I'm reading von Neumanns 1931 book on Quantum Mechanics. Here he presents spectral theory for unbounded operators and quite a bit of the notation and naming has changed in the mean time.

Specifically for him a symmetric operator (for all $x,y\in D(A)$ $\langle Ax,y\rangle = \langle x,Ay\rangle$) is called hermitian and an symmetric operator is called maximal if it admits no proper symmetric extensions.

He characterises maximality via the Cayley Transform, an operator $A$ is maximal iff the Cayley Transform $U(A)$ has image or domain the entire Hilbert space $H$. An operator is defined to be hyper-maximal if the Cayley Transform is unitary, ie has both image and domain all of $H$. He shows that the hyper-maximal symmetric operators are precisely those that admit a spectral decomposition.

For example it is clear that self-adjoint implies maximal: If $A=A^*$ and $B$ is symmetric so that $B\lvert_{D(A)}=A$, if $y\in D(B)$ then for every $x\in D(A)$ you've got: $$|\langle Ax,y\rangle| =| \langle x,By\rangle| ≤\|x\|\,\|By\|$$ it follows $y\in D(A^*)=D(A)$ and $A^*y = By$.

My questions:

What is the modern terminology for maximal?


Is hyper-maximal the same as self-adjoint?


The usual term for "maximal" is now "maximally symmetric," meaning that $A$ is symmetric and there are no proper symmetric extensions. von Neumann's work gives the decomposition $$ \mathcal{D}(A^*)=\mathcal{D}(A)\oplus\mathcal{N}(A^*-iI)\oplus\mathcal{N}(A^*+iI), $$ which is orthogonal with respect to the following graph inner product $$ \langle x,y\rangle_{A^*}=\langle x,y\rangle_H+\langle A^*x,A^*y\rangle_H. $$ If neither of the deficiency spaces $\mathcal{N}(A^*\pm iI)$ is $\{0\}$, then you can always extend $A$ in such a way that the extension remains symmetric. (You can write down such an extension.) So maximally symmetric operators $A$ have either $\mathcal{N}(A^*-iI)=\{0\}$ or $\mathcal{N}(A^*+iI)=\{0\}$. These are maximal, as you have described them. If both are $\{0\}$, then $A^*=A$ because they have the same domain and $A^*$ extends $A$, and this is what you have referred to as hyper-maximal.

This can be recast without too much trouble in terms of the Cayley transform.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.