I am currently dealing with an operator-valued function $f(\hat{T})$ of the following kind: $$f(\hat{T}) =\sqrt{1 + b\hat{T}^2} $$ where $b$ $\in \mathbb{R}$ and $\hat{T}$ is a linear operator acting on the usual Hilbert space of quantum mechanics, i.e. $L^2(\mathbb{R})$. What I am interested in is to define its action on a generic physical state $|\Psi\rangle$ in the representation of its own eigenbasis (of course I am supposing that this exists).
If I am correct, I could proceed with a series expansion of the previous operator. This leads me to write: $$\sqrt{1 + b\hat{T}^2}=\sum_{n=0}^{\infty} c_n (b\hat{T}^2)^n $$ where $c_n$ is the proper generalized binomial coefficent and the series should converge for $\left\|b\hat{T}^2\right\|\leq 1$.
This should not be a problem in principle, nevertheless, it turns out that, in my framework, the operator $\hat{T}$ is the usual momentum operator of quantum mechanics $\hat{p}$, which - to my knowledge - is unbounded with respect its domain of definition (which should be a dense subset of the Hilbert space $L^2(\mathbb{R})$). From this, I deduce that the previous procedure is not suitable for the present case. Hence what can be done? My question can be exposed from two different points of view:
1)How can I find the action of this operator if the series expansion fails to be a proper instrument?
2)Is it possible to "impose" that the momentum operator $\hat{p}$ be bounded - like some kind of constraint, which I imagine could lead to a redefinition of the operator itself? From a mathematical and physical point of view is this even possible?
Thank you all in advance and sorry for my possible lack of mathematical rigour.
Having such a property satisfied indeed requires a redefinition of the operator. This can be done by carefully choosing the domain of the operator. For example, take: $\hat{T}^2 : H^{2}(\mathbb{R}) \to L^2(\mathbb{R})$, i.e., define the domain of this operator to be $H^{2} (\mathbb{R})$, a Sobolev space containing functions that are $L^2$ together with their weak derivatives up to second order. In such a case, $\hat{T}^{2}$ can be shown to be a bounded operator since the Laplacian is bounded, see, e.g., here. Note that it's common to define Hamiltonians on $H^2$ in quantum mechanics, see, e.g., from Quantum to Classical Molecular Dynamics by Christian Lubich.