1. Spectral Madness - Widening the Scope of a
Partial Dynamic Symmetry
Wesley Pereira, Ricardo Garcia Larry Zamick and Alberto Escuderos
Department of Physics and Astronomy, Rutgers University, Piscataway, New Jersey 08854
Kai Neergård
Fjordtoften 17, 4700 Næstved, Denmark
August 24, 2015
Abstract
If one examines two-body matrix elements from experiment, one no-
tices that not only J=0 T=1 lies low, but also J=1 T=0 and J=Jmax =2j
T=0. It is sometimes thought that one needs both T=1 and T=0 two-
body matrix elements to get equally spaced spectra of even I states, i.e.,
vibrational spectra. We here attempt to get equally spaced levels with
only those that have T=1 (even J). As an example, we perform single-j
calculations (f7/2 ) in 44
Ti and 46
Ti. We then shift gears and decide to
play around with the input two particle matrix elements (not worrying
about experiment) to generate interesting spectra, e.g., rotational spectra,
with and then without T=0 two-body matrix elements. We also consider
simple interactions (e.g. "123",”1234” and “12345”) and find an expanded
partial dynamical symmetry.
1 Introduction
In a Nature article by B. Cederwall et al.[1], they report findings of equally
spaced levels J=0, 2, 4, 6 in the 8 hole nucleus 92
Pd. Their calculated B(E2)’s
[1.2] are not consistent with a simple vibrational interpretation. In the “sup-
plements material” of the Nature article they feel that they have an isoscalar
spin aligned coupling scheme and emphasize configurations in which a neutron
and proton couple to J=Jmax=9 in the g9/2 shell. Cleary, they emphasize the
importance of odd-J T=0 two-body matrix elements. Robinson et al.[3] noted
that in a large space calculation the static quadrupole moment of the lowest 2+
state of 92
Pd was very small, consistent with the vibrational picture. On the
other hand, 96
Cd turned out to be prolate, and 88
Ru, oblate.
Previous to this, Robinson et al.[4] made a study of doing full fp calculations
of another 8 particle system, 48
Cr. They compared the results of including
all two-body matrix elements with those in which all T=0 two-body matrix
1
arXiv:1506.07209v11[nucl-th]21Aug2015

2. elements were set to zero. The qualitative discussions in the 2 cases were quite
different. In a lanl preprint by Robinson et al.[5], a figure is shown which
confirms the implications of refs [1,2] that dire consequences occur when T=0
two-body matrix elements are set to zero. With one of the interactions used, the
first 6+
and 8+
states are nearly degenerate and the B(E2)’s 8 to 6 and 6 to 4
are very small. On the other hand, for 48
Cr the emphasis was that it was hard to
tell which of the calculations agreed better with experiment – full interaction of
T=1 only. Sure, there were some differences, but both spectra looked similar –
sort of vibrational, but with a tendency to rotation. The B(E2)’s were larger in
the full calculation, but this could be accomodated to a large extent by changing
the effective charges.
Although the above works serve as a stimulus for we are about to do, we
would like to separate ourselves from discussions and arguments about the rel-
ative merits therein. Rather, we start anew and address the problem of how
two-body matrix elements affect the spectra of more complex nuclei.
2 Attempt to generate near equally spaced spec-
tra with only T=1 two-body matrix elements
-44
Ti
In 1963 and 1964, calculations were performed to obtain wave functions and
energy levels in the f7/2 shell by Bayman et al. [6], McCullen et al. [7] and
French et al. [8]. At that time, the T=1 two-body matrix elements were well
known but not so T=0. In 1985, the T=0 matrix elements were better known
and the calculations were repeated by Escuderos et al. [9]. The two-particle
matrix elements, obtained mainly from the spectrum of 42
Sc, are shown in the
table. Note that not only J=0 T=1, but also J=1 T=0 and J=7 T=0, are low
lying .
We next make a search for a set of T=1 two-body matrix elements which
will give close to equal spaces spectra for even I states in 44
Ti, and for which all
the T=0 (odd-J) matrix elements are set equal to zero. We find a good choice
for J=0, 2, 4, 6 are, respectively, 0.00, 1.00, 1.60, 2.00.
Table I: Spectra of 42
Sc
J MBZE T=0 MBZE T=1 only MeV
0 0.0000 1 0.6111 0 0.0000
2 1.5865 3 1.4904 2 1.0000
4 2.8153 5 1.5101 4 1.6000
6 3.2420 7 0.6163 6 2.0000
3 Spectra (MeV) of 44
Ti and 46
Ti
Table II: Spectra of 44
Ti and 46
Ti
2

