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I am permanently confused about the distinction between Hermitian and self-adjoint operators in an infinite-dimensional space. The preceding statement may even be ill-defined. My confusion is due to consulting Wikipedia, upon which action I have the following notion.

Let $H$ be a pre-Hilbert space equipped with an inner product ${\langle}.,.{\rangle}$ and $T:D(T){\subset}H{\longmapsto}H$ a linear operator. Then

  1. If ${\langle}Tx,y{\rangle}$=${\langle}x,Ty{\rangle}$ for all $x,y{\in}D(T)$ then $T$ is symmetric.

  2. If $T$ is symmetric and also bounded then it is Hermitian.

  3. If $T$ is symmetric and $D(T)=H$ then $T$ is self-adjoint.

As a corollary, if the above is true then a symmetric and self-adjoint operator must be Hermitian since a symmetric operator defined on all of $H$ must be bounded. On the other hand, a Hermitian operator need not be self-adjoint: it would not be if its domain were a strict subset of $H$.

Would people agree with this? I always see the second and third of these treated as equivalent, hence my confusion.

Many thanks.

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The terminology probably differs depending on whether you're talking to a physicist or a functional analyst. –  Qiaochu Yuan May 11 '11 at 3:56
    
Thanks. I think I'd prefer to talk to a functional analyst. I guess the former would resent the apparent disjointness of the two groups, or perhaps that's just what I'm reading and you require at least one of the identities to hold true. –  Josef K. May 11 '11 at 5:12
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In the 1960s Friedrichs met Heisenberg and used the occasion to express to him the deep gratitude of mathematicians for having created quantum mechanics, which gave birth to the beautiful theory of operators on Hilbert space. Heisenberg allowed that this was so; Friedrichs then added that the mathematicians have, in some measure, returned the favor. Heisenberg looked noncommittal, so Friedrichs pointed out that it was a mathematician, von Neumann, who clarified the difference between a self-adjoint operator and one that is merely symmetric. "What's the difference," said Heisenberg. –  leslie townes May 8 '12 at 8:17
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- story from Peter Lax, Functional Analysis (slightly edited for length) –  leslie townes May 8 '12 at 8:17
    
@leslietownes thanks for the reference! ;) So it was Friedrichs who met Heisenberg, not von Neumann himself, and this story is not an anecdote. –  Wildcat Jun 3 '13 at 11:20

2 Answers 2

up vote 22 down vote accepted

These are not the usual definitions as I know them.$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

First, I am only familiar with the situation that $H$ is a Hilbert space and $D(T)$ is dense in $H$ (which entails no loss of generality, as we can replace $H$ with the completion of $D(T)$.)

I would say:

  1. $T$ is symmetric if $\inner{Tx}{y} = \inner{x}{Ty}$ for all $x,y \in D(T)$. (Note your definition doesn't make sense, because you are applying $T$ to vectors that may not be in $D(T)$.)

  2. $T$ is Hermitian if it is symmetric and bounded. (If $T$ is bounded then it has a unique bounded extension to all of $H$, so we may as well assume $D(T) = H$ in this case.) Since a symmetric operator is always closable, the closed graph theorem implies that a symmetric operator with $D(T) = H$ is automatically bounded.

  3. $T$ is self-adjoint if the following, more complicated condition holds. Let $D(T^*)$ be the set of all $y \in H$ such that $|\inner{Tx}{y}| \le C_y ||x||$ for all $x \in D(T)$, where $C_y$ is some constant depending on $y$. If $T$ is symmetric, one can show that $D(T) \subset D(T^*)$; $T$ is said to be self-adjoint if it is symmetric and $D(T) = D(T^*)$.

With these definitions, we have Hermitian implies self-adjoint implies symmetric, but all converse implications are false.

The definition of self-adjoint is rather subtle and this may not be the place for an extended discussion. However, I'd recommend a textbook such as Reed and Simon Vol. I. Perhaps I'll just say that symmetric operators, although the definition is simple, turn out not to be good for much, per se. One needs at least self-adjointness to prove useful theorems.

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+1: I agree with all the points you're making (you beat me to it again). I made a small correction in 3. One minor TeX-point on this site: If you're using \newcommand or something similar this results in strange spacing at the point you're making the declaration. I therefore put these things at the end of the first paragraph when I need them, then this can't be seen. –  t.b. May 11 '11 at 4:39
    
Sorry on point 1: it was an omission which I have corrected in my question now. This is very helpful. I would urge you to consider editing the Wikipedia article! –  Josef K. May 11 '11 at 5:10
    
I should say I added "pre-" to "Hilbert space" retrospectively as a slight provocation, since I can't see why the definitions can't apply for any inner product space. However, in light of your intro to self-adjointness I think I prefer to stick to Hilbert spaces. –  Josef K. May 11 '11 at 5:20
    
@Josef K.: The definitions apply, but they don't have quite the same meaning. For instance, without completeness a symmetric operator which is everywhere defined (i.e. $D(T) = H$) need not be bounded. –  Nate Eldredge May 11 '11 at 12:05
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@kemiisto: I clarified the definitions a little; in particular, a bounded operator may as well be taken to be everywhere defined ($D(T) = H$). Then it's easy to see that $D(T^*) = H$ as well, since $|\langle Tx, y \rangle| \le \|T\| \|x\| \|y\|$. –  Nate Eldredge Jun 3 '13 at 14:24

The definition is quite simple when you realize it. But it takes some time to realize the difference. There are some contradictions with Nate answer, but this just a matter of terminology.

  • $\mathrm T$ is Hermitian if $\forall x,y \in D(\mathrm T) (\mathrm Tx,y) = (x,\mathrm T y)$
  • $\mathrm T$ is symmetric if $\mathrm T$ is Hermitian and densely defined. As far as i understand the only advantage of symmetric op over Hermitian is guaranted exitance of $\mathrm T$'s closure.
  • $\mathrm T$ is self-adjoint if $\mathrm T^* == \mathrm T$, where $\mathrm T^*$ defined as from following relation$\forall x \in D(\mathrm T) \exists y,z \in \mathbb H: (\mathrm Tx,y) = (x,z)$. The operator $\mathrm T^*: z = T^*y$ and is called adjoint.

    For finite-dimensional spaces all this definitions turn to be the same. Bounded symmetric operators are essentially self-adjoint (closure is self-adjoint).

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