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The moment of inertia in the interacting boson model
Nuclear Physics A452 (1986) l-29 @ North-Holland Publishing Company
THE MOMENT OF INERTIA IN THE INTERACTING BOSON MODEL H. SCHAASER Department
and
D.M.
BRINK
of Theoretical Physics, 1 Keble Road, Oxford OXI 3NP, UK Received
27 August
1985
We perform a self-consistent cranking calculation of intrinsic states of IBM-l and IBM-2 for three reasons: (a) to study the effect of Coriolis and centrifugal forces in the IBM; (b) to derive approximate analytic formulae for the energy in some cases where the IBM hamiltonian does not have dynamical symmetry; (c) to study the dependence of the moments of inertia on the variables of the intrinsic IBM states. Choosing a coordinate system where the quadrupole tensor is diagonal reduces the number of variables considerably. Among other applications of our formalism we study the transition between the symmetries SU(3) and SU*(3) in the IBM-2 hamiltonian H = -(Q, + Q,) (Q,, + Q.). We find that the moments of inertia and the asymmetry variables show a kind of phase transition, as a function of the parameter x, occurring in Q..
Abstract.
1. Introduction During the past years the interacting boson model (IBM) 1*2) has had a lot of success in providing a phenomenological description of the spectroscopic data of nuclei. In this model the valence nucleons of even nuclei are described as a collection of bosons of angular momentum L= 0 and L = 2. Two versions of the IBM are popular: the IBM-l which deals only with one sort of nucleon, and the IBM-2 which takes account of both neutrons and protons. In this paper we study how Coriolis and centrifugal effects arise in the IBM-l and IBM-2. We have outlined our procedure for IBM-1 in ref. ‘). The energy expectation values and the electromagnetic transition probabilities are influenced in an important way by Coriolis effects. We make a self-consistent cranking calculation of the intrinsic IBM wave function to study the inclusion of Coriolis and centrifugal effects. We also investigate the dependence of the moment of inertia of the nucleus on the variables of the wave function which describe the deformation of the nuclear shape. Except in the three limiting cases of dynamical symmetry SU(5), SU(3), O(6) (and SU*(3) in the case of IBM-2) one always had to perform a numerical calculation in order to diagonalise the IBM hamiltonian. This becomes increasingly difficult (or impossible) for very large numbers of bosons. These difficulties do not occur in the intrinsic-state formalism 3X5V11*‘2). In the course of studying the moments of inertia, Coriolis and centrifugal forces, we derive approximate formulae for the energy of some of the states of the IBM-l 1 April 1986
H. Schaaser, D.M. Brink / Moment
2
and the IBM-2. These formulae not hold but rotational the number of variables in terms of the collective practical
of inertia
are valid in cases where dynamical
symmetry
does
features dominate. A special advantage of our work is that is small and they can be given a simple physical interpretation model.
This makes our formulae
especially
attractive
for
applications. 2. The cranked intrinsic state in the IBM-l
The IBM-l simulates band we use normalised
2N valence nucleons intrinsic states
by N bosons.
For the ground-state
IN; a) = JV&J+)~(O)
(2.1)
with 1
B+= s’+C a&d;,
JN=(N!(l+&,
P
(2.2)
+,)N)‘/*
and with p = -2, -1, . . . ,2. s+ and d: are the creation operators for a monopole and a quadrupole boson (with magnetic projection number p). IO) represents the core. By normalising the wave function one can always remove a possible coefficient of the s+ operator in (2.2). (2.1) are not eigenstates of the IBM-l hamiltonian. Being coherent states they are also over-complete and non-orthogonal “). Nevertheless one can use them to estimate physical quantities. The states (2.1), which are generalisations of the states used by Ginocchio and Kirson 5), are introduced because we want to study non-static effects of the nucleus. As we are interested in rotational features, it is important that states of type (2.1) can have large quadrupole moments ‘). By going to an orthogonal coordinate system x, y, z where the axes are the principal axes of the quadrupole tensor we can reduce the number of independent variables in B+ and arrive at a simpler description of the intrinsic state. We have the usual problem of finding principal axes for the matrix elements of a quadrupole tensor, which for the IBM is defined as Q,=~~~~+dd=s+x(d+d)ZT’, L?, = (-l)*d_,
(ab)Z” = We abbreviate the matrix elements ing (Q,) means that
C (LCm,mb~~~)~,abm,. m,,mb of Q, between
(QJ = (Q-d = 0, The diagonalization
of (Q,)
,
is discussed
(2.3)
the states (2.1) to (Q,). Diagonalis-
(Q2> =
(Q-J.
in more detail in appendix
(2.4) A.
H. Schaaser, D. M. Brink / Moment of inertia
Choosing
a coordinate
equivalent to requiring mations A,
system
3
where the axes are the principal
that the intrinsic
states are symmetric
under
axes of (0,) certain
IN; a’) = A/N; a) = (N; a),
is
transfor-
(2.5)
which we now discuss. It is instructive to look first at the static case. Imposing the geometrical symmetries R,(r), RY( 7r), R,(V) implies that the coordinate axes are the principal axes of (0,) (appendix A). (R,(r) is the rotation of the system through the angle r around the x-axis.) In the static case it is usual to require also that the intrinsic state is time-reversal invariant. This in conjunction with the other symmetries leads to real a, and the intrinsic states are identical to the ones used by Ginocchio and Kirson ‘). As combinations between time reversal T and the geometrical symmetry operations are possible, we have the following abelian group of transformations: S = {K(r), where
Ry(r),
R,(n),
T, TUT),
TRY(r),
T%(m),
El,
(2.6)
E is the unit element.
One usually determines the moment of inertia by cranking the nucleus. The introduction of cranking around the x-axis (this is the usual choice) destroys some of the symmetries of S. We choose an appropriate subgroup S, of S, which still diagonalises (0,). A sketch (fig. 1) of the problem helps one to see which symmetries remain. Certainly T will no longer be a symmetry; it changes the sense of the cranking movement. But some of the combinations between T and rotations are
Fig. 1. Principal-axis
system of a triaxial
nucleus.
Here the x-axis
is taken as the cranking
axis.
4
H. Schaaser, D.M. Brink / Moment of inertia
still symmetries: S,= {K(V), Example:
T&(r),
%(r),
R,,(T) alone is no longer a symmetry
(2.7)
El.
operation,
because
it changes
the
cranking sense. But the combination T&,(P) leaves the cranking sense and the configuration as a whole unaffected. S, is isomorphic to the group Dz. We prove in appendix A that (Q,) is still diagonal with this choice of S, in the cranked case. The operations of S, give us relations between the a,. In appendix A it is shown that these relations are found by solving AB+A-’
(for all A from S,) .
= B+
We consider R,(V) first. From the table in appendix R,(r) and TRY(r) imply
A we know that the symmetries
a, = a-+,
(2.8a)
a,=aE.
(2.8b)
TR,( n) is redundant and gives us no new relations between the a, in addition to (2.8). S, diagonalises (Q,) for all values of the parameter x of Q,, and gives real a,. This case should be general enough for studying nuclei in which rotational features dominate. Complex intrinsic variables a, have been used by some authors “). In their case their variables are associated with a special symmetry (O(6)) and the energy surface is -y-independent ‘). In comparison with the states used by Ginocchio and Kirson ‘) our states have in general a, = a-, # 0. In the next section we see that a, is closely associated with the Coriolis and centrifugal forces. Hence in the non-static case, B+ is equal to B+=s++a,d,++a,(d:+d’,)+a,(d:+d’,)
with a,, real.
Using [A, BN] = NBN-‘[A,
B]+fN(N-
+&N(N-l)(N-2)B”-‘[[[A, we can easily calculate
the overlap
(N; aIN; a’) =
l)BN-‘[[A,
B], B]
B], B,], B]+.
-.,
(2.9)
of the states (2.1):
(1+x
(1 +C &a,l” a,a,)N’2(1+C aLaL)N’2’
If we call A =da,a,, B=JaLaL and cos (Y=(aLa,)/(AB), then (N; a IN; a’) in the limit N + ~0 is approximately gaussian in (A-B)
the overlap and (Y.
H. Schaaser,
D. M. Brink / Moment
of inertia
5
3. The moment of inertia for the IBM-1 A first step towards calculating depend on the cranking frequency cranking
problem
is formulated
the moment of inertia is to find out how the a,, o. We crank around the x-axis. The self-consistent as
6(N;
aJHIBM- wl,lN;
(3.1)
a) = 0
where 1, is the x-component of the angular momentum operator and the expectation value of HIBM- w!, is minimised with respect to the a,. We use the states (2.1) as trial wave functions for the ground-state band. The term -01, comes about by making the transition from the laboratory to the intrinsic frame of the nucleus ‘); it incorporates the Coriolis and centrifugal forces in the intrinsic frame.
3.1. DEPENDENCE
OF THE
ENERGY
We use the most general
IBM-l
ON THE
INTRINSIC
hamiltonian
VARIABLES
1,2) with two boson
interactions:
H = E,S+S + &,d+ . d+~,(s+)*s*+u~~+d+~&++~[d+~d+s*+(s+)~d”~d] +~*[(d+d+)(~).
&+++d+
- (c?c?)‘~‘]+
C
C,(d+d+)“’
- (J&“’
(3.2)
1=0,2,4
with a - b = 1, a,b_,(-l)@. All terms conserve the total number of bosons. Using (2.8) and the g-band states (2.1) we can easily calculate the matrix elements of (3.2) and EN = (N; a[H,,,IN; a), the energy in the laboratory system, is
+ C4&[4(9a:+
10&a0a2+22a~+21u~)a~+9(a~+2a~)*]
11
(3.4)
with C=l+~uZ,=l+u;+2u:+2u:. P By setting function:
a, = 0 we can recover
the formula
EN = &(a,,
for EN in ref. ‘). We see that EN is a
a:, a*).
(3.5)
Eq. (3.5) will considerably simplify our calculations later in this section. The dependence on the absolute value of a, is explained by the time-reversal invariance of EN. It is related to the fact that the energy cannot depend on a sign change in the cranking frequency w (see appendix B).
6
H. Schaaser, D.M. Brink / Moment
3.2. CRANKING
When
of inertia
CALCULATION
w = 0, (3.1) reduces
to the variation
problem
@Iv; alH,,,IN;
a) = 0.
This is the static case and R,,( rr) and R,(n) 2). Their action on the operator B+ results appendix A). So we have
(3.6)
are also symmetry operations (see sect. in the additional relation a, = -a, (see
a,(w=O)=O.
(3.7)
In order to solve (3.1) we need an expression C. The matrix element of 1, is L,=(N;
a\/,jN;
a)=2Nu,[l+~
for 1,. This is found
u,uJ’(&,+2u,)
That Lx has a, as factor is explained in appendix The extremum condition for (3.1) is aEN --
au,
(3.8)
B. So we have L,.(w = 0) = 0.
aL, o-=0.
(3.9)
au,
When o is small we solve (3.9) by expanding the solution of the static variational problem
EN and Lx about (3.6):
a, = a:‘,
~,=a~“++~, EN = E$+
which is
(3.10a) ..,
c ~A,A,E~~~,+~ cI.y
L, = c A,L~;+ fi As EN has an extremum
in appendix
c $A,A,L!$+. P’.Y
(3.10b) ...
(3.1Oc)
at a,(‘) , there is no term linear in A, in (3.10b). The constant
term in the expansion of Lx vanishes at o = 0. Before inserting (3.10) into (3.9) we note that a Taylor expansion of A, in terms of o has no constant term: A, = WA,,, + &J~A,_, The term w C APL!&
is then of second C A,E$!$, P
+. . . .
order in o. So the expansion = coL$‘i +O(02)
.
of (3.9) is (3.11)
Taking into account (3.5) and (3.7) we see that terms like E$jF, vanish for all p it 1. of (3.7). So (3.11) simplifies Also terms like LEL vanish for (Y# 1 because considerably: AoE$!!,~oo+A,E~~20=0, AoE$$,2+A2E$;22=0, A 1E ‘;jl . 1= o Lp,’3 .
(3.12)
H. Schaaser,
D.M. Brink / Moment
of inertia
The determinant E N,OO
EN,,,
E N,O2
E~,22
('I because of the extremum does not vanish at a, two lines of (3.12) have only a trivial solution:
of EN at this location.
Ao,A2=O+O(w2),
(3.13a)
d,=a,=wL!,fi/E$!$,+O(~~).
(3.13b)
(3.13) shows how the values of a,,which are the solution of the variation (3.1), depend on the cranking frequency w. Going back to (3.10~) we easily find the moment of inertia I: L,=A,L~,‘+O(W~)=O(L~;)~/E~~~,+O(W~). Comparison
with the classical
formula
a,=pcosy,
problem
(3.14)
Lx = Iw +O(w2) yields (3.15)
r,=(L~1)2/E~!11+O(W). The equilibrium state is specified shape-defining variables by ‘)
So the first
by a, and
a, which
are related
to Bohr’s
a,=J&/3siny.
An extension of IBM-l in the static case (w = 0) [ref. 8)], which we here, can describe triaxial features. For the IBM hamiltonian (3.2), y = 0 in the equilibrium state provided p is allowed to take negative we have a,= p, a2= 0.Because we have a symmetrical rotor I = 1,
do not consider one can choose values “). Then = 1, is given by
1=6@(1+/32)/A+O(W)) A=(E/N)(~+~~)+(~-~/N)[(~+~~)A~-~A~],
A,=u2-2vo-~~hv2/?-~Co/32+~C4~2, A2= ~,+(~,+2v,)/3*-2$/~ The value of p, which minimises root of
v2P3+($CO+fC2+~Cq)P4.
