Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative of an operator valued function

  1. Feb 11, 2012 #1
    If I have a function
    [tex]
    f:R\rightarrow L
    [/tex]
    where L is the space of linear operators from an hilbert space to itself, how can i define the derivative of f at a particular point of R? I mean, it is "obvious" that one should try:

    [tex]
    f'(s_0)=lim_{\Delta s\rightarrow0}\frac{|f(s_0 + \Delta s) -f(s_0)|}{\Delta s}
    [/tex]

    but i'm concerned with the fact that when in the numerator i use ||, i mean the norm on L, which as i said is the space of linear operators from an hilbert space to itself. Since in physics one often deals with operators that have infinite operator norm, I wanted to know if the norm of that difference of operators is finite or not, even if we take unbounded operators.

    It is a mathematical question, but since in QM one encounters these things all the time, I wanted some clarifications..

    thanks!
     
  2. jcsd
  3. Feb 11, 2012 #2

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Yea I was going to suggest to define the norm on the space of operators by the following inner product of operators (A,B) = Trace (Abar B) - here Abar means the adjoint which reduces to Trace (AB) for the usual Hermtian operators in QM.

    The above is called the Frobenius norm (older texts refer to it as the Hilbert–Schmidt norm or the Schur norm).

    But ouch - even the identity operator has an infinite norm.

    Its probably better to use the operator norm:
    http://en.wikipedia.org/wiki/Operator_norm

    But even then I don't know how to guarantee it being finite.

    Thanks
    Bill
     
    Last edited: Feb 12, 2012
  4. Feb 12, 2012 #3

    Bill_K

    User Avatar
    Science Advisor

    I would think the "obvious" thing to do is to take the ||'s off and just let f'(s) be an operator.
     
  5. Feb 12, 2012 #4
    Ok, one doesn't need to put || in THAT definition, and I know that f'(s_0) is an operator, what I meant was that when i use the usual [tex]\epsilon,\delta[/tex]

    definition of limit, the norm on L will surely pop out: I mean that f'(s_0) is the operator that does this thing:

    [tex]\forall\epsilon>0\exists\delta>0:if\,\,0<|s-s_0|<\delta\Rightarrow\bigg|\frac{f(s) - f(s_0) - (s-s_0)f'(s_0)}{s-s_0}\bigg|_L<\epsilon[/tex]

    where the [tex]||_L[/tex] is the norm on L. Here I wanted to know if there are any problems if i'm dealing with unbounded operators. I admit I made a mistake, and in my first post the || were not needed, but they surely are in this definition (which is that of the differential of a function from a vector space to another, i think)
     
  6. Feb 12, 2012 #5
    In the original post, things should look like this:

    [tex]f'(s_0)[/tex]

    is an operator such that:

    [tex]lim_{\Delta s\rightarrow 0}\frac{|f(s_0+\Delta s) - f(s_0) - \Delta s\,f'(s_0)|_L}{\Delta s} = 0[/tex]

    The oroginal post was wrong since it compared a number to an operator :smile:
     
  7. Feb 12, 2012 #6

    strangerep

    User Avatar
    Science Advisor

    For bounded operators, one can use the supremum norm (aka operator norm), which bhobba mentioned already: http://en.wikipedia.org/wiki/Operator_norm

    See also: http://en.wikipedia.org/wiki/Topologies_on_the_set_of_operators_on_a_Hilbert_space

    For unbounded operators, one can talk in terms of densely-defined operators on an ordinary Hilbert space, or else pass to a rigged Hilbert space and use the theory of generalized functions (distributions). But for that, one must first understand how operators on (say) the Schwartz space are extended to operators on its dual space of tempered distributions via the dual pairing. (If that all sounds like gobble-de-gook, I could probably post some introductory references. It depends what you're really trying to do.)
     
  8. Feb 13, 2012 #7
    My knowledge about this is very limited.. So i'd be glad if you could post some references. I just started a course on quantum field theory and since (but it also happens in the mathematical formalism of quantum mechanics) we're dealing with operator valued functions, I wanted to know more about the mathematics behind that..

    So, if you know an introductory book about the mathematics of QFT, feel free to tell me about that, too :smile::smile:
     
  9. Feb 13, 2012 #8

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you want to learn this stuff, you have to start with topology and functional analysis. Kreyszig's book on functional analysis covers the topology you need as well, so it's much easier to read than many of the others. (I haven't read it myself, but I know how crazy hard some other books are).
     
