values of j1 and j2
j1 = 3
j2 = 3
J = 2
Jp = 2
M = 2
Mp = 3
sum = 0.0
for m1 in range(-j1, j1+2, 2):
for m2 in range(-j2, j2+2, 2):
M = (m1+m2)/2.
""" Call j1, j2, J, m1, m2, m1+m2 """
sum += clebsch_gordan(S(j1)/2, S(j2)/2, J, S(m1)/2, S(m2)/2, M)*clebsch_gordan(S(j1)/2, S(j2)/2, Jp, S(m1)/2, S(m2)/2, Mp)
print sum
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
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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
We will write these functions in a more compact form hereafter, namely,
and
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
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
and due to the symmetry properties of the Clebsch-Gordan coefficient we have
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
and the uncoupled basis
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
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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
which we can now extend to
with the isospin spinors defined as
and
We can then define the proton state function as
and similarly for neutrons as
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and
and operating with $\hat{\tau}_z$ on the proton state function we have
and for neutrons we have
which results in
and
and its corresponding isospin projection as
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
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
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
and finally for a two-particle-two-hole state we
The anti-symmetrized matrix element is detailed as
and note that anti-symmetrization means
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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
with
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
Furthermore, using the property of the Clebsch-Gordan coefficient
which can be used to show that
is equal to
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
we do refer to the angular momenta only. This means that another way of writing the last state is
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
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the anti-symmetrized matrix elements need now to obey the following relations
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and
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
from left in
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we obtain
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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. We would obviously like to reduce this to one only by rewriting our equations in $j$-scheme.
Our Hartree-Fock basis is orthogonal by definition, meaning that we have
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
as
where we have suppressed the dependence on $n_p$ and $t_z$ in the matrix element. Using the definition
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
can be rewritten 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 need to define the so-called $6j$ and $9j$ symbols
as well as the the Wigner-Eckart theorem
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 ($\hbar=1$)
and
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
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
and assuming that states with different $J$ and $M$ are orthonormal we arrive at
is independent of $M$. To show that
but this is also equal to
meaning that
The double bars indicate that this expectation value is independent of the projection $M$. We can manipulate the Clebsch-Gordan coefficients using the relations
and
together with the so-called $3j$ symbols. It is then normal to encounter the Wigner-Eckart theorem in the form
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
and
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
Multiplying from the left with the same $3j$ symbol and summing over $M,\mu,M'$ we obtain the equivalent relation
where we used
we have that
With the Pauli spin matrices $\sigma$ and a state with $J=1/2$, the reduced matrix element
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
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
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
and
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
and
and
do not behave as a spherical tensor. However, the modified quantity
we can then define the expectation value
since the Clebsch-Gordan coefficients are real. The rhs is equivalent with
which is equal to
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 $r_{ij}=|{\bf r}_i-{\bf r}_j|$,
with $P_{\nu}$ being a Legendre polynomials
We can rewrite the Wigner-Eckart theorem as
and for $\theta=0$, the spherical harmonic
which results in
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 ${}^{16}\mbox{O}$),
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 $J$.
Now, nothing hinders us from recoupling this state by coupling $j_b$ to $j_c$, yielding an intermediate angular momentum $J_{bc}$ and then couple this angular momentum to $j_a$, resulting in the final angular momentum $J'$.
That is, we can have
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
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
The $6j$ symbol is also invariant if upper and lower arguments are interchanged in any two columns
The following program tests this relation for the case of $j_1=3/2$, $j_2=3/2$, $j_3=3$, $j_4=1/2$, $j_5=1/2$, $j_6=1$
In [2]:
from sympy import S
from sympy.physics.wigner import wigner_6j
# Twice the values of all js
j1 = 3
j2 = 5
j3 = 2
j4 = 3
j5 = 5
j6 = 1
""" The triangular relation has to be fulfilled """
print wigner_6j(S(j1)/2, S(j2)/2, j3, S(j4)/2, S(j5)/2, j6)
""" Swapping columns 1 <==> 2 """
print wigner_6j(S(j2)/2, S(j1)/2, j3, S(j5)/2, S(j4)/2, j6)
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
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
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 $J$
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
is equal to
The permutation of any two rows or any two columns yields a phase factor $(-1)^S$, where
As an example we have
and using the orthogonality properties of the Clebsch-Gordan coefficients we can rewrite the above as
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,
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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
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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)
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which can be rewritten in terms of a $9j$ symbol as
and obtain
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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
Using this expression we get
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and inserting these expressions for the two reduced matrix elements we get
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
with $n$ and $N$ the principal quantum numbers of the relative and center-of-mass motion, respectively.
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,
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2 2 1
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The label $a$ represents here all the single particle quantum numbers
$n_{a}l_{a}j_{a}$.