(3.16)
(3.4) in the static case (a,= a2= 0),is given as the
O=~(I+~~)+(~-l){(~~+2v~-2u~)-3&~~, +(2W-U2-2vo)~2+J3v2~3}
(P + 0))
(3.17)
with
Eqs. (3.16) and (3.17) should be used only if v2 # 0 (or u2 # 0; see eqs. (3.18), (3.20)). This corresponds to a well-pronounced minimum in the energy surface of a deformed nucleus. For large N the l/N terms in eq. (3.10) can be neglected and the moment of inertia tends to a limiting value which is independent of N.
8
H. Schaaser, D.M. Blink / Moment of inertia
3.3. APPLICATIONS
If we omit terms which (3.2) can be expressed
contribute
only to the binding
energy,
the hamiltonian
l+azQ,.
T3+crbT4.
in the form lo)
H = E”q,+ uoPf . P+a,l-
Q3+a,T,-
T4
(3.18)
with QXr from (2.3): QXP = Q,(x For the operators
=
-#) .
in (3.18) we have used the definition
(3.18) and (3.2) are related
(3.19) of ref. lo). The hamiltonians
by
ES=h2,
E~=E”+~u,+~~~+~~~+~u~,
1 uo=zuo,
u2=2u2,
I u,=u,-~u~,
u2=-J?uT2,
Co=~uo-6u,+.&72--gcr3+~u~, C2=-3u,-~a,+$r,+~u,,
C,=4u,+&r,++ju~+$7,,
&=&d-us.
(3.20)
There are three subchains of groups for which the eigenvalues of H can be given analytically. They are named after the largest group in each chain: SU(3), O(6), SU(5). We are especially interested in the case in which H has a rotational spectrum. The SU(3) symmetry is an example of such a case. For SU(3) the eigenvalues of (3.18) are lo)
E([W(A, P)KLM) = ((+*-bduL+ where
L is the angular
momentum
l)+tu*C(h PI 9
and C(A, CL) is the quadratic
Casimir
(3.21) operator
of SU(3): C(A,~_L)=A~+~~+A~+~(A+/.L). We see that the moment
of inertia
has the same value for all different
(3.22) representations
(A, CL), namely z=
4 8u, -3u,
*
(3.23)
Solving (3.17) for N 9 1 gives p = 42. Inserting this value for p and the appropriate to (3.20) gives values for the parameters E,, .Q, uo, u2, uo, v2, Co, C,, C, according the moment of inertia (3.23). Naively one would expect that the moment of inertia should increase as the number of bosons increases, but this is not the case. We notice that formulae (3.16), (3.17) exhibit the usual phase ambiguity of SU(3), which allows A to have the values A = *+(J7). It is easy to prove that the sign choice to a prolate A = *IAl corresponds to v2= flu21. If u2<0 then A 0 gives an oblate shape ‘).
H. Schaaser, D.M. Brink / Moment of inertia
In order to improve Dukelsky
et al. “)
their results
obtained
used the random-phase
thus obtained become important (for example for O(6) symmetry)
9
with the intrinsic-state method.
only in some regions and for the isolation
formalism
The higher-order
I’),
corrections
away from SU(3) symmetry of spurious states ‘*). In the
case of exact SU(3) symmetry, our result (3.16) for the moment of inertia and that of ref. 12) both agree with the exact result (3.23) to the same degree of accuracy (in the leading order of l/N). We also note that for large angular momentum the IBM-l predicts triaxial nuclei. If we insert Eq. (3.8) gives a, as a function of LIP, a2 or of Z.3, y, respectively. Ui = ai(P, y) (i = 0, 1,2) we get a complicated energy surface in ~3, y. For L # 0 the minimum does not occur at y = 0”, nor at y = 60”. Triaxial shapes are then possible because a, is included in the generalised wave function (2.1), (2.2). More restricted wave functions used previously ‘) do not have stable triaxial minima in the energy surface. Investigations of this possibility are in progress. Earlier ‘) we studied an application of the formulae (3.16), (3.17) of our cranking formalism to a hamiltonian, which contains only QJ . Q3, 1. 1 and P6 = Pf . P. The operators Q3 . Q3 and 1. 1 can be expressed by the Casimir operators of the group chain SU(3) and the operator P6 belongs to the group chain O(6). For this special hamiltonian our formula (3.16) gives results nearly as accurate as those of the IBM-l computer program PHINT 13) and agree just as well as PHINT with the experimental data of ‘68Er. So for sufficiently large boson number N the rotational structure in the formula (3.21) can be understood by the picture of cranking an intrinsic state (3.16), (3.17).
4. The p- and y-bands One uses as normalised
intrinsic
states of the /3- and y-bands
l”*14) (4.1)
IyrJ=d:IN-l).
(4.2)
Together with the ground state IN) from (2.1) they form a set of three orthonormal states. We can use these states to calculate the excitation energy of the p- and y-bands above the O+ energy level of the g-band. We shall also show that the moments of inertia Zp, Z, of these bands are approximately equal to the moment of inertia of the g-band. As discussed in sect. 3 we can choose y = 0 in the equilibrium state and set a,=0
where /3 is the Bohr variable.
9
The operator
ao=P, B+ of IN - 1) then has the form
B+=s++Pdof+a,(d:+d’,).
(4.3)
10 4.1.
H. Schonser, D.M. Brink / Moment THE
of inertia
P-BAND
We look first at the P-band
ground-state
energy
EP,=(PP(H]!P&=(N-l]CHC+]N-l). With the commutation
(4.4)
relations [C, B+]=O,
[c,c+]=l,
(4.5)
we arrive easily at CMC+=[[C,M],C+]+C+[C,M]+MC+C+M, with M a given operator,
(4.6)
and
Ec=(N-l][[C,
H], C+]]N-l)+EN_r
(4.7)
with EN-, of (3.4). For the first term in (4.7) we use (1 +P’)KC,
HI, C’l= P2[b, HI, s+1 - PU[s, HI,
(h.c. = hermitian
conjugate).
A straightforward
d,+l + h.c.I+ [[do,HI, calculation
&I
(4.8)
gives
[[d,, H], d;] = Ed+ u,s+s+2u2(22~o~2~>(d,‘s+s+d,) +
;
1
v=-2
EpN=-
1 1+p2
&d+p2ES+
(4.9)
,
N-l 1+p2+2a: Uo-U2-4~o+2(fCo+~C2+~C4)}
X(U2-4U2&$+2p2{2
+4V2p3G+
C,4(22/~, v-~.JIv)*d:_,d,_,
1=0,2,4
U2p4+a:{2U2~2+4~2pJ~+$(~c,+3c,)})
1 +
EN-1 . (4.10)
The ground state of the p-band has L = 0. Hence a, = 0 and the excitation of the lowest-lying state of the p-band above the ground state EN is
1-p’ N-1 = el+p2+(l+p2)* +8$ with W of (3.17).
[u2-2uo-4zr2~~+4~2(uo-u2-3~o+
v2p3+(u2-2W)p4]
energy
w) (4.11)
11
H. Schaaser, D.M. Brink / Moment of inertia
To calculate
the moment
of (4.10) with respect
of inertia with (3.15) we need the second partial derivative
to a,: EC!:; = A+ ~Yfj),,,~ .
Inspection of this proportional to (N to (N - 1)2. So for (4.12). In order to element
(4.12)
formula with the help of (4.10) and (3.4) shows that A is - 1) only, while in E”’N 1,11there are terms which are proportional a sufficiently high number of bosons, A will not contribute in derive the moment of inertia we need to calculate the matrix
of 1, (for 1, see appendix
C)
L~=(~~lI~I~~)=(N-ll[[C,
(4.13)
LJ, C’]IN-l)+(N-lII,IN-1)
with the help of (4.6). A short calculation shows that [[C, 1,], C’] = 0. This in conjunction with (4.12) gives for the moment of inertia of the P-band; (4.14) This is approximately equal to the moment of inertia I of the g-band as the leading terms in (3.16) do not depend on N. As an illustration we study the band-head energy of the P-band for the case of SU(3) symmetry; only the leading terms in N are retained. We can therefore set p=J2: E,p,‘,‘A(SU(3)) = -6Na2. We compare this formula with the excitation using (3.21). The ground-state band belongs the p- and y-bands to (2N-4,2) [ref. “)I:
(4.15)
energy derived from group theory by to the representation (A, CL)= (2N, 0),
E~~,“~(SU(3))=E([N](2N-4,2)000)-E([N](2N,0)000) = -6N(lObviously
1/(2N))a,.
(4.15) and (4.16) coincide
(4.16)
in their leading
terms in N.
4.2. THE y-BAND
A similar
calculation
for the y-band
using
HI, GIIN-~)+EN-~
EY,=(~~IHI~~)=(N-~I[[~z, =&d+
N-1 1 +p2+2af
+$~:[6C2+ The matrix
element
(4.5), (4.6) and (4.9) gives
{u2+4u2pJ3+~p2[4c2+3c4]
Cal}+ EN_,.
(4.17)
of 1, is LZ=(~~II,IIYy)=(N-llIxlN-l)
because
[[d2, &I, d:] = 0. Similar
arguments
IY = (L;,yy/E~,y)1=
as for the P-band 10 f I,
result in (4.18)
12
H. Schaaser, D.M. Brink / Moment of inertia
which gives the general result that for large boson numbers the moments of inertia for the g-, p- and y-bands are the same. The excitation energy of the y-band ground state is
1 =-+ 1+p2
N-l (1+p’)’
[u,-~u,+~u,&-[~~+~V~-$C~--~C‘&?~
+8~~&?‘-2~~[4C,,-~C~+&C~]]+(3/1)~
(4.19)
with L = 2. In the SU(3) limit (4.19) gives E,Y,,,(SU(3)) Analogously
to (4.16) the group-theoretical
- -6N~z.
(4.20)
result with L = 2 is
E~~,,(SU(3))=-%a,+E~~,~(SU(3))=-6N(I-l/8N)~2, which shows equality
with (4.20) in the leading
terms in N.
For comparison of the formulae (4.11), (4.18), (4.19) with the program and experiment, we refer to ref. ‘).
PHINT
5. The moment of inertia in the IBM-Z In the IBM-2 one distinguishes between protons and neutrons. We study a particular IBM-2 hamiltonian which contains all terms which have been used in the comparisons with experiments. In these comparisons the program NPBOS 16) has mostly been used ‘*2). In sect. 6 we shall compare our results with NPBOS and experiment, and also with the analytical formulae for the limiting cases of SU(3) and SU*(3) symmetry “). We also study the transition between the limiting cases SU(3) and SU*(3). Iachello ‘) discussed the different group chain reductions for the IBM-2. In IBM-l in the static case, the intrinsic states are always axially symmetric. In IBM-2 they may be axially symmetric, but not in all cases. For example, the limit of SU*(3) symmetry “), which does not occur in IBM-l, describes a triaxial rotor I*). In the first part of this section we study a general intrinsic state for an IBM-2 hamiltonian. We include in particular that (i) the intrinsic state for each sort of nucleon can be triaxial, (ii) the principal-axis systems of the neutron and the proton distributions may be different. We shall derive formulae for the moments of inertia about the x, y, z-axes of the principal-axis system. In general all three moments of inertia are needed for the calculation of spectra I’). 5.1. THE HAMILTONIAN,
Our choice
INTRINSIC
for the hamiltonian
STATES
AND
MATRIX
ELEMENTS
is
H ,e~=%G+~,~,+‘Grv~,’
Qv+ v,r,+ vw+M.
(5.1)
H. Schaaser, D. M. Brink / Moment of inertia
13
We reserve the index 7r (v) for the proton (neulron) degree of freedom, and p, p’ for +rror V. nr and n, are the number operators for the d-bosons. Analogously to (2.3) we have for the quadrupole
operator
Q,,, = G+&,p+ d;,s, +xp[d;L?pl:’ . For the interaction
between
like nucleons
(5.2)
we take
V,, = ‘$,Q,. Q, . The Majorana
operator
(5.3)
is
M = -2
C
&(d:df,)‘k’
. (&,(?,)(k)
k=1,3
(&s,
+&(d;sf,-s;d+,)‘2’.
-s~&)‘~‘.
(5.4)
The notation for these operators is the same as in ref. lo). We proceed in a similar way to our method for IBM-l. Our wave function IN,N,)=[N,!N,!(C,)N-(CY)N”]-“2(B~)Nn(B~)N”IO),
is (5.5)
with
c, = 1+c 4,,a,, II
B; = s;+C
up,+dp++. I”
(5.5)
The variables up,@ are taken in the principal-axis system of the nucleons of sort p. From sect. 2 it is clear that the quadrupole tensors of the neutrons and protons are diagonal
if the following
generalisations a P.P = qr,.
of (2.8) hold: %,P = CA .
First we study the static case (w = 0). In this case we have in addition (see appendix
(5.6) to (5.6)
A) a, = (-l)“a,.
So there are four independent real variables up,+ (p = r or v and t.~= 0,2) and three system of the protons with Euler angles LY,,(Ye, a3 which connect the principal-axis that of the neutrons:
where D is a rotation matrix. If both nucleon distributions have axial symmetry only one Euler angle (Ye is necessary to account for the relative angle between the axes of symmetry. For our hamiltonian the equilibrium value of CY~can only be a2 = 0” or cz2= 90”. If we allow the proton and neutron mass distributions each to be triaxially deformed, two more
14
H. Schaaser, D.M. Brink / Moment of inertia
Euler angles
(Y,, (Yeare required.
six combinations
Analogously
we have an equilibrium
only for the
of Euler angles: (ff,, a29(Yj) = (0°, O”,00), (90”, O”, 00), (0°, 90”, 0°) ) (90”, 90”, 00), (0°, 90”, 90”), (90”, 90”, 90°) .
The combinations
(O’, O”, 90’) and (90”, 0”, 90’) are redundant.
In practice
this means
that the principal-axis systems of protons and neutrons are the same, but may differ in the labels x, y, z of the axes. Minimising E for a, = 0 provides us with values for the uP and (Yi in the static case. We write E explicitly for w = 0: Q(v)+K,,Q~(T)+K,,Q~(v)
E”‘=E,~(~)+E,~(~)+K,,Q(~).