3. I MBZE 44
Ti MBZE 46
Ti T=1 only 44
T T=1 only 46
Ti
0 0.0000 0.0000 0.0000 0.0000
2 1.1631 1.1483 0.8302 0.8397
4 2.7900 2.2225 1.5723 1.5535
6 4.0618 3.1575 2.1508 1.9492
8 6.0842 4.8720 3.4643 3.1042
10 7.3839 6.3334 4.2524 4.0223
12 7.7022 8.0257 4.8524 4.9490
We have not made an exhaustive search the optimum T=1 matrix elements
so as to obtain an equally spaced spectrum of 44
Ti, but it is more than sufficient
to get the point across. With a perfect vibrator the I=12+
state would be at
6×0.8302=4.9812 MeV. In fact, it is at 4.8524 MeV. The 8 → 6 splitting is
significantly dfferent for the 2 → 0 splitting of 0.8397 MeV. It is, in fact, 1.3105
MeV.
4 Rotational Spectra.
We can also play the game of obtaining rotational spectra. This is easier. We
consider 3 cases:
1. Set all two-particle matrix elements to J(J+1). Then it is easy to show
that for 44
Ti, one also gets a perfect I(I+1) spectrum with the I=2 state at 6
MeV and the I=12 at 156 MeV.
Table III: Perfect rotator spectrum for 44
Ti
I E T=1 only scaled
0 0 0
2 6 6
4 20 19.58
6 42 41.16
8 72 70.16
10 110 107.80
12 156 150.61
3

4. 2. We next refer to Table III. In the first spectral column we have the perfect
rotator spectrum. We obtain this from a J(J+1) 2-body spectrum where both
T=0 and T=1 2-body matrix elements are included. In the next column we have
a spectrum in which only the T=1 matrix elements are included but we rescale
them so the first 2+
state comes at 6 MeV, the same as in the first column.
Without rescaling the 2+
state would be at 4.625 MeV and the 12+
at 116.10
MeV. We do not obtain a perfest rotational spectrum but still a fairly good one.
3. If we set all two-body T=1 matrix elements to zero and set the T=0 ones
to J(J+1), we get a most fascinating spectrum. The even I states with I=0, 2, 4,
6 and 8 are at zero energy whilst the 10+
and 12+
states are at 94 MeV and 156
MeV respectively. While at first glance surprising, there is an easy explanaton
for the multiple ground state zeros. All states of zero energy have isospin T=2.
They are therefore double analogs of states in
44
Ca, which in our model space
consists of 4 valence neutrons. The neutron-neutron interacton occurs only in
the T=1 channel, but in this case all three of the T=1 matrix elemnts are set
to zero. This explains why the T=2 states in 44
Ti lie at zero energy. A closer
examination shows that there are two I=2+
and two I=4+
states at zero energy.
But it is well known that in 44
Ca such states occur twice. One I=2+
state has
seniority v=2 and the other v=4; likewise for I=4+
. Note that there are no
I=10+
or 12+
states of the (f7/2)4
configuration in 44
Ca. This explains why
these states do not appear at zero energy in 44
Ti.
In summary, with a J(J+1) two-body interaction one gets a perfect rotations
spectrum in more complex nuclei, e.g., 44
Ti. If this interaction is only in the
T=1 channel, one gets an imperfect, but still fairly good rotational spectrum.
If this interacton is only in the T=0 channel one gets a spectrum with multiple
T= 2 degeracies at zero energy and the spectum does not at all resemble a
rotational spectrum.
4

