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For a general unitary symplectic invariant operator AΓ AJ, although it can happen [see Eq. 62].
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Note that P+ P+J =0= P+J P+, since Φ+ (1J) Φ+ =0, which implies [P+, P+J] =0. Also [P+J, FJ] = [P+, F]J =0, because F Φ+ = Φ+ . Finally, [F, FJ] ik = Ji,Jk - Ji,Jk =0 for any basis vector ik.
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Observe that (P+ A A′ FB B′) T A′ B′ = (dB / dA) FA A′ P+ B B′ and analogously (P+ A A′ 1B B′) T A′ B′ = (1/ dA) FA A′ 1B B′, and (1A A′ FB B′) T A′ B′ = dB 1A A′ P+ B B′.
-
Observe that (P+ A A′ FB B′) T A′ B′ = (dB / dA) FA A′ P+ B B′ and analogously (P+ A A′ 1B B′) T A′ B′ = (1/ dA) FA A′ 1B B′, and (1A A′ FB B′) T A′ B′ = dB 1A A′ P+ B B′.
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The other possibilities give nothing new as (J1) P+ (J† 1) = P+J and, trivially, JJF J† J† =F.
-
The other possibilities give nothing new as (J1) P+ (J† 1) = P+J and, trivially, JJF J† J† =F.
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