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The following is stated in (among others) the articles

The three discrete symmetries Time reversal symmetry ($T$), Particle-hole symmetry ($C$) and Chiral symmetry ($S$) all commute (are real symmetries) of the full second quantized Hamiltonian, $$\hat{H}=\sum_{A,B}\Psi_A^\dagger \mathcal{H}_{AB}\Psi_B. $$ That is, $$[\hat{T},\hat{H}]=[\hat{C},\hat{H}]=[\hat{S},\hat{H}]=0.\;\;\;\;\;\;\; (1)$$ They go on to claim that, $\hat{C}$ and $\hat{S}$ are not real symmetries in the "classical" sense of the first quantized Hamiltonian $\mathcal{H}$. In fact, the first quantized Hamiltonians obey,

$$[T,\mathcal{H}]=\{C,\mathcal{H}\}=\{S,\mathcal{H}\}=0. \;\;\;\;\;\; (2)$$ How can I prove that imposing Eq. (1) implies Eq. (2)?

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marked as duplicate by Jens Roderus, Community Apr 4 '18 at 13:08

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