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.)


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.)


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.)

  • $\begingroup$ Nice one, I guess that I messed up directions of inclusions. We have that if $S \subset T$ then $T^* \subset S^*$, which here doesn't do the job, and the part "then" doesn't work in other direction of inclusion. $\endgroup$ – m_gnacik Jan 13 '15 at 18:16
  • $\begingroup$ @mgn: Yes, that was the problem: The inclusion reverses. $\endgroup$ – C-Star-W-Star Jan 13 '15 at 18:19
  • $\begingroup$ @T.A.E.: I already expected such an example. Thanks!! :) $\endgroup$ – C-Star-W-Star Jan 13 '15 at 18:21
  • $\begingroup$ Sorry for that it is not the first time when I messed up with this inclusion. $\endgroup$ – m_gnacik Jan 13 '15 at 18:21
  • 1
    $\begingroup$ @mgn : Operator domain inclusions always seem to trip me up, too. I think the notation is not good, but I don't have an alternative. $\endgroup$ – DisintegratingByParts Jan 13 '15 at 18:27

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.


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