(5.7)
with
The other terms are n(p)=(n,)‘O’=~(a~,o+2aZ,,), P
Q(pO)=(~~)bO)=~[2a,,o+x~(2a~,2-a~,o)~l, P
QW)
2% Qp,2[1+XpQp,OJ?l 3 P
= (Q&A” = 7
02b)=(Qp-
0,)‘“‘=5N,+~[(x2,-4)CyKp P
+(N,-1)[4&+44& with K, = at,,+ 2az,,. The matrix case
(,)(O)+!
~~a,o(6& element
- a;,~) +fx’,K’,ll
of the Majorana
(253[%,2%,0-
5-rY
+ &[%,o - %,J
operator
(5.9a) is in the static
G,o%,212
+ 252[%,2- %,21’1.
(5.9b)
We note that (M)(O) does not depend on the parameter 5,. [The second derivative of(M) with respect to a,, or uv,,, however, depends on all three parameters 6 (see eq. (5.18)).]
15
H. Schaaser, D.M. Brink / Moment of inertia 5.2. CRANKING
FORMALISM
OF THE
IBM-2
We discuss the dynamic case (w # 0) now. In this case it is easier to evaluate all matrix elements in. the coordinate system of one type of nucleon. We choose arbitrarily the coordinate system of the protons to be the standard system. We use this approach because the relations (5.6) are not necessarily the same in a rotated coordinate system. If, for example, we assume that the proton coordinate system and the neutron system are related by a rotation around the y-axis about CY~ = 90”, then in the rotated system the equivalent equations to (5.6) are 4.m = a:,, .
G,m = a,,(-1)“)
In this rotated coordinate system in the dynamic case, a,, and a,_, vanish and u”,-~ are real and independent variables. In general in each of the rotated a v.2, %,o. coordinate systems we may have relations between the up,+, which are different from (5.6). We find that it is much easier to discuss the dynamical case once and for all in a standard coordinate system, for which we have chosen the proton coordinate system. From appendix A and (5.6) it is easy to see that in this standard system the expectation value of the energy in the laboratory system is a function of six real variables a,, :
=
Ha,,,,
G.2,
a,,,
a”,29
(%,,a”,,),
a:,,,
&,,I .
(5.10)
Analogously to (3.5), E depends only on terms which are homogeneous functions of second degree in up,,. This particular dependence on ap,, will help to keep the system of equations (5.13) simple. Similar to the variable a, of sect. 3, the up.1 are related system a sign The
to the cranking frequency o. As we have chosen the same standard coordinate for the proton and neutron variables, a,, and av,, will both be affected by change of w. So the energy does not depend on the sign of w. expectation value of the angular momentum, LX, is equal to
Lx = Lx,, + LX.“, So the obvious
generalisation aPIW
=
E =
with Lx,p = Wap,,(CJ1~J~
(5.11)
of (3.10) is
42 +A,,
(5.12a)
3
+..., ,,a’E,‘,“’
E(O)+1 c i~~,,ft,.,, P.P,
Lx=
aP,of2@P,2).
id
I3L(O) C Ap,+~+-.. P,W PIP
The variation problem 6( N,NVIHrBM - ol,(N,N,) equations in A,, and w, if we omit terms in 0(02).
P,cI
(5.12b)
P’A
(5.12~) = 0 gives us a system of linear We take into account eqs. (5.10)
16
H. Schaaser, D.M. Brink / Moment of inertia
and (5.11). Then we have a system which consists to (3.12):
C p.i
of six equations
AmoE$‘J,pi + Am,E’,$,pi + AvoEZ,pi + Av2Etfpi =0 , 4,Ef?,p,
+
4,E!?,p,
=
p=r,
wL$‘;~
v;
of a type similar
i=O,2,
(5.13a) (5.13b)
.
E pk,olmis the second derivative of E with respect to ap,k and am,,,. Because we evaluate the derivatives at w = 0 and the energy has its minimum there, the determinant of the coefficients Es,;, of the first four equations (5.13a) does not vanish and the solutions are Ap,o=A,,z=O+O(02). The same applies
to the determinant L(o) A,,
Inserting
=
w
E’O’_
of the coefficients _ L’o’
E;;; p’*p’ E(o) Tr1,Tr1 v,,v,
X.Pl
-
nl,vl
(P # PI >
b%,vJ*
(5.13) into (5.12~) gives us the moment
It is instructive
to discuss
in (5.13b). The solutions
are
E(o)
It can easily be shown that the mixed derivatives
similarity
(5.13a)
(5.13b)
of inertia:
of E come from the Majorana
term:
(5.13b) for A4 = 0; then (5.13b) and (5.14) have a strong
to (3.13b) and (3.15): A,, =
wL$‘,,,lE$‘,p,+O(m*) , (5.15)
L = L/w =C [L!.t,hlz/E~~,p,, P
with
LAY:, =2Np(Cp)~1[ap,oJ~+2ap,2], E (0) p,,p,
=
“p”%‘)+
c k=0.2
(5.16)
2k’2KppQ($+?“(pk)
+Kpp[Q- Ql”(P)+(@?,p,
(P#P),
(5.17)
In (5.17) and (5.18) the double prime denotes the second derivatives of n(p), Q(pk), Q * Q(p) with respect to ap,l taken at w = 0. The terms occurring in (5.17) are listed
17
H. Schaaser, D.M. Brink / Moment of inertia
below:
n”(p) =
3,
(5.18a)
P
(5.18b)
(5.18~)
(5.18d)
+
&mb,* - 4J1+5Jx
+ &MQ,.&
%,& +~,&I’
q&1’+
For K,, see eq. (5.9). In order to complete
+
253
2 [
l-I
p=lr,v
h%,&
&I + &I *
(5.18e)
eq. (5.13b) and (5.14) we also note
~poJ%+a,2~v~ ,
,
,
11 +&
,
(5.18f)
where ~ denotes the product symbol. Eq. (5.15) for the moment of inertia is similar to eq. (3.15) for the moment of inertia of the IBM-l. It is a sum of two terms p = r and p = V, corresponding to neutrons and protons, but there is a coupling and each of these terms depends on the neutron and proton interaction. It is not surprising that such coupling terms exist in the IBM-2 because the IBM is not a geometrical model and our usual intuition, which is based on geometrical pictures, does not apply. We also have to determine what the moments of inertia around the y- and z-axes are. We have already studied the case of cranking around the x-axis in the standard system. Subsequently we crank around the y- or z-axis by rotating the wave function (5.5) so that the old x-axis is in place of the axis under study. This results in the formulae 1, = mp,o, -$,J 9 1, = L(%,O, %,Z) , 1, = 4 (- ~%I + %,*4 , tqLl,, + a,&)
,
p=~
or V.
According to the triaxial rotor model of Davidov and Filippov 18), one has to use all three moments of inertia for the calculation of the spectra. The CZ~,~are related
18
H. Schaaser, D.M. Brink / Moment of inertia
to the Bohr variables
by up.0 = P, cos 3$,
6. Applications We study the limiting
of the IBM-2
cranking
cases of SU(3) and SU*(3) H
These symmetries
ap,2 = &&sin
IBM=K3(Q~+C~)
are realised
yP.
(5.19)
formalism
for
’ (&+QY)*
with the following
(6.1)
sets of parameters
17) (parameters
in (5.1) to (5.4) not listed here are set equal to zero): K,,
=
SU(3): su*(3):
Kw
=
K3,
K ,ru = 2K3
Xw=&=T;J7
3
x,=-xy=r;J5.
(6.2)
We see in eq. (6.2) that the distinction between the SU(3) and SU*(3) limits comes from (x,/xy) = 1 or -1, while in both limits x,, can take both values. This possibility of choosing the sign of xP in IBM-2 is analogous to the possibility of choosing the sign of x in IBM-l (see sect. 3). For SU(3) and SU*(3) the energy spectrum of the hamiltonian (6.1) is given by eqs. (3.21), (3.22) with (+, = 0, 02= K~. The allowed values for (A, p) are given in ref. I’). Thus the formula for the moment of inertia (3.23) is valid for IBM-2 in the limiting cases of SU(3) and SU*(3) if (+, = 0, o2 = K~.
6.1. THE W(3)
SYMMETRY
First we study the symmetry SU(3). In the limit of N,, minimising E = (HtBM) in the static case are “) SU(3):
%=[;;c,]
3
Yv=[;;~],
&=py=d2,
N, > 1, the results
C=O”
of
(6.3)
The upper line& yP corresponds to x,, = xy = -$/?, while the lower line is assigned to x,, = xy = 4d7. G is the angle between the symmetry axis of the neutrons and the protons. For large N, the moments of inertia 1, and 1, are equal and have the value I = -4/3K3, which is the value of I in eq. (3.23). 1, vanishes. If we study (6.1) for more physical values of N,, NV than in the limit of large N,,, i.e. O< N,, NV < 10, we notice that in the SU(3) limit the variables differ slightly from the values mentioned in eq. (6.3). We have ym = yy = O”, & = &, < 42 with Z, = 1, and 1, =O. This is very similar to IBM-l, where eq. (3.17) also gives p
H. Schaaser, D.M. Brink / Moment of inertia 6.2. THE SU*(3)
SYMMETRY
In the case of SU*(3) symmetry we also study first the limit results of minimising E = (H,,,) in the static case are 17) su*(3):
19
YT=(;;~),
y”=(T),
p,=py=&
N,,, N, 4 1. The
(Y=90°,
(6.4)
It is easy to show that in both cases for (‘y,, yy) we get the moment of inertia of (3.23): Z, = Z, = Z, = -4/3~~. We see that for SU*(3) the nucleus is triaxial, because Cu= 90” and &, py # 0 [ref. “)I. We can also easily understand that we get an SU*(3) spectrum if all moments of inertia are equal: In the case of Z, = Z, = Z we have a formula
for the rotational
energy
E=w+l) -+
(+-& >
K2.
21
In this case the usual combinations
z
of K and L are possible:
K=O,
L = 0,2,4,
...)
K=2,
L=2,3,4,5
)...)
K=4,
L = 4,5,6,7,
...
...,
... 3
If Z = Z, = Z, = Z, then the energy states with different K but same L are degenerate. This is the SU*(3) spectrum 17). Because in the limit of large N,, N, all three moments of inertia are equal we have a spherical top even though our nucleus is triaxial. A rotation Z = -4/(3Kj).
around
As a more physical
any axis through
example
the centre
of the nucleus
for (6.1) with SJJ”(3) symmetry
would
have
we study the case
N,, = 10, N, = 2, K~= -1 MeV, xrr = -xy = -iJ7. We get the results & = 1.36, py = 1.16, ym = O.l”, yv = 53X’, leading to the moments of inertia (in MeV-‘) Z, = 1.39, Z, = 1.40, Z, = 1.59. This shows that for SU*(3) the neutron and proton distributions, or both, may be triaxial, which was not noted in ref. r7). This tendency increases for both neutron and proton distributions if (N,, + N,) is small. Our example shows that if only one of these N, decreases, then only the distribution of nucleons of type p will tend to become triaxial.
6.3. TRANSITION
BETWEEN
SU(3)
AND
SU*(3)
We study the hamiltonian (6.1). One possible way to move continuously from SU(3)toSU*(3)istovaryx,andtokeepx,fixedtothevaluex,=-~J?(seeeq. (6.2)). We shall find three modes: an SU(3) mode, an SU*(3) mode and a mode of large xy. In the SU(3) mode the distributions of protons and neutrons are each axially
20
symmetric distribution
H. Schaaser, D.M. Brink / Moment of inertia
and have the same symmetry axis. In the other two modes the proton remains almost axially symmetric around the z-axis. In the SU*(3)
mode the neutron to the proton
distribution
symmetry
tends to be axially symmetric
axis. For large xy we have a separate
about an axis orthogonal mode. In this mode the
neutron distribution has yy = 30” and & is large. The large values of &, do not necessarily pose problems in the interpretation of this mode, because by using the collective model one can show that in the case of IBM-l some large p correspond only to small physical deformation strengths ‘) and probably this also holds for the IBM-2 In the mode with large xy the neutron mass distribution is a triaxial ellipsoid, both thin and elongated. There is no longer any axial symmetry and the longest axis is orthogonal to the proton symmetry axis. It is relatively complicated to discuss the Euler angles ((Ye, Q*, (Ye) in conjunction with yV, yy. For example if yy = 60” the neutron distribution is axially symmetric with respect to the y-axis of the neutron coordinate system ‘O). So in order to find out about the relative orientation of the proton and neutron mass distributions one always has to look at (a,, CQ, CX~)and y,, yy as well. It is much easier to study the asymmetry variable 7, which is defined as 20)
(6.5) This relation is valid only for small xP and small up+. One easily finds that by reducing (6.5) to the case of one type of nucleon only and inserting (5.9a) and (5.19) into (6.5). 7 corresponds to the variable y of IBM-l (see sect. 3), where no distinction is made between protons and neutrons. 6.3.1 Symmetrical system: N, = N,, = 3. We see in fig. 2 that p,, is almost constant, while By varies substantially.