5. 5 Fun Spectra
We here look around for interesting relations between input 2-particle spectra
and more complicated systems. As before, the latter will be 44
Ti as 2 protons
and 2 neutrons in the f7/2 shell. We use trial and error. We have found one
interesting case which we here mention. The two-body interacton matrix ele-
ments are 0, 0, 1, 0, 2, 0, 3, 0. We call this the 123 interaction. In other words,
all the T=0 matrix elements are set to zero and the T=1’s are set to J/2. The
resulting spectra are shown in table IV. Of special interest is that the levels
from I=6 to I=12 equally spaced with a separation of 1.5 MeV, almost twice the
2 → 0 splitting. Thus, as shown in table IV, we have achieved a “vibrational
spectrum” from I=6 to I=12, but not from I=0 to I=6.
Of even more interest are the structures of the I=6 and I=8 wave functions
in tables VI and VII. That the (2,4) and (4,2) configurations have the same
value is not a surprise. It follows from charge symmetry, and the same sign
from the fact that we are dealing with a T=0 state. What is a surprise is that
for I=6 we have a multitude of zeros – (2,6), (4,4), (6,2), (6,4) and (6,6). There
are corresponding zeros for I=8. We note one common feature – things seem
to separate into classes such that configuratons with the same value of the sum
of the proton-proton and neutron-neutron angular momenta act the same. For
I=6 the (6,0) and (4,2) coefficients are non-zero. The Jp+Jn sum is 6. For (2,6)
and (4,4) the sum is 8 and for this class the coefficients are all zero. And (6,6)
gives us 12 and here we also get a zero coefficient.
A more complete picture is afforded if we include the odd angular momenta.
This is shown in Table XXIX. We show all the equally spaced levels and all the
angular momenta of degenerate T=0 states. We note that for a given energy, all
the states have the same value of (Jp+ Jn). That quantity is listed in the first
column. Next comes the energy E, a bit rounded off to show more clearly that
the spacing is 1.5 MeV. Then we list, for a given energy, the angular momenta
of the states with that energy. For example, for 4.65 MeV, all states have Jp+
Jn = 8. There are 3 degenerate states in this case with I=6, 7, 8.
If we go to the g9/2 shell we get a similar behaviour (now with the 1234
interaction), but starting from I=8 and ending at I=14. The spacings are still
1.5 MeV and the I=8 and beyond wave functions are very strange looking.
Some of the results have been known before. In the f7/2 case there are several
states at the same energy 6.15 MeV. They have angular momenta 3, 7, 9 and 10.
But this is not specific to the 123 interaction – it applies to any interaction acting
only for T=1. This has been noted and explained [10,11,12,13] as an example of
a partial dynamical symmetry. They noted that all these T=0 states have the
same dual quantum numbers (Jp, Jn), not just the sum (Jp +Jn). A common
feature of these angular momenta is that they cannot occur or a system of 4
identical nucleons in the f7/2 shell e.g. 44
Ca. This can be seen in the tables of
identical-particle fractional parentage coefficients, B.F. Bayman and A. Lande
[14].The conditions that are imposed by the non-existence of these states can
be used to show that indeed the states 3, 7, 9, and 10 with the same (Jp, Jn)
are degenerate. What is new in this work is that with the specific 123 (T=1)
5

6. interaction, two other angular momenta I=6 and 8 enter the game. Note that
we are considering only T=0 states. There are no T=0 states with I=1 or 11 in
the (f7/2)4
configuration.
We also show more briefly results in table XXX for the 135 interaction 0,
0, 1, 0, 3, 0, 5, 0. The spacings are now 3 MeV, double those for the 123
interaction. We note that the results for (Jp+Jn) = 10 and 12 are the same as
in table XXIX. This is to be expected, since they involve angular momenta not
present for 4 identical nucleons in the f7/2 shell and are therefore true for any
T=1 interaction. The main difference is that with 135 there is only one special
I=6+
state, whereas with 123 there were 2 special 6+
states.
If we go to the g9/2 shell we get a similar behavour, (now with the 1234
interaction), but starting from I=7 and ending at I=14. The results are shown
in table XXXI. The spacings are still 1.5 MeV and the I=7 and beyond wave
functions have fixed (Jp+Jn). As an example there are degerate states at 7.29
MeV with I=10, 11, and 12, as well as states at 8.79 MeV with I=11, 13, and 14.
In the (g9/2)4
configuration of identical particles, e.g. neutrons, the following
angular momenta cannot occur: 1, 11, 13, 14, 15, 16. There are no T=0 states
with I=1, so this angular momentum is not under consideration.
In the next section we will be showing wave funcitons as column vectors
with ammplitudes DI
(Jp,Jn), such that for a i state of tota angular momentum
I |DI
(Jp,Jn)|2
is the probaility that the protons couple to Jp and the neutrons
to Jn. Note that for an N=Z nucleus e.g. 44
Ti one has the following relation
for a state of isospin T:
D(Jp,Jn)= (-1)(I+T )
D(Jn,Jp).
5.1 123 Tables in f7/2
44
Ti
Table IV: Fun Spectra and Differences
I E Diff.
0 0.0000
2 0.7552 0.7552
4 1.8338 1.0786
6 3.1498 1.3160
8 4.6498 1.5000
10 6.1498 1.5000
12 7.6498 1.5000
Table V: Wave functions for I=3
6

