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Unbounded operator and expansion of commutator

  1. Oct 11, 2015 #1

    ShayanJ

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    Consider two self-adjoint operators A and B with commutator [A,B]=C such that [A,C]=0.
    Now I consider an operator which is a function of A and is defined by the series ## F(A)=\sum_n a_n A^n ## and try to calculate its commutator with B:

    ## [F(A),B]=[\sum_n a_n A^n,B]= \\ \sum_n a_n [A^n,B]=\sum_n a_n \left[ A^{n-1}[A,B]+A^{n-2}[A,B]A+\dots+A[A,B]A^{n-2}+[A,B]A^{n-1} \right]=\\ \sum_n a_n \left[ A^{n-1}C+A^{n-2}CA+\dots+ACA^{n-2}+CA^{n-1} \right]=\sum_n n a_n A^{n-1} C##

    Now if I formally write ## \sum_n n a_n A^{n-1}=\frac{dF}{dA} ##, I can have ##[F(A),B]=\frac{dF}{dA} C ##.

    But what happens if A is unbounded?

    Thanks
     
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  3. Oct 11, 2015 #2

    Krylov

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    If ##A## is unbounded, then without additional information I cannot say whether ##F(A)## is even defined in the first place. Do you have a functional calculus for ##A##?
     
  4. Oct 11, 2015 #3

    ShayanJ

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    Just assume that F can be defined. Also I didn't assume any functional calculus because as I said, equation ## \sum_n n a_n A^{n-1} =\frac{dF}{dA} ## is only formal, only a recipe with no math behind it.
     
  5. Oct 11, 2015 #4

    Krylov

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    Sorry, I should have written ##F(A)##, not just ##F##.

    I suppose that if ##A## is bounded, I could possibly make mathematical sense out of ##\frac{dF}{dA}##, but if ##A## is unbounded, I'm already stuck at the definition of ##F(A)## itself.
     
  6. Oct 11, 2015 #5

    ShayanJ

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    I see it now. Because A is unbounded, even a function of it with a finite number of terms in its expansion will diverge somewhere. So we can't define functions of A, whether the series is finite or infinite. So its just meaningless to define an operator e.g. ## e^{iA} ## for an unbounded A.

    But something comes to my mind. We don't abandon ordinary functions which diverge at some point of their definition. So why do we abandon functions of operators that diverge for some vectors that they act on?
     
  7. Oct 11, 2015 #6

    Krylov

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    Indeed, whether you can define ##F(A)## meaningfully depends in subtle ways on the characteristics of both ##F## and ##A##. As you said, there is no need to discard the idea of defining a function of an unbounded operator entirely, just because it does not work for certain ##A##.

    In fact, there exists a good functional calculus for closed operators on Banach spaces, see e.g. Section V.8 of Taylor & Lay, Introduction to Functional Analysis, Wiley, 1980. The specific case of the exponential is the subject of the theory of operator semigroups, see e.g. the book by Engel & Nagel, One-Parameter Semigroups for Linear Evolution Equations, Springer, 2000. For example, Stone's theorem II.3.24 is of importance in quantum mechanics.

    So in summary: You can still do a lot when the operator is unbounded, but the mathematical treatment is more involved.
     
  8. Oct 11, 2015 #7

    micromass

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    No, it isn't. The operator ##e^{iA}## is well-defined for unbounded operators. But you can't just take any function of an unbounded operator. And you definitely can not define these functions using Taylor series.
     
  9. Oct 11, 2015 #8

    Krylov

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    I would say that there exist unbounded operators ##A## for which ##e^{iA}## may be well-defined. The way you put it suggests that you can take any unbounded ##A##.
     
  10. Oct 11, 2015 #9

    ShayanJ

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    Do you mean that the exponential operator can't be defined using Taylor series? So how is it defined?
     
  11. Oct 11, 2015 #10

    micromass

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    See Reed and Simon, volume 1.
     
  12. Oct 11, 2015 #11

    Krylov

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    I just gave you two specific references: Taylor & Lay and Engel & Nagel.
    Reed & Simon is good, too.
     
  13. Oct 11, 2015 #12

    ShayanJ

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    Actually I was expecting a summary when I asked. But I'll read the books.
    Thanks guys.
     
  14. Oct 11, 2015 #13

    micromass

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    We get that. But I'm not going through the effort describing the unbounded spectral theorem when you could just read it yourself and ask whenever something is unclear.
     
  15. Oct 11, 2015 #14

    ShayanJ

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    I wasn't protesting anything. I just explained that I didn't just ignore Krylov's references.
    And I understand what you say.
     
  16. Oct 11, 2015 #15

    ShayanJ

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    I tried reading those books but they're too advanced for me. So let me just ask one question. Whatever the definition of ## e^{iA} ## is for an unbounded operator, is the following formula true for it under the conditions ## [A,B]=C ## and ## [A,C]=0 ##?:
    ## [e^{iA},B]=iCe^{iA} ##

    I prefer the answer for general C but if its too general to answer, then specialize to ##C \propto I##.
    Thanks again.
     
  17. Oct 12, 2015 #16

    Krylov

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    Without wishing to sound pedantic, I don't believe it's possible to prove a mathematical identity involving an object that is not defined, and in order to know more about how it could possibly be defined, it would help to know more about ##A## and the space on which it acts. For example, is ##iA## perhaps the generator of a (strongly continuous) group or semigroup (i.e. is ##e^{t i A}## a "time evolution operator")? Your desire to take the exponential suggests this.
     
  18. Oct 12, 2015 #17

    ShayanJ

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    Of course without defining an object, you can't prove theorems about it. I just meant that I couldn't understand the references. So just tell me, with the usual definition of exponential of an operator, is that relation correct or not?

    But now that you also need information about A, so it seems I should give more context. Actually I was reading this paper. I understand it. The only problem is that, it says to ensure that the equation 2.2(essentially the formula in the OP) is correct, its necessary to consider a bounded operator as the exponent. I just want to find a theorem about this.
     
  19. Oct 13, 2015 #18

    Krylov

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    I don't know, that's up to you to find out. I don't have the appetite to delve into that paper, but I will repeat that it is important to know more about ##A## or ##iA##. You cannot reasonably define the exponent of just any unbounded operator. Sometimes it's possible, but which of the definitions is used (I know at least two) depends on the case. As I asked before: Do you perhaps know if ##iA## is an evolution operator?
     
  20. Oct 13, 2015 #19

    ShayanJ

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    No, its not an evolution operator in the sense used in QM.
    OK, So I just put this question for the time I know enough about functional analysis.
    Thanks guys.
     
  21. Oct 13, 2015 #20

    Krylov

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    I should of course have asked: "Do you know if ##iA## generates a group of evolution operators?" but I suppose that the answer would have been the same. With that settled, you should follow micromass's suggestion and look into the functional calculus for self-adjoint operators. Also for a physicist, getting to know the book by Reed & Simon (particularly the first part, as it is a series) is probably a very good investment of his time.

    EDIT: Hold on... I realise now that ##iA## is skew-adjoint, so (assuming ##A## is also densely defined) by Stone's theorem ##iA## does generate a strongly continuous group of unitary operators. Of course, that was silly of me, but it has been a long time that I did anything on a Hilbert space. So, yes, you can assume that ##(e^{tiA})_{t \in \mathbb{R}}## is properly defined. Now you may be able to work something out by using that, on the domain of ##A##, the operator and the exponentials commute.
     
    Last edited: Oct 13, 2015
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