Hello,
I read that if $A\subset D$ is the heart of a bounded t structure, D a triangulated category, then D=D^b(A).
Have not been able to find a proper reference for this. Can anyone confirm this result?
Thanks!
Hello, I read that if $A\subset D$ is the heart of a bounded t structure, D a triangulated category, then D=D^b(A). Have not been able to find a proper reference for this. Can anyone confirm this result? Thanks! 


You're never going to find a reference because it's false. Just take $D$ to be the stable homotopy category of Postnikov pieces, i.e. spectra with finitely many nontrivial homotopy groups. Take $D_{\geq 0}$ (resp. $D_{\leq 0}$) the full subcategory of spectra with nontrivial homotopy groups concentrated in nonnegative (resp. nonpositive) degrees. This is a $t$structure whose heart $D_{\geq 0}\cap D_{\leq 0}$ is equivalent to the category $Ab$ of abelian groups. There's even a functor $D^b(Ab)\rightarrow D$ preserving the $t$structure (take the canonical $t$structure on the left) and inducing an equivalence between the hearts, but $D$ cannot be equivalent to $D^b(Ab)$. Actually $D$ has no algebraic model. 


To add to what Fernando Muro said, there are examples where this is true. Beilinson showed in his paper "On the derived category of perverse sheaves" (Springer LNM 1289 pp. 2741) that it is true when you are talking about the middle perverse tstructure in the constructible derived category of sheaves on schemes. If you look through Chapter 3 of BBD ("Faisceaux pervers") you will see how subtle it is even to define a functor from $D^b(A)$ to D, and you can also see from Beilinson's solo article how different that task is from the task of showing that such a map is an equivalence. That it works at all depends quite a bit on the fact that the perverse tstructure is used. 

