m***x
发帖数: 492 | 1
先求解：A |an> = an |an>
if an is not degenerated, A B |an> = B A|an>=an B|an>, then B|an> is also
eigenvector of A so B|an> is only different from |an> by a constant, so just
calculate which is the eigenvalue for B's eigenvector |an>
if an 简并(degree as s)，AB|an(i)>=BA|an(i)>=anB|an(i)> so B|an(i)> is still
eigenvector of A, B|an(i)> = \sum_{i}{ Bii'|an(i')}> where
Bii'=
solve this problem as degenerated problem, you will get answer. |
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