Self-Adjoint versus Hermitian
rick1138 said:
What is the difference between SA and Hermitian operators? I have heard it mentioned a few times, but have never heard anyone state it explicitly.
When the Hilbert space is infinite-dimensional, a difficulty arises in that many linear operators cannot be defined on the
entire space. For example, think of the position operator Q for a particle moving in one dimension. Since Q takes the function f(q) to qf(q), there will be many square-integrable functions f(q) such that qf(q) is not square-integrable. This means that Q
cannot be defined on the
entire Hilbert space. On account of this sort of phenomenon, the mathematician is forced to take into consideration the
domain (a linear subspace of the Hilbert space) on which a linear operator is defined.
So, for example, to say that A = B means two things:
(i) Domain(A) = Domain(B);
and
(ii) |Af> = |Bf>, for all |f> in the common domain.
In a similar vein, A + B can only be defined on the
intersection of Domain(A) and Domain(B).
Also, AB can only be defined on those elements |f> of Domain(B) such that |Bf> is an element of Domain(A).
Physicists normally do not worry about such matters.
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Now, what does all of this have to do with "self-adjoint" and "hermitian" operators?
Well, first of all, let A and B be two linear operators acting
in a Hilbert space H. Then, A and B are said to be
adjoint to one another iff:
<Bg|f> = <g|Af> for all |f> in Domain(A) and all |g> in Domain(B).
Note that B is
"an" adjoint of A, and not necessarily
"the" adjoint of A.
"The" adjoint of A (call it A'), if it exists, is a
maximal extension of all possible B which are adjoint to A. A' exists iff Domain(A) is "big enough" (specifically, "big enough" means: the (set theoretic) closure of Domain(A) equals the entire Hilbert space H).
I'm not going to give any more details than this on A' (... but if anyone would like more, feel free to ask).
We are now in a position to display the definitions of "hermitian" and "self-adjoint".
A linear operator A is
hermitian iff <g|Af> = <Ag|f> for all f,g in Domain(A).
A linear operator A is
self-adjoint iff: (i) A is hermitian, (ii) A' exists, and (iii) Domain(A) = Domain(A').
As it turns out, there exist
hermitian operators A, such that A' exist
but Domain(A) is only a
proper subspace of Domain(A').
The upshot of all of this is that a
hermitian operator will
not in general guarantee the so called "closure" property with regard to its "eigenkets" and corresponding "bras"; i.e. a relation like:
Sigma_n { |n><n| } and/or Integral {ds |s><s| }
= Identity operator on H .
A
self-adjoint operator will
always guarantee this sort of relation.
Mathematicians generalize and rigorously formulate this notion of "closure" with what is called a "spectral family" (or "spectral function"), arriving at what is called "The Spectral Theorem" for
self-adjoint operators.
********************************************
*
A note on NOTATION: Let C be
any linear operator
* such that the adjoint C' exists.
*
* Then, for all |f> in Domain(C) and all |g> in Domain(C'),
*
* <g|Cf> = <g|(C|f>) = <g|C|f> = (<g|C)|f> = <C'g|f> .
********************************************
_____________________
Note that "self-adjoint" is a
stricter condition than "hermitian", "self-adjoint" being "hermitian with the two additional properties mentioned above".
Somehow Turin got it the other way around:
turin said:
I thought that Hermiticity was the more strict condition: being self-adjoint-ness with the additional requirement on the boundary.
And Robphy found someone else who, in an example, (just how?) also gets it backwards:
robphy said:
http://mathworld.wolfram.com/Self-Adjoint.html
Near the end of this description, there is a comment confirming what turin said above.
_______________________
Reilly brings out another point:
reilly said:
I'm very puzzled. A standard definition of a Hermitian operator/matrix H, is that H is Hermitian if and only if H is self adjoint ... [and then in a later post] ... In fact, von Neumann in his Mathematical Foundations of Quantum Mechanics gives the definition Hermitian if and only if self adjoint, in Hilbert Space.
I haven't checked von Neumann's book ... but you are right, there do exist sources which define "hermitian" as you say. In that case, any time I mentioned the word "hermitian" above, you should replace it with the word "symmetric" to denote the corresponding property.
But be careful regarding what you found on the Conjugate Transpose page (
http://mathworld.wolfram.com/ConjugateTranspose.html):
reilly said:
I checked out the Wolfram page ... If you go to the Conjugate Transpose page via the given link, you'll find my definition. So I'll stick with von Neumann's definition.
On that page, the Hilbert space is
finite-dimensional, for which the above mentioned "domain problems" do not exist. At the outset one can talk about linear operators acting
on the
entire space, and none of the above distinctions need to be considered.
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Slyboy offers us a complication.
slyboy said:
Just to complicate the discussion further, an operator does not have to be self-adjoint in order to have real eigenvalues. It need only be normal, i.e. commute with its adjoint.
What (I think) he wishes to say is:
An operator does not have to be "self-adjoint" in order to
satisfy "The Spectral Theorem".
Note that a
normal operator can, in general, have
complex eigenvalues.
This is the definition:
A linear operator N is
normal iff: (i) the adjoint N' exists, (ii) Domain(N) = Domain(N'), (iii) [N,N'] = 0.
This is a
generalization of "self-adjoint" and "unitary" operators.