7. Jp Jn D(Jp,Jn)
2 2 0.0000
2 4 0.0000
4 2 0.0000
4 4 0.0000
4 6 -0.7071
6 4 0.7071
6 6 0.0000
Table VI: Wave functions for I=6
Jp Jn D(Jp,Jn) D(Jp,Jn)
0 6 0.3953 0.0000
2 4 0.5863 0.0000
2 6 0.0000 -0.4743
4 2 0.5863 0.0000
4 4 0.0000 0.7416
4 6 0.0000 0.0000
6 0 0.3953 0.0000
6 2 0.0000 -0.4743
6 4 0.0000 0.0000
6 6 0.0000 0.0000
Table VII: Wave functions for I=8
Jp Jn D(Jp,Jn)
2 6 0.4882
4 4 0.7234
4 6 0.0000
6 2 0.4882
6 4 0.0000
6 6 0.0000
Table VIII: Wave functions for I=9
Jp Jn D(Jp,Jn)
4 6 -0.7071
6 4 0.7071
6 6 0.0000
Table IX: Wave functions for I=10
Jp Jn D(Jp,Jn)
4 6 0.7071
6 4 0.7071
6 6 0.0000
7

8. Table X: Wave functions for I=12
Jp Jn D(Jp,Jn)
6 6 1.0000
5.2 135 Tables in f7/2
Table XI: Wave functions for I=3
Jp Jn D(Jp,Jn) D(Jp,Jn)
2 2 0.0000 0.0000
2 4 -0.7071 0.0000
4 2 0.7071 0.0000
4 4 0.0000 0.0000
4 6 0.0000 -0.7071
6 4 0.0000 0.7071
6 6 0.0000 0.0000
Table XII: Wave functions for I=6
Jp Jn D(Jp,Jn)
0 6 0.0000
2 4 0.0000
2 6 -0.4743
4 2 0.0000
4 4 0.7416
4 6 0.0000
6 0 0.0000
6 2 -0.4743
6 4 0.0000
6 6 0.0000
Table XIII: Wave functions for I=7
Jp Jn D(Jp,Jn)
2 6 0.0000
4 4 0.0000
4 6 -0.7071
6 2 0.0000
6 4 0.7071
6 6 0.0000
Table XIV: Wave functions for I=8
8

9. Jp Jn D(Jp,Jn)
2 6 0.4882
4 4 0.7234
4 6 0.0000
6 2 0.4882
6 4 0.0000
6 6 0.0000
Table XV: Wave functions for I=9
Jp Jn D(Jp,Jn)
4 6 -0.7071
6 4 0.7071
6 6 0.0000
Table XVI: Wave functions for I=10
Jp Jn D(Jp,Jn)
4 6 0.7071
6 4 0.7071
6 6 0.0000
Table XVII: Wave functions for I=12
Jp Jn D(Jp,Jn)
6 6 1.0000
5.3 1234 Tables in g9/2
96
Cd
Table XVIII: Wave functions for I=3
Jp Jn D(Jp,Jn)
2 2 0.0000
2 4 -0.6598
4 2 0.6598
4 4 0.0000
4 6 -0.2449
6 4 0.2449
6 6 0.0000
6 8 0.0690
8 6 -0.0690
8 8 0.0000
Table XIX: Wave functions for I=7
9

10. Jp Jn D(Jp,Jn)
2 6 0.0000
2 8 -0.3459
4 4 0.0000
4 6 0.6167
4 8 0.0000
6 2 0.0000
6 4 -0.6167
6 6 0.0000
6 8 0.0000
8 2 0.3459
8 4 0.0000
8 6 0.0000
8 8 0.0000
Table XX: Wave functions for I=8
Jp Jn D(Jp,Jn) D(Jp,Jn)
0 8 0.2792 -0.2173
2 6 0.4949 0.1322
2 8 0.0000 0.4214
4 4 0.5951 0.0000
4 6 0.0000 0.0000
4 8 0.0000 -0.4985
6 2 0.4949 -0.1322
6 4 0.0000 0.0000
6 6 0.0000 0.0000
6 8 0.0000 0.0962
8 0 0.2792 0.2173
8 2 0.0000 -0.4214
8 4 0.0000 0.4985
8 6 0.0000 -0.0962
8 8 0.0000 0.0000
Table XXI: Wave functions for I=9
Jp Jn D(Jp,Jn)
2 8 0.6281
4 6 0.3248
4 8 0.0000
6 4 -0.3248
6 6 0.0000
6 8 0.0000
8 2 -0.6281
8 4 0.0000
8 6 0.0000
8 8 0.0000
10

11. Table XXII: Wave functions for I=10
Jp Jn D(Jp,Jn) D(Jp,Jn)
2 8 0.3293 0.0000
4 6 0.6258 0.0000
4 8 0.0000 -0.5126
6 4 0.6258 0.0000
6 6 0.0000 0.6888
6 8 0.0000 0.0000
8 2 0.3293 0.0000
8 4 0.0000 -0.5126
8 6 0.0000 0.0000
8 8 0.0000 0.0000
Table XXIII: Wave functions for I=11
Jp Jn D(Jp,Jn) D(Jp,Jn)
4 8 0.7071 0.0000
6 6 0.0000 0.0000
6 8 0.0000 0.7071
8 4 -0.7071 0.0000
8 6 0.0000 -0.7071
8 8 0.0000 0.0000
Table XXIV: Wave functions for I=12
Jp Jn D(Jp,Jn)
4 8 0.4715
6 6 0.7452
6 8 0.0000
8 4 0.4715
8 6 0.0000
8 8 0.0000
Table XXV: Wave functions for I=13
Jp Jn D(Jp,Jn)
6 8 0.7071
8 6 -0.7071
8 8 0.0000
Table XXVI: Wave functions for I=14
Jp Jn D(Jp,Jn)
6 8 0.7071
8 6 0.7071
8 8 0.0000
11

12. Table XXVII: Wave functions for I=15
Jp Jn D(Jp,Jn)
8 8 1.0000
Table XXVIII: Wave functions for I=16
Jp Jn D(Jp,Jn)
8 8 1.0000
5.4 Two protons and two neutrons in the h11/2 with the
12345 interaction
In this section we will be more brief. We will just show in table XXXII the
special states with (Jp+Jn) constant for 2 protons and 2 neutrons in the h11/2
shell. We use the 12345 interaction 0, 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0. The angular
momenta that cannot occur in this case are 1, 15, 17, 18, 19, and 20. Since we
are considering only T=0 states, we do not consider I=1 or I=19.
Table XXIX: Special states in the f7/2shell (123 interaction)
Jp+Jn E (MeV) I
6 3.15 3, 6
8 4.65 6, 7, 8
10 6.15 3, 7, 9, 10
12 7.65 10, 12
Table XXX: Special states in the f7/2shell (135 interaction)
Jp+ Jn E(MeV) I
6 4.42 3
8 7.42 6, 7, 8
10 10.42 3, 7, 9, 10
12 13.42 10, 12
12

13. Table XXXI: Special states in the g9/2shell (1234 interaction)
Jp+Jn E(MeV) I
8 4.29 8
10 5.79 7, 9,10
12 7.29 10, 11, 12
14 8.79 11, 13, 14
16 10.29 14, 16
Table: XXXII: Special states in the h11/2shell (12345 interaction)
Jp+Jn E (MeV) I
10 5.46 10
12 6.96 11, 12
14 8.46 11, 13, 14
16 9.96 14, 15, 16
18 11.46 15, 17, 18
20 12.96 18, 20
5.5 Explanation
As noted above, all the states in tables XXIX–XXII are composed of basic
states |JpJn with Jp + Jn = k for some constant k. To understand how such a
state can be an eigenstate of the Hamiltonians H and why the eigenvalues are
degenerate and equidistant, notice that for a given j and I, where j is single
nucleon angular momentum, the matrix elements of H are
JpJn|H|JpJn = δJpJp
δJnJn
(EJp
+ EJn
) + 4
J1J2
JpJn J1J2 EJ1
J1J2 JpJn .
(1)
Here, EJ
is the energy of two nucleons with total angular momentum J, and
JpJn J1J2 is shorthand for a unitary 9-j symbol,
JpJn J1J2 = (jj)Jp(jj)Jn|(jj)J1(jj)J2
I
.
13

