Chapter 4
The 4N Bound State
The 4-body problem is richer in structure than 3 bodies. It has dierent types of substructures, 3+1 partitions and 2+2 partitions. Furthermore the 3 body subclusters can
again be partitioned into 2+1 as we already know. The strategy of Yakubovsky (O.A.
Yakubovsky, Sov. J. Nucl. Phys. 5 (1967) 971) is, to decompose the total state in so
many components as there are sequences of partitions:
8
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1234 ! >
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8
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:
(
12
13
123 + 4 !
23
12
14
124 + 3 !
24
13
14
134 + 2 !
34
23
24
234 + 1 !
34
12
12 + 34 !
( 34
13
13 + 24 !
( 24
14
14 + 23 !
23
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
3
2
1
4
2
1
4
3
1
4
3
2
3
1
2
1
2
1
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
4
4
4
3
3
3
2
2
2
1
1
1
4
2
4
3
3
4
At the end in each chain we end up with one specic pair. There are altogether
90
4x3 + 3x2 = 18 chains. Correspondingly one has 18 components of , which nowadays
are called Yakubovsky components.
The 4N system is very tightly bound in the -particle. It has a high nuclear density. It
is an important test ground to study nuclear dynamics and deserves a strong eort to
solve that system precisely for any type of nuclear force. Moreover it has "excited states"
(in fact resonances), which still pose a real challenge to be determined theoretically by
rigorous few-body techniques.
4.1 The Yakubovsky Equations for the 4N Bound
State
We start again from the 4N Schrodinger equation
X
(H +
Vij ) = E Now there are 6 pair forces Vij . In integral form (4.1) reads
X
X
Vij G
Vij = E ,1 H
ij
The rst step is the Faddeev decomposition
X
ij
(4.1)
0
(4.2)
0
0
(4.3)
ij
with
Separating the term
ij
ij
= G Vij = G Vij
0
X
0
on the right hand side and solving for
X
(1 , G Vij ) ij = G Vij
kl
kl6 ij
X
ij = G tij
kl
0
(4.4)
kl
kl
ij
yields
(4.5)
0
=
0
(4.6)
kl6 ij
=
which is a set of 6 coupled Faddeev equations. Here we use (1 , G Vij ), Vij = tij . Let
us look into the physics of this coupled set of equations. Again we iterate this set many
times which leads to
X
X
G tkl
G tmn : : :
(4.7)
ij = G tij
0
0
kl6 ij
0
=
mn6 kl
1
0
=
Note that the last tmn can again carry the indices ij . Graphically (4.7) looks like
91
1
2
3
4
We recognize dierent possible patterns: t-matrices within a 3-body subcluster, tmatrices within two dierent 2-body subclusters and t-matrices in sequence between all 4
particles. All that can come in arbitrary order. The only requirement is, that sequential
t-matrices are never within the same pair. The rst two groups separated by the vertical
lines have spectators: the rst group has a noninteracting particle number 4 and in the
second case the two two-body subclusters do not interact. It appears natural to rst sum
up all interactions within each 3-body subcluster to innite order and also within the 2+2
noninteracting subclusters. This is achieved according to Yakubovsky in the following
way. Let us consider for instance the equation for . Applying (4.6) gives
12
12
= Gt (
0 12
13
+
23
+
14
+
24
+
34
)
(4.8)
As
in (4.8) is driven by t , so is
driven by t ; by t , etc. Thus, processes
which are restricted to the three-body subsystem (1 2 3), are generated by the components
; and . Therefore, we can split into various parts
12
12
12
13
13
23
13
23
23
12
12
= Gt ( +
= Gt ( +
+ Gt
:
0
12
13
23
0
12
14
24
0
12
)
)
34
(4.9)
The way (4.9) is written shows that the rst two components on the right-hand side do
not contain an interaction ti , i.e., with particle 4. The next two components (second row)
do not contain an interaction ti , i.e., particle 3 is the operator. The last item (last row)
contains t and through the interaction t , which corresponds to a 2 + 2 clustering.
4
3
12
Let us introduce
34
34
Gt (
1
0 12
92
13
+
23
):
(4.10)
Then for identical particles one has with P = ,, since the full wave function is
antisymmetric,
34
P 34
= Gt P ( + )
= G t [P G V + P G V ]
= ,G t (G V + G V ) = ,G t (
1
0 12
34
0 12
0
13
34
12
0
0
23
13
34
14
0
0
23
24
0
12
14
+
24
)
(4.11)
Further we introduce a second amplitude
Gt
(4.12)
The amplitudes and are called Yakubovsky components. Using their denition, the
amplitude can be written as
= , P + (4.13)
All other amplitudes ij can be decomposed similarity, just permuting particles. Obviously there can be no new structure.
