what is eigenvalue of $P^{1/n}$ operator if we know eigenvalue equation of $P$ ? If $P$ is an operator and $PΨ=pΨ$ ( $p$ as the eigenvalue ) then is it true to say $P^{1/n}Ψ=p^{1/n}Ψ$ ( n is an integer and positive number )
 A: In general, the operator $P^{1/n}$ is relatively hard to define: we have sort of the same problems we have with complex numbers (there are in general multiple $n$th roots of any nonzero $z\in\mathbb C$) except that we have them separately on every dimension, which sort of mostly kills the whole thing. 
However, if $P$ is positive semidefinite, and self-adjoint, then there are good chances that it will have a unique self-adjoint positive semidefinite $n$th root. Such an operator $Q$ is then characterized by


*

*having all-nonnegative eigenvalues, and

*$Q^n\psi=P\psi$ for all $\psi$.


If $P$ has a discrete eigenbasis $\{\psi_k\}$ with eigenvalues $\{p_k\}$, then the operator $Q$ such that
$$Q\psi_k=p_k^{1/n}\psi_k  \tag{$\ast$} $$
obviously satisfies these criteria, and therefore that's the operator you want. In general, it's very hard for an operator that doesn't satisfy $(\ast)$ to make any sense as an $n$th root of $P$ (although I can't discount some pathological case outright) and generally you can always assume such a relationship.
