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The free particle wave function <x|p>?

  1. May 5, 2015 #1
    I am following the math of scattering theory in Sakurai, Revised Edition pp.380-381

    For a free particle, one can find that the solution is a plane wave that can be written (in position space) as,

    [tex]<x|\phi>=\frac{e^{ip \cdot x}}{(2 \pi \hbar)^{3/2}}[/tex]

    However, how does one obtain ##<x|p>?## In the book it has..

    [tex]<x|p>=\frac{e^{ip \cdot x}}{(2 \pi \hbar)^{3/2}}[/tex]

    Which is identical to ##<x|\phi>##. Why are these two expressions the same? I also don't know what ##<x|p>## means physically. Momentum in the position basis?
     
    Last edited: May 5, 2015
  2. jcsd
  3. May 5, 2015 #2

    fzero

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    The state ##|p\rangle## is the eigenstate of the momentum operator:

    $$ \hat{p} | p\rangle = p |p\rangle.$$

    The free particle state ##|\phi\rangle## is also a momentum eigenstate, so it should be identified with ##|p\rangle##.

    In order to compute ##\langle x|p\rangle##, we should consider how to expand ##|p\rangle## in the position basis:

    $$ | p \rangle = \int dx f(p,x) |x\rangle.$$

    In the position basis, ##\hat{p}=-i\hbar d/dx##, so it must be that

    $$ \hat{p} | p \rangle = -i\hbar \int dx \frac{df(p,x)}{dx} |x\rangle .$$

    But this is an eigenvector with eigenvalue ##p##, so it is also the case that

    $$ \hat{p}| p \rangle = p | p \rangle = p\int dx f(p,x) |x\rangle.$$

    Comparing these leads to an equation:

    $$-i\hbar \frac{df(p,x)}{dx} =p f(p,x)$$

    which has solution

    $$f(p,x) = e^{i px/\hbar}.$$

    Computing ##\langle x |p \rangle ## is then just a matter of recognizing a delta function in

    $$\int dx' e^{i px'/\hbar} \langle x |x'\rangle.$$
     
  4. May 5, 2015 #3
    Thank you for the reply! This answers my question!
     
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