2
1
0 12
34
2
12
12
1
34
1
2
Remembering that the Faddeev amplitude for the 3N system was given by (3.17)
= G tP
:
0
(4.14)
From this we see that the part (1 , G t P ) = 0 represents a 3-body amplitude.
We also remember that P P
=
and P P
= . With the denition
P = P P + P P , we can rewrite (4.10) as follows
0
12
12
23
13
23
12
12
1
23
13
23
12
13
23
=
=
=
=
for the last row, (4.13) was used.
1
Gt
Gt
Gt
Gt
0 12
0 12
0 12
0 12
( + )
(P P + P P )
P
( , P + )
13
23
13
23
12
23
12
12
1
34
1
(4.15)
2
If the last two terms on the right hand side would be absent, we would just have the
3-body problem for particles 123 with particle number 4 as a spectator (it occurs only
as part of the kinetic energy H in G ). Since we want to sum up all forces within the
3-body subcluster 123, we bring the rst term of the right hand side to the left:
0
0
(1 , G t P ) = G t P (,P + )
0
12
1
0
93
12
34
1
2
(4.16)
Now we can invert with the help of a 3-body T-operator
= (1 , G t P ), G t P (,P + )
G TP (,P + )
1
1
0
12
0
34
0
12
1
34
1
2
2
(4.17)
where T obeys the 3-body Faddeev equation
T = t + TPG t
(4.18)
For the verication of (4.18) and the subsequent considerations, we drop the pair index
12 from t:
(1 , G tP ), G t G T
G t = (1 , G tP ) G T = G (T , tPG T )
(4.19)
or
T = t + tPG T = t + tPG t + tPG tPG t + : : :
= t + TPG t = t + tPG t + tPG tPG t + : : :
(4.20)
Similarly we treat :
12
0 12
1
0
0
0
0
0
0
0
0
0
0
0
0
!
0
0
0
0
2
= Gt
= Gt P P
(4.21)
The permutations P~ P P permute the two two-body subclusters 12 and 34. With
(4.13), we obtain
2
13
0 12
34
0 12
13
24
12
24
= G t P~ ( , P + )
2
0 12
1
34
1
(4.22)
2
We can now separate the subcluster problem for 12 and 34
(1 , G t P~ ) = G t P~ ( , P )
Inverting yields
= (1 , G t P~ ), G t P~ ( , P )
G T~P~ ( , P )
0
12
2
0
12
1
34
1
(4.23)
1
2
0 12
0
0 12
1
34
94
1
1
34
1
(4.24)
where T~ obeys the operator equation
~ t :
T~ = t + T~PG
In order to see the action of T~, we iterate (4.25)
12
0
(4.25)
12
~ t + t PG
~ t PG
~ t +
T~ = t + t PG
~ + t G t P~ PG
~ t +
= t + t G Pt
= t + t G t P~ + t G t G t + 12
12
0
12
12
0
12
12
0 34
12
12
12
12
0 12
0
12
0 12
34
0
0 34
12
0 12
(4.26)
This can be graphically represented as
1
2
3
4
Thus we end up with two coupled Yakubovsky equations for the Yakubovsky components
and and two accompanying equations for the subcluster problems:
1
2
1
= ,G TPP + G TP
(4.27)
= G T~P~ (1 , P ) (4.28)
0
2
34
0
1
0
34
2
1
T = t + TPG t
(4.29)
~ t
T~ = t + T~PG
(4.30)
0
0
Here we dropped the pair index 12 from the t operator.
The set of equations (4.27) - (4.30) is the set of equations to be solved for the 4N bound
state. The total wave function for the 4N bound state can be constructed from the
dierent components. According to the decomposition (4.3), we have
95
=
X
ij
ij
=
+| +
{z } + | +
{z } +
=
+P
,P P +P P
= (1 + P , P P + P~ )
= (1 + P , P P + P~ )((1 , P ) + ) :
12
23
12
14
31
12
34
24
34
12
34
13
24
12
12
34
34
1
2
(4.31)
To arrive at the last row, (4.15) was employed. Counting the number of terms yields 18
pieces, the 18 Yakubovsky components representing all possible chains of partitions.
4.2 Momentum Space Representations
Now we have to introduce two sets of Jacobi momenta
2
u1
u2
~u
3
1
1
u3
~u
4
3
2
4
v1
v2
1
3
~v
2
(4.32)
2
(4.33)
= 43 (~k , 13 (~k + ~k + ~k ) )
(4.34)
= 21 (~k , ~k )
(4.35)
3
1
4
~v
v3
1
= 32 (~k , 12 (~k + ~k ) )
2
~u
= 21 (~k , ~k )
1
2
1
1
2
3
2
= 12 ((~k + ~k ) , 12 (~k + ~k ))
(4.36)
= 21 (~k , ~k )
(4.37)
1
~v
3
2
3
3
96
4
4
As an example we consider the kinetic energy term in the above coordinates. The free
Hamiltonian is given as
X ki
H =
:
(4.38)
i 2m
Using the rst set of coordinates (4.32) - (4.34), one obtains
(4.39)
H = 2KM + 2u + 2uM + u^
2M`
`
with M = 4m; = m ; M` = m and M^ ` = m. This leads to (when taking the c.m.
