# Positive but not self-adjoint operator

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?

• 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). Jun 9, 2021 at 15:42
• Why is it not self adjoint ? Jun 9, 2021 at 15:53
• Does $\langle \psi , T \psi \rangle$ even need to be real ? I'm confused ... Jun 9, 2021 at 15:59

$$-\frac{d^2}{dx^2}$$ 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

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).

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 $$T^2$$.