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Given ${\frak L}({\scr H}) \equiv \{ A : {\scr H} \to {\scr H} \ \vert \ A$ linear$ \}$, the operator $ T \in {\frak L}({\scr H})$ is said to be positive if:

$\langle \psi, T \psi \rangle \geq 0 \quad \forall \ \psi \in \scr H $, where $ {\displaystyle}{ \langle . , . \!\rangle }$ denotes the hermitian inner product.

It can be shown that every positive bounded operator is self-adjoint.


Is there a counterexample for unbounded ($ {\rm dom} \ T \subsetneq \scr H $) positive operators?

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    $\begingroup$ How about $-\frac{d^2}{dx^2}$ defined on $C_c^{\infty}(\mathbb{R})$? This is positive and symmetric but not self-adjoint (though it is essentially self-adjoint I believe). $\endgroup$ Jun 9, 2021 at 15:42
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    $\begingroup$ Why is it not self adjoint ? $\endgroup$ Jun 9, 2021 at 15:53
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    $\begingroup$ Does $\langle \psi , T \psi \rangle$ even need to be real ? I'm confused ... $\endgroup$ Jun 9, 2021 at 15:59

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$-\frac{d^2}{dx^2}$ on $C_c^\infty((a,b))$ has different self adjoint extensions, with different eigenvalues, for different choices of boundary conditions at $a, b$. Thus, it cannot be self adjoint because:

A symmetric operator is called maximal symmetric if it has no symmetric extensions, except for itself.

Every self-adjoint operator is maximal symmetric.

Wikipedia

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As already noted, even a very nice and natural symmetric operator (such as the classic $-d^2/dx^2$ on $C^\infty_c(a,b)$, giving simple Sturm-Liouville problems), can have several/many self-adjoint (=maximal symmetric) extensions.

But/and in the land of unbounded operators, there is an interesting complementary question, namely, does a symmetric unbounded (densely-defined) operator have any self-adjoint extension(s)?

In general, there answer is "no": J. von Neumann's ideas about deficiency indices clarifies this. An easy example, amenable to explicit computation, is $Tf=x^3f'+(x^3f)'$ on Schwartz functions $f$, which has no extension to a self-adjoint operator on $L^2(\mathbb R)$.

But von Neumann knew, and K. Friedrichs emphasized this construction in 1934/5, that positive (or, more generally, semi-bounded) symmetric operators do always have at least one self-adjoint extension (with some good properties).

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Let $\mathcal{H}=L^2[0,\infty)$, and define $T : \mathcal{D}(T)\subset\mathcal{H}\rightarrow\mathcal{H}$ by $$ Tf=if', $$ where $\mathcal{D}(T)$ consists of all absolutely continuous functions $f\in L^2[0,\infty)$ for which $f(0)=0$. This is a closed, symmetric operator, but it is not self-adjoint because the adjoint domain is the same as $\mathcal{D}(T)$, except that the restriction $f(0)=0$ is not present. The graph of $T$ is of co-dimension $1$ in the graph of $T^*$, which prevents there from being a self-adjoint extension of $T$. This operator is related to the Laplace transform calculus.

A positive version is $T^2$.

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