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.

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    $\begingroup$ The terminology probably differs depending on whether you're talking to a physicist or a functional analyst. $\endgroup$ Commented May 11, 2011 at 3:56
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    $\begingroup$ 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. $\endgroup$
    – Josef K.
    Commented May 11, 2011 at 5:12
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    $\begingroup$ 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. $\endgroup$ Commented May 8, 2012 at 8:17
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    $\begingroup$ - story from Peter Lax, Functional Analysis (slightly edited for length) $\endgroup$ Commented May 8, 2012 at 8:17
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    $\begingroup$ Related Phys.SE question: physics.stackexchange.com/q/68826/2451 $\endgroup$
    – Qmechanic
    Commented Jun 22, 2013 at 18:15

3 Answers 3


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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    $\begingroup$ +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. $\endgroup$
    – t.b.
    Commented May 11, 2011 at 4:39
  • $\begingroup$ 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! $\endgroup$
    – Josef K.
    Commented May 11, 2011 at 5:10
  • $\begingroup$ 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. $\endgroup$
    – Josef K.
    Commented May 11, 2011 at 5:20
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    $\begingroup$ @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 $|\langle Tx, y \rangle| \le \|T\| \|x\| \|y\|$. $\endgroup$ Commented Jun 3, 2013 at 14:24
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    $\begingroup$ @ErikPillon: Whichever you prefer. The statement is known as the Hellinger-Toeplitz theorem but you can prove it from the closed graph theorem in about two lines. $\endgroup$ Commented Oct 7, 2017 at 17:54

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).


Although a reasonable part of the question is about the commonly-accepted senses of the terminology, another part of it is surely about facts (rather than terminological conventions).

And I am sympathetic to bad reactions to the (fluctuating per source, and over time!) terminology. E.g., in other contexts, "symmetric" (over $\mathbb C$) would mean $\mathbb C$-bilinear, while "hermitian" would mean ... um, hermitian? ($\mathbb C$-sesquilinear?).

Ok, we can agree that the terminology is not self-explanatory? In my experience, disparities among usage and implied properties are as much due to differences in conception of (apparently official, but in different ways in different peoples' minds) what these words mean.

Nevertheless, yes, there are actual mathematical questions here that do not depend on what we declare (or, worse, tacitly presume) words to mean.

So, for example, "symmetric" does not refer to the complex-bilinear versus complex-sesquilinear issue, but to the issue of, if the operator is unbounded (hence, not defined, nor definable everywhere, by the closed graph theorem), whether or not it "appears to be self-adjoint" on its domain.

And there's the important-yet-hard-to-promote issue of whether some random unbounded operator is densely defined. Things tend to screw up if not. E.g., no adjoint. But/and giving things a name that does not refer to this feature is ... unhelpful. So, really, we should say that we care about densely-defined, symmetric operators. (Or maybe call them "potentially self-adjoint", though there's the problem that (cf. von Neumann) not all symmetric operators have self-adjoint extensions!)

E.g., sure, continuous (a.k.a. "bounded") operators don't have any of these issues... In any reasonable sense, symmetric=hermitian=self-adjoint.

For genuinely unbounded operators, symmetric does not imply self-adjoint, and, unless the thing is already self-adjoint, its adjoint is definitely not symmetric. (Crazy, right?)

But, in practical situations, all these seeming bait-and-switch or faux-paradoxical things are actually sensible. Abstractly, it's hard to see the forest for the trees, or vice-versa. In my opinion, this is yet another instance where too-aggressive generalization/abstraction makes things incomprehensible, if it goes so far as to disconnect from the (oh-so-) tangible examples that gave rise to the ideas.


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