In the SU(3) limit (xy = -id?)
and in the SU*(3) limit
(xy = ;JJ?), the deformation strength variables &, /3” have the same value. For xy % 1, py becomes large. We note that the region of largest curvature in the curve for & is near to xy = 0.6 which indicates that this point is somehow special. We see in fig. 3 that for xy s 0.64 we have yT = xy = 0. For xy > 0.64 yT and y,, diverge suddenly: y,, remains always near zero while y,, increases to yV = 60” at xy = 1.4. For xy > 1.4, yy decreases and approaches yy = 30” asymptotically, while y_ approaches zero. In fig. 4 we see a sudden rise of 7 at xy = 0.64. At the SU*(3) value of xy we have 7 = 30” exactly 17), and this holds for all nuclei with N,, = NV. At 7 = 30” the nucleus has its largest triaxial asymmetry 20). For xy > 2.5, formula (6.5) can no longer be applied. A sudden transition also occurs in the moments of inertia (fig. 5). For ,Y”< 0.64 the neutron and proton bosons both have axial symmetry about the same axis. At xy =0.64 there is a bifurcation in Z, and Z,, and Z, increases steeply. At the SU*(3) value xy = 48 the moments of inertia Z,. Z, are exactly equal and Z, is slightly smaller. For this value of xy the inertia tensor is axially symmetric about the y-axis. The
H. Schaaser,
D.M.
Brink / Moment
of inertia
21
1.8 1.6 -
NV=3 0.6-
I I sum I
0.4 -
I I I I I
I
I
I 1
1
0.6
1
I
0.8
1.0
1.2
I
1.4
1.6
1.8
X,
Fig. 2. Transition from SU(3) to S_U*(3). We plot the deformation variables p,, and p, as functions x, and keep x,, fixed to x, = -g7 = 1.32. x, and x, are parameters in the quadrupole operators and Q,, respectively. H = -(Q, + 0,) (0, + 0,) (in units of MeV).
of Q,
Y [degrees] 1
SU’(3) I
60
I
0.7 Fig. 3. Transition
0.8 from
0.9
SU(3)
1.0
1.1
1.2
1.3
N,=3
1.4
to SU*(3). We plot the deformation and x, are as in fig. 2.
1.5
16
variables
1.7
1.8
X,
‘y, and y, against
x,. H
22
H. Schoaser, D.M. Brink / Moment of inertia [degreesI''
I
\
Nn:3,
\
NV=3
IO-
5-
0.5
I
1.0
; SP(3) I II 1.5
I
I
20
2.5
\ \ \ \ \ \ \ \ \ \ \ \
\ I \ I IL
I
3.0
3.5
x,
Fig. 4. The asymmetry variable 7 plotted against the parameter x,,. H and x,, are as in fig. 2. 7 takes into account the deformation of both protons and neutrons. For x, > 2.5,eq. (6.5)for 7 can no longer be applied, which is denoted by a dashed line.
1.6 1.4
0.6
L_Ll ISU'(3)
,-,, V.-t
I
0.2
I
I
0.6
0.8
1.0
1.2
I
I
1.4
1.6
1.8
X"
Fig. 5. The moments of inertia in the transition from SU(3) and SU*(3) symmetry. H and x,, are as in fig. 2. At x, = 0.64 there is a bifurcation in Z, and ZYand there is a steep rise in Z,.
H. Schaaser, D.M. Brink / Moment of inerlia
Fig. 6. The analogue as function
23
of fig. 2 for an asymmetrical system (N, f N,). Deformation variables p,, and p” of x, in the transition from SU(3) to SU*(3). H and x, are as in fig. 2.
transition at xy = 0.64 is something like a phase transition between the SU(3) mode and the SU*(3) mode. 6.3.2. Asymmetric system N, = 4, N, = 2. This case has the same total number of bosons as in subsect. 6.3.1. Figs. 6-9 showing the deformation variables and moments of inertia as a function of xy are similar to those for the case N,, = N, = 3. In the present case there are also two phase transitions which happen at xY--.0.99 1.4 which is slightly larger than and at xy - 2.19. The pV and p,, curves cross at xy = the SU”(3) value. For xy > 1.4, yy approaches the values yy = 30” asymptotically. In fig. 9 the bifurcation between I, and ZYis less pronounced than in the symmetrical system N, = N, = 3 discussed before. At the SU*(3) value of xy all three moments of inertia are approximately equal, which is a sufficient condition to have an SU*(3)-like spectrum.
6.4. COMPARISON
WITH
NPBOS
AND
EXPERIMENT
In fig. 10 we show the ground-band energy levels of 16’Yh calculated with the cranking formalism (CF) and NPBOS and compared to experiment. We have taken the NPBOS parameters of ref. 19) (see fig. 10.) The corresponding hamiltonian is
H. Schaaser, D.M. Brink / Moment of inertia
24
Y
[degrees]
su731
4
I
60 50 -
Fig. 7. Deformation variables y, and y, for N, # N,. In analogy to fig. 3 we perform the transition from SU(3) to SU*(3) by varying x,. H and x,, are as in fig. 2. We note that the deviation from y, =60” at the SU*(3) value of x,, is stronger for the asymmetrical system than for the symmetrical system (N,, = N,). For x, + co the variable y, approaches y, = 30” as limit.
[degreeslAY 30-
25NTT=4,Nv=2 20 -
15-
IO-
5-
II 1.0
0.8 Fig. 8. The asymmetry
variable
7 plotted
transitional between SU(5) the cranking formalism is we see from fig. 10 this is in the ground-state band simple formula.
I
I1
1.2
against
1.4
L
I
I
1.6
1.8
2.0
2.2
x,
x,. H and x, are as in fig. 6. For x. < 0.99, 7 vanishes.
and SU(3), so there is no triaxiality. Fig. 10 shows that less accurate compared to experiment than NPBOS. As due to the deviation from constant angular momentum in 16”Yb, which is included in NPBOS, but not in our
H. Schaaser, D.M. Brink / Moment of inertia
25
system (IV,, # N,). Fig. 9. Moments of inertia in the transition from SU(3) to SU*(3) for an asymmetrical H and x,, are as in fig. 2. There is a bifurcation in I, and I, and a steep rise in I, at xv = 0.99, which is similar to fig. 4.
7. Summary In this paper we have studied the cranking of intrinsic states for IBM-l and IBM-2. In the first part on IBM-l we described in detail the method used in ref. ‘). It was useful to use symmetry arguments to reduce the number of intrinsic variables to three or less. We give a formula for the moment of inertia when the cranking angular velocity is small. We also extended the formalism to discuss the excitation energies and moments of inertia of the /3- and y-bands. In the second part on IBM-2 we also gave a general formula for the moment of inertia. We considered the SU(3) and SU*(3) limits and studied the transition between them. We found an interesting phase-transition-like behaviour. One moment of inertia (1,) is zero up to the transition point and then rises steeply, while the other two show a bifurcation at the transition point. A comparison with experiment lying between the SU(3) and was made for the nucleus 16’Yb using parameters SU(5) limits. The results presented in this paper hold if the cranking angular velocity w is small. It would be interesting to extend them to larger values of w.
26
H. Schaaser, D.M. Brink / Moment of inertia E [M&l A
,/ CF 1'
3.0-
2.5 -
2.0-
Fig. 10. Comparison between the cranking formalism (CF), NPBOS and experiment for the energy levels of the g-band of $FYb,,. The parameters of the hamiltonian are E, = E, = 0.67 MeV, K,, = -0.12 MeV, x, = 0.49, x, = -1.2, N,, = 6, N, = 4. This is a hamiltonian transitional between the dynamical symmetries SU(5) and SU(3). It gives an axially-symmetric nucleus. The values of the geometric variables found by our intrinsic-state formalism are p,, = 0.51, p, = 0.75, y, = y, = 0, 6 = 0”. 1, = I, = 16.78 MeV-‘, 1; = 0.
The authors thank M. Zirnbauer, P. Halse, J. Lomnitz-Adler, P. Van Isacker, A. Frank and J. Dukelsky for valuable discussions. One of us (H.S.) would like to thank the Studienstiftung des Deutschen Volkes, Dr. Carl Duisberg - Stiftung, and Deutscher Akademischer Austauschdienst for financial support.
Appendix A DIAGONALISATION
OF (0,)
(0,) is diagonalised by states IN; a) which are solutions of (2.4). Initially taking the transformations A to be those of S, we notice that transformations containing T change complex numbers into their complex conjugates *“):
Ta,T-’
= a*W’
The a, do not transform under the rotations of S because S leaves the coordinate frame unaffected (principal-axis system). As NN of (2.2) is a real number it is left unaffected by all A and we can write (2.5) (AB+A-‘)NIO)=
(B+)NIO).
(‘4.1)
H. Schaaser, D.M. Brink / Moment of inertia
This equation
27
is solved by B+, which fulfills AB+A-’
= eiPB+ ,
(A.2)
where (Y is an appropriate real number. The effect of the transformations states with good angular momentum is well known *‘). s+ is invariant these transformations. So we can set a equal to zero in (A.2):
of S on under all
AB+A-'=B+.
(A.3)
We give a table of how dz transforms and which relations from (A.3). Using the conventional phasing *‘)
between
the a,, result
and the property that any of the Ri(r)of S can be expressed as a product of the two others, one can derive the entire table from (for example) R,(r) and R,(r):
Rx(v) d’,
R,(n) (-)*d+, (-Y”a-,
a-,
R,(r)
T%(n)
TR,(r)
T&(r)
T
(-Yd;
(-)pd;
d;
d’,
( -)pd’,
(-Y”a,
(-)%I;
a:
a?,
(-)‘“a!,
The parity P is an identity transformation with respect to B+[ref. ‘“)I. From the table we note that the same is valid for T*. The matrix elements of Q, from (2.3) with the states (2.1) are
(Q,>=
1 +I
N a;a,
{(-)‘a_,+a~+~(a*a”)~‘}.
(2.8) has been used to derive this expression.
(A-4)
It is easy to see that
(Q,) = (-)“(o-,)*.
(A.5)
(9,) and (Qz) are, explicitly, (QJ=
N l+Ca:a,
(Q*)=
{aT-a_,+~x[JS(a_,a*,-ait;a,)+~(ata_,-a,aT)]>,
N
1 -t-c a;a,
{az*+a_,+&[h(
We prove first that the geometrical (0,). From the table it follows that
R,(n)+ ~~=a-,,
(A.6)
a$a_,+a$a,)+&
symmetries
aTa_,]}.
(A.7)
R,(n),R,(n),R,(r) diagonalise
R,(r)+ a, = (-)fia_,
.
H. Schaaser,
28
D.M. Brink / Moment
of inertia
This implies a, = -a, = 0 and a2 = a_z. R,(r) is redundant. Inserting a, = 0 into (A.6) gives (Q,) = (Q-J = 0 and with the help of (AS) we see that ( Q2) = (Q-2). As {R,(n), (0,)
R,,(n),
R,(n),
E} is a subgroup
is also diagonalised
of S, S also diagonalises
(0,).
by S,:
R,(r)
TR,(r)+a,=a;.
+ a, = a_, ,
(TR,( T) is redundant.) These relations between reduces to a real function in the a,. Then (AS)
the a, make gives
(9,) vanish
and
(QA
(‘4.8)
(QJ=(Q-J=o, (Qd=(Q-2). Appendix B
DEPENDENCE
OF E,
As T commutes
AND
L, ON THE
VARIABLE
a,
with HiBM, we have al&&V;
EN =(N;
We note that IN; a) is symmetric
a)=(N; under
alT-‘H,,,TIN;
R,(T)
a).
we have
TIN; a) = R,(r)lN;
a) = IN; a,, a_,, a*).
Inserting this into (B.l) we see that EN can only depend on [a,[. We now study the matrix element LX8of the angular momentum angular
momentum
changes
its direction
under
LX,= (N; alT-‘l,TIN; IN; a) is invariant
under
L,,=(N;
Comparison
operator
I,. The
time reversal:
a) = -(N;
all,lN;
a).
03.2)
S,:
aj(R,(n)T-‘)T-‘l,T(R,(r)T)jN;
=(N;
(B-1)
S, so that
TIN; a)=R,(vr)[TR,(r)]TIN; As R,(n)R,,(r)=
a).
alR,(4lxRx(~)lN
a)
a)=(N;
ao, -a,, a,llxk
with (B.2) shows that Lx is proportional
a,, -a,,
az).
to al.
Appendix C CONSTRUCTION
OF ANGULAR
MOMENTUM
OPERATORS
and then determine the A,, by demanding which are the usual angular momentum [I+, d:l=C,, (6-~b++1))-~'*d;+,, formulae. By taking the complex conjugate of 1, we find I_. For 1, = $(I++ I_) the result is We take the ansatz
1, = C,,, A,,dLd,
lx=;
i a=-*
(6-(~(~~+1))“*{d~+,d~+d~d,+~}.
H. Schaaser, D.M. Brink 1 Moment of inertia
When
going
interchanged
from the laboratory
to the intrinsic
29
frame the roles of 1, and I- are
‘$. But as 1, is the sum of 1, and I- it is not affected by this transition.
For 1, we find 1, =&6(d+d)l”.
