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The dimensionality of an operator?

  1. Sep 5, 2005 #1
    I have some basic questions concerning operators. What is actually meant by the following:

    1) The dimensionality of an operator? E.g., what does it mean to say that the operator K has the dimension of 1/length (an example from Sakurai's book)? Operators act on abstract mathematical states to produce other states - how can you ascribe a dimension to such a quantity?

    2) The derivative of an operator? Like dA/dt, where A is an operator. Can anyone offer an intuitive explanation?

    3) An arbitrary function applied to an operator? Like exp(A) where A is an operator. In this case we can write exp(A) = 1 + A + A^2/2 + A^3/3 + ... - is this how you define a function of an operator in the general case, by using the taylor expansion?
     
    Last edited: Sep 5, 2005
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  3. Sep 5, 2005 #2

    Tom Mattson

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    When he says "dimension" there he's not talking about the dimension of a vector space, he's talking about units. K has to be an inverse length for the units to work out right.

    If an operator A is time-dependent, then it's eigenvalues are time-dependent. dA/dt is a measure of the rate of change of that time dependence.

    Yes.
     
  4. Sep 5, 2005 #3

    Hurkyl

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    It means exactly what it "should" mean: for example, I have an operator-valued function A whose domain is the reals, and whose range is operators, then we have:

    [tex]
    \frac{d}{dx}A(x) = \lim_{h \rightarrow 0} \frac{A(x + h) - A(x)}{h}
    [/tex]

    This definition makes sense because operators have norms, so the meaning of the limit is just like any other metric space. (such as R³)


    For the sufficiently insane, there's also ridiculously abstract way to make sense of feeding an operator into certain complex valued functions. (And it even lets you work with things that aren't analytic... things that don't have power series expansions)
     
  5. Sep 5, 2005 #4
    A function [tex]f(\hat{A})[/tex] of an operator [tex] \hat{A}[/tex] can be calculated by:

    [tex] f(\hat{A}) = \sum_{n} f(a_{n}) |n \rangle \langle n| [/tex]

    where you have the eigenvalue equation:

    [tex] \hat{A} |n \rangle = a_{n} |n \rangle [/tex]

    So if you want to calculate [tex]exp(\hat{A})[/tex] you can do this by:

    [tex]exp(\hat{A}) = \sum_{n} e^{a_{n}} |n \rangle \langle n|[/tex]

    with
    [tex] f(\hat{A}) = exp(\hat{A}) [/tex] and
    [tex] f(a_{n}) = e^{a_{n}} [/tex]

    I think this is called the spectral decomposition (or spectral theorem?). Someone else maybe can tell you what this 'technique' is called.

    -----------------

    I just found in the Preskill lecture notes:
    http://www.theory.caltech.edu/people/preskill/ph229/notes/book.ps
    See page 38 and 39:

    A self-adjoint operator in a Hilbert space H has a spectral representation - it's eigenstates form a complete orthonormal basis in H. We can express a self-adjoint operator [tex]\hat{A}[/tex] as

    [tex] \hat{A} = \sum_{n} a_{n} \hat{P_{n}}[/tex]

    with [tex] \hat{P_{n}} = |n \rangle \langle n| [/tex]
     
    Last edited: Sep 5, 2005
  6. Sep 5, 2005 #5
    Thanks, guys. I understand 2) and 3) now. There's still the issue about 1), though...
    I know he means units, but as far as I know only numbers can have units. We need to extend the concept of units if we are to apply to operators, aren't we? Or am I missing something?
     
  7. Sep 5, 2005 #6

    Tom Mattson

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    If we maintain that the dimensions of both sides of any equation in physics be the same then it naturally follows that operators have dimensions.

    And in any case, you learned in Physics I that objects other than numbers can have dimension. Take a vector such as velocity, for example.
     
  8. Sep 6, 2005 #7

    Hurkyl

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    Bleh, what I had posted is about bounded operators... to apply what I said exactly as I said it, you have to either restrict attention to an appropriate subspace of the state space or to an appropriate subspace of the operators.
     
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