Can one amplify P=NP beyond P=PH?
From  Russell Impagliazzo's comment:


  As a way of formalizing 
  what languages are in $\mathsf{P}$ if $\mathsf{P}=\mathsf{NP}$, 
  Regan introduced the complexity class $\mathsf{H}$. 
  A language $L$ is in $\mathsf{H}$ if and only if $L$ is in $\mathsf{P}^O$
  relative to every oracle $O$ so that $\mathsf{P}^O=\mathsf{NP}^O$. 
  Thus, $L$ is in $\mathsf{H}$ if the statement 
  $\mathsf{P}=\mathsf{NP} \implies L\in\mathsf{P}$ relativizes. 
  $\mathsf{PH} \subseteq \mathsf{H} \subseteq \mathsf{AltTime}(O(\lg\lg n),\mathsf{poly})$. 
  From Toda's theorem, and some of the lemmas in Toda's theorem, 
  it is also true that $\mathsf{H} \subseteq \mathsf{P}^{\mathsf{mod}_q \mathsf{P}}$ for every $q$. 
  Basically, any oracle satisfying $\mathsf{P}^O=\mathsf{NP}^O$ 
  gives a new upper bound on $\mathsf{H}$. 
  It is open whether $\mathsf{H}=\mathsf{PH}$.


And from Lance Fortnow's comment:


  Let $f(n)$ be any unbounded function. 
  $\mathsf{H}$ is not contained in $\mathsf{AltTime}(f(n),\mathsf{poly})$ and 
  if you could prove 
  $\mathsf{P}=\mathsf{NP}$ implies $\mathsf{P}=\mathsf{AltTime}(f(n),\mathsf{poly})$ then $\mathsf{NP}$ is different than $\mathsf{L}$.


For definition of $\mathsf{H}$ see definition 6.3 in


Kenneth W. Regan, "Index sets and presentations of complexity classes", 1999