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Confusion related to momentum eigenstates

  1. Jan 31, 2016 #1
    When we deal with an infinite-dimensional basis,the normalization condition of this basis becomes <x|x'>=δ(x-x')(here for example the position basis).Same thing for momentum eigenstates <p|p'>=δ(p-p').
    Lets look now on the eigenvalue problem of the momentum operator:
    [itex]\hat p | p \rangle =p | p\rangle [/itex]
    projecting [itex]\langle x |[/itex] on both sides this will yield to a differential equation of the form :

    [itex]-iħ\frac{d\psi_{p}(x)}{dx}=p\psi_{p}(x)[/itex]

    where [itex]\psi_{p}(x)=\langle x | p\rangle[/itex]

    this will finally yield that [itex]\psi_{p}(x)=\frac{1}{√(2πħ)}e^{ipx/ħ}[/itex]

    so [itex] | p \rangle \leftrightarrow \frac{1}{√2πħ} (e^{ ipx_{1} } | x_{1} \rangle +e^{ ipx_{2} } |x_{2} \rangle+ . . . . . +e^{ ipx_{n} } | x_{n} \rangle)[/itex] where n goes to infinity and [itex]x_{n}=x_{n-1}+dx[/itex]
    but doesn't this violate<x|x'>=δ(x-x')? since if we project,for example,<x1| on |p> and we take by consideration δ(x-x')=<x|x'>, the result will be <x|p>=∞ and not [itex]\frac{1}{√2πħ}e^{ ipx_{1} }[/itex] ???
     
  2. jcsd
  3. Jan 31, 2016 #2

    vanhees71

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    I do not understand what you mean with the last paragraph. The consistency between the generalized position and momentum eigen bases is given by a well-known theorem in Fourier analysis:
    $$\langle x|x' \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|x' \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{2 \pi \hbar} \exp[\mathrm{i} p(x-x')/\hbar]=\delta(x-x').$$
     
  4. Jan 31, 2016 #3
    I mean multiply <x1| by |p> , the result will be 1/√(2πħ) *exp(ipx1)*<x1|x1> and as <x1|x1>=∞ then <x1|p>=∞ and not 1/√(2πħ)*exp(ipx1).
     
  5. Jan 31, 2016 #4

    blue_leaf77

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    That's where the mistake lies because that's not the proper way to expand a state into position basis. You should instead operate the identity operator ##1 = \int |x\rangle \langle x| dx## to the ket ##|p\rangle##.
    $$
    |p\rangle = \int |x\rangle \langle x|p\rangle dx
    $$
    You can easily see if you project this onto certain position basis ##|x'\rangle##, you will obtain what you want.
     
  6. Feb 1, 2016 #5

    vanhees71

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    I don't understand your math at all. As you have derived correctly yourself, the generalized momentum eigenstates in the position representation read
    $$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x)$$
    without any undetermined (diverging) factors.
     
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