3

I have heard some of my physics professors mention that hypermaximal means to have an infinite number of self-adjoint extensions. However, this was only mentioned during lectures and I could find no mention of this in math resources while scouting the internet, which makes me think that this is some local hearsay perpetuated by mistranslations of English books in the past. (It might not be just a local thing, I think I heard this in a foreing lecture on YouTube also, but can't find the exact timestamp).

On top of this, I have found this book: https://books.google.ro/books?id=4PR1-WRz87gC&pg=PA13&lpg=PA13&dq=hypermaximal+operator+definition&source=bl&ots=ZZm4I1mTGh&sig=ACfU3U2mT5ZfwhOTEstyTjLCLYpIlVwZ1Q&hl=en&sa=X&ved=2ahUKEwibgteQzNb2AhUTuKQKHSGUAz0Q6AF6BAgUEAM#v=onepage&q=hypermaximal%20operator%20definition&f=false , which mentions (bottom of page 13) that von Neumann himself was using hypermaximal to mean self-adjoint (this confuses me even more, since the same people who told me that hypermaximal stands for having an infinite number of SA extensions are also big fans of von Neumann's old book).

Now, in recent math texts I can't even find discussions on having infinite SA extensions (there is only talk about having at least one, i.e. possibly more than one). So it makes me wonder if that is even a thing.

Does anyone have some insight on this?

Mirod the best
  • 377

2 Answers2

2

A symmetric operator may or may not be self-adjoint. Furthermore, there are symmetric operators that have no self-adjoint extensions. Of course, every self-adjoint operator is symmetric.

A symmetric operator may or may not have a self-adjoint extension. A self-adjoint operator seems to have been called hypermaximal by von Neumann, at least according to the reference link you posted. And this terminology would make sense because of how a self-adjoint operator has no proper symmetric extension.

The key identity that von Neumann established for a closed symmetric operator $A:\mathcal{D}(A)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ on a complex Hilbert space $\mathcal{H}$ was the following orthogonal decomposition in the $\mathcal{H}\times\mathcal{H}$: $$ \mathcal{D}(A^*)=\mathcal{D}(A)\oplus\mathcal{N}(A^*-iI)\oplus\mathcal{N}(A^*+iI) $$ And he showed that every self-adjoint extension of $A$ comes about through a unitary map $U : \mathcal{N}(A^*-iI)\rightarrow\mathcal{N}(A^*+iI)$. Any such unitary map gives rise to a self-adjoint extension of $A^*$. If these "deficiency" spaces (as von Neumann called them,) are not isomorphic, then there are no self-adjoint extensions. That's why an operator such as $A=\frac{1}{i}\frac{d}{dx}$ has no self-adjoint extension: $$ A : \mathcal{D}(A)\subset L^2[0,\infty)\rightarrow L^2[0,\infty), $$ where the domain $\mathcal{D}(A)$ consists of all absolutely continuous $f\in L^2[0,\infty)$ such that $f'\in L^2[0,\infty)$ and $f(0)=0$. The adjoint $A^*$ has the same domain, without the condition that $f(0)=0$. $A^*$ is not symmetric, and, indeed, $A$ has no proper symmetric (or self-adjoint) extensions.

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149
0

In his 'Mathematical Foundations of Quantum Mechanics,' Von Neumann says that hypermaximal operators R, S, ... correspond to physical quantities (p. 161, for instance). By physical quantities, he means observables, so 'hypermaximal' simply means 'Hermitian.'

Lory
  • 203
  • 1
  • 6