References 1) A. Arima and F. Iachello, Phys. Rev. Lett. 35 (1975) 1069; F. Iachello, Prog. Part. Nucl. Phys. 9 (1983), and references therein; R.F. Casten and D.D. Warner, ibid., and references therein 2) A. Arima and F. Iachello, in Advances in Nuclear Physics, vol. 13, ed. J.W. Negele, (Plenum, NY, 1984), and references therein 3) H. Schaaser and D.M. Brink, Phys. Lett. 143B (1984) 269 4) A.E.L. Dieperink, Prog. Part. Nucl. Phys. 9 (1983) 121 5) J.N. Ginocchio and M.W. Kirson, Nucl. Phys. A350 (1980) 31; Phys. Rev. Lett. 44 (1980) 1744 6) P. Van Isacker, A. Frank and J. Dukelsky, Phys. Rev. 31C (1985) 671; 0. Castanos et al., Phys. Rev. 24C (1981) 1367 7) P. Ring and P. Schtick, The nuclear many-body problem (Springer, NY, 1980), pp. 128-130 8) R.F. Casten, A. Aprahamian and D.D. Warner, Phys. Rev. 29C (1984) 356 9) J.M. Eisenberg and W. Greiner, Nuclear theory, vol. 1, 2nd ed. (North-Holland, Amsterdam, 1975) pp. 107-115 10) F. Iachello, in Lecture Notes in Physics, ed. J.S. Dehesa, vol. 161, (Springer, Berlin, 1982); F. [achello, in Nuclear Structure, ed. K. Abrahams et al. (Plenum, NY, 1981) 11) J. Dukelsky et al., Phys. Lett. 130B (1983) 123 12) J. Dukelsky et al, Nucl. Phys. A425 (1984) 93; Phys. Lett. 129B (1983) 1 13) 0. Scholten, program PHINT, KVI Groningen, Holland (1980) 14) A.E.L. Dieperink and 0. Scholten, Nucl. Phys. A346 (1980) 125; R. Bijker and A.E.L. Dieperink, Phys. Rev. C26 (1982) 2688 15) A. Arima and F. Iachello, Ann. of Phys. 111 (1978) 201 16) T. Otsuka, program NPBOS, University of Tokyo, Japan (1977) 17) A.E.L. Dieperink and R. Bijker, Phys. Lett. 116B (1982) 77; A.E.L. Dieperink, Nucl. Phys. A421 (1984) 189 18) AS. Davidov and B.F. Filippov, Nucl. Phys. 8 (1958) 237; J.P. Davidson, Rev. Mod. Phys. 37 (1965) 105 19) C. Garrett et al., Phys. Lett. 118B (1982) 292 20) A. Bohr and B.R. Mottelson, Nuclear structure, ~01s. 1 and II (Benjamin, NY, 1969 and 1975)
THE MOMENT OF INERTIA IN THE INTERACTING BOSON MODEL H. SCHAASER Department
and
D.M.
BRINK
of Theoretical Physics, 1 Keble Road, Oxford OXI 3NP, UK Received
27 August
1985
We perform a self-consistent cranking calculation of intrinsic states of IBM-l and IBM-2 for three reasons: (a) to study the effect of Coriolis and centrifugal forces in the IBM; (b) to derive approximate analytic formulae for the energy in some cases where the IBM hamiltonian does not have dynamical symmetry; (c) to study the dependence of the moments of inertia on the variables of the intrinsic IBM states. Choosing a coordinate system where the quadrupole tensor is diagonal reduces the number of variables considerably. Among other applications of our formalism we study the transition between the symmetries SU(3) and SU*(3) in the IBM-2 hamiltonian H = -(Q, + Q,) (Q,, + Q.). We find that the moments of inertia and the asymmetry variables show a kind of phase transition, as a function of the parameter x, occurring in Q..
Abstract.
1. Introduction During the past years the interacting boson model (IBM) 1*2) has had a lot of success in providing a phenomenological description of the spectroscopic data of nuclei. In this model the valence nucleons of even nuclei are described as a collection of bosons of angular momentum L= 0 and L = 2. Two versions of the IBM are popular: the IBM-l which deals only with one sort of nucleon, and the IBM-2 which takes account of both neutrons and protons. In this paper we study how Coriolis and centrifugal effects arise in the IBM-l and IBM-2. We have outlined our procedure for IBM-1 in ref. ‘). The energy expectation values and the electromagnetic transition probabilities are influenced in an important way by Coriolis effects. We make a self-consistent cranking calculation of the intrinsic IBM wave function to study the inclusion of Coriolis and centrifugal effects. We also investigate the dependence of the moment of inertia of the nucleus on the variables of the wave function which describe the deformation of the nuclear shape. Except in the three limiting cases of dynamical symmetry SU(5), SU(3), O(6) (and SU*(3) in the case of IBM-2) one always had to perform a numerical calculation in order to diagonalise the IBM hamiltonian. This becomes increasingly difficult (or impossible) for very large numbers of bosons. These difficulties do not occur in the intrinsic-state formalism 3X5V11*‘2). In the course of studying the moments of inertia, Coriolis and centrifugal forces, we derive approximate formulae for the energy of some of the states of the IBM-l 1 April 1986
H. Schaaser, D.M. Brink / Moment
2
and the IBM-2. These formulae not hold but rotational the number of variables in terms of the collective practical
of inertia
are valid in cases where dynamical
symmetry
does
features dominate. A special advantage of our work is that is small and they can be given a simple physical interpretation model.
This makes our formulae
especially
attractive
for
applications. 2. The cranked intrinsic state in the IBM-l
The IBM-l simulates band we use normalised
2N valence nucleons intrinsic states
by N bosons.
For the ground-state
IN; a) = JV&J+)~(O)
(2.1)
with 1
B+= s’+C a&d;,
JN=(N!(l+&,
P
(2.2)
+,)N)‘/*
and with p = -2, -1, . . . ,2. s+ and d: are the creation operators for a monopole and a quadrupole boson (with magnetic projection number p). IO) represents the core. By normalising the wave function one can always remove a possible coefficient of the s+ operator in (2.2). (2.1) are not eigenstates of the IBM-l hamiltonian. Being coherent states they are also over-complete and non-orthogonal “). Nevertheless one can use them to estimate physical quantities. The states (2.1), which are generalisations of the states used by Ginocchio and Kirson 5), are introduced because we want to study non-static effects of the nucleus. As we are interested in rotational features, it is important that states of type (2.1) can have large quadrupole moments ‘). By going to an orthogonal coordinate system x, y, z where the axes are the principal axes of the quadrupole tensor we can reduce the number of independent variables in B+ and arrive at a simpler description of the intrinsic state. We have the usual problem of finding principal axes for the matrix elements of a quadrupole tensor, which for the IBM is defined as Q,=~~~~+dd=s+x(d+d)ZT’, L?, = (-l)*d_,
(ab)Z” = We abbreviate the matrix elements ing (Q,) means that
C (LCm,mb~~~)~,abm,. m,,mb of Q, between
(QJ = (Q-d = 0, The diagonalization
of (Q,)
,
is discussed
(2.3)
the states (2.1) to (Q,). Diagonalis-
(Q2> =
(Q-J.
in more detail in appendix
(2.4) A.
H. Schaaser, D. M. Brink / Moment of inertia
Choosing
a coordinate
equivalent to requiring mations A,
system
3
where the axes are the principal
that the intrinsic
states are symmetric
under
axes of (0,) certain
IN; a’) = A/N; a) = (N; a),
is
transfor-
(2.5)
which we now discuss. It is instructive to look first at the static case. Imposing the geometrical symmetries R,(r), RY( 7r), R,(V) implies that the coordinate axes are the principal axes of (0,) (appendix A). (R,(r) is the rotation of the system through the angle r around the x-axis.) In the static case it is usual to require also that the intrinsic state is time-reversal invariant. This in conjunction with the other symmetries leads to real a, and the intrinsic states are identical to the ones used by Ginocchio and Kirson ‘). As combinations between time reversal T and the geometrical symmetry operations are possible, we have the following abelian group of transformations: S = {K(r), where
Ry(r),
R,(n),
T, TUT),
TRY(r),
T%(m),
El,
(2.6)
E is the unit element.
One usually determines the moment of inertia by cranking the nucleus. The introduction of cranking around the x-axis (this is the usual choice) destroys some of the symmetries of S. We choose an appropriate subgroup S, of S, which still diagonalises (0,). A sketch (fig. 1) of the problem helps one to see which symmetries remain. Certainly T will no longer be a symmetry; it changes the sense of the cranking movement. But some of the combinations between T and rotations are
Fig. 1. Principal-axis
system of a triaxial
nucleus.
Here the x-axis
is taken as the cranking
axis.
4
H. Schaaser, D.M. Brink / Moment of inertia
still symmetries: S,= {K(V), Example:
T&(r),
%(r),
R,,(T) alone is no longer a symmetry
(2.7)
El.
operation,
because
it changes
the
cranking sense. But the combination T&,(P) leaves the cranking sense and the configuration as a whole unaffected. S, is isomorphic to the group Dz. We prove in appendix A that (Q,) is still diagonal with this choice of S, in the cranked case. The operations of S, give us relations between the a,. In appendix A it is shown that these relations are found by solving AB+A-’
(for all A from S,) .
= B+
We consider R,(V) first. From the table in appendix R,(r) and TRY(r) imply
A we know that the symmetries
a, = a-+,
(2.8a)
a,=aE.
(2.8b)
TR,( n) is redundant and gives us no new relations between the a, in addition to (2.8). S, diagonalises (Q,) for all values of the parameter x of Q,, and gives real a,. This case should be general enough for studying nuclei in which rotational features dominate. Complex intrinsic variables a, have been used by some authors “). In their case their variables are associated with a special symmetry (O(6)) and the energy surface is -y-independent ‘). In comparison with the states used by Ginocchio and Kirson ‘) our states have in general a, = a-, # 0. In the next section we see that a, is closely associated with the Coriolis and centrifugal forces. Hence in the non-static case, B+ is equal to B+=s++a,d,++a,(d:+d’,)+a,(d:+d’,)
with a,, real.
Using [A, BN] = NBN-‘[A,
B]+fN(N-
+&N(N-l)(N-2)B”-‘[[[A, we can easily calculate
the overlap
(N; aIN; a’) =
l)BN-‘[[A,
B], B]
B], B,], B]+.
-.,
(2.9)
of the states (2.1):
(1+x
(1 +C &a,l” a,a,)N’2(1+C aLaL)N’2’
If we call A =da,a,, B=JaLaL and cos (Y=(aLa,)/(AB), then (N; a IN; a’) in the limit N + ~0 is approximately gaussian in (A-B)
the overlap and (Y.
H. Schaaser,
D. M. Brink / Moment
of inertia
5
3. The moment of inertia for the IBM-1 A first step towards calculating depend on the cranking frequency cranking
problem
is formulated
the moment of inertia is to find out how the a,, o. We crank around the x-axis. The self-consistent as
6(N;
aJHIBM- wl,lN;
(3.1)
a) = 0
where 1, is the x-component of the angular momentum operator and the expectation value of HIBM- w!, is minimised with respect to the a,. We use the states (2.1) as trial wave functions for the ground-state band. The term -01, comes about by making the transition from the laboratory to the intrinsic frame of the nucleus ‘); it incorporates the Coriolis and centrifugal forces in the intrinsic frame.
3.1. DEPENDENCE
OF THE
ENERGY
We use the most general
IBM-l
ON THE
INTRINSIC
hamiltonian
VARIABLES
1,2) with two boson
interactions:
H = E,S+S + &,d+ . d+~,(s+)*s*+u~~+d+~&++~[d+~d+s*+(s+)~d”~d] +~*[(d+d+)(~).
&+++d+
- (c?c?)‘~‘]+
C
C,(d+d+)“’
- (J&“’
(3.2)
1=0,2,4
with a - b = 1, a,b_,(-l)@. All terms conserve the total number of bosons. Using (2.8) and the g-band states (2.1) we can easily calculate the matrix elements of (3.2) and EN = (N; a[H,,,IN; a), the energy in the laboratory system, is
+ C4&[4(9a:+
10&a0a2+22a~+21u~)a~+9(a~+2a~)*]
11
(3.4)
with C=l+~uZ,=l+u;+2u:+2u:. P By setting function:
a, = 0 we can recover
the formula
EN = &(a,,
for EN in ref. ‘). We see that EN is a
a:, a*).
(3.5)
Eq. (3.5) will considerably simplify our calculations later in this section. The dependence on the absolute value of a, is explained by the time-reversal invariance of EN. It is related to the fact that the energy cannot depend on a sign change in the cranking frequency w (see appendix B).
6
H. Schaaser, D.M. Brink / Moment
3.2. CRANKING
When
of inertia
CALCULATION
w = 0, (3.1) reduces
to the variation
problem
@Iv; alH,,,IN;
a) = 0.
This is the static case and R,,( rr) and R,(n) 2). Their action on the operator B+ results appendix A). So we have
(3.6)
are also symmetry operations (see sect. in the additional relation a, = -a, (see
a,(w=O)=O.
(3.7)
In order to solve (3.1) we need an expression C. The matrix element of 1, is L,=(N;
a\/,jN;
a)=2Nu,[l+~
for 1,. This is found
u,uJ’(&,+2u,)
That Lx has a, as factor is explained in appendix The extremum condition for (3.1) is aEN --
au,
(3.8)
B. So we have L,.(w = 0) = 0.
aL, o-=0.
(3.9)
au,
When o is small we solve (3.9) by expanding the solution of the static variational problem
EN and Lx about (3.6):
a, = a:‘,
~,=a~“++~, EN = E$+
which is
(3.10a) ..,
c ~A,A,E~~~,+~ cI.y
L, = c A,L~;+ fi As EN has an extremum
in appendix
c $A,A,L!$+. P’.Y
(3.10b) ...
(3.1Oc)
at a,(‘) , there is no term linear in A, in (3.10b). The constant
term in the expansion of Lx vanishes at o = 0. Before inserting (3.10) into (3.9) we note that a Taylor expansion of A, in terms of o has no constant term: A, = WA,,, + &J~A,_, The term w C APL!&
is then of second C A,E$!$, P
+. . . .
order in o. So the expansion = coL$‘i +O(02)
.
of (3.9) is (3.11)
Taking into account (3.5) and (3.7) we see that terms like E$jF, vanish for all p it 1. of (3.7). So (3.11) simplifies Also terms like LEL vanish for (Y# 1 because considerably: AoE$!!,~oo+A,E~~20=0, AoE$$,2+A2E$;22=0, A 1E ‘;jl . 1= o Lp,’3 .
(3.12)
H. Schaaser,
D.M. Brink / Moment
of inertia
The determinant E N,OO
EN,,,
E N,O2
E~,22
('I because of the extremum does not vanish at a, two lines of (3.12) have only a trivial solution:
of EN at this location.
Ao,A2=O+O(w2),
(3.13a)
d,=a,=wL!,fi/E$!$,+O(~~).
