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Separation of Variables in Spherical Schrodinger Equation

  1. May 12, 2012 #1
    The normalization condition is:

    ∫|ψ|[itex]^{2}[/itex]d[itex]^{3}[/itex]r=1

    In spherical coordinates:

    d[itex]^{3}[/itex]r=r[itex]^{2}[/itex]sinθdrdθd[itex]\phi[/itex]

    Separating variables:

    ∫|ψ|[itex]^{2}[/itex]r[itex]^{2}[/itex]sinθdrdθd[itex]\phi[/itex]=∫|R|[itex]^{2}[/itex]r[itex]^{2}[/itex]dr∫|Y|[itex]^{2}[/itex]sinθdθd[itex]\phi[/itex]=1

    The next step is the part I don't understand. It says:

    ∫[itex]^{∞}_{0}[/itex]|R|[itex]^{2}[/itex]r[itex]^{2}[/itex]dr=1 and ∫[itex]^{2\pi}_{0}[/itex]∫[itex]^{\pi}_{0}[/itex]|Y|[itex]^{2}[/itex]sinθdθd[itex]\phi[/itex]=1

    I don't understand why they both have to be one. I remember learning that if a function dependent on one variable equals the function dependent on another variable then they both must equal a constant, which makes sense to me. Given the equations here, however, why cant the R part of ψ in Equation 3 equal, say 0.5, and the Y part = 2?
     
  2. jcsd
  3. May 12, 2012 #2

    cepheid

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    Well here's a heuristic argument that I can think of for why they must be separately normalized to 1.

    You can think of R as a 1D position wavefunction for the radial position. If the radial position is a random variable, then |R|^2 is its probability density function, and like any pdf, the integral over all possible r values of this pdf must equal 1, since the probability of finding the electron *at some radius between 0 and infinity* is 100%

    Similary, Y is a 2D position wavefunction for the angular position coordinates theta and phi. Hence |Y|^2 is a joint probability density function for theta and phi, and like any joint pdf, the integral over all possible values of all of the variables must be 1, because the probability of finding the electron at *some* azimuthal angle and *some* polar angle on the sphere is 100%.

    Another argument: It would be very fortuitous indeed if the function R(r) and the function Y(θ, ##\phi## ) just so happened to be of the right shape that the integral over the radial function was x (for x < 1) and the integral over the angular function was (1-x). I don't think that there is any reason why this would have to be true for every R and Y that satisfied the Schrodinger equation.
     
  4. May 13, 2012 #3

    dextercioby

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    Incidentally, the Y's are the spherical harmonics which are an orthonormal basis in the Hilbert space [itex] \displaystyle{\mathcal{L}^2\left(S^2,d\Omega\right)} [/itex], so the second integral in the product is necessarily 1. Then it follows that the first integral should be equal to 1 as well.
     
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