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Very elementary notation question

  1. Jul 18, 2009 #1
    Shankar p68-69 gives a mathematical "derivation" of the action of the X (position) operator, the summary of which is as follows:
    I followed the logic without a problem, since it only involves using the matrix elements of X in the basis of eigenfunctions of X. However, the next paragraph reads:
    Similarly, he writes, of the action of X in the K basis
    Now, to me

    [tex]f(x) = \left\langle x | f \right\rangle[/tex]

    and

    [tex]g(k) = \left\langle k | g \right\rangle[/tex]

    are scalars. Hence I cannot comprehend what is intended by Shankar's notation. Any insight?
     
    Last edited: Jul 18, 2009
  2. jcsd
  3. Jul 18, 2009 #2
    I am wondering if the next computation gives some insight: Shankar then computes the matrix elements of X in the K basis as:
    where, again, to me

    [tex]\left\langle x|k\right\rangle \propto e^{ikx}[/tex]

    and not

    [tex]\left|k\right\rangle \propto e^{ikx}[/tex]
     
    Last edited: Jul 18, 2009
  4. Jul 19, 2009 #3
    Inside the integral you have

    [tex]e^{-ikx} x e^{ik'x}[/tex]

    [tex]= e^{-ikx} \frac{1}i \frac{d}{dk'} e^{ik'x}[/tex]

    which is what leads him to write that [tex]\textbf{X} \left| g(k) \right\rangle = \left|i\frac{dg(k)}{dk}\right\rangle[/tex].


    The notation [tex]|f(x)\rangle[/tex] as the ket corresponding to f(x) (which he says near the top of pg. 69) is sloppy, but I don't think there is anything wrong with it. He's not saying that [tex]|k\rangle \propto e^{ikx}[/tex], he's just using it as a notation to express the ket that comes out of the operation [tex]X|f\rangle[/tex].
     
    Last edited: Jul 19, 2009
  5. Jul 19, 2009 #4
    This is correct. To go from left to right, simply insert
    [tex]1=\int_{-\infty}^{\infty}\text{d}x\left|x\right\rangle\left\langle x\right|[/tex]
     
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