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I have this wave function: $$\Psi=C e^{-\rho / 2} \rho^{l} L_{1}^{3} Y_{l, m} $$ To normalize the function I have tried to express the polynomial in the function as follows. If: $$L_{1}^{3}(x)=(-1)^{1} \frac{d^{1}}{d x^{1}} L_{3+1}(x)=-4 x^{3}+48 x^{2}-144 x+96$$ Then: $$\Psi=A e^{-\rho / 2} \rho^{l}\left(-4 \rho^{3}+48 \rho^{2}-144 \rho+96\right) Y_{l, m}$$ Using the normalization condition: $$ 1=\int|\Psi| \rho^{2} d \rho d \Omega = C^2 \int_0^{\infty} e^{-\rho} \rho^{2(l+1)} L_{1}^{3} d\rho \int|Y_{l,m}|^2 d \Omega $$ where: $\int|Y_{l,m}|^2 d \Omega = 1$. Therefore $$C^2 \int_{0}^{\infty} e^{-\rho} \rho^{2(l+1)} (-4 \rho^{3}+48 \rho^{2}-144 \rho+96) d \rho = 1$$ The integral diverges. What am I wrong?

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    $\begingroup$ With the vanishing exponential inside the integral, are you sure it diverges? In any case, it'd help if you gave the context for this calculation. $\endgroup$
    – Miyase
    Jun 18 at 8:43
  • $\begingroup$ It is a wave function in spherical coordinates. I just noticed that term $\rho^{2l}$ should have entered expression $\int\left|Y_{l, m}\right|^{2} d \Omega=1$ . So $\int \rho^{2l} \left|Y_{l, m}\right|^{2} d \Omega=1$? I'm sorry, but I don't deal much with the properties of spherical harmonics. $\endgroup$
    – Dayzk
    Jun 18 at 17:23

1 Answer 1

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The integral won't diverge since

\begin{equation} \int_{0}^{\infty} \rho^{x}e^{-\rho} \text{d}\rho = \Gamma(x+1) = x! \quad \text{for} \quad x \in \mathbb{N} \end{equation}

Just plug in the appropriate values for $x$ in order to evaluate each term of the integral and you're sorted.

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