/sci/ - Science & Math

>>11307598
>I'm dealing with QM observables, so A and B have to self-adjoint
Careful, fermion creation/annihilation operators [math]c,c^\dagger[/math] are [math]not[/math] themselves observable, if they're what you're studying. Gauge non-invariant quantities aren't typically observables, unless some quantum phase transition occurs (like in a superconductor).
>Does the restriction to self-adjoint A, B change anything
If [math]C=A,B[/math] are in addition self-adjoint (SA) then [math]\operatorname{Spec}C\subset \mathbb{R}[/math] and [math]C^2 = 1[/math] implies that [math]\operatorname{Spec}C \in \{\pm 1\}[/math]. Using the spectral theorem you can see that all matrix elements of [math]A+B[/math] are bounded by 2.
If on the other hand you mean to restrict [math]C[/math] to the domain on which it is SA, then
1. [math]\{A,B\} = 0[/math] may not make any sense as the domains of their SA restrictions may be disjoint; in particular, their domains may not even intersect [math]P\mathcal{H}^*[/math].
2. Since [math]A,B[/math] aren't necessarily SA, they do not form SA extensions of their SA restrictions, so the spectral theorem only holds on the SA domains, not all of [math]\mathcal{H}[/math].
By considering SA restrictions you are basically throwing away the "bad vectors" so of course you can find tighter bounds on the quantities. But then the question becomes "how much do you lose with this restriction?"