14. The angular momenta Jp, Jn, Jp and Jn are even, while J1 and J2 take all values
allowed by the triangle inequalities. It is convenient to define an operator X
such that
J1J2|X|J1J2 = J1J2 J1J2 .
The space of states with even isospin T is spanned by the orthonormal states
|JpJn e = 2−
1+δJpJn
2 (|JpJn + (−)I
|JnJp )
with Jp ≥ Jn for even I and Jp > Jn for odd I. By the symmetry of JpJn J1J2 ,
the matrix element JpJn|e X|J1J2 vanishes unless J1 and J2 have equal pari-
ties. Therefore, in the even T space, when EJ
= 0 for odd J, only even J1 and
J2 contribute to the sum in eq. (1), and we have
H = Ω + 2NΩN
with operators Ω and N acting within the even T spaces and defined by
JpJn|Ω|JpJn e = δJpJp
δJnJn
(EJp
+ EJn
),
JpJn|N|JpJn e = JpJn|X|JpJn e.
Here the subscript e indicates a matrix element taken between states |JpJn e.
Because the two-proton and two-neutron states are antisymmetric, one can
apply in the last equation the relation
4X = (13) + (14) + (23) + (24),
where the operator (ik) interchanges the states of the ith and kth nucleons. It is
here understood that in a state |JpJn , the 1st and 2nd nucleons are the protons
and the 3rd and 4th nucleons the neutrons. By eq. (4) of ref. [15] we have
1≤i<k≤4
(ik) = 4 − 42
/4 − T(T + 1) = −T(T + 1),
and due to the antisymmetry of the two-proton and two-neutron states, (12)
and (34) equal −1. As a result, the matrix N has eigenvalue (−T(T + 1) − 2 ×
(−1))/4 = 1/2 for T = 0. In particular, if some T = 0 state is an eigenstate of
Ω, it is an eigenstate of H with eigenvalue 1 + 2 × (1/2)2
= 3/2 times that of Ω.
With EJ
= J/2 for even J and EJ
= 0 for odd J, if some T = 0 state is a
linear combination of states |JpJn with Jp + Jn = k, it is an eigenstate of Ω
with eigenvalue k/2 and therefore an eigenvalue of H with eigenvalue 3k/4. Such
states are degererate if they have the same k, and the energies for different k
are spaced by the constant 3/2. These energy relations persist when the ground
state energy is subtracted. The 135 interaction is related to the 123 interaction.
To display this relation we indicate E0
by writing the interaction symbols as
0135 and 0123. Then, in an obvious notation, we have
0135 = 2 × 0123 − 1111 + 1000.
14

15. Table: XXXIII: Special states not shown in previous tables
j Jp + Jn I
1/2 0 0
3/2 2 2
4 2, 4
5/2 4 4
6 3, 5, 6
8 6, 8
13/2 12 12
14 13, 14
16 15, 16
18 15, 17, 18
20 18, 19, 20
22 19, 21, 22
24 22, 24
15/2 14 14
16 15, 16
18 17, 18
20 19, 20
22 19, 21, 22
24 22, 23, 24
26 23, 25, 26
28 26, 28
The EJ
of the interaction 1111 depend only on the parity of J. It is well known
that adding such an interaction has no other effect than shifting all levels with
the same T by the same amount. This does not change the relations between
T = 0 levels. For j = 7/2 and Jp +Jn > 6 neither Jp nor Jn can be zero. Neither
can Jp or Jn be zero for I = 3. As these states in table XXIX are eigenstates of
N, they are then unaffected by the interaction 1000. This explains table XXX
by table XXIX.
It only remains to understand the existence of T = 0 states composed of
basic states |JpJn with Jp + Jn = k for some constant k. We call such a
state a special state. Before explaining how such states can occur we present
in table XXXIII the result of a complete search of special states for j ≤ 15/2.
Cases already included in previous tables are omitted. From tables XXIX and
XXXI–XXXIII one infers the following systematics.
1. For a given j there is a special state for any I from 2j − 1 to 4j − 2 except
4j − 3 (which I is impossible for j = 1/2 and accommodates for j ≥ 3/2
just a single T = 1 state). These states have k = Jp + Jn = I for even I
and k = I + 1 for odd I.
15

16. 2. Besides, there are special states with (I, k) = (4j − 11, 4j − 8),
(4j − 8, 4j − 6), (4j − 7, 4j − 4), (4j − 4, 4j − 2) provided this I is not
negative.
Only two deviations from these rule are found, both of them for fairly low j:
First, there is no (I, k) = (4j − 8, 4j − 6) = (2, 4) special state for j = 5/2.
Second, there is an additional (I, k) = (3, 10) = (4j − 11, 4j − 4) special state
for j = 7/2. We shall prove that for j ≥ 9/2, special states occur systematically
according to the above rules. The proofs are different for the two rules.
5.5.1 Rule 1
We discuss the cases of even and odd I separately. Throughout, k is even.
Even I Let
|ψ =
Jp+Jn=k
cJp
|(JpJn)kk ,
with some set of coefficients cJ , where |(JpJn)IMI is the basic state with
magnetic quantum number MI. The above state evidently has I = k. We
assume k ≥ 2j − 1, so the range S of Jp in the summation is the set of even
integers J with k − 2j + 1 ≤ J ≤ 2j − 1. Using formulas for vector coupling
coefficients [16] we get
m1m2m3m4|ψ := ( jm1| × jm2| × jm3| × jm4|) |ψ
= δm1+m2+m3+m4,k(−)m1−m3
a(m1)a(m2)a(m3)a(m4)f(m1 + m2),
where
a(m) =
(j + m)!
(j − m)!
,
f(M) =
b(M)b(k − M)cM , M ∈ S,
0, otherwise,
b(J) =
(2J + 1)!(2j − J)!
(2j + J + 1)!(J!)2
.
The state |ψ has T = 0 when it belongs to the kernel of
K = (13) + (14) + (23) + (24) − 2,
which condition is seen to be equivalent to
(−)m3−m1
f(m3 + m2) + (−)m4−m3
f(m4 + m2)
+ (−)m1−m2
f(m1 + m3) + (−)m1−m3
f(m1 + m4)
− 2(−)m1−m3
f(m1 + m2) = 0
16

