$p^4$ in radial coordinates not Hermitian Griffiths' quantum textbook claims in question 6.15 that "$p^2$ is Hermitian, but $p^4$ is not, for hydrogen states with $l=0$."  
First off, I am puzzled at his use of terminology.  An operator is either Hermitian or it is not.  It seems wrong to say that an operator is "Hermitian for certain states."  
Nevertheless, even allowing this abuse of terminology, the wording is still confusing: is he implying that there are certain states (i.e. $l \neq 0$) "for which $p^2$ is not Hermitian"?  
Depending on the answer to the above question, there are two possibilities, neither of which seem possible:
1) $p^2$ is not a Hermitian operator in general, which would make the Hamiltonian non-Hermitian.  
2) $p^2$ is a Hermitian opeartor ("for all states"), which I believe then requires $p^4$ to also be Hermitian, by an obvious derivation: 
$ <f|p^4g> = <f|p^2p^2g> = <p^2f|p^2g> = <p^2p^2f|g> = <p^4f|g>$
But this contradicts what the question asks you to show.  
I have not yet gone through the question myself as it seems like a lengthy derivation, and I would first like to know what, if anything, is amiss with my contradictory results above.  
 A: Without specifying the domains of the involved operators all the discussion has not much sense. 
Let me say that, if $A :D(A) \to H$ with $D(A)\subset H$ a linear dense subspace of the Hilbert space $H$, $A$ is self-adjoint if $D(A)=D(A^\dagger)$ and $A=A^\dagger$.
Notice that consequently (I stress that the converse is false) self-adjointness implies
$\langle f| Ag\rangle = \langle Af| g \rangle \mbox{ for all } f,g \in D(A) \tag{1}$  
Relaxing the hypotesis of density od $D(A)$ and $D(A)=D(A^\dagger)$, $A$ is Hermitian, by definition, if (1) holds true.
The squared momentum operator is self-adjoint (not only Hermitian) when the domain is  $$D(P^2)=\{\psi \in L^2(\mathbb R^3, dx)\:|\: \mathbb R^3 \ni p \mapsto \: p^2\:\hat{\psi}(p)\: \mbox{belongs to } L^2(\mathbb R^3, dp)\}$$
 where $\hat{\psi}$ is the Fourier-Plancherel transform of $\psi$. $(P^2)^n$ is similarly self-adjoint in the natural domain $D(P^{2n})$ constructed inductively $D(P^{2n}) := D(P^2(P^2)^{n-1})$ where $D(AB) := \{\psi \in D(B)\:|\: B\psi \in D(A)\}$. In the examined particular case these domains are standard Sobolev spaces.  
If referring to self-adjointness (including the definition of natural domains) instead of Hermiticity and using the natural domain of $P^4$, Griffiths' statement is false. Referring to Hermiticity, the statement is false as well or at least ambiguous, because the notion of a Hermitian operator with respect to a vector is not a standard definition.  What presumably Griffiths means is that (1) is false if $A=P^4$ is viewed as a differential operator and $\psi$ belongs to an, actually not completely defined, domain $X$ of differentiable functions in the eigenspace of $L^2$ with $\ell =0$. Obviously in the intersection of $D(P^4)$ and the eigenspace with $\ell=0$ of $L^2$ Griffiths' claim is false again, due to (1) which holds for $P^4$ thereon.
A: It was an incorrect statement, as it is explained here by Griffiths himself.
Anyways, the mathematical explanation is straightforward: given a self-adjoint operator $A$ with domain $D(A)$, any sufficiently regular real function $f(A)$ of it (and the square is perfectly ok for $-\Delta=p^2$) is self-adjoint on some domain $D(f(A))$ by the spectral theorem (in its functional calculus form).
In particular, $p^4$ is self-adjoint on the Sobolev space $H^4(\mathbb{R}^3)$ defined by: $$H^s(\mathbb{R}^d)=\{f\in L^2(\mathbb{R}^d), (1+\lvert k\rvert^2)^{s/2}\hat{f}\in L^2(\mathbb{R}^d)\}\; .$$
The only thing that I expect to be possible (but I am not sure) is that some eigenfunction of the hydrogen atom is in the domain $H^2(\mathbb{R}^3)$ of $p^2$ but not in the domain $H^4(\mathbb{R}^3)$ of $p^4$.
