# triple integral (quantum mechanics)

I recently started a quantum mechanics course after a long time with no serious maths and I'm having some problems with the most basic maths operations.

Please, help me solve this triple integral (it's a non-graded book exercise and I know that the result should be: $1 \over \sqrt{2\pi}$)

Data: \begin{align*} \Psi_{2p1}(r,\theta,\phi) &= \sqrt{1 \over{64\pi a^5}}re^{-r \over 2a} \sin\theta ·e^{i\phi}\\ \Psi_{2px}(r,\theta,\phi) &= \sqrt{1 \over{32\pi a^5}}re^{-r \over 2a} \sin\theta \cos\phi\end{align*} Demonstrate that: $$\left\langle\Psi_{2p1} | \Psi_{2px}\right\rangle= \frac{1}{\sqrt{2\pi}}$$

The actual integral to solve is: $$\int\limits_{0}^\infty \int\limits_{0}^\pi \int\limits_{0}^{2\pi}{\Psi_{2p1}^*\Psi_{2px}r^2\sin\theta\,\mathrm d\phi\,\mathrm d\theta\,\mathrm dr}$$

Thanks!

My try: $$\left\langle\Psi_{2p1} | \Psi_{2px}\right\rangle = \sqrt{1 \over{2^{11}\pi^2 a^{10}}}\int\limits_{0}^\infty \int\limits_{0}^\pi \int\limits_{0}^{2\pi}r^4e^{-r \over a} \sin^2\theta \cos\phi e^{i\phi} \,\mathrm d\phi\,\mathrm d\theta\,\mathrm dr$$ $$={1 \over{2^{5}\pi a^{5}\sqrt{2}}}\int\limits_{0}^\pi\int\limits_{0}^{2\pi} [(-a /5r)r^5e^{-r \over a}]_o^\infty \sin^2\theta \cos\phi e^{i\phi} \,\mathrm d\phi\,\mathrm d\theta$$

Is it right so far? How do I continue?

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It's not clear from the question which part you're having trouble with. Do you know how to set up the integral for that matrix element? Do you know how to carry out the angular integration? The radial integration? – joriki Mar 29 '12 at 5:20
I'm rapidly losing interest in helping you. I wrote twice that there seems to be a problem with the wave functions, and you ask whether there are more things wrong that you missed without addressing those comments? Are you actually reading the comments? – joriki Mar 29 '12 at 5:45
@joriki I don't see what else is wrong with the wave functions (I am copying them from a book). Could you be more explicit, please? maybe say exactly what the issue is? – margaritam Mar 29 '12 at 5:53
b) You had the integrations in the wrong order. They should nest so that the innermost differential symbol corresponds to the innermost integral symbol. Also it's unusual to repeat the variable names in the integral limits, and it's not necessary for clarity if you get the order right. c) You'd missed a \rho. d) It's rather unusual to have explicit multiplication dots everywhere, and I don't think it makes for easier reading, rather the contrary. e) I put the d's in the differential symbols in Roman, but that's personal preference; some people put them in italics. – joriki Mar 29 '12 at 6:28
Sorry, I give up. I suggest you consider going back to learning more elementary things before tackling quantum mechanics. Your attempt at solving the integral again contains more than one very basic error per equation. You're missing two factors of $\sin\theta$ and all factors containing $\phi$; you dropped the $r$ differential but the $\phi$ integral; and it's unclear which of the two you were actually trying to carry out because the result would be correct for neither. I won't be commenting anymore; best of luck with this, and do consider learning more basic things first. – joriki Mar 29 '12 at 7:01

Start from

$$\left\langle\Psi_{2p1} | \Psi_{2px}\right\rangle = \sqrt{1 \over{2^{11}\pi^2 a^{10}}}\int\limits_{0}^\infty \int\limits_{0}^\pi \int\limits_{0}^{2\pi}r^4e^{-r \over a} \sin^3\theta \cos\phi e^{i\phi}~ \mathrm d\phi~\mathrm d\theta~\mathrm dr$$ (which, note, is not the same as the expression in the question statement; thanks @joriki.)

1. Expand the $e^{i\phi}$ term using Euler's formula to be $\cos\phi + i \sin\phi$. If you multiply this against the $\cos\phi$ factor already in the expression, and integrate from 0 to $2\pi$, you see that the term $\sin\phi\cos\phi$ integrate to zero (why?) and that $\cos^2\phi$ integrate to some constant (why? And I'll leave it to you to compute that constant yourself).

2. Now the innermost integral is taken care of, you can integrate the term $\sin^3\theta$ from $0$ to $\pi$. This gives you another constant (what is it?).

3. Lastly, you need to evaluate the $r$ integral. Note that after the previous two steps you are left with something that looks like $$\text{Constant}\cdot \frac{1}{a^5} \int_0^\infty r^4 e^{-\frac{r}{a}} \mathrm{d}r$$ Now, you can rewrite it as $$\text{Constant}\cdot \int_0^\infty \left(\frac{r}{a}\right)^4 e^{-\frac{r}{a}} \mathrm d \left(\frac{r}{a}\right)$$ so doing the change of variables $\rho = r/a$, your integral becomes $$\text{Constant}\cdot \int_0^\infty \rho^4 e^{-\rho