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Normalization condition for free & spherically symmetric

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data

    I think, to normalize a wavefunction, we integrate over the solid angle ##r^2 dr d\theta d\phi##. Typically we have ## R(r)Y(\theta, \phi) ## as solutions. If ##Y## is properly normalized, then the normalization condition for ##R(r)## ought to be

    $$ \int_0^\infty dr r^2 |R(r)|^2 = 1$$

    2. Relevant equations

    Now I am trying to develop a general normalization condition for the spherical Bessel functions, which are solutions to the free particle in free space, but solved using spherical symmetry. For example the 0th solution is

    $$ R(r) = \sin (kr)/r $$

    where ## k = \sqrt{2mE/\hbar} ##.

    Now if I were to normalize with integration over ## r^2 dr ##, the normalization integral will not converge. Google searching has been unhelpful so far, but I've tended to see just integration over ## dr ##. So I'm conceptually unsure why this would be the case.

    3. The attempt at a solution

    The end goal is to have a general normalization factor for the spherical Bessel functions. If I normalize

    $$ \int_0^\infty dr | A R_0(r)|^2 = 1 $$

    In this case, ## A = \sqrt{2/\pi k} ##.

    In fact I found by just doing the normalization for the first few functions, sqrt(2l+1) works as the additional normalization factor.
    But the current sticking point is the integration over dr or r^2 dr.

    Thanks.
     
    Last edited: Nov 22, 2015
  2. jcsd
  3. Nov 22, 2015 #2

    king vitamin

    User Avatar
    Gold Member

    Free particle states aren't normalizable. You can see this pretty quickly in cartesian coordinates (showing one dimension for simplicity):
    [tex]
    \psi_k(x) = Ne^{i k x}
    [/tex]
    where [itex]k[/itex] is a quantum number related to energy [itex]E = \hbar^2 k^2/2m[/itex]. Since [itex]|\psi(x)|^2 = 1[/itex], clearly you can't normalize this state.

    Instead, it's conventional to normalize the states as
    [tex]
    \int dx \psi_{k'}^*(x) \psi_{k}(x) = \delta(k - k')
    [/tex]
    with the dirac delta function. Then you can show that [itex]N = 1/\sqrt{2 \pi}[/itex]. This normalization is useful because you can then write down general normalizable states as
    [tex]
    \Psi(x) = \int dk A(k) \psi_k(x)
    [/tex]
    and now the normalization condition on [itex]\Psi[/itex] becomes
    [tex]
    \int dx |\Psi(x)|^2 = \int dk A(k) = 1.
    [/tex]
    So [itex]A(k)[/itex] is a normalized distribution over which [itex]k[/itex]'s contribute to the wave function. So a good choice for normalizing for your spherical problem would be to generalize the delta function normalization above.
     
  4. Nov 23, 2015 #3
    I guess i have
    $$
    |A|^2 \int_0^\infty dr r^2 \psi^*_{0,k'}(r)\psi_{0,k}(r) = \delta(k-k')
    $$
    and ## \psi_0(r) = \sin(kr)/r = \frac{i}{r}( \exp(ikr) - \exp(-ikr))##, and the r^2 will cancel, and then one does the integral...
     
    Last edited: Nov 23, 2015
  5. Nov 23, 2015 #4

    king vitamin

    User Avatar
    Gold Member

    Looks good. You'll need to use the integral I used above,
    [tex]
    \int_{-\infty}^{\infty} dx e^{ikx} = 2 \pi \delta(k),
    [/tex]
    and be careful with all of the algebra. A few points: you should actually have [itex]j_{0}(kr) = \sin(kr)/(kr)[/itex], and you're missing some factors for the exponential form for [itex]\sin[/itex].
     
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