Fock Space: Formal Adjoints Problem
Given a pre-Hilbert space $\mathcal{H}$.
Consider unbounded operators:
$$S,T:\mathcal{H}\to\mathcal{H}$$
Suppose they're formal adjoints:
$$\langle S\varphi,\psi\rangle=\langle\varphi,T\psi\rangle$$
Regard the completion $\hat{\mathcal{H}}$.
Here they're partial adjoints:
$$S\subseteq T^*\quad T\subseteq S^*$$
In particular, both are closable:
$$\hat{S}:=\overline{S}\quad\hat{T}:=\overline{T}$$

But they don't need to be adjoints, or?
  $$\hat{S}^*=\hat{T}\quad\hat{T}^*=\hat{S}$$
  (I highly doubt it but miss a counterexample.)

Application
Given the pre-Fock space $\mathcal{F}_0(\mathcal{h})$.
The ladder operators are pre-defined by:
$$a(\eta)\bigotimes_{i=1}^k\sigma_i:=\langle\eta,\sigma_k\rangle\bigotimes_{i=1}^{k-1}\sigma_i\quad a^*(\eta)\bigotimes_{i=1}^k\sigma_i:=\bigotimes_{i=1}^k\sigma_i\otimes\eta$$
and extended via closure:
$$\overline{a}(\eta):=\overline{a(\eta)}\quad\overline{a}^*(\eta):=\overline{a^*(\eta)}$$
regarding the full Fock space $\mathcal{F}(\mathcal{h})$.
They are not only formally:
$$\langle a(\eta)\varphi,\psi\rangle=\langle\varphi,a^*(\eta)\psi,\rangle$$
but really adjoint to eachother:
$$\overline{a}(\eta)^*=\overline{a}^*(\eta)\quad\overline{a}^*(\eta)=\overline{a}(\eta)^*$$
(The usual proof relies on Nelson's theorem, afaik.)
 A: Let $S=\frac{d}{dx}$ and $T=-\frac{d}{dx}$ on the linear subspace $\mathcal{H}=\mathcal{C}_{0}^{\infty}(0,2\pi)\subset \hat{\mathcal{H}}=L^{2}[0,2\pi]$ consisting of infinitely differentiable functions on $[0,2\pi]$ which vanish outside some compact subset of $(0,2\pi)$. Then
$$
             (Sf,g) = (f,Tg),\;\;\; f,g\in\mathcal{C}_{0}^{\infty}.
$$
Both operators $S$ and $T$ are closable in $L^{2}$ and the domains consist of all $f \in L^{2}$ which are absolutely continuous on $[0,2\pi]$ with $f'\in L^{2}$ and $f(0)=f(2\pi)=0$. So $S^{\star} \ne \overline{T}$  and $T^{\star} \ne\overline{S}$ because the domains of the two adjoints are also equal and consist of absolutely continuous $f \in L^{2}$ with $f' \in L^{2}$ (no endpoint conditions.)
A: Let $S_0$, $T_0$ be densely defined linear operators (this is required for adjoints to exists) on a Hilbert space such that 
$$(1) \ \qquad S_0 \subseteq T_0^* \quad \& \quad T_0 \subseteq S_0^*. $$
Hence, we can conclude that $S_0$, $T_0$ are closable ($S_0^*$, $T_0^*$ are closed) and since the closure is the smalled closed extension we obtain that
$$(2) \ \qquad \overline{S_0} \subseteq T_0^* \quad \& \quad \overline{T_0} \subseteq S_0^*. $$
Also, the fact that $S_0$ and $T_0$ are closable implies that $S_0^*$ and $T_0^*$ are densely defined, and in particular $\overline{S_0}={S_0}^{**}$ and so $\overline{T_0}={T_0}^{**}$.
Now for simplicity denote $S:=\overline{S_0}$ and $T:= \overline{T_0}$. 
Note that
$$(3) \ \qquad S \subseteq T^* \quad \& \quad T\subseteq S^*. $$
The converse (reversed inclusion) is not true and the counterexample was delivered by @T.A.E.
