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Representation of QM

  1. Mar 11, 2013 #1
    Coordinate representation ##\psi=\psi(x)##
    Momentum representation ##\psi=\psi(p)##
    Fourier transform
    [tex]\psi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}\psi(x)e^{-\frac{ipx}{\hbar}}dx[/tex]
    I'm confused with this ##\hbar##? Why not
    [tex]\psi(p)=\frac{1}{\sqrt{2\pi}}\frac{1}{\hbar}\int^{\infty}_{-\infty}\psi(x)e^{-\frac{ipx}{\hbar}}dx[/tex]?
     
  2. jcsd
  3. Mar 11, 2013 #2

    Bill_K

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    Science Advisor

    Because ∫exp(ikx) dk ≡ 2π δ(x). And therefore ∫exp(ipx/ħ) dp = 2π δ(x/ħ) = 2πħ δ(x).

    Look at the normalization integral, with the 1/√2πħ factors included:

    1 = ∫ψ*(p) ψ(p) dp = 1/(2πħ) ∫∫∫ψ*(x) exp(ipx/ħ) ψ*(x') exp(-ipx'/ħ) dp dx dx'

    Do the p integral, using line 1:

    = 1/(2πħ) ∫∫ψ*(x) ψ*(x') (2πħ) δ(x - x') dx dx'

    Next do the x' integral, using the δ-fn. See we again get the correct normalization:

    1 = ∫ψ*(x) ψ*(x) dx

    Note that the factors of 2πħ canceled, showing that the 1/√2πħ was necessary!

    EDIT: Hold on! I see you asked the exact same question on two consecutive threads!
     
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