AlbertYang的甘苦滋味

http://en.wikipedia.org/wiki/Triplet_state

問題是這個樣子的

在上面的Wiki link中

提及了一些關於電子的組態

可是也不曉得為啥 就assign某3個組合為triplet

另一個組合就是singlet

想了一陣子

這邊是我的想法 也請網友多指教囉

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

The spin multiplicity (total spin momentum) for 2e^- can

be regarded as "2S+1". For S = 1 (2e- are arranged in parallel

configuration), the sub-configurations (combination) of e- pairs

are 1, 0, and -1 which can be regarded as the vector, S, projects

on to an imaginary z-axis (as the external magnetic field

orientation). Easily, we know that the S=0 is an anti-parallel

configuration, is paired.

Graphically, we can "draw" that the combination of 2e- are

1(↑)2(↑), 1(↓)2(↓), and 1(↑)2(↓). But, because 2e- are

fundamentally indistinguishable, the 1(↓)2(↑) configuration

also must take into account. That is, we must consider the

linear combination of a*1(↑)2(↓)+b*1(↓)2(↑) as a possible

solution under operation of spin operator (this also can be

regarded as Zeeman or Stark effect). a and b are arbitrary

coefficient. We define the combination as

[1(↑)2(↓)+/-1(↓)2(↑)]/(2)^0.5 to use.

So we got 4 spin combination for all possible e- orientations:

1(↑)2(↑)

1(↓)2(↓)

[1(↑)2(↓)+/-1(↓)2(↑)]/((2)^0.5)

The first 3 spin wavefunction is symmetric, and the last one

is anti-symmetric.

When starting considering e- configuration of H2, the

Pauli exclusion principle must be taken. That is, the interchange

of 2e- must obey the antisymmetric requirement. This can be

accomplished by putting on the spatial part of H2 (這邊沒有解).

So the first 3 wavefunction is corresponding to higher energy

(excited state) which is solved by spatial part, the last one

corresponds to lower energy which is named ground state. The

excited state of H2 is triple degeneracy (triplet) due to

its possible spin configurations, and the ground state is singlet.