Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am a physicist. I always heard physicists used the terminology "symmetric", "Hermitian", "self-adjoint", and "essentially self-adjoint" operators interchangeably.

Actually what is the difference between all those operators? Presumably understandable by a physicist.

share|improve this question
1  
The case of unbounded operators (like position and momentum operators) is more complicated, since these operators can't be defined on the hole Hilbert space. See en.wikipedia.org/wiki/Self-adjoint_operator –  jjcale Jun 22 '13 at 18:06
1  
Related math.SE question: math.stackexchange.com/q/38387/11127 –  Qmechanic Jun 22 '13 at 18:14

3 Answers 3

Yes, for physicists, it is the same thing. I try to summarize differences from math sources:

An operator A verifying $<Au,v> = <u,AV>$ is called symmetric.

In this case, the definitions domain verify $D(A) \subset D(A^*)$. So you have, in general, no equality between $A$ and $A^*$, because the domains of definition are different.

An operator is hermitian if it is bounded and symmetric.

A self-adjoint operator is by definition symmetric and everywhere defined, the domains of definition of $A$ and $A^*$ are equals,$D(A) = D(A^*)$, so in fact $A = A^*$ .

A theorem (Hellinger-Toeplitz theorem) states that an everywhere defined symmetric operator is bounded.

There is also a subtelty, that is, for every hermitian operator, you can construct an extension of this operator, which is self-adjoint.

share|improve this answer
    
What do you mean by "extension"? That's also relates to my confusing point in wiki, "An operator which has a unique self-adjoint extension is said to be essentially self-adjoint." What is extension? –  user26143 Jun 22 '13 at 19:35
    
This is a math subtelty and I am not at all a specialist. But the idea is that an hermitian operator is defined on a dense set of the Hilbert space, and not on the whole Hilbert space. So apparently, one can find an extension of this Hermitian operator defined on the whole Hilbert space, and so this extension is self-adjoint. –  Trimok Jun 22 '13 at 19:47
    
thank you very much –  user26143 Jun 22 '13 at 19:52

Here's a mathematician's stab at an answer. The tl;dr version: "symmetric" and "self-adjoint" are the same thing for bounded operators, whilst they are the same thing for unbounded operators only insofar as the Aharonov--Bohm effect doesn't exist!

  1. A densely defined operator $D$ on your Hilbert space $H$ is symmetric if for any vectors $\xi$ and $\eta$ in the domain $D(A)$ of $A$, $\langle \xi, A\eta \rangle = \langle A\xi, \eta \rangle$. This, in particular, implies that $D(A) \subseteq D(A^\ast)$, where $D(A^\ast)$ denotes the domain of $A^\ast$.
  2. A symmetric operator $A$ is self-adjoint (or equivalently Hermitian, though some apparently define a Hermitian operator to be a bounded self-adjoint operator) if $A = A^\ast$ as unbounded operators, i.e., $D(A) = D(A^\ast)$ and for any $\xi \in D(A) = D(A^\ast)$, $A^\ast \xi = A \xi$. This is equivalent to saying that $A$ is symmetric and $D(A) = D(A^\ast)$. Note that self-adjoint operators absolutely need not be everywhere-defined!
  3. A self-adjoint operator $A^\prime$ is called a self-adjoint extension of a given symmetric operator $A$ if $D(A) \subseteq D(A^\prime)$ and if $A^\prime \xi = A\xi$ for any $\xi \in D(A)$. Then, $A$ is called essentially self-adjoint if it admits a unique self-adjoint extension, in which case even we mathematicians are usually quite happy to conflate $A$ with its unique self-adjoint extension.

OK, so why on earth should theoretical physicists ever care about all this business, given that so many of the operators they use (e.g., pretty much any Hamiltonian you care about on the line or on all of $3$-space, if I recall correctly) can be shown to be essentially self-adjoint? Well, roughly speaking, any time you need to fuss about imposing different boundary conditions, you're quite possibly actually fussing with different self-adjoint extensions without knowing it.

For example, the momentum operator $i\tfrac{d}{dx}$ on the interval $[0,1]$ (or your finite interval of choice) is symmetric but not essentially self-adjoint; you have a full circle's worth of different self-adjoint extensions, each coming precisely from imposing a quasi-periodic boundary condition of the form $f(0) = e^{2\pi i \beta}f(1)$. These, in turn, are no idle mathematical curiosities, but rather are the mathematical upshot/manifestation of the Aharonov--Bohm effect; for details see, for instance, http://homepages.physik.uni-muenchen.de/~helling/pextension.pdf, though I'm sure the relevant section of Reed--Simon will also discuss this.

Of course, the moment you're dealing rather with bounded operators (e.g., discrete Schroedinger operators), then there's no difference whatsoever between being symmetric and being self-adjoint.

share|improve this answer

Using bra-ket notation, a symmetric operator $S$ verifies: $\langle i|S|j \rangle =\langle j|S|i\rangle$ for every $i$ and $j$. In matrix form, it would be equal to its transpose ($S=S^T$).

An Hermitian or self-adjoint operator (it's the same) $H$ verifies: $\langle i|H^\dagger|j \rangle =\langle i|H|j\rangle$. If you use only $\mathbb{R}$, instead of $\mathbb{C}$, then the Hermitian operator is the same as the symmetric ($^\dagger$ becomes $^T$).

I've never heard of essential self-adjoint operators.

There are some mathematical subtleties, you better check here.

share|improve this answer
1  
According to wiki, link "Bounded symmetric operators are also called Hermitian." The symmetric operator includes \dagger operation –  user26143 Jun 22 '13 at 17:35
    
I think there are several definitions of symmetric. Usually, it would mean "equal" to the transpose, but in functional analysis it's different. See: math.stackexchange.com/q/38387/… –  jinawee Jun 22 '13 at 17:46
    
thanks a lot for the link! –  user26143 Jun 22 '13 at 17:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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