(3.13b)
(3.13) shows how the values of a,,which are the solution of the variation (3.1), depend on the cranking frequency w. Going back to (3.10~) we easily find the moment of inertia I: L,=A,L~,‘+O(W~)=O(L~;)~/E~~~,+O(W~). Comparison
with the classical
formula
a,=pcosy,
problem
(3.14)
Lx = Iw +O(w2) yields (3.15)
r,=(L~1)2/E~!11+O(W). The equilibrium state is specified shape-defining variables by ‘)
So the first
by a, and
a, which
are related
to Bohr’s
a,=J&/3siny.
An extension of IBM-l in the static case (w = 0) [ref. 8)], which we here, can describe triaxial features. For the IBM hamiltonian (3.2), y = 0 in the equilibrium state provided p is allowed to take negative we have a,= p, a2= 0.Because we have a symmetrical rotor I = 1,
do not consider one can choose values “). Then = 1, is given by
1=6@(1+/32)/A+O(W)) A=(E/N)(~+~~)+(~-~/N)[(~+~~)A~-~A~],
A,=u2-2vo-~~hv2/?-~Co/32+~C4~2, A2= ~,+(~,+2v,)/3*-2$/~ The value of p, which minimises root of
v2P3+($CO+fC2+~Cq)P4.
(3.16)
(3.4) in the static case (a,= a2= 0),is given as the
O=~(I+~~)+(~-l){(~~+2v~-2u~)-3&~~, +(2W-U2-2vo)~2+J3v2~3}
(P + 0))
(3.17)
with
Eqs. (3.16) and (3.17) should be used only if v2 # 0 (or u2 # 0; see eqs. (3.18), (3.20)). This corresponds to a well-pronounced minimum in the energy surface of a deformed nucleus. For large N the l/N terms in eq. (3.10) can be neglected and the moment of inertia tends to a limiting value which is independent of N.
8
H. Schaaser, D.M. Blink / Moment of inertia
3.3. APPLICATIONS
If we omit terms which (3.2) can be expressed
contribute
only to the binding
energy,
the hamiltonian
l+azQ,.
T3+crbT4.
in the form lo)
H = E”q,+ uoPf . P+a,l-
Q3+a,T,-
T4
(3.18)
with QXr from (2.3): QXP = Q,(x For the operators
=
-#) .
in (3.18) we have used the definition
(3.18) and (3.2) are related
(3.19) of ref. lo). The hamiltonians
by
ES=h2,
E~=E”+~u,+~~~+~~~+~u~,
1 uo=zuo,
u2=2u2,
I u,=u,-~u~,
u2=-J?uT2,
Co=~uo-6u,+.&72--gcr3+~u~, C2=-3u,-~a,+$r,+~u,,
C,=4u,+&r,++ju~+$7,,
&=&d-us.
(3.20)
There are three subchains of groups for which the eigenvalues of H can be given analytically. They are named after the largest group in each chain: SU(3), O(6), SU(5). We are especially interested in the case in which H has a rotational spectrum. The SU(3) symmetry is an example of such a case. For SU(3) the eigenvalues of (3.18) are lo)
E([W(A, P)KLM) = ((+*-bduL+ where
L is the angular
momentum
l)+tu*C(h PI 9
and C(A, CL) is the quadratic
Casimir
(3.21) operator
of SU(3): C(A,~_L)=A~+~~+A~+~(A+/.L). We see that the moment
of inertia
has the same value for all different
(3.22) representations
(A, CL), namely z=
4 8u, -3u,
*
(3.23)
Solving (3.17) for N 9 1 gives p = 42. Inserting this value for p and the appropriate to (3.20) gives values for the parameters E,, .Q, uo, u2, uo, v2, Co, C,, C, according the moment of inertia (3.23). Naively one would expect that the moment of inertia should increase as the number of bosons increases, but this is not the case. We notice that formulae (3.16), (3.17) exhibit the usual phase ambiguity of SU(3), which allows A to have the values A = *+(J7). It is easy to prove that the sign choice to a prolate A = *IAl corresponds to v2= flu21. If u2<0 then A
H. Schaaser, D.M. Brink / Moment of inertia
In order to improve Dukelsky
et al. “)
their results
obtained
used the random-phase
thus obtained become important (for example for O(6) symmetry)
9
with the intrinsic-state method.
only in some regions and for the isolation
formalism
The higher-order
I’),
corrections
away from SU(3) symmetry of spurious states ‘*). In the
case of exact SU(3) symmetry, our result (3.16) for the moment of inertia and that of ref. 12) both agree with the exact result (3.23) to the same degree of accuracy (in the leading order of l/N). We also note that for large angular momentum the IBM-l predicts triaxial nuclei. If we insert Eq. (3.8) gives a, as a function of LIP, a2 or of Z.3, y, respectively. Ui = ai(P, y) (i = 0, 1,2) we get a complicated energy surface in ~3, y. For L # 0 the minimum does not occur at y = 0”, nor at y = 60”. Triaxial shapes are then possible because a, is included in the generalised wave function (2.1), (2.2). More restricted wave functions used previously ‘) do not have stable triaxial minima in the energy surface. Investigations of this possibility are in progress. Earlier ‘) we studied an application of the formulae (3.16), (3.17) of our cranking formalism to a hamiltonian, which contains only QJ . Q3, 1. 1 and P6 = Pf . P. The operators Q3 . Q3 and 1. 1 can be expressed by the Casimir operators of the group chain SU(3) and the operator P6 belongs to the group chain O(6). For this special hamiltonian our formula (3.16) gives results nearly as accurate as those of the IBM-l computer program PHINT 13) and agree just as well as PHINT with the experimental data of ‘68Er. So for sufficiently large boson number N the rotational structure in the formula (3.21) can be understood by the picture of cranking an intrinsic state (3.16), (3.17).
4. The p- and y-bands One uses as normalised
intrinsic
states of the /3- and y-bands
l”*14) (4.1)
IyrJ=d:IN-l).
(4.2)
Together with the ground state IN) from (2.1) they form a set of three orthonormal states. We can use these states to calculate the excitation energy of the p- and y-bands above the O+ energy level of the g-band. We shall also show that the moments of inertia Zp, Z, of these bands are approximately equal to the moment of inertia of the g-band. As discussed in sect. 3 we can choose y = 0 in the equilibrium state and set a,=0
where /3 is the Bohr variable.
9
The operator
ao=P, B+ of IN - 1) then has the form
B+=s++Pdof+a,(d:+d’,).
(4.3)
10 4.1.
H. Schonser, D.M. Brink / Moment THE
of inertia
P-BAND
We look first at the P-band
ground-state
energy
EP,=(PP(H]!P&=(N-l]CHC+]N-l). With the commutation
(4.4)
relations [C, B+]=O,
[c,c+]=l,
(4.5)
we arrive easily at CMC+=[[C,M],C+]+C+[C,M]+MC+C+M, with M a given operator,
(4.6)
and
Ec=(N-l][[C,
H], C+]]N-l)+EN_r
(4.7)
with EN-, of (3.4). For the first term in (4.7) we use (1 +P’)KC,
HI, C’l= P2[b, HI, s+1 - PU[s, HI,
(h.c. = hermitian
conjugate).
A straightforward
d,+l + h.c.I+ [[do,HI, calculation
&I
(4.8)
gives
[[d,, H], d;] = Ed+ u,s+s+2u2(22~o~2~>(d,‘s+s+d,) +
;
1
v=-2
EpN=-
1 1+p2
&d+p2ES+
(4.9)
,
N-l 1+p2+2a: Uo-U2-4~o+2(fCo+~C2+~C4)}
X(U2-4U2&$+2p2{2
+4V2p3G+
C,4(22/~, v-~.JIv)*d:_,d,_,
1=0,2,4
U2p4+a:{2U2~2+4~2pJ~+$(~c,+3c,)})
1 +
EN-1 . (4.10)
The ground state of the p-band has L = 0. Hence a, = 0 and the excitation of the lowest-lying state of the p-band above the ground state EN is
1-p’ N-1 = el+p2+(l+p2)* +8$ with W of (3.17).
[u2-2uo-4zr2~~+4~2(uo-u2-3~o+
v2p3+(u2-2W)p4]
energy
w) (4.11)
11
H. Schaaser, D.M. Brink / Moment of inertia
To calculate
the moment
of (4.10) with respect
of inertia with (3.15) we need the second partial derivative
to a,: EC!:; = A+ ~Yfj),,,~ .
Inspection of this proportional to (N to (N - 1)2. So for (4.12). In order to element
(4.12)
formula with the help of (4.10) and (3.4) shows that A is - 1) only, while in E”’N 1,11there are terms which are proportional a sufficiently high number of bosons, A will not contribute in derive the moment of inertia we need to calculate the matrix
of 1, (for 1, see appendix
C)
L~=(~~lI~I~~)=(N-ll[[C,
(4.13)
LJ, C’]IN-l)+(N-lII,IN-1)
with the help of (4.6). A short calculation shows that [[C, 1,], C’] = 0. This in conjunction with (4.12) gives for the moment of inertia of the P-band; (4.14) This is approximately equal to the moment of inertia I of the g-band as the leading terms in (3.16) do not depend on N. As an illustration we study the band-head energy of the P-band for the case of SU(3) symmetry; only the leading terms in N are retained. We can therefore set p=J2: E,p,‘,‘A(SU(3)) = -6Na2. We compare this formula with the excitation using (3.21). The ground-state band belongs the p- and y-bands to (2N-4,2) [ref. “)I:
(4.15)
energy derived from group theory by to the representation (A, CL)= (2N, 0),
E~~,“~(SU(3))=E([N](2N-4,2)000)-E([N](2N,0)000) = -6N(lObviously
1/(2N))a,.
(4.15) and (4.16) coincide
(4.16)
in their leading
terms in N.
4.2. THE y-BAND
A similar
calculation
for the y-band
using
HI, GIIN-~)+EN-~
EY,=(~~IHI~~)=(N-~I[[~z, =&d+
N-1 1 +p2+2af
+$~:[6C2+ The matrix
element
(4.5), (4.6) and (4.9) gives
{u2+4u2pJ3+~p2[4c2+3c4]
Cal}+ EN_,.
(4.17)
of 1, is LZ=(~~II,IIYy)=(N-llIxlN-l)
because
[[d2, &I, d:] = 0. Similar
arguments
IY = (L;,yy/E~,y)1=
as for the P-band 10 f I,
result in (4.18)
12
H. Schaaser, D.M. Brink / Moment of inertia
which gives the general result that for large boson numbers the moments of inertia for the g-, p- and y-bands are the same. The excitation energy of the y-band ground state is
1 =-+ 1+p2
N-l (1+p’)’
[u,-~u,+~u,&-[~~+~V~-$C~--~C‘&?~
+8~~&?‘-2~~[4C,,-~C~+&C~]]+(3/1)~
(4.19)
with L = 2. In the SU(3) limit (4.19) gives E,Y,,,(SU(3)) Analogously
to (4.16) the group-theoretical
- -6N~z.
(4.20)
result with L = 2 is
E~~,,(SU(3))=-%a,+E~~,~(SU(3))=-6N(I-l/8N)~2, which shows equality
with (4.20) in the leading
terms in N.
For comparison of the formulae (4.11), (4.18), (4.19) with the program and experiment, we refer to ref. ‘).
PHINT
5. The moment of inertia in the IBM-Z In the IBM-2 one distinguishes between protons and neutrons. We study a particular IBM-2 hamiltonian which contains all terms which have been used in the comparisons with experiments. In these comparisons the program NPBOS 16) has mostly been used ‘*2). In sect. 6 we shall compare our results with NPBOS and experiment, and also with the analytical formulae for the limiting cases of SU(3) and SU*(3) symmetry “). We also study the transition between the limiting cases SU(3) and SU*(3). Iachello ‘) discussed the different group chain reductions for the IBM-2. In IBM-l in the static case, the intrinsic states are always axially symmetric. In IBM-2 they may be axially symmetric, but not in all cases. For example, the limit of SU*(3) symmetry “), which does not occur in IBM-l, describes a triaxial rotor I*). In the first part of this section we study a general intrinsic state for an IBM-2 hamiltonian. We include in particular that (i) the intrinsic state for each sort of nucleon can be triaxial, (ii) the principal-axis systems of the neutron and the proton distributions may be different. We shall derive formulae for the moments of inertia about the x, y, z-axes of the principal-axis system. In general all three moments of inertia are needed for the calculation of spectra I’). 5.1. THE HAMILTONIAN,
Our choice
INTRINSIC
for the hamiltonian
STATES
AND
MATRIX
ELEMENTS
is
H ,e~=%G+~,~,+‘Grv~,’
Qv+ v,r,+ vw+M.
(5.1)
H. Schaaser, D. M. Brink / Moment of inertia
13
We reserve the index 7r (v) for the proton (neulron) degree of freedom, and p, p’ for +rror V. nr and n, are the number operators for the d-bosons. Analogously to (2.3) we have for the quadrupole
operator
Q,,, = G+&,p+ d;,s, +xp[d;L?pl:’ . For the interaction
between
like nucleons
(5.2)
we take
V,, = ‘$,Q,. Q, . The Majorana
operator
(5.3)
is
M = -2
C
&(d:df,)‘k’
. (&,(?,)(k)
k=1,3
(&s,
+&(d;sf,-s;d+,)‘2’.
-s~&)‘~‘.
(5.4)
The notation for these operators is the same as in ref. lo). We proceed in a similar way to our method for IBM-l. Our wave function IN,N,)=[N,!N,!(C,)N-(CY)N”]-“2(B~)Nn(B~)N”IO),
is (5.5)
with
c, = 1+c 4,,a,, II
B; = s;+C
up,+dp++. I”
(5.5)
The variables up,@ are taken in the principal-axis system of the nucleons of sort p. From sect. 2 it is clear that the quadrupole tensors of the neutrons and protons are diagonal
if the following
generalisations a P.P = qr,.
of (2.8) hold: %,P = CA .
First we study the static case (w = 0). In this case we have in addition (see appendix
(5.6) to (5.6)
A) a, = (-l)“a,.
So there are four independent real variables up,+ (p = r or v and t.~= 0,2) and three system of the protons with Euler angles LY,,(Ye, a3 which connect the principal-axis that of the neutrons:
where D is a rotation matrix. If both nucleon distributions have axial symmetry only one Euler angle (Ye is necessary to account for the relative angle between the axes of symmetry. For our hamiltonian the equilibrium value of CY~can only be a2 = 0” or cz2= 90”. If we allow the proton and neutron mass distributions each to be triaxially deformed, two more
14
H. Schaaser, D.M. Brink / Moment of inertia
Euler angles
(Y,, (Yeare required.
six combinations
Analogously
we have an equilibrium
only for the
of Euler angles: (ff,, a29(Yj) = (0°, O”,00), (90”, O”, 00), (0°, 90”, 0°) ) (90”, 90”, 00), (0°, 90”, 90”), (90”, 90”, 90°) .
The combinations
(O’, O”, 90’) and (90”, 0”, 90’) are redundant.
In practice
this means
that the principal-axis systems of protons and neutrons are the same, but may differ in the labels x, y, z of the axes. Minimising E for a, = 0 provides us with values for the uP and (Yi in the static case. We write E explicitly for w = 0: Q(v)+K,,Q~(T)+K,,Q~(v)
E”‘=E,~(~)+E,~(~)+K,,Q(~).
(5.7)
with
The other terms are n(p)=(n,)‘O’=~(a~,o+2aZ,,), P
Q(pO)=(~~)bO)=~[2a,,o+x~(2a~,2-a~,o)~l, P
QW)
2% Qp,2[1+XpQp,OJ?l 3 P
= (Q&A” = 7
02b)=(Qp-
0,)‘“‘=5N,+~[(x2,-4)CyKp P
+(N,-1)[4&+44& with K, = at,,+ 2az,,. The matrix case
(,)(O)+!
~~a,o(6& element
- a;,~) +fx’,K’,ll
of the Majorana
(253[%,2%,0-
5-rY
+ &[%,o - %,J
operator
(5.9a) is in the static
G,o%,212
+ 252[%,2- %,21’1.
(5.9b)
We note that (M)(O) does not depend on the parameter 5,. [The second derivative of(M) with respect to a,, or uv,,, however, depends on all three parameters 6 (see eq. (5.18)).]
15
H. Schaaser, D.M. Brink / Moment of inertia 5.2. CRANKING
FORMALISM
OF THE
IBM-2
We discuss the dynamic case (w # 0) now. In this case it is easier to evaluate all matrix elements in. the coordinate system of one type of nucleon. We choose arbitrarily the coordinate system of the protons to be the standard system. We use this approach because the relations (5.6) are not necessarily the same in a rotated coordinate system. If, for example, we assume that the proton coordinate system and the neutron system are related by a rotation around the y-axis about CY~ = 90”, then in the rotated system the equivalent equations to (5.6) are 4.m = a:,, .
G,m = a,,(-1)“)
In this rotated coordinate system in the dynamic case, a,, and a,_, vanish and u”,-~ are real and independent variables. In general in each of the rotated a v.2, %,o. coordinate systems we may have relations between the up,+, which are different from (5.6). We find that it is much easier to discuss the dynamical case once and for all in a standard coordinate system, for which we have chosen the proton coordinate system. From appendix A and (5.6) it is easy to see that in this standard system the expectation value of the energy in the laboratory system is a function of six real variables a,, :
=
Ha,,,,
G.2,
a,,,
a”,29
(%,,a”,,),
a:,,,
&,,I .
(5.10)
Analogously to (3.5), E depends only on terms which are homogeneous functions of second degree in up,,. This particular dependence on ap,, will help to keep the system of equations (5.13) simple. Similar to the variable a, of sect. 3, the up.1 are related system a sign The
to the cranking frequency o. As we have chosen the same standard coordinate for the proton and neutron variables, a,, and av,, will both be affected by change of w. So the energy does not depend on the sign of w. expectation value of the angular momentum, LX, is equal to
Lx = Lx,, + LX.“, So the obvious
generalisation aPIW
=
E =
with Lx,p = Wap,,(CJ1~J~
(5.11)
of (3.10) is
42 +A,,
(5.12a)
3
+..., ,,a’E,‘,“’
E(O)+1 c i~~,,ft,.,, P.P,
Lx=
aP,of2@P,2).
id
I3L(O) C Ap,+~+-.. P,W PIP
The variation problem 6( N,NVIHrBM - ol,(N,N,) equations in A,, and w, if we omit terms in 0(02).
P,cI
(5.12b)
P’A
(5.12~) = 0 gives us a system of linear We take into account eqs. (5.10)
16
H. Schaaser, D.M. Brink / Moment of inertia
and (5.11). Then we have a system which consists to (3.12):
C p.i
of six equations
AmoE$‘J,pi + Am,E’,$,pi + AvoEZ,pi + Av2Etfpi =0 , 4,Ef?,p,
+
4,E!?,p,
=
p=r,
wL$‘;~
v;
of a type similar
i=O,2,
(5.13a) (5.13b)
.
E pk,olmis the second derivative of E with respect to ap,k and am,,,. Because we evaluate the derivatives at w = 0 and the energy has its minimum there, the determinant of the coefficients Es,;, of the first four equations (5.13a) does not vanish and the solutions are Ap,o=A,,z=O+O(02). The same applies
to the determinant L(o) A,,
Inserting
=
w
E’O’_
of the coefficients _ L’o’
E;;; p’*p’ E(o) Tr1,Tr1 v,,v,
X.Pl
-
nl,vl
(P # PI >
b%,vJ*
(5.13) into (5.12~) gives us the moment
It is instructive
to discuss
in (5.13b). The solutions
are
E(o)
It can easily be shown that the mixed derivatives
similarity
(5.13a)
(5.13b)
of inertia:
of E come from the Majorana
term:
(5.13b) for A4 = 0; then (5.13b) and (5.14) have a strong
to (3.13b) and (3.15): A,, =
wL$‘,,,lE$‘,p,+O(m*) , (5.15)
L = L/w =C [L!.t,hlz/E~~,p,, P
with
LAY:, =2Np(Cp)~1[ap,oJ~+2ap,2], E (0) p,,p,
=
“p”%‘)+
c k=0.2
(5.16)
2k’2KppQ($+?“(pk)
+Kpp[Q- Ql”(P)+(@?,p,
(P#P),
(5.17)
In (5.17) and (5.18) the double prime denotes the second derivatives of n(p), Q(pk), Q * Q(p) with respect to ap,l taken at w = 0. The terms occurring in (5.17) are listed
17
H. Schaaser, D.M. Brink / Moment of inertia
below:
n”(p) =
3,
(5.18a)
P
(5.18b)
(5.18~)
(5.18d)
+
&mb,* - 4J1+5Jx
+ &MQ,.&
%,& +~,&I’
q&1’+
For K,, see eq. (5.9). In order to complete
+
253
2 [
l-I
p=lr,v
h%,&
&I + &I *
(5.18e)
eq. (5.13b) and (5.14) we also note
~poJ%+a,2~v~ ,
,
,
11 +&
,
(5.18f)
where ~ denotes the product symbol. Eq. (5.15) for the moment of inertia is similar to eq. (3.15) for the moment of inertia of the IBM-l. It is a sum of two terms p = r and p = V, corresponding to neutrons and protons, but there is a coupling and each of these terms depends on the neutron and proton interaction. It is not surprising that such coupling terms exist in the IBM-2 because the IBM is not a geometrical model and our usual intuition, which is based on geometrical pictures, does not apply. We also have to determine what the moments of inertia around the y- and z-axes are. We have already studied the case of cranking around the x-axis in the standard system. Subsequently we crank around the y- or z-axis by rotating the wave function (5.5) so that the old x-axis is in place of the axis under study. This results in the formulae 1, = mp,o, -$,J 9 1, = L(%,O, %,Z) , 1, = 4 (- ~%I + %,*4 , tqLl,, + a,&)
,
p=~
or V.
According to the triaxial rotor model of Davidov and Filippov 18), one has to use all three moments of inertia for the calculation of the spectra. The CZ~,~are related
18
H. Schaaser, D.M. Brink / Moment of inertia
to the Bohr variables
by up.0 = P, cos 3$,
6. Applications We study the limiting
of the IBM-2
cranking
cases of SU(3) and SU*(3) H
These symmetries
ap,2 = &&sin
IBM=K3(Q~+C~)
are realised
yP.
(5.19)
formalism
for
’ (&+QY)*
with the following
(6.1)
sets of parameters
17) (parameters
in (5.1) to (5.4) not listed here are set equal to zero): K,,
=
SU(3): su*(3):
Kw
=
K3,
K ,ru = 2K3
Xw=&=T;J7
3
x,=-xy=r;J5.
(6.2)
We see in eq. (6.2) that the distinction between the SU(3) and SU*(3) limits comes from (x,/xy) = 1 or -1, while in both limits x,, can take both values. This possibility of choosing the sign of xP in IBM-2 is analogous to the possibility of choosing the sign of x in IBM-l (see sect. 3). For SU(3) and SU*(3) the energy spectrum of the hamiltonian (6.1) is given by eqs. (3.21), (3.22) with (+, = 0, 02= K~. The allowed values for (A, p) are given in ref. I’). Thus the formula for the moment of inertia (3.23) is valid for IBM-2 in the limiting cases of SU(3) and SU*(3) if (+, = 0, o2 = K~.
6.1. THE W(3)
SYMMETRY
First we study the symmetry SU(3). In the limit of N,, minimising E = (HtBM) in the static case are “) SU(3):
%=[;;c,]
3
Yv=[;;~],
&=py=d2,
N, > 1, the results
C=O”
of
(6.3)
The upper line& yP corresponds to x,, = xy = -$/?, while the lower line is assigned to x,, = xy = 4d7. G is the angle between the symmetry axis of the neutrons and the protons. For large N, the moments of inertia 1, and 1, are equal and have the value I = -4/3K3, which is the value of I in eq. (3.23). 1, vanishes. If we study (6.1) for more physical values of N,, NV than in the limit of large N,,, i.e. O< N,, NV < 10, we notice that in the SU(3) limit the variables differ slightly from the values mentioned in eq. (6.3). We have ym = yy = O”, & = &, < 42 with Z, = 1, and 1, =O. This is very similar to IBM-l, where eq. (3.17) also gives p
H. Schaaser, D.M. Brink / Moment of inertia 6.2. THE SU*(3)
SYMMETRY
In the case of SU*(3) symmetry we also study first the limit results of minimising E = (H,,,) in the static case are 17) su*(3):
19
YT=(;;~),
y”=(T),
p,=py=&
N,,, N, 4 1. The
(Y=90°,
(6.4)
It is easy to show that in both cases for (‘y,, yy) we get the moment of inertia of (3.23): Z, = Z, = Z, = -4/3~~. We see that for SU*(3) the nucleus is triaxial, because Cu= 90” and &, py # 0 [ref. “)I. We can also easily understand that we get an SU*(3) spectrum if all moments of inertia are equal: In the case of Z, = Z, = Z we have a formula
for the rotational
energy
E=w+l) -+
(+-& >
K2.
21
In this case the usual combinations
z
of K and L are possible:
K=O,
L = 0,2,4,
...)
K=2,
L=2,3,4,5
)...)
K=4,
L = 4,5,6,7,
...
...,
... 3
If Z = Z, = Z, = Z, then the energy states with different K but same L are degenerate. This is the SU*(3) spectrum 17). Because in the limit of large N,, N, all three moments of inertia are equal we have a spherical top even though our nucleus is triaxial. A rotation Z = -4/(3Kj).
around
As a more physical
any axis through
example
the centre
of the nucleus
for (6.1) with SJJ”(3) symmetry
would
have
we study the case
N,, = 10, N, = 2, K~= -1 MeV, xrr = -xy = -iJ7. We get the results & = 1.36, py = 1.16, ym = O.l”, yv = 53X’, leading to the moments of inertia (in MeV-‘) Z, = 1.39, Z, = 1.40, Z, = 1.59. This shows that for SU*(3) the neutron and proton distributions, or both, may be triaxial, which was not noted in ref. r7). This tendency increases for both neutron and proton distributions if (N,, + N,) is small. Our example shows that if only one of these N, decreases, then only the distribution of nucleons of type p will tend to become triaxial.
6.3. TRANSITION
BETWEEN
SU(3)
AND
SU*(3)
We study the hamiltonian (6.1). One possible way to move continuously from SU(3)toSU*(3)istovaryx,andtokeepx,fixedtothevaluex,=-~J?(seeeq. (6.2)). We shall find three modes: an SU(3) mode, an SU*(3) mode and a mode of large xy. In the SU(3) mode the distributions of protons and neutrons are each axially
20
symmetric distribution
H. Schaaser, D.M. Brink / Moment of inertia
and have the same symmetry axis. In the other two modes the proton remains almost axially symmetric around the z-axis. In the SU*(3)
mode the neutron to the proton
distribution
symmetry
tends to be axially symmetric
axis. For large xy we have a separate
about an axis orthogonal mode. In this mode the
neutron distribution has yy = 30” and & is large. The large values of &, do not necessarily pose problems in the interpretation of this mode, because by using the collective model one can show that in the case of IBM-l some large p correspond only to small physical deformation strengths ‘) and probably this also holds for the IBM-2 In the mode with large xy the neutron mass distribution is a triaxial ellipsoid, both thin and elongated. There is no longer any axial symmetry and the longest axis is orthogonal to the proton symmetry axis. It is relatively complicated to discuss the Euler angles ((Ye, Q*, (Ye) in conjunction with yV, yy. For example if yy = 60” the neutron distribution is axially symmetric with respect to the y-axis of the neutron coordinate system ‘O). So in order to find out about the relative orientation of the proton and neutron mass distributions one always has to look at (a,, CQ, CX~)and y,, yy as well. It is much easier to study the asymmetry variable 7, which is defined as 20)
(6.5) This relation is valid only for small xP and small up+. One easily finds that by reducing (6.5) to the case of one type of nucleon only and inserting (5.9a) and (5.19) into (6.5). 7 corresponds to the variable y of IBM-l (see sect. 3), where no distinction is made between protons and neutrons. 6.3.1 Symmetrical system: N, = N,, = 3. We see in fig. 2 that p,, is almost constant, while By varies substantially.