mass momentum K~ = 0)
(4.40)
H = um + 43m u + 32m u :
Using the second set of coordinates (4.35) - (4.37), one obtains
(4.41)
H = 2v + 2v + v~
2M`
where = m and M~ ` = m. (The latter results from the reduced mass M = m + m =
m ), or
(4.42)
H = vm + vm + 2vm :
2
0
2
2
2
2
1
2
3
0
2
2
3
3
4
2
1
0
2
2
2
3
2
2
2
1
2
3
0
1
1
~
2
1
2
2
2
1
2
3
2
`
1
2
0
In addition one needs the relative orbital, spin and total angular momenta and isospins.
Thus we get two sets of partial wave projected momentum space basis states
j u u u ; ai
(4.43)
j v v v ; bi
(4.44)
where a,b comprise all discrete quantum numbers. It is natural to present and as
hu u u ; a j i
(4.45)
hv v v ; b j i ;
(4.46)
respectively. The detailed and explicit representations of all the operators in that basis
are given in H. Kamada et al, Nucl. Phys. A548 (1992) 205
1
2
3
1
2
3
1
1
1
2
2
3
2
1
3
2
For a realistic calculation, allowing the NN force to act in two-nucleon states up to jmax =
4 one needs about 400 sets of discrete quantum numbers a,b. More physically spoken one
has to take into account that the 3N subclusters are in the states of total angular momenta
and parities 1=2; 3=2 and 5=2 and the two-body subclusters in the states 0 up to 4.
The resulting discretized 4-body kernel is huge, of the order of 10 10 , and the huge
eigenvalue problem is handled by a Lanczos type method. For all that see the Reference
given above.
+
7
97
7
4.3 Results for the -Particle Ground State
Fully converged four nucleon binding energies are shown now for various NN forces
potential
Bonn B
Nijm 78
Paris
Ruhrpot
AV14
pd
4.99
5.39
5.77
5.85
6.08
E3 H
8.14
7.63
7.46
7.64
7.68
E4 He
27.07
25.10
24.32
24.60
24.73
Also triton binding energies and deuteron d-states probabilities pd are given. We see
that the -particle is theoretically underbound. Since we do not include the Coulomb
force one has to compare to the Coulomb corrected "experimental" value 29 MeV. The
underbinding is about 4-5 MeV except for Bonn B, where it is only 2 MeV. This looks big
but again in relation to the total potential energy of about -120 MeV the defect is only a
few percent. Nevertheless 3NF's are needed in addition to the present day NN forces.
In applications one does not access the wave function directly but rather only certain
matrix elements thereof. On example, which can be "measured" at least approximately
in electron scattering, is the momentum distribution, the probability to nd a nucleon
with momentum p in the nucleus.
The nucleon momentum distribution n(p) is shown for the lightest systems. The calculations are based on AV 14.
Another quantity to characterize the wave function is the NN correlation function
C (r) = h j (~r , ~rij ) j i :
(4.47)
h j i
It represents the probability to nd two nucleons at a distance ~r in d; H and He.
3
4
The calculations are based on AV 14. The correlation function for He is remarkably close
to the one of H and the deuteron for r 1fm, a feature already seen in H in relation
to the deuteron.
4
3
3
In the following, the state-dependent correlation functions
C`sjt(r) h j (~r , ~rij ) P`sjt j i
98
(4.48)
are shown for the largest partial wave projected functions. Calculations are based on the
AV 14 potential.
99
Again we see a strong similarity.
Finally we show the most probable states for the four nucleons in the -particle ground
state. It is a tetrahedron with approximate pair distances of 1 fm.
100
The inclusion of 3NF's into the Yakubovsky scheme is interesting. There are obviously
4 3NF's, where each one is split naturally into 3 pieces as shown in chapter 2. One can
nd an optimal distribution of the 12 pieces over the 18 Yakubovsky components, which
makes the equations as simple as possible see (W. Glockle et al, Nucl. Phys. A 560 (1993)
541).
Only calculations with a strongly truncated set of channels exist up to now.
As an open problem sticks out the study of the "excited states" of the -particle (from
D.R. Tilley, H.R. Weller, G.M. Hale, Nucl. Phys. A541, 1 (1992):
101
The states are all resonances and it would be highly interesting to locate the S -matrix
pole positions in the complex energy plane for all these states. It would be important to
determine their widths. Also since the spatial arrangements of the four nucleons in the
various excited states might be quite dierent, the action of possible 3 NF's can be quite
dierent from state to state.
Therefore that spectrum would be an ideal testing ground for the nuclear Hamiltonian.
102