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Operator is it function?

  1. Aug 22, 2004 #1
    Hi
    i am tyring to self study some QM.
    I am referring to set of notes by BrianD.Serot and feynman lectures.

    What is exactly a operator ?Isnt it a fancy name for function excepting that we write it independant of the variable.
    How are operators represented by matrix method? :confused:

    Thanks in advance
    poolwin2001
     
  2. jcsd
  3. Aug 22, 2004 #2

    jcsd

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    An operator is an operator, it operates on a function to produce a new function:

    One of the most simple examples would be the differential operator D:

    if:

    [tex]f(x) = x^2[/tex]

    then:

    [tex]Df = \frac{d}{dx} f(x) = 2x[/tex]

    [tex]D^2f = \frac{d^2}{dx^2}f(x) = 2[/tex]
     
  4. Aug 22, 2004 #3

    HallsofIvy

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    One can think of an "operator" as a function from one set of functions to another. In otherwords, the "independent variable" and "dependent variable" of operators are functions, not numbers.
     
  5. Aug 22, 2004 #4

    arildno

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    An operator can be seen as a function(al) whose domain is a function space, and whose range is the same function space.
    An operator is therefore a special case of a functional.
     
  6. Aug 23, 2004 #5
    Could you briefly state the difference between a functional and an operator?
     
  7. Aug 23, 2004 #6

    arildno

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    Wel from what I know, a functional is a function whose domain is a function space.
    No restrictions are laid upon the range.
    For example, consider the function space F of continuous, integrable functions on the interval (a,b). (That it is a function space should be obvious).
    Form the funcional on F, I(f), which relates the scalar [tex]\int_{a}^{b}fdx[/tex] to each f.
    I is therefore a functional from F into R (the real numbers).
     
  8. Aug 23, 2004 #7

    selfAdjoint

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    Yes, in the most general view, a functional is a function and a matrix is a function, it's only a question of what are the range and domain. But that is not a useful definition; in QM it is important that there are ordinary functions of a complex variables and ALSO things which are different from that, operators which act on a function space.
     
  9. Aug 23, 2004 #8

    According to wikipedia, arildno has it backwards.

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

    This says the functional is a special case of operator.

    Furthermore (as I am also learning QM on my own), My notes from Introductory Quantum Chemistry by Victor S. Bastista, state that an operator tranforms a function into another, and a linear functional transforms a function into a number. if you want the more general term, see link above.

    As for the matrix, there's an example if you look "hermitian" up on wikipedia.
    It also has a bit on the Feynman integrals.
     
  10. Aug 23, 2004 #9

    arildno

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    Not QUITE backwards, qouting from the source:
    "The initial meaning is a function that takes functions as its argument; that is, a function !!whose domain is a set of functions!!. This was how the word was used initially, in the calculus of variations, where the integrand to be minimized should be a functional.."

    This is how I know it; however, it may well be that I don't know the proper definition of
    of an operator. :wink:
    Thx for the link.
     
  11. Aug 23, 2004 #10
    Hey, I'm just learning this myself--I had to review all of the way back to linear algebra, so I've been spending a lot of time looking these things up.

    The problem is, since I'm new to the in-depth mathematics of it, I'm not sure which definitions or terms poolwin needs.
     
  12. Aug 23, 2004 #11

    Fredrik

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    The relationship between linear transformations and matrices is quite simple.

    Suppose U and V are vector spaces, and that

    [tex]T:U\rightarrow V[/tex]

    is linear. Any vector in either of the two vector spaces can be expressed as a linear combination of basis vectors. We can use this, and the linearity of T, to express y=Tx in two different ways.

    [tex]y=\sum_{i=1}^m y_i v_i[/tex]
    [tex]y=Tx=T\bigl( \sum_{j=1}^n x_j u_j\bigr) =\sum_{j=1}^n x_j Tu_j=\sum_{j=1}^n x_j \sum_{i=1}^{m} [Tu_j]_i v_i=\sum_{i=1}^m \bigl( \sum_{j=1}^n [Tu_j]_i x_j \bigr) v_i[/tex]

    I hope the notation is easy enough to understand.

    Since basis vectors are linearly independent, we must have

    [tex]y_i=\sum_{j=1}^n [Tu_j]_i x_j[/tex]

    and this can be interpreted as a matrix equation:

    [tex]\begin{pmatrix}y_1\\ \vdots\\ y_m\end{pmatrix}=

    \begin{pmatrix}[Tu_1]_1& \dots& [Tu_n]_1\\
    \vdots& \ddots& \vdots& \\
    [Tu_1]_m& \dots& [Tu_n]_m\end{pmatrix}

    \begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix}[/tex]

    The column matrices "represent" the vectors x and y, and the m×n matrix "represents" the operator T.

    The term "linear operator" is usually reserved for the special case U=V. (This is why I called T a linear transformation).

    In quantum mechanics we're dealing with linear operators that map a Hilbert space H into itself (onto itself, in the case of observables). The bases are orthonormal. Because of this, we can use the inner product to write the matrix equation in a different way:

    [tex]\begin{pmatrix}\langle u_1|y\rangle\\ \vdots\\ \langle u_n|y\rangle\end{pmatrix}=

    \begin{pmatrix}\langle u_1|Tu_1\rangle& \dots& \langle u_1|Tu_n\rangle\\
    \vdots& \ddots& \vdots& \\
    \langle u_n|Tu_1\rangle& \dots& \langle u_n|Tu_n\rangle\end{pmatrix}

    \begin{pmatrix}\langle u_1|x\rangle\\ \vdots\\ \langle u_n|x\rangle\end{pmatrix}[/tex]
     
  13. Aug 24, 2004 #12

    arildno

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    Hmm..I can't agree with this.
    As long as one regards the domain/range as crucial components of the function (and not only the "assignment rule"), then it follows that knowledge gained about functions having one type of domain cannot necessarily be regarded as knowledge about functions having a completely different type of domain.
    For example, by saying that the domain is a function space will in general mean that we're having an infinite-dimensional domain. This facet alone should tell us that, for example, questions of basis, norms and convergence must be treated with greater care, and with other tools than those which suffice for functions with finite-dimesional domains.
     
  14. Aug 24, 2004 #13

    selfAdjoint

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    Quite so, but bases. norms, etc. are questions about a particular type of function, and the function concept is much more general than that. Generally a function preserves something in addition to doing its mapping. See Category theory for a rgorous treatment of this. The idea that all functions are homeopmorphisms is as restrictive as the notion that all manifods are differentiable.
     
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