- #1
BillKet
- 313
- 29
Hello! I have a question about this paper. They claim that the hyperfine and spin-rotational terms can be treated perturbatively (they do perform a full diagonalization, too, but they claim that perturbation theory is good to get an estimate of the effect). I agree with that for most of the terms, except for ##c(I\cdot n)(S\cdot n)##. For example, let's assume we use as the level crossing the states (using the notation in order N, S, I):
$$\psi_+ = |0,0>|1/2,1/2>|1/2,1/2>$$
and
$$\psi_- = |1,1>|1/2,-1/2>|1/2,1/2>$$
The operator ##c(I\cdot n)(S\cdot n)## can be expanded (I will ignore some constants, I will write just the operators) as ##c(I_zY_1^0+I_+Y_1^{-1}+I_-Y_1^1)(S_zY_1^0+S_+Y_1^{-1}+S_-Y_1^1)## and among these, the term ##cI_zY_1^0S_-Y_1^1## doesn't seem to vanish when calculated between ##\psi_+## and ##\psi_-## i.e.
$$<\psi_-|cI_zY_1^0S_-Y_1^1|\psi_+> \neq 0$$
which is about equal to c. But when doing perturbation theory, the effect of this term would be about ##\frac{c}{E_+-E_-}## and while c is very small, ##E_+-E_-## is much smaller (ideally as small as the parity violation effect), so I don't see how this can be treated pertubatively. Am I missing something? Is that matrix element actually vanishing? Thank you!
$$\psi_+ = |0,0>|1/2,1/2>|1/2,1/2>$$
and
$$\psi_- = |1,1>|1/2,-1/2>|1/2,1/2>$$
The operator ##c(I\cdot n)(S\cdot n)## can be expanded (I will ignore some constants, I will write just the operators) as ##c(I_zY_1^0+I_+Y_1^{-1}+I_-Y_1^1)(S_zY_1^0+S_+Y_1^{-1}+S_-Y_1^1)## and among these, the term ##cI_zY_1^0S_-Y_1^1## doesn't seem to vanish when calculated between ##\psi_+## and ##\psi_-## i.e.
$$<\psi_-|cI_zY_1^0S_-Y_1^1|\psi_+> \neq 0$$
which is about equal to c. But when doing perturbation theory, the effect of this term would be about ##\frac{c}{E_+-E_-}## and while c is very small, ##E_+-E_-## is much smaller (ideally as small as the parity violation effect), so I don't see how this can be treated pertubatively. Am I missing something? Is that matrix element actually vanishing? Thank you!