Schur マニュアルの問題を解く　その３

6-6節 Exercises 4-6

問４

4. In the simple $SU_3$ quark model of baryons and mesons the $(u,d,s)$ quarks span the threedimensional $SU_3$ irreducible representation $\{1\}$ while the corresponding antiquarks $(u,d,s)$ span the three-dimensional $SU_3$ irreducible representation $\{1^2\}$ . The $SU_3$ group is sometimes referred to as the flavour group $SU^{fl}_3$ .
(a) Show that combining quarks with antiquarks leads to an octet and singlet of mesons.
(b) Show that combining a triple product of quarks leads to a baryon singlet and two baryon octets.

素粒子クオーク模型

REP> gr su3
Group is SU(3)

REP> prod 1, 1^2
{21} + {0}
REP> dim 21
dimension = 8
REP> dim 0
dimension = 1

REP> prod prod 1,1,1
{3} + 2{21} + {0}
REP> std last 
{3} + 2{21} + {0}
REP> dim 3
dimension = 10
REP>

(b) で {3} も出てきてよく分からない。　

問５

5. A student tries to unite the lowest lying baryons into a single irreducible representation of a Lie group G. Noting that the baryon octet has spin $J_p = {1\over2}^+$ and the decuplet spin $J_p = {3\over2}^+$ identify G and the relevant irreducible representation she found.

よく分からんが･･･

spin SU(2) * quark SU(3) なので Lie 群的に SU(6)→SU(2)*SU(3) か？
spin3/2, decuplet → 4*10=40 次元, spin1/2, octet → 2*8=16 次元となる。
問４で SU(3) の {3} は１０重項、{21} は八重項と分かっているので、求める規約表現は {3}{3} + {1}{21} となるはず。

以下の計算で、SU(2) $\{\lambda_1, \lambda_2\}=\{\lambda_1-\lambda_2,0\}$ であることを考えると、SU(6) の規約表現 {3} が SU(2)xSU(3) では {3}{3}+{21}{21} = {3}{3}+{1}{21} と見なせるので、これが求める答えだと思ふ。

BRM> stop
enter branching & rule numbers> 5 2,3
U(6) to U(2) * U(3) 
BRM> 0
{0}{0}
BRM> dim 0
Irrep modifies to null
BRM> 0
{0}{0}
BRM> 1
{1}{1}
BRM> 2
{2}{2} + {1^2}{1^2}
BRM> 3
{3}{3} + {21}{21}
BRM> 1^2
{2}{1^2} + {1^2}{2}
BRM> 1^3
{3}{1^3} + {21}{21}
BRM> quit

問６

6. Assume that you have the group $SU^{fl}_3$ , as in exercise 5. Your quarks are now also endowed with spin $(SU^s_2)$ and color $(SU^c_3)$ . The total of 18 single quark states span the vector irreducible representation $\{1\}$ of $SU_{18}$ and you have the chain of groups
$SU_{18}\supset SU^{fl}_3\times(SU^{cs}_6\supset SU^{s}_2\times SU^{c}_3)$
Assume that each quark is in an $S-$ state and that the states of the six quark configuration
$q^6$ , span the totally antisymmetric irreducible representation $\{1^6\}$ of $SU_{18}$ . In terms of QCD (Quantum ChromoDynamics) physical states correspond to color singlets i.e. states transforming as $\{0\}$ under $SU^c_3$ . With respect to the color-spin group $SU^{cs}_6$ this can only happen if the weight $w_\lambda$ of the partition $\lambda$ labelling the irreducible representation $\lambda$ of $SU^{cs}_6$ is divisible by three. That is for irreducible representations of null triality.
(a) Determine the irreducible representations of $SU^{fl}_3\times SU^{cs}_6$ contained in $\{1^6\}$ of $SU_{18}$ that have null triality.
(b) Determine the spin content of each $SU_6^{cs}$ found in (a).
(c) Which of the multi-quark configurations $q^7, q^8$ and $q^9$ might you expect to contain color singlet states?
(d) In the MIT bag model of multi-quark states there is an energy term proportional to the
eigenvalues of the second-order Casimir operator of the color-spin group. Show that
the eigenvalues for the irreducible representations $\{21^4\}, \{2^3\}$ and $\{3^2\}$ are in the ratio 1:2:3.

謎が多いが･･･

(a)
U(18) ⊃ U(3) * U(6) で、U(6) の λi がみな３の倍数になっていて null triality の条件をみたしているのは、{2^3}{3^2} の項だけ。

DP> brm
Branch Mode
enter branching & rule numbers> 5 3,6
U(18) to U(3) * U(6) 
BRM> 1^6
{6}{1^6} + {51}{21^4} + {42}{2^2 1^2} + {41^2}{31^3} + {3^2}{2^3} + {321}{321} + {2^3}{3^2}

(b)
SU(6) ⊃ SU(2) * SU(3) で、U(2) のスピン成分は $\{\lambda_1, \lambda_2\}=\{\lambda_1-\lambda_2,0\}$ であるから、{6}={6}, {51}={4}, {42}={2}, {3^2}={0} となる。

BRM> stop
enter branching & rule numbers> 5 2, 3
U(6) to U(2) * U(3) 
BRM> 3^2
{6}{3^2} + {51}{42} + {51}{321} + {42}{51} + {42}{41^2} + {42}{3^2} + {42}{321} + {3^2}{6} + {3^2}{42} + {3^2}{2^3}

SU(3) の方は {321}={21}, {41^2}={3}, {2^3}={0} となるので、{3^2}{2^3}={0}{0} が color singlet と思われる。（が、よく分からんｗ　３の倍数でアホになります）

BRM> stop
enter branching & rule numbers> 5 3 6
U(18) to U(3) * U(6) 
BRM> 1^7
{61}{21^5} + {52}{2^2 1^3} + {51^2}{31^4} + {43}{2^3 1} + {421}{321^2} + {3^2 1}{32^2} + {32^2}{3^2 1}
BRM> 1^8
{62}{2^2 1^4} + {61^2}{31^5} + {53}{2^3 1^2} + {521}{321^3} + {4^2}{2^4} + {431}{32^2 1} + {42^2}{3^2 1^2} + {3^2 2}{3^2 2}
BRM> 1^9
{63}{2^3 1^3} + {621}{321^4} + {54}{2^4 1} + {531}{32^2 1^2} + {52^2}{3^2 1^3} + {4^2 1}{32^3} + {432}{3^2 21} + {3^3}{3^3}
BRM> exit

(d)
SU(6)群で規約表現 {21^4}, {2^3}, {3^2} の性質を書かせて、２次のカシミール演算子のところを見ると、72:144:216＝1:2:3 になっている。

DP> gr su6
Group is SU(6)
DP> rep
REP mode       
Group is SU(6)
REP> prop 21^4
<dynkin label> (10001)
dimension = 35   72*2nd-casimir=72
2nd-dynkin = 12

REP> prop 2^3
<dynkin label> (00200)
dimension = 175   72*2nd-casimir=144
2nd-dynkin = 120

REP> prop 3^2
<dynkin label> (03000)
dimension = 490   72*2nd-casimir=216
2nd-dynkin = 504

REP>

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Schur マニュアルの問題を解く　その３

6-6節 Exercises 4-6

問４

問５

問６