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Hauke Reddmann
Hauke Reddmann
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Assume these are general 6j symbols with multiplicity labels and "spins" which are not necessarily selfconjugate (say $SU_3$). From a programming viewpoint, in some use cases it would be far more elegant (especially as if the triad is not "legit", the CG symbol is 0) to write

Clebsch$[l1,l2,l3,l4,j1,j2,j3,j4,j5,j6]=$

Triad$[l1,j1,j2,j3,$

$l2,j1',j5,j6',$

$l3,j2',j4',j6,$

$l4,j3',j4,j5']$

=Triad$[t1,t2,t3,t4]$

where j are spins, l multiplicity labels and ' means conjugation. The triads should be sortable: If $t1=[l1,j3,j2,j1]$ is used instead, the 6j symbol should still be reconstructable uniquely from $t1,t2,t3,t4$. This is surely true for six different spins, but...

Give four triads with them and the spins sorted in any order, but correct conjugation (remember: $t1\neq t1'=[l1,j1',j2',j3']$). (Quick check: Does each spin occur as a pair $j,j'$ in the twelve spin arguments of the triads?) Can you uniquely reconstruct the 6j symbol (modulo its symmetries) defined by them? Or is there a "bad" case where two principially different 6j symbols have the same set of sorted triads?

(There are only a finite number of cases from six equal to six different spins, but the conjugation problem made me shriek back from solving it by brute force. EDIT: I wrote a program, which is not yet complete, but the number of different "spin content" of principially different 6j symbols is around ~1000. Multiply with 5 label sets and 30*24 permutations and you get somewhere between 10000 and 100000 triad sets. This seems manageable for a brute force duplicate check.)

Assume these are general 6j symbols with multiplicity labels and "spins" which are not necessarily selfconjugate (say $SU_3$). From a programming viewpoint, in some use cases it would be far more elegant (especially as if the triad is not "legit", the CG symbol is 0) to write

Clebsch$[l1,l2,l3,l4,j1,j2,j3,j4,j5,j6]=$

Triad$[l1,j1,j2,j3,$

$l2,j1',j5,j6',$

$l3,j2',j4',j6,$

$l4,j3',j4,j5']$

=Triad$[t1,t2,t3,t4]$

where j are spins, l multiplicity labels and ' means conjugation. The triads should be sortable: If $t1=[l1,j3,j2,j1]$ is used instead, the 6j symbol should still be reconstructable uniquely from $t1,t2,t3,t4$. This is surely true for six different spins, but...

Give four triads with them and the spins sorted in any order, but correct conjugation (remember: $t1\neq t1'=[l1,j1',j2',j3']$). (Quick check: Does each spin occur as a pair $j,j'$ in the twelve spin arguments of the triads?) Can you uniquely reconstruct the 6j symbol (modulo its symmetries) defined by them? Or is there a "bad" case where two principially different 6j symbols have the same set of sorted triads?

(There are only a finite number of cases from six equal to six different spins, but the conjugation problem made me shriek back from solving it by brute force.)

Assume these are general 6j symbols with multiplicity labels and "spins" which are not necessarily selfconjugate (say $SU_3$). From a programming viewpoint, in some use cases it would be far more elegant (especially as if the triad is not "legit", the CG symbol is 0) to write

Clebsch$[l1,l2,l3,l4,j1,j2,j3,j4,j5,j6]=$

Triad$[l1,j1,j2,j3,$

$l2,j1',j5,j6',$

$l3,j2',j4',j6,$

$l4,j3',j4,j5']$

=Triad$[t1,t2,t3,t4]$

where j are spins, l multiplicity labels and ' means conjugation. The triads should be sortable: If $t1=[l1,j3,j2,j1]$ is used instead, the 6j symbol should still be reconstructable uniquely from $t1,t2,t3,t4$. This is surely true for six different spins, but...

Give four triads with them and the spins sorted in any order, but correct conjugation (remember: $t1\neq t1'=[l1,j1',j2',j3']$). (Quick check: Does each spin occur as a pair $j,j'$ in the twelve spin arguments of the triads?) Can you uniquely reconstruct the 6j symbol (modulo its symmetries) defined by them? Or is there a "bad" case where two principially different 6j symbols have the same set of sorted triads?

(There are only a finite number of cases from six equal to six different spins, but the conjugation problem made me shriek back from solving it by brute force. EDIT: I wrote a program, which is not yet complete, but the number of different "spin content" of principially different 6j symbols is around ~1000. Multiply with 5 label sets and 30*24 permutations and you get somewhere between 10000 and 100000 triad sets. This seems manageable for a brute force duplicate check.)

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Hauke Reddmann
Hauke Reddmann
  • 4.8k
  • 1
  • 18
  • 24

"Uniqueness" of 6j symbols via triads

Assume these are general 6j symbols with multiplicity labels and "spins" which are not necessarily selfconjugate (say $SU_3$). From a programming viewpoint, in some use cases it would be far more elegant (especially as if the triad is not "legit", the CG symbol is 0) to write

Clebsch$[l1,l2,l3,l4,j1,j2,j3,j4,j5,j6]=$

Triad$[l1,j1,j2,j3,$

$l2,j1',j5,j6',$

$l3,j2',j4',j6,$

$l4,j3',j4,j5']$

=Triad$[t1,t2,t3,t4]$

where j are spins, l multiplicity labels and ' means conjugation. The triads should be sortable: If $t1=[l1,j3,j2,j1]$ is used instead, the 6j symbol should still be reconstructable uniquely from $t1,t2,t3,t4$. This is surely true for six different spins, but...

Give four triads with them and the spins sorted in any order, but correct conjugation (remember: $t1\neq t1'=[l1,j1',j2',j3']$). (Quick check: Does each spin occur as a pair $j,j'$ in the twelve spin arguments of the triads?) Can you uniquely reconstruct the 6j symbol (modulo its symmetries) defined by them? Or is there a "bad" case where two principially different 6j symbols have the same set of sorted triads?

(There are only a finite number of cases from six equal to six different spins, but the conjugation problem made me shriek back from solving it by brute force.)