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Hermitian vs. self-adjoint operators

  1. Sep 25, 2008 #1
    Hello,

    what's the difference between Hermitian and self-adjoint operators? Our professor in Group Theory made a comment once that the two are very similar, but with a subtle distinction (which, of course, he failed to mention :smile: )

    Thanks!
     
  2. jcsd
  3. Sep 25, 2008 #2
    Oops... I found the answer. Sorry :redface:
     
  4. Sep 25, 2008 #3

    morphism

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    Well, what's this subtle distinction?
     
  5. Sep 25, 2008 #4
  6. Sep 25, 2008 #5

    morphism

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    Just make sure you're aware that it's a matter of convention. Most people use the two words to mean the same thing.
     
  7. Sep 25, 2008 #6

    George Jones

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    While this is true for bounded operators, all the books I have checked on my shelf make a distinction for unbounded operators. According to the books that I have checked, a Hermitian operator is a bounded self-adjoint operator, i.e., all Hermitian operators are self-adjoint, but not all self-adjoint operators are Hermitian.
     
  8. Sep 25, 2008 #7

    atyy

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    Is this an example of the difference you are talking about, or is it yet another difference?
    http://mathworld.wolfram.com/BerryConjecture.html:"...physicists define this operator to be Hermitian, mathematicians do not."
     
  9. Sep 26, 2008 #8

    George Jones

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    Let [itex]A[/itex] be a densely defined linear operator on a Hilbert space. One of the defining properties, *, for all cases is

    [tex]\left<Ax,y\right> = \left<x,Ay\right>, \ *[/tex]

    where [itex]x[/itex] and [itex]y[/itex] are elements of the Hilbert space.

    The text that was used for the functional analysis course that I took as a student makes the following definitions:

    [itex]A[/itex] is Hermitian if [itex]A[/itex] is bounded and * is true for every [itex]x[/itex] and [itex]y[/itex] in the Hilbert space;

    [itex]A[/itex] is symmetric if * holds for for every [itex]x[/itex] and [itex]y[/itex] in the domain of [itex]A[/itex];

    [itex]A[/itex] is self-adjoint if [itex]A[/itex] is symmetric and the domain of [itex]A[/itex] equals the domain of [itex]A^\dagger[/itex].

    According to these defintions, every Hermitian operator is and self-adjoint, but not all self-adjoint operators are Hemitian. Some books leave off the first definition and call symmetric operators Hermitian. Then, every self-adjoint operator is Hermitian, but not all Hermitian operators are self-adjoint.
    A symmetric (or Hemitian, depending on the terminology used) operator is self-adjoint iff it has trivial deficiency indices. Some physicists and physics references do not distinguish between the above defintions, and use the terms Hermitian and self-adjoint interchangeably to refer to all cases.

    If [itex]A[/itex] is unbounded, then the Hilbert space is necessarily infinite-dimensional, and the domain of [itex]A[/itex] need not be all of the Hilbert space. If this is the case and if [itex]A[/itex] is symmetric, then the domain of [itex]A[/itex] is a subset of the domain of [itex]A^\dagger[/itex], which I have not defined.

    In physics, the canonical commutation relation is important. If self-adjoint operators [itex]A[/itex] and [itex]B[/itex] satisfy such a relation, then it is easy to show that at least one of the operators must be unbounded. Suppose it is [itex]A[/itex]. The Hellinger-Toeplitz theorem states any linear operator that satisfies * for all elements of the Hilbert space must bounded. Since [itex]A[/itex] is self-adjoint and unbounded, the domain of physical observable [itex]A[/itex] cannot be all of Hilbert space!

    For one consequence of these concepts, see

    https://www.physicsforums.com/showthread.php?t=122063&highlight=dirac.
     
    Last edited: Dec 17, 2013
  10. Sep 26, 2008 #9

    morphism

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  11. Sep 27, 2008 #10

    atyy

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  12. Dec 16, 2008 #11
    I know, George you have to be the most interesting person i know, gosh gosh

    "what are you going to do today napoleon?"
    "whatever the frick i wanna do gosh! gosh"
     
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