17. for m1 + m2 + m3 + m4 = k. The above equation holds when f(M) is constant
for M ∈ S. Indeed, when m1 + m2 + m3 + m4 = k, no sum M of two of the
m’s is greater that 2j or less than k − 2j, so M ∈ S if M is even. If m1 + m2 is
even, because m3 + m4 is also even and the m’s are half-integral, the numbers
m1 +m3 and m1 +m4 have opposite parities. For the one that is even, the term
in the equation with f of this argument has the sign opposite to that of the
last term on the left. Similarly for the numbers m3 + m2 and m4 + m2. So the
equation holds. If m1 +m2 is odd, because m3 +m4 is also odd, all of m3 +m2,
m4 + m2, m1 + m3, m1 + m4 have the same parities. If all of them are even,
m1 + m4, in particular, is even, so (−)m3−m1
(−)m4−m3
= (−)m4−m1
= −1.
Similarly, because m2 +m3 is even, (−)m3−m1
(−)m1−m2
= (−)m3−m2
= −1. So
again the equation holds.
Thus |ψ is special when
cJ ∝
1
b(J)b(k − J)
.
Odd I We now consider a state
|ψ =
Jp+Jn=k
cJn
|(JpJn)(k − 1)(k − 1) ,
which has I = k − 1, and so far assume again k ≥ 2j − 1. This assumption is
sharpened later. We further assume
cJ = −ck−J . (2)
as required for even T. This rules out k = 4j − 2, because in that case S has
only one element J = 2j − 1, whose cJ would then vanish. (It was actually
noted already that T cannot be even for I = k − 1 = 4j − 3 because only
Jp = Jn = 2j − 1 is allowed by the triangle inequalities.) We then get
m1m2m3m4|ψ = δm1+m2+m3+m4,k−1(−)m1−m3
1
2k
a(m1)a(m2)a(m3)a(m4)f(m1 + m2) ×
m3 − m4, even m1 + m2,
m1 − m2, odd m1 + m2,
with
f(M) =



d(M)d(k − M)cM , M ∈ S,
f(k − 1 − M), k − 1 − M ∈ S,
0, otherwise,
d(J) =
√
J b(J).
As m1m2m3m4|ψ vanishes unless m1+m2+m3+m4 = k−1, we understand
this relation in the following. The expression for m1m2m3m4|ψ is even under
17

