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

Position and Momentum Operators

  1. Nov 25, 2011 #1
    I would just like some clarification and some assertion that I've got the right idea. Please correct everything I say!

    For any observable [itex]A[/itex] over a finite-dimensional vector space with orthonormal basis kets [itex]\{|a_i\rangle\}_{i=1}^n[/itex] we can write
    [tex]A = IAI = \left(\sum_{i=1}^n |a_i\rangle\langle a_i|\right) A \left(\sum_{j=1}^n |a_j\rangle\langle a_j|\right) = \sum_{i=1}^n |a_i\rangle\langle a_i| \sum_{j=1}^n A|a_j\rangle\langle a_j| = \sum_{i=1}^n \sum_{j=1}^n |a_i\rangle\langle a_i|A|a_j\rangle\langle a_j| = \sum_{i,j} |a_i\rangle\langle a_j| \: \langle a_i|A|a_j\rangle \; \; (1)[/tex]
    and we say that [itex]\langle a_i|A|a_j\rangle[/itex] is the coordinate representation of A.

    My professor commonly talks about the coordinate representation of the momentum and position operators. I know these live in an infinite-dimensional vector space. Is this a Hilbert space? A rigged Hilbert space?

    We know that [itex]\hat{X}[/itex] acts on vectors [itex]|x\rangle[/itex] such that [itex]\hat{X}|x\rangle = x|x\rangle[/itex]. Why can we write the following? What does it mean?
    [tex]\int_{-\infty}^{\infty} dx \; x \;|x\rangle\langle x| \; \; (2)[/tex]
    For an arbitrary operator [itex]A[/itex] in this infinite-dimensional space, (suppressing the bounds of integration and writing x' to denote another variable and not the derivative of x),
    [tex]A = IAI = \left(\int dx' \; |x'\rangle\langle x'|\right) A \left(\int dx \; |x\rangle\langle x|\right) = \int\int dx' dx \; |x'\rangle\langle x'|A|x\rangle\langle x| = \int\int dx' dx \; |x'\rangle\langle x| \; \langle x'|A|x\rangle \; \; (3)[/tex]
    This is analogous to the finite-dimensional case in the sense that the coordinate representation of [itex]A[/itex] is given by [itex]\langle x'|A|x\rangle[/itex].

    Now if [itex]A = \hat{X}[/itex] then [itex]\langle x'|\hat{X}|x\rangle = \langle x'|x|x\rangle = x \langle x'|x\rangle = x \delta(x' - x)[/itex], and with this, (3) simplifies into (2) (by the sifting property of the Dirac delta function).

    I don't know how to formally proceed if [itex]A = \hat{P}[/itex] because it seems a little bit more tricky. Can anybody help there? Also, why do most textbooks avoid this integration business and simply state that [itex]\hat{X} = x[/itex] and [itex]\hat{P} = -i\hbar \frac{\partial}{\partial x}[/itex]?

    Thanks!
     
  2. jcsd
  3. Nov 25, 2011 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I would just call those numbers the components (or matrix elements) of A in the basis [itex]\{|a_i\rangle\}_{i=1}^n[/itex]. See post #3 here (the part before the quote) if you need to refresh your memory about the relationship between linear operators and matrices.

    The position and momentum operators are linear maps. Their codomains are Hilbert spaces (the same one actually). Their domains are dense subsets of that Hilbert space. They don't have any eigenvectors, so to make sense of what you wrote as [itex]\hat X|x\rangle=x|x\rangle[/itex], you would have to replace the Hibert space with a rigged Hilbert space.

    This will be a disappointing answer, but I don't think there's a complete answer that can be understood by a typical physics student that covers less than 200 pages. I still don't get it myself. It's been one of my long term goals for a long time. I think the short article http://www.math.neu.edu/~king_chris/GenEf.pdf [Broken] by Mustafa Kesir proves it, but it relies on the spectral theorem for not necessarily bounded normal operators on a Hilbert space, which is extremely hard. You need to be good at topology, measure theory and functional analysis before you can even begin to study its proof.

    I need to go and do something else for a while, so I don't have time to think about the rest of it right now.
     
    Last edited by a moderator: May 5, 2017
  4. Nov 26, 2011 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    That's just that: formally proceed, since most books do so. Indeed, to make full sense of is formally written in most books starting with Dirac's text of 1930, you need to understand, jokingly, be fluent in speaking functional analysis, or better, be one of the PhD students of Arno Böhm.

    So <most textbooks> use formal manipulations of vectors and distributions, simply because it's usually beyond the reader's interest to know the maths behind it. It's an identical situation with Feynman's path integrals. 99.9% of textbook readers don't CARE why they work and 99% of textbook authors don't KNOW why they work, if they ever work at all...
     
  5. Nov 26, 2011 #4

    strangerep

    User Avatar
    Science Advisor

    Rigged Hilbert space. (Although people sometimes try to work with an ordinary
    Hilbert space on which these operators are only densely defined.)

    Its meaning is very similar to your eqn (1). Since your [itex]a_i[/itex] are (I presume) eigenvalues associated with A, what happens if you evaluate your eq(1) one step further?
    I think this happens:
    [tex]
    \def\<{\langle}
    \def\>{\rangle}
    A ~=~ \dots ~=~ \sum_{i,j} |a_i\> \< a_j| ~ \<a_j| A |a_i\>
    ~=~ \sum_{i,j} |a_i\> \< a_j| ~ a_i \delta_{ij}
    ~=~ \sum_i |a_i\> \< a_i| ~ a_i
    [/tex]
    Get it?

    They don't really "avoid this integration business". Those operators are used because they satisfy the canonical commutation relations between position and momentum when represented as operators on a space of functions. (Strictly speaking, it's a space of generalized functions.)

    Try this introductory paper if you haven't already seen it:

    Rafael de la Madrid,
    "The role of the rigged Hilbert space in Quantum",
    Available as: arXiv:quant-ph/0502053
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook