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And while I'm at it: eigenfunction of position

  1. Jul 7, 2004 #1
    I've yet to see a decent argument as to why the eigenfunctions of the position operator are delta functions. (Griffith's argues this, but oh so weakly). Could someone provide one or a couple dozen

    Kevin
     
  2. jcsd
  3. Jul 7, 2004 #2

    reilly

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    Science Advisor

    Tricky indeed. If I recall correctly, Dirac's argument goes roughly:

    In momentum space, x=id/dp (up to factors or h and +/-1)

    <x|p> = exp(+iKX) (apart from normalizing factors)

    So <p|x> ~exp(-iKX)

    thus <X | X'> ~Integral over p of <X|P><P|X'> ~ delta(X-X'), the wave function in space of a position eigenstate with eigenvalue of X'.

    A more complicated but physically understandable approach is to work with the delta function as a limit of Gaussians, wave packets and all that.

    Remember, it took quite a while for mathematicians to catch up with Dirac.

    Regards,
    Reilly Atkinson
     
  4. Jul 8, 2004 #3

    turin

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    I suppose that you mean "... in the position basis?"

    The way I've understood it is that the components of the matrix that is used to represent the position operator in the position basis are the "values" of the Dirac delta functional: &delta;(x,x'). When the matrix operates on a function, in the position basis, it is an integration over one of the two variables, x or x' (conventionally x'), analogous to the summation over one of the two indices of a matrix component when a discrete matrix multiplies a discrete vector. Since the matrix is diagonal in the position basis, each row or column of the matrix is a Dirac delta function: &delta;(x - x'), where one of the two variables, x or x' (conventionally x'), is held fixed and treated as a constant. Since it is the row or column of a diagonal matrix (which I believe is also Hermitian), then it is an eigenvector of that matrix.
     
  5. Jul 8, 2004 #4
    Again, check out

    http://www.math.sunysb.edu/~leontak/book.pdf

    pages 46 and 47 (... maybe starting at chapter 2, page 35).

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

    Or, in more simplistic terms:

    Interpret <x|x'> as a distribution. We have, for any "test function" f(x)

    f(x) = <x|f> = Integral { <x|x'><x'|f> dx' } = Integral { <x|x'> f(x') dx' } .

    But, we also have

    f(x) = Integral { delta(x,x') f(x') dx' } .

    So, subtract to get

    Integral { [<x|x'> - delta(x,x')] f(x') dx' } .

    Since f(x) is an arbitrary "test function", it follows that

    <x|x'> - delta(x,x') = 0 .

    [The above is (more or less) Turin's words in symbolic form.]
     
    Last edited: Jul 8, 2004
  6. Jul 8, 2004 #5
    First, thank you all for helping.

    reilly wrote
    Hmmm, why is it the case that <x|p>=exp(iKX)?

    with regards to turin

    The matrix description is interesting, I'll chew on that.

    Eye_in_the_Sky wrote

    I must be confused (but perhaps close to understanding) since you've just proved that free states aren't normalizable, i.e. eigenfunctions corresponding to a continuous spectrum of eigenvalues can't be normalized like those of a discrete spectrum can be. How does this prove that the eigenfunction of position (in position basis) is the delta function? (By the way thanks for the website, i'll check it out).

    Once again, thanks for your help.

    Kevin
     
  7. Jul 8, 2004 #6
    <x|x'> - delta(x,x') = 0 implies <x|x'> = delta(x,x') , and <x|x'> is the position-representation of |x'> which is the eigenket of X with corresponding eigenvalue x'.


    By theorem <x|k> ~ exp(ikx) iff k ~ -i(d/dx).

    (... think of Fourier transforms)
     
  8. Jul 9, 2004 #7
    Oh I think I'm getting it.

    So while driving to my folks house for a bar-b-que a light opened up upon me, it was the sun. Half an hour later though, another light began to shine and it was about position eigenfunctions.

    It makes a great deal of sense now that I think about it. If |x> is an eigenfunction of x then <x|f>=f(x), I just didn't see that of course this reeks of delta functions. A lot of thanks for you folks for helping out.

    Kevin
     
  9. Jul 10, 2004 #8
    more answers for your "deeper" mathematical questions about qm

    While Takhtajan's e-book is great ( i had the pleasure of taking that course with him some years ago, while simultaneously taking a course in QM using Von Neumann's text and his was so much clearer; his exposition is always fantastic) it may be that you will want to work out the rigorous matheamtical justifications and the details of QM all along as you go. If so, you will find the road leads to more and more functional analysis but it can become quite a consuming passion. If this is the case you can do no better than Reed and Simon's 4 volume fantastic work on the subject titled simply Functional Analysis.

    Reed Simon Functional Analysis

    This book isn't enclyclopedic like say, Yoshida's, or terse and written to impress like Rudin's books on the same topic. In fact, these books take quantum mechanics as the starting point (and the whole motivation, really) for functional analysis. They assume nothing (really nothing...they even do some set theory in chapt one vol 1) but they go all the way (for non- relativisitc spinless, stuff) including the only detailed exposition i have ever seen on why the schrodinger equation shouldn't have any solutions and then why it does in some cases that turn out to be important. (compactness of resolvent and all that stuff)
    If you are interested in QM and love math,open these books...every decent university library will have them....they are a rarely mentioned gem for sure.
     
    Last edited: Jul 10, 2004
  10. Jul 10, 2004 #9
    Just be sure you keep track of "where" each operation is taking place and where each function, distribution, operator, etc "lives". The dirac notation doesn't make this clear at all. In fact despite its convenience for self adoint operaotrs, it often obfuscates the issues associated with domain and image and makes it hard to visualize where the operators take place.
    But its is all rigorously worked out.
    Of course when you get to quantum field theory, well, that's another story...
     
    Last edited: Jul 10, 2004
  11. Jul 10, 2004 #10
    Thanks for your response. I've always sort of pegged myself as an aspiring mathematical-physicist. I really can't do one without the other. As I have been looking for the rigorous foundations of QM I have begun to notice that functional analysis is key. Lucky for me, this coming year is all measure theory and functional analysis, which should help me come to grips with all of this stuff.

    I hope to read that one as well...someday :)

    Thanks,

    Kevin
     
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