18. the permutation (13)(24) and odd under (12), both of which commute with
K. Because the m’s are half-integral, we can therefore assume without loss of
generality that m1 + m2 and m1 + m3 are even. Then m3 + m4, m2 + m4,
m2 + m3 are odd and m1 + m4 is even. A sufficient condition for |ψ belonging
to the kernel of K then is
(−)m3−m1
(m3 − m2)f(m3 + m2) + (−)m4−m3
(m4 − m2)f(m4 + m2)
+ (−)m1−m2
(m2 − m4)f(m1 + m3) + (−)m1−m3
(m3 − m2)f(m1 + m4)
− 2(−)m1−m3
(m3 − m4)f(m1 + m2) = 0.
This can be reduced by (−)m3−m1
(−)m4−m3
= (−)m4−m1
= −1,
(−)m3−m1
(−)m1−m2
= (−)m3−m2
= 1 and f(M) = f(k − 1 − M) to
(m3 − m2)f(m3 + m2) + (m2 − m4)f(m2 + m4) + (m4 − m3)f(m4 + m3) = 0.
An odd sum M of two of the m’s cannot be greater than 2j or less than
k−1−(2j−1) = k−2j. If M = 2j, both m’s equal j, so their difference vanishes
and f(M) does not matter in the above equation. For k − 2j ≤ M ≤ 2j − 2
the number k − 1 − M belongs to S. The equation is seen to hold if f(M) is
a first degree polynomial in M for the odd sums M of two of the m’s, and it
is by the preceding remark sufficient that this holds for k − 1 − M ∈ S. By
f(M) = f(k−1−M), this is equivalent to f(M) being a first degree polynomial
in M for M ∈ S.
Choosing f(M) ∝ k − 2M, we get
cJ ∝
k − 2J
d(J)d(k − J)
,
which satisfies eq. (2). The state |ψ is then special. In order that the denomi-
nator does not vanish in the above expression, we must have 0 /∈ S, which rules
out k = 2j − 1. The final scope of the proof is thus 2j + 1 ≤ k ≤ 4j − 4,
corresponding to odd I with 2j ≤ I ≤ 4j − 5.
5.5.2 Rule 2
We call the number of basic states |JpJn e with Jp+Jn = k the k dimension. For
a state to be a linear combination of only such basic states states, x coefficients
of basic states |JpJn e must vanish, where x is the difference between the even
T dimension and the k dimension. This is realized within the T = 0 space
when its dimension is greater than x or, equivalently, the k dimension is greater
than the T = 2 dimension. That the k dimension is greater than the T = 2
dimension is thus a sufficient condition for the occurrence of a special state. In
particular, a special state occurs when there is an even T state but no T = 2
state. Then every basic state |JpJn e is special and thus an eigenstate of the
Hamiltonian with eigenvalue 3(EJp
+ EJn
)/2 only provided EJ
= 0 for odd J.
This is discussed in detail in refs. [10,11].
18

19. The k and T = 2 dimensions can be calculated by combinatorics, and for
k ≥ I ≥ 2j −1 they are functions of y = 4j −k and z = 4j −I, respectively. The
first few values of these functions are given below. The k dimension vanishes
unless y ≥ 2 and the T = 2 dimension vanishes unless z ≥ 6.
y 2 4 6 8 10 12 14 16
k dim., even z 1 1 2 2 3 3 4 4
k dim., odd z 0 1 1 2 2 3 3 4
z 6 7 8 9 10 11 12 13 14 15 16 17
T = 2 dim. 1 0 1 1 2 1 3 2 4 3 5 4
The k dimension is seen to be greater than the T = 2 dimension in the four
cases of rule 2: (z, y) = (11, 8), (8, 6), (7, 4), (4, 2). Other cases are those covered
by rule 1 (z = y or z = y + 1 except z = y + 1 = 3) with even z ≤ 10 or odd
z ≤ 13.
The present calculation requires I ≥ 2j − 1. For z ≤ 13 this holds when
j ≥ 13/2. The number of combinations of j, I, k with j ≤ 11/2 is finite, so these
cases can be examined individually. Our complete seach for j ≤ 15/2 shows that
the rules 1 and 2 predict all special states for such j with the two exceptions
mentioned. It turns out that all special states with even z ≤ 10 or odd z ≤ 13
have the k dimension greater than the T = 2 dimension. The k dimension never
exceeds the T = 2 dimension by more than one. This can be turned around to
say that the dimension of the configuration space of four identical nuclons for
given j and I is never less than the maximal k dimension minus one.
5.6 Final Comment
Preiviously we had found a partial dynamical symmetry when we set all T=0
two-body amtrix elemnts to zero in a single j shell calculation for 2 protons and
2 neutrons [10.11.1,12,13]. The angular momenta involved were those that could
not occur for 4 identical particles. For the states in question one had (Jp, Jn) as
good dual quantum numbers. When we consider more restricitve interactions,
still with T=0 interactions set to zero e.g. “123” in f7/2, “1234” in g9/2 ,”12345”
in h11/2 we get some selected equally spaced levels which are usually multi-
degenerate. As well as the old we get new angular momenta as part of the
eaually spaced spectra.These can occur for systems of identical particles. The
wave functions have the constraint that ( Jp + Jn) is a constant. This is less
constrictive than the previous condition.
5.7 Acknowledgments
Wesley Pereira is a student at Essex College, Newark, New Jersey, 07102. His
research at Rutgers is funded by a Garden State Stokes Alliance for Minorities
Participation (G.S.L.S.A.M.P.) internship.
Ricardo Garcia has two institutional affiliations: Rutgers University, and
the University of Puerto Rico, Rio Piedras Campus. The permament address
19

20. associated with the UPR-RP is University of Puerto Rico, San Juan, Puerto Rico
00931. He acknowledges that to carry out this work he has received support via
the Research Undergraduate Experience program (REU) from the U.S. National
Science Foundation through grant PHY-1263280, and thanks the REU Physics
program at Rutgers University for their support.
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21