  10. Feb 13, 2012 #9
    I second the recommendation of Kreyszig. It's probably the only functional analysis book that assumes so little background and yet covers so much ground.
     
  11. Feb 13, 2012 #10

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    I also recommend Kreyszig and I am thinking of actually purchasing a copy.

    I also recommend the following book on Distribution Theory that I have in my library. It helped me a lot with this sort of stuff:
    https://www.amazon.com/Theory-Distr...=sr_1_1?s=books&ie=UTF8&qid=1329180440&sr=1-1

    You can usually define what you want on a test space then use distribution theory extend it. And if you are doing QFT then the above IMHO is a must - they are really distributions rather than functions.

    Thanks
    Bill
     
  12. Feb 14, 2012 #11

    strangerep

    User Avatar
    Science Advisor

    Yeah, I was going to say something about how basic stuff on distributions and generalized functions is probably more important to understand clearly (for QM/QFT purposes) before tackling Functional Analysis books (which usually don't cover generalized functions anyway). I didn't know a good basic book on distributions and generalized functions, so I'll take at look at the one you suggested (Richards & Youn) and see what it's like.

    Cheers.
     
  13. Feb 14, 2012 #12
    The space of BOUNDED operators on a Hilbert space is itself a Banach space, so it is easy to define the usual Fréchet derivative:

    http://en.wikipedia.org/wiki/Fréchet_derivative

    I wouldn't know how to define this derivative, on the other hand, if you map into some unbounded operators, typical in physics.
     
  14. Feb 14, 2012 #13

    strangerep

    User Avatar
    Science Advisor

    Gel'fand triples, i.e., using topological duals of a suitable nuclear space.
     
  15. Feb 15, 2012 #14
    Can you spell this out in more detail? Are you taking the Frechet derivative of an unbounded operator on the dense nuclear space, and then doing ... something with the dual pairing between the nuclear space and the space of continuous antilinear functionals?
     
  16. Feb 15, 2012 #15

    strangerep

    User Avatar
    Science Advisor

    Well, it can't be the standard Frechet derivative, since that's for Banach spaces, but the small nuclear space [itex]\Omega[/itex] in a Gel'fand triple
    [tex]
    \Omega ~\subset~ \bar{\Omega} ~\subset~ \Omega' ~,~~~~
    \Big(\mbox{where}~ \Omega' ~\mbox{is the topological dual of}~ \Omega\Big) ~.
    [/tex]
    is not complete, hence not Banach.

    But here's my (quick) understanding of what's going on. (Warning: I could be wrong about parts of this, in which case I hope someone more knowledgeable will say so.)

    Let [itex]\phi[/itex], [itex]\Psi[/itex] be arbitrary vectors in [itex]\Omega, \Omega'[/itex] respectively. Let A be an operator defined everywhere on [itex]\Omega[/itex]. Then by standard theorems (cf. Gelfand & Vilenkin vol 4) it can be extended uniquely to an operator A' on [itex]\Omega'[/itex] via the dual pairing
    [tex]
    (\phi, A'\Psi) ~:=~ (A\phi, \Psi)
    [/tex]
    such that A' coincides with A when acting on elements of [itex]\Omega[/itex].

    Now let A(t) be an operator-valued function of a real parameter t such that A(t) is defined everywhere on [itex]\Omega[/itex]. Then, by the above construction it can be extended to [itex]\Omega'[/itex].

    If, in addition, there is an operator B such that
    [tex]
    \frac{A(t+h) - A(t) - B}{h} ~\rightarrow~ 0 ~~~~~~~~ (1)
    [/tex]
    in weak topology, we can call B a derivative of A at t. By duality, we can extend B to all of [itex]\Omega'[/itex].

    The subtleties lie in what it means for the LHS of (1) to approach 0 weakly. The standard meaning for weak operator topology is that a sequence of operators [itex]\{T_n\}[/itex] converges weakly to [itex]T[/itex] iff [itex]T_n \phi \to T \phi[/itex] for all [itex]\phi[/itex].

    (Once again: caveat emptor.)
     
  17. Feb 15, 2012 #16

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    He is talking about Rigged Hilbert Spaces which is an extension of distribution theory - probably best to get the book I recommended on distribution theory first but do check out:
    http://www.abhidg.net/RHSclassreport.pdf [Broken]

    Thanks
    Bill
     
    Last edited by a moderator: May 5, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook