我看了书有一段不是很了解 他说nucleon的total spin J = 1/2 而不管是uud还是udd u和d都有J = 1/2 我想问的第一个问题是从nucleon的total spin J = 1/2 可以因此知道uud或是udd是能量最低的束缚态吗? 这两种波函数是否symmetric还是anti-symmetric? 怎麽看出来的? 另外书本又提到Δ++ Its spin, J = 3/2, is obtained by combing three identical J = 1/2 u quarks in their "ground state". That is, the quark scheme forces us to combine three identical fermions u in a "completely symmetric ground state" uuu in order to accommodate the known properties of the Δ-particle. Such a state is of course forbidden by Fermi statistics. 我不懂为什麽从uuu就知道是completely symmetric ground state? 对称波函数能量不是都比较低吗? 又存在有不完全对称或反对称这种组合吗? 我猜测书中的意思是说J = 3/2 = 1/2 + 1/2 + 1/2 所以态是(up, up, up)所以是对称?这样对吗? 所以如果按照Fermi-Dirac统计 qqq的组合是不可能发生的 是吗? 谢谢回答 --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 128.220.147.133