Functions of operators Consider the function $e^{\Omega}$ where $\Omega$ is a Hermitian operator. We can show that this function is well defined by going to the eigenbasis of $\Omega$ and studying the convergence of the power series involving the eigenvalues of the operator.
In page 55 of R. Shankar's Principles of Quantum Mechanics are a few lines:

$e^{\Omega}$ is indeed well defined by the power series in this basis (and therefore in any other). 

Could one give me hints on how to proceed with the mathematical reasoning for the function being well defined in any basis other than the eigenbasis? 
 A: One should use the spectral theorem to properly define such exponential. Let $\Omega$ be a (possibly unbounded) self-adjoint operator, with domain $D(\Omega)$. Then by the spectral theorem there exists a projection valued measure $\mathrm{d}P_{\lambda}$ such that
$$\Omega = \int_{\mathbb{R}}\lambda \mathrm{d}P_{\lambda}\; .$$
More precisely, the action of $\Omega$ on any $\phi,\psi\in D(\Omega)$ can be written as
$$\langle\phi, \Omega\psi\rangle=\int_{\mathbb{R}} \lambda \mathrm{d}(\langle\phi,P_\lambda \psi\rangle)<\infty\; ,$$
where $\mathrm{d}(\langle\phi,P_\lambda \psi\rangle)$ is the spectral measure evaluated on $\phi,\psi$ (and it is a complex measure on the reals).
The operator $e^{\Omega}$ is a densely defined operator acting as follows:
$$e^{\Omega}=\int_{\mathbb{R}}e^{\lambda} \mathrm{d}P_{\lambda}\; .$$
One can prove that it is indeed densely defined (it is just another form of the spectral theorem), and the domain of definition consists of such vectors $\psi$ for which
$$\lVert e^\Omega\psi\rVert^2=\int_{\mathbb{R}} e^{2\lambda} \mathrm{d}(\langle\psi,P_\lambda \psi\rangle)<\infty\; .$$
As it is easy to see, such definition works well in any basis, and if the operator has continuous spectrum. If $\Omega$ has compact resolvent, then there exists an orthonormal basis of eigenvectors $\{\psi_n\}_{n\in \mathbb{N}}$. In that case, the spectral representation is much easier, and it is given by
$$\Omega= \sum_{n\in \mathbb{N}}\lambda_n \lvert\psi_n\rangle\langle\psi_n\rvert\; .$$
It then follows that the exponential is defined by
$$e^{\Omega}=\sum_{n\in \mathbb{N}}e^{\lambda_n} \lvert\psi_n\rangle\langle\psi_n\rvert\; ,$$
and the domain of definition (the vectors on which the "series converges") are those vectors $\psi$ such that
$$\sum_{n\in \mathbb{N}}e^{2\lambda_n}\lvert\langle\psi_n,\psi\rangle\rvert^2<\infty\; .$$
A: I'm a bit rusty on this, but I suspect that if H is the said hermitian operator, B is your new space phase, then there's an invertible matrix T such that H' in B can be written $H' = T^{-1}HT$ (property of hermitian operators)
So all the terms of type $\frac{H'^{n}}{n!}$ in $e^{H}$ will be written as  $T^{-1}H^{n}T$ (because of the product $TT^{-1} = 1$)
Hence $e^{H}$ is well defined (and once can demonstrate that $<\varphi | e^{H} | \varphi> = <\varphi | e^{H'} | \varphi>$ for all wave functions using the same argument)
Edit: formatting...
