Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 19I / a / i

Table of contents
- Solution i

Solution

Applying the Clebsch-Gordan decomposition for SU2 gives

V_{4} \otimes V_{3} ≅ V_{7} \oplus V_{5} \oplus V_{3} \oplus V_{1},

and

V_{3} \otimes V_{3} ≅ V_{6} \oplus V_{4} \oplus V_{2} \oplus V_{0} .

On the summand

V_{6 - 2 j}

, interchange of the two tensor factors acts by

(- 1)^{j}

; this is the flip parity in the SU2 tensor square. The even-

j

summands make the symmetric square, so

S^{2} V_{3} ≅ V_{6} \oplus V_{2} .

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version