ABSTRACT
Odd tensor operators play an important role in the seniority scheme. In the previous section we showed that components of irreducible tensor operators with odd ranks are quasi-spin scalars, s = 0. This follows from the fact that if k is odd, ( a j + × a ˜ j ) κ ( k ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203739716/ff034a11-a369-4bc6-a470-f8cf1794dcb5/content/eq1748.tif"/> commutes with S j + https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203739716/ff034a11-a369-4bc6-a470-f8cf1794dcb5/content/eq1749.tif"/> (or S j − https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203739716/ff034a11-a369-4bc6-a470-f8cf1794dcb5/content/eq1750.tif"/> ). In a subsequent section we will see how odd tensor operators may be used to define the seniority scheme and to derive additional properties. Here we just point out that two-body interactions including in their expansion only odd tensors have special properties in the seniority scheme. Looking at (19.46) we see that such odd tensor interactions V odd are a sum of a quasi-spin scalar and a term proportional to n. In order to determine the proportionality factor of the latter term, we apply V odd to the state with seniority v = 0 (and J = 0) in the j 2 configuration. The eigenvalue of Vs =0 for this state is equal to that of the state with v = 0 in any jn configuration with even n and in particular, to that of the vacuum state |0〉. Hence, the eigenvalue of Vs =0 is zero and the term in V odd proportional to n must have the eigenvalue V 0 = Vj =0(jjjj). Hence, for an odd tensor interaction we 378deduce from (19.46) the relation () − 1 2 j + 1 ∑ k odd F k = V 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203739716/ff034a11-a369-4bc6-a470-f8cf1794dcb5/content/eq1751.tif"/>
