Self-adjoint operators and Hermitian operators

[spaghetti3451]

I was wondering what the difference is between the two. Would be nice if someone could explain the difference in simple terms, because it appears to be essential to my quantum mechanics course.

 

[economicsnerd]

Basically, they're interchangeable.

- Every operator $T$ on a (complex) Hilbert space $\mathcal H$ has a unique adjoint operator $T^*$, pinned down by the property that$\langle Tx,y\rangle=\langle x,T^*y\rangle$ for every $x,y\in\mathcal H$. We say $T$ is a self-adjoint operator if $T^*=T$.
- Every square (compex) matrix $A$ has a unique conjugate transpose $A^*$. We say $A$ is a Hermitian matrix if $A^*=A$.
- If $\mathcal H$ is finite-dimensional, and $A$ is the matrix which represents $T$ with respect to some given orthonormal basis (e.g. the standard basis if $\mathcal H = \mathbb C^n$), then it's straightforward to verify that the conjugate transpose matrix $A^*$ represents the adjoint operator $T^*$. In particular, $T$ is self-adjoint if and only if $A$ is Hermitian.

So if we're careful (e.g. only using orthonormal bases, for instance), then "self-adjoint" and "Hermitian" mean the same thing in the finite dimensional world. I would suppose they're also being used interchangeably for operators on an arbitrary Hilbert space.

 

[dextercioby]

For a general complex separable Hilbert space, Hermitean is an old-fashioned notion for what mathematicians call symmetric operator. A symmetric operator has an unique adjoint who's nothing but an extension of it. Self-adjoint is the operator which is equal to its adjoint (in von Neumann's words: hypermaximal symmetric). Quantum Mechanics needs self-adjoint operators to depict observables, not symmetric ones.

 

[Terandol]

Things are slightly more complicated in the infinite dimensional case when unbounded operators are considered. In fact when physicists say Hermitian, the definition they often give is something like A is Hermitian if \langle u | A | v \rangle=\langle v|A|u\rangle^* for all u,v\in \mathcal{H} however they also want the spectral theorem and all it's consequences to apply to A. Strictly speaking this definition merely says that A is what mathematicians would call symmetric and the spectral theorem does NOT necessarily apply to such operators.

To give the precise definition for unbounded operators, it is necessary to pay close attention to the domains of the operators in question. So, suppose that an operator A is defined on the domain \mathcal{D}(A) which is dense in \mathcal{H}. Then the adjoint A^\dagger of A is defined on the domain consisting of all u\in \mathcal{H} such that the map v\mapsto \langle u,Av \rangle for v\in \mathcal{D}(A) extends to a bounded linear functional on all of \mathcal{H}. (Note that this extension is unique since the domain is dense.) In this case, the adjoint A^\dagger is defined to be the element of \mathcal{H} corresponding to this linear functional via the Riesz-representation theorem so that the equality \langle A^\dagger u,v\rangle=\langle u,Av\rangle as expected.

The definition of symmetric (or Hermitian as physicists often call it) is simply that \langle Au,v\rangle=\langle u,Av\rangle for all u,v\in \mathcal{D}(A). The definition of self-adjoint is that A^\dagger =A. The key point here is that in general, \mathcal{D}(A^\dagger) \supseteq \mathcal{D}(A) so a self-adjoint operator has the extra condition that it's adjoint has the same domain as the operator itself which a merely symmetric operator need not satisfy. It is often possible to extend a symmetric operator to a self-adjoint operator by extending it's domain but, contrary to what a previous poster stated, even if this is possible I believe there is no guarantee that you get a unique extension. A related class of operators, the essentially self-adjoint ones, are precisely those which have a unique extension to a self-adjoint operator but this does not include all symmetric operators.

Since the majority of physics books I've read anyway (at least the more introductory ones) do not pay very close attention to the precise domain on which unbounded operators are defined, the above distinction between self-adjoint and symmetric seems to get blurred or not mentioned at all but as I said, the spectral theorem only applies to self-adjoint operators not Hermitian(symmetric) ones so these are what are really meant if you want to be precise.

 

[WWGD]

Are we making use here somewhere that for a continuous (i.e., linear + bounded ) map L to extend from a dense subspace into the space (when the target space is a complete normed space), that L must be uniformly-continuous?