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Problem with normalization wave function position/momentum space

  1. Jan 22, 2014 #1
    1. The problem statement, all variables and given/known data
    We start with a pure state at t=0 of an electron is
    [itex]C e^{- a^2 x^2} \left(\begin{array}{c}
    1\\
    i
    \end{array}\right)[/itex]

    Probability density of measuring momentun [itex]p_0[/itex] and third component of spin [itex]- \frac{\hbar}{2}[/itex]

    And probability of measuring a state with momentum between 0 and [itex]p_0[/itex]


    2. Relevant equations

    Fourier Transform

    3. The attempt at a solution

    I have a wave function with the spatial and spinor (Z basis),i know the normalization for the spinor is [itex]\frac{1}{\sqrt{2}}[/itex] and for the spatial part i just have to solve this
    [itex] \int_{-\infty}^{\infty}\psi^{*}\psi dx = 1[/itex]

    Wich gives me
    [itex] C = (\frac{2}{\pi})^{1/4} \sqrt{a} [/itex]

    Now i have a spatial normalized wave function wich is

    [itex] \psi = (\frac{2}{\pi})^{1/4} \sqrt {a} e^{- a^2 x^2} [/itex]

    Since the momentum is not well defined in the spatial basis i cant obtain the probability of measuring [itex]p_0[/itex] right away, i have to use the momentum basis wave function, so i must do a fourier transform of the function like this

    [itex] \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty}\psi e^{\frac{-i p x}{\hbar}} dx = \frac{1}{\sqrt{a \hbar}} \frac{1}{(2 \pi)^{1/4}} e^-{\frac{p^2}{4 a^2 \hbar^2}}=\phi [/itex]

    Now of the probability to obtain p0 [itex] <p_0|\phi> [/itex]

    Wich i assume is the integral from p0 to p0 wich is zero, and the probability between 0 and [itex] p_0[/itex] its the integral from 0 to [itex] p_0[/itex]

    [itex] \int_{0}^{p_0} \phi p dp = \frac{2^{3/4} (a \hbar)^{3/2}}{\pi^{1/4}} (1-e^{\frac{-p_0^2}{4 a^2 \hbar^2}}) [/itex]


    But this has to be wrong, because if i take [itex] p_0 = \infty [/itex] it must be equal to 1 wich is not, whats wrong with the normalization ? or is something else ?

    Thanks !
     
  2. jcsd
  3. Jan 22, 2014 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The idea with the Fourier trnsform to get the momentum wave function is correct. Now just remember how to get the probability density (momentum distribution) from the momentum wave function.
     
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