In the SU(3) limit (xy = -id?)
and in the SU*(3) limit
(xy = ;JJ?), the deformation strength variables &, /3” have the same value. For xy % 1, py becomes large. We note that the region of largest curvature in the curve for & is near to xy = 0.6 which indicates that this point is somehow special. We see in fig. 3 that for xy s 0.64 we have yT = xy = 0. For xy > 0.64 yT and y,, diverge suddenly: y,, remains always near zero while y,, increases to yV = 60” at xy = 1.4. For xy > 1.4, yy decreases and approaches yy = 30” asymptotically, while y_ approaches zero. In fig. 4 we see a sudden rise of 7 at xy = 0.64. At the SU*(3) value of xy we have 7 = 30” exactly 17), and this holds for all nuclei with N,, = NV. At 7 = 30” the nucleus has its largest triaxial asymmetry 20). For xy > 2.5, formula (6.5) can no longer be applied. A sudden transition also occurs in the moments of inertia (fig. 5). For ,Y”< 0.64 the neutron and proton bosons both have axial symmetry about the same axis. At xy =0.64 there is a bifurcation in Z, and Z,, and Z, increases steeply. At the SU*(3) value xy = 48 the moments of inertia Z,. Z, are exactly equal and Z, is slightly smaller. For this value of xy the inertia tensor is axially symmetric about the y-axis. The
H. Schaaser,
D.M.
Brink / Moment
of inertia
21
1.8 1.6 -
NV=3 0.6-
I I sum I
0.4 -
I I I I I
I
I
I 1
1
0.6
1
I
0.8
1.0
1.2
I
1.4
1.6
1.8
X,
Fig. 2. Transition from SU(3) to S_U*(3). We plot the deformation variables p,, and p, as functions x, and keep x,, fixed to x, = -g7 = 1.32. x, and x, are parameters in the quadrupole operators and Q,, respectively. H = -(Q, + 0,) (0, + 0,) (in units of MeV).
of Q,
Y [degrees] 1
SU’(3) I
60
I
0.7 Fig. 3. Transition
0.8 from
0.9
SU(3)
1.0
1.1
1.2
1.3
N,=3
1.4
to SU*(3). We plot the deformation and x, are as in fig. 2.
1.5
16
variables
1.7
1.8
X,
‘y, and y, against
x,. H
22
H. Schoaser, D.M. Brink / Moment of inertia [degreesI''
I
\
Nn:3,
\
NV=3
IO-
5-
0.5
I
1.0
; SP(3) I II 1.5
I
I
20
2.5
\ \ \ \ \ \ \ \ \ \ \ \
\ I \ I IL
I
3.0
3.5
x,
Fig. 4. The asymmetry variable 7 plotted against the parameter x,,. H and x,, are as in fig. 2. 7 takes into account the deformation of both protons and neutrons. For x, > 2.5,eq. (6.5)for 7 can no longer be applied, which is denoted by a dashed line.
1.6 1.4
0.6
L_Ll ISU'(3)
,-,, V.-t
I
0.2
I
I
0.6
0.8
1.0
1.2
I
I
1.4
1.6
1.8
X"
Fig. 5. The moments of inertia in the transition from SU(3) and SU*(3) symmetry. H and x,, are as in fig. 2. At x, = 0.64 there is a bifurcation in Z, and ZYand there is a steep rise in Z,.
H. Schaaser, D.M. Brink / Moment of inerlia
Fig. 6. The analogue as function
23
of fig. 2 for an asymmetrical system (N, f N,). Deformation variables p,, and p” of x, in the transition from SU(3) to SU*(3). H and x, are as in fig. 2.
transition at xy = 0.64 is something like a phase transition between the SU(3) mode and the SU*(3) mode. 6.3.2. Asymmetric system N, = 4, N, = 2. This case has the same total number of bosons as in subsect. 6.3.1. Figs. 6-9 showing the deformation variables and moments of inertia as a function of xy are similar to those for the case N,, = N, = 3. In the present case there are also two phase transitions which happen at xY--.0.99 1.4 which is slightly larger than and at xy - 2.19. The pV and p,, curves cross at xy = the SU”(3) value. For xy > 1.4, yy approaches the values yy = 30” asymptotically. In fig. 9 the bifurcation between I, and ZYis less pronounced than in the symmetrical system N, = N, = 3 discussed before. At the SU*(3) value of xy all three moments of inertia are approximately equal, which is a sufficient condition to have an SU*(3)-like spectrum.
6.4. COMPARISON
WITH
NPBOS
AND
EXPERIMENT
In fig. 10 we show the ground-band energy levels of 16’Yh calculated with the cranking formalism (CF) and NPBOS and compared to experiment. We have taken the NPBOS parameters of ref. 19) (see fig. 10.) The corresponding hamiltonian is
H. Schaaser, D.M. Brink / Moment of inertia
24
Y
[degrees]
su731
4
I
60 50 -
Fig. 7. Deformation variables y, and y, for N, # N,. In analogy to fig. 3 we perform the transition from SU(3) to SU*(3) by varying x,. H and x,, are as in fig. 2. We note that the deviation from y, =60” at the SU*(3) value of x,, is stronger for the asymmetrical system than for the symmetrical system (N,, = N,). For x, + co the variable y, approaches y, = 30” as limit.
[degreeslAY 30-
25NTT=4,Nv=2 20 -
15-
IO-
5-
II 1.0
0.8 Fig. 8. The asymmetry
variable
7 plotted
transitional between SU(5) the cranking formalism is we see from fig. 10 this is in the ground-state band simple formula.
I
I1
1.2
against
1.4
L
I
I
1.6
1.8
2.0
2.2
x,
x,. H and x, are as in fig. 6. For x. < 0.99, 7 vanishes.
and SU(3), so there is no triaxiality. Fig. 10 shows that less accurate compared to experiment than NPBOS. As due to the deviation from constant angular momentum in 16”Yb, which is included in NPBOS, but not in our
H. Schaaser, D.M. Brink / Moment of inertia
25
system (IV,, # N,). Fig. 9. Moments of inertia in the transition from SU(3) to SU*(3) for an asymmetrical H and x,, are as in fig. 2. There is a bifurcation in I, and I, and a steep rise in I, at xv = 0.99, which is similar to fig. 4.
7. Summary In this paper we have studied the cranking of intrinsic states for IBM-l and IBM-2. In the first part on IBM-l we described in detail the method used in ref. ‘). It was useful to use symmetry arguments to reduce the number of intrinsic variables to three or less. We give a formula for the moment of inertia when the cranking angular velocity is small. We also extended the formalism to discuss the excitation energies and moments of inertia of the /3- and y-bands. In the second part on IBM-2 we also gave a general formula for the moment of inertia. We considered the SU(3) and SU*(3) limits and studied the transition between them. We found an interesting phase-transition-like behaviour. One moment of inertia (1,) is zero up to the transition point and then rises steeply, while the other two show a bifurcation at the transition point. A comparison with experiment lying between the SU(3) and was made for the nucleus 16’Yb using parameters SU(5) limits. The results presented in this paper hold if the cranking angular velocity w is small. It would be interesting to extend them to larger values of w.
26
H. Schaaser, D.M. Brink / Moment of inertia E [M&l A
,/ CF 1'
3.0-
2.5 -
2.0-
Fig. 10. Comparison between the cranking formalism (CF), NPBOS and experiment for the energy levels of the g-band of $FYb,,. The parameters of the hamiltonian are E, = E, = 0.67 MeV, K,, = -0.12 MeV, x, = 0.49, x, = -1.2, N,, = 6, N, = 4. This is a hamiltonian transitional between the dynamical symmetries SU(5) and SU(3). It gives an axially-symmetric nucleus. The values of the geometric variables found by our intrinsic-state formalism are p,, = 0.51, p, = 0.75, y, = y, = 0, 6 = 0”. 1, = I, = 16.78 MeV-‘, 1; = 0.
The authors thank M. Zirnbauer, P. Halse, J. Lomnitz-Adler, P. Van Isacker, A. Frank and J. Dukelsky for valuable discussions. One of us (H.S.) would like to thank the Studienstiftung des Deutschen Volkes, Dr. Carl Duisberg - Stiftung, and Deutscher Akademischer Austauschdienst for financial support.
Appendix A DIAGONALISATION
OF (0,)
(0,) is diagonalised by states IN; a) which are solutions of (2.4). Initially taking the transformations A to be those of S, we notice that transformations containing T change complex numbers into their complex conjugates *“):
Ta,T-’
= a*W’
The a, do not transform under the rotations of S because S leaves the coordinate frame unaffected (principal-axis system). As NN of (2.2) is a real number it is left unaffected by all A and we can write (2.5) (AB+A-‘)NIO)=
(B+)NIO).
(‘4.1)
H. Schaaser, D.M. Brink / Moment of inertia
This equation
27
is solved by B+, which fulfills AB+A-’
= eiPB+ ,
(A.2)
where (Y is an appropriate real number. The effect of the transformations states with good angular momentum is well known *‘). s+ is invariant these transformations. So we can set a equal to zero in (A.2):
of S on under all
AB+A-'=B+.
(A.3)
We give a table of how dz transforms and which relations from (A.3). Using the conventional phasing *‘)
between
the a,, result
and the property that any of the Ri(r)of S can be expressed as a product of the two others, one can derive the entire table from (for example) R,(r) and R,(r):
Rx(v) d’,
R,(n) (-)*d+, (-Y”a-,
a-,
R,(r)
T%(n)
TR,(r)
T&(r)
T
(-Yd;
(-)pd;
d;
d’,
( -)pd’,
(-Y”a,
(-)%I;
a:
a?,
(-)‘“a!,
The parity P is an identity transformation with respect to B+[ref. ‘“)I. From the table we note that the same is valid for T*. The matrix elements of Q, from (2.3) with the states (2.1) are
(Q,>=
1 +I
N a;a,
{(-)‘a_,+a~+~(a*a”)~‘}.
(2.8) has been used to derive this expression.
(A-4)
It is easy to see that
(Q,) = (-)“(o-,)*.
(A.5)
(9,) and (Qz) are, explicitly, (QJ=
N l+Ca:a,
(Q*)=
{aT-a_,+~x[JS(a_,a*,-ait;a,)+~(ata_,-a,aT)]>,
N
1 -t-c a;a,
{az*+a_,+&[h(
We prove first that the geometrical (0,). From the table it follows that
R,(n)+ ~~=a-,,
(A.6)
a$a_,+a$a,)+&
symmetries
aTa_,]}.
(A.7)
R,(n),R,(n),R,(r) diagonalise
R,(r)+ a, = (-)fia_,
.
H. Schaaser,
28
D.M. Brink / Moment
of inertia
This implies a, = -a, = 0 and a2 = a_z. R,(r) is redundant. Inserting a, = 0 into (A.6) gives (Q,) = (Q-J = 0 and with the help of (AS) we see that ( Q2) = (Q-2). As {R,(n), (0,)
R,,(n),
R,(n),
E} is a subgroup
is also diagonalised
of S, S also diagonalises
(0,).
by S,:
R,(r)
TR,(r)+a,=a;.
+ a, = a_, ,
(TR,( T) is redundant.) These relations between reduces to a real function in the a,. Then (AS)
the a, make gives
(9,) vanish
and
(QA
(‘4.8)
(QJ=(Q-J=o, (Qd=(Q-2). Appendix B
DEPENDENCE
OF E,
As T commutes
AND
L, ON THE
VARIABLE
a,
with HiBM, we have al&&V;
EN =(N;
We note that IN; a) is symmetric
a)=(N; under
alT-‘H,,,TIN;
R,(T)
a).
we have
TIN; a) = R,(r)lN;
a) = IN; a,, a_,, a*).
Inserting this into (B.l) we see that EN can only depend on [a,[. We now study the matrix element LX8of the angular momentum angular
momentum
changes
its direction
under
LX,= (N; alT-‘l,TIN; IN; a) is invariant
under
L,,=(N;
Comparison
operator
I,. The
time reversal:
a) = -(N;
all,lN;
a).
03.2)
S,:
aj(R,(n)T-‘)T-‘l,T(R,(r)T)jN;
=(N;
(B-1)
S, so that
TIN; a)=R,(vr)[TR,(r)]TIN; As R,(n)R,,(r)=
a).
alR,(4lxRx(~)lN
a)
a)=(N;
ao, -a,, a,llxk
with (B.2) shows that Lx is proportional
a,, -a,,
az).
to al.
Appendix C CONSTRUCTION
OF ANGULAR
MOMENTUM
OPERATORS
and then determine the A,, by demanding which are the usual angular momentum [I+, d:l=C,, (6-~b++1))-~'*d;+,, formulae. By taking the complex conjugate of 1, we find I_. For 1, = $(I++ I_) the result is We take the ansatz
1, = C,,, A,,dLd,
lx=;
i a=-*
(6-(~(~~+1))“*{d~+,d~+d~d,+~}.
H. Schaaser, D.M. Brink 1 Moment of inertia
When
going
interchanged
from the laboratory
to the intrinsic
29
frame the roles of 1, and I- are
‘$. But as 1, is the sum of 1, and I- it is not affected by this transition.
For 1, we find 1, =&6(d+d)l”.
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