Baker Campbell Hausdorff formula for unbounded operators Baker Campbell Hausdorff formula says that for elements $X,Y$ of a Lie algebra we have
$$e^Xe^Y=\exp(X+Y+\frac12[X,Y]+...),$$
which for $[X,Y]$ being central reduces to
$$e^Xe^Y=\exp(X+Y+\frac12[X,Y]).$$
As I am not an expert in Lie algebras, I was wondering if this setup is suitable to say that thus a formula like this holds when $X,Y$ are unbounded operators with common core such that $[X,Y]=i$. If so, I would be grateful for the reference to a proof in that particular case.
Motivation: An important identity in quantum mechanics is the decomposition of the Weyl operator as an exponential of position and momentum operators:
$$W(z_1,...,z_n)=\exp(-i\sqrt{2}\sum_{j}(x_jp_j-y_jq_j)).$$
When I was trying to prove this in an explicit way (straight from the definition of the Weyl operator), there comes a point when I need to use the BCH formula. Since for unbounded operators I cannot use the usual expressions for exponential (as a sum of a convergent series), this worries me slightly.
 A: Let $\mathfrak{g}$ be a Lie algebra, and let $G$ be the (unique up to isomorphism) simply-connected group with Lie algebra $\mathfrak{g}$. Here, $\mathfrak{g}$ and $G$ should be thought as "abstract", with the exponential map $\exp : \mathfrak{g} \rightarrow G$ defined in terms of the 1-parameter sub-groups in $G$. The BCH formula is fulfilled for this "abstract" exponential map, although the completly general proof is a bit lengthy. It involves deriving a formula for the derivative of the exponential mapping, then proving $\exp \big( [X,\cdot] \big) Y = \exp (X) Y \exp (-X)$ and finally deducing the BCH formula. A concise presentation can be found eg. in chapter 3 of Brian Hall's "Lie Groups, Lie Algebras, and Representations". This reference aims at being fairly accessible, at the price of restricting itself to those Lie goups $G$ which can be constructed as matrix subgroups: this is the case of most groups used in practice (in particular the Heisenberg group generated by the position/momentum operators of quantum mechanics, together with the identity operator, can be constructed as a subgroup of 3x3 real matrices).
Next, let $X \mapsto i\hat{X}$ be a representation of this Lie algebra by unbounded symmetric operators on a complex Hilbert space $\mathcal{H}$ (note that we do not even need to assume that $\hat{X}$ be essentially self-adjoint, just symmetric).
Since we are dealing with unbounded operators, we have to be a bit careful as to what we mean by a representation, because a priori $\hat{X}\hat{Y}$, $\hat{Y}\hat{X}$, and therefore $[\hat{X}, \hat{Y}]$ may fail to be well-defined. Specifically, we demand that there exists a common invariant dense domain $\mathcal{D} \subseteq \mathcal{H}$, such that, for all $X \in \mathfrak{g}$, $\hat{X}$ is defined on $\mathcal{D}$ and stabilizes $\mathcal{D}$ (ie. $\hat{X} \left\langle \mathcal{D} \right\rangle \subseteq \mathcal{D}$). Then, we can define arbitrary products of the $\hat{X}$'s on $\mathcal{D}$, and in particular commutators.
The question is then whether this representation can be exponentiated, ie:


*

*for all $X \in \mathfrak{g}$, $\hat{X}$ is actually essentially self-adjoint;

*and there exists a unitary representation $U \mapsto \hat{U}$ of the group $G$ on $\mathcal{H}$ such that, for any $X \in \mathfrak{g}$, $\widehat{\exp X} = \widehat{\exp} (i\hat{X})$. Here, I denote by $\widehat{\exp}$ the "operator" exponential map, which is defined by spectral resolution of the essentially self-adjoint operator $\hat{X}$: as you observed, in the case of unbounded operators, the exponential cannot be defined in terms of the exponential series (note that the notation $\widehat{\exp}$ is non-standard, I use it here only to prevent confusion between the various notions of exponential maps).
If this holds, the BCH formula satisfied by the "abstract" exponential map $\exp$ will be inherited by the "operator" exponential map $\widehat{\exp}$.
So, with all these preliminaries in place: when can a Lie algebra representation by symmetric unbounded operators be exponentiated? A sufficient condition (useful in practice, albeit not a necessary condition) is the Nelson criterion (lemma 9.1 of Edward Nelson, "Analytic Vectors", Annals of Mathematics, Second Series, Vol. 70, No. 3 (Nov., 1959), pp. 572-615): it demands that there exists a basis $X_1,\dots,X_n$ of $\mathfrak{g}$, a dense subdomain $\mathcal{D}_o \subseteq \mathcal{D}$, and a real $s > 0$, such that:
$$
\forall \psi \in \mathcal{D}_o,\; \sum_{m=0}^{\infty} \frac{s^m}{m!} \sum_{k_1,\dots,k_m} \left\| X_{k_1} \dots X_{k_m} \psi \right\| < \infty.
$$
This is a fairly technical result, but the intuition behind it is that this condition is precisely what you need to define $\widehat{\exp} (i\hat{X})$ over $\mathcal{D}_o$ via the exponential series, and, from there, deduce that $\hat{X}$ is indeed essentially self-adjoint (using the Stone's theorem discussed below), that this definition of $\widehat{\exp} (i\hat{X})$ coincides over $\mathcal{D}_o$ with the spectral one, and finally, that this indeed gives you a unitary representation of $G$ (matching the "abstract" BCH-formula with the one that can be proven directly using the exponential series).
In the case of the position/momentum operators of quantum mechanics, one can for example take $X_1 = \text{id}$, $X_2 = q$, $X_3 = p$ and take $\mathcal{D} = \mathcal{D}_o$ to be spanned by finite linear combination of the harmonic osciallator energy eigenstates. Then, the condition can be proven using the expression of the position/momentum operators in terms of ladder operators.
However, coming back to your specific motivation, I do not honestly think that you need all this machinery. Instead, you can use the explicit expression of the Weyl operators to prove that $t \mapsto W(tz)$ is a strongly continuous one-parameter unitary group: ie. $W(sz) W(tz) = W\big((s+t)z\big)$ and, for any $\psi \in \mathcal{H}$, $t \mapsto W(tz) \psi$ is continuous (with respect to the norm of $\mathcal{H}$; in this case, the $L_2$-norm). Then, Stone's theorem (theorem VIII.8 of Reed and Simon, "Methods of Modern Mathematical Physics", volume 1) tells you that $W(tz) = \widehat{\exp} (it\hat{X})$ with $\hat{X}$ the self-adjoint operator defined by:
$$i \hat{X} \psi = \left. \frac{d}{dt} W(tz) \psi \right|_{t=0}$$
for any $\psi$ in the dense domain $\mathcal{D}$ where this derivative exists. Using again the explicit expression of $W(tz)$ you can then check that $\hat{X} = \sqrt{2} (y\hat{q} - x\hat{p})$.
