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The solution to Schrodinger's equation are wave functions $\Psi (r,\theta ,\phi )$ of the form, $\Psi (r,\theta ,\phi )= R(r)\Theta(\theta)\Phi(\phi)$ Where, the probability of finding an electron in a region, V:$0<r<\infty , 0<\theta <\pi , 0<\phi<2\pi $, is found using the volume integral;

$$N^2\int_{0}^{\infty }r^2R^2(r)dr\int_{0}^{2\pi }\Phi^2(\phi)d\phi\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta =1$$

Now, consider the $2p_{y}$ orbital; $\Psi _{2p_{y}}=\frac{1}{4\sqrt{2\pi a_{0}^{5}}}re^{-\frac{r}{2a_{0}}}sin \theta sin \phi$

And now the question itself; Evaluate the three integrals,

One. $\int_{0}^{\infty }r^2R^2(r)dr$

Two. $\int_{0}^{2\pi }\Phi^2(\phi)d\phi$

Three. $\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta$

and show $\int_{V}^{}\Psi ^2dV=1$

I understand this may be alot but any help would be greatly appreciated. Integration is something I really need to work on. Thanks guys.

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The tag complex-integration does not seem to fit. –  Fabian Sep 22 '12 at 6:18
    
@Fabian It's a complex problem about integration. –  Austin Mohr Sep 22 '12 at 6:51
1  
The solutions to the Schrödinger equation are linear combinations of such things. –  Qiaochu Yuan Sep 22 '12 at 6:53
2  
Note that in many languages that use diacritical marks (including German) the letters with and without them bear only a historical relationship and don't represent similar phonemes. Thus, leaving them off is about as bad as replacing the letter by a completely different one; that is, when you write "Schrodinger", you might as well write "Schridinger". If you can't produce letters with diacritics on your keyboard, you can always copy them from the Web, e.g. in this case from the Wikipedia article on Schrödinger (which you can find by searching for "Schrodinger" :-). –  joriki Sep 22 '12 at 7:40

1 Answer 1

From your expression for $\Psi$ one can deduce

\begin{align} R(r) &= re^{-r/(2a_0)} \\ \Phi(\phi) &= \sin\phi \\ \Theta(\theta) &= \sin\theta \\ N &= \frac{1}{4\sqrt{2\pi a_0^5}} \end{align}

so that

\begin{align} &\int_0^{\infty}r^2R^2(r)dr = \int_0^{\infty}r^4e^{-r/a_0}dr = \\ &\quad= -a_0^5\left.\left[ \left(\frac{r}{a_0}\right)^4 +4\left(\frac{r}{a_0}\right)^3 +12\left(\frac{r}{a_0}\right)^2 +24\left(\frac{r}{a_0}\right) +24 \right]e^{-r/a_0}\right|_0^{\infty}=24a_0^5\\ &\int_0^{2\pi}\Phi^2(\phi)d\phi=\int_0^{2\pi}\sin^2\phi d\phi=\\ &\quad=\left.\frac{1}{2}(\phi-\sin\phi\cos\phi)\right|_0^{2\pi}=\pi\\ &\int_0^{\pi}\Theta^2(\theta)\sin\theta d\theta=\int_0^{\pi}\sin^3\theta d\theta=\\ &\quad=\left.\left(-\cos\theta+\frac{1}{3}\cos^3\theta\right)\right|_0^{\pi}=\frac{4}{3} \end{align}

Putting all together:

$$ N^2\int_0^{\infty}r^2R^2(r)dr\int_0^{2\pi}\Phi^2(\phi)d\phi\int_0^{\pi}\Theta^2(\theta)\sin\theta d\theta=\frac{1}{16(2\pi a_0^5)}\cdot24a_0^5\cdot\pi\cdot\frac{4}{3}=1 $$

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Damn, I was just warming up mu wxmaxima. –  copper.hat Sep 22 '12 at 7:40
    
The first two integrals don't really require antiderivatives -- $\int_0^\infty r^n\mathrm e^{-r}\mathrm dr=n!$ (by $n$-fold integration by parts, or the definition of the gamma function), and $\sin^2\phi$ averages to $\frac12$ over any quarter period. –  joriki Sep 22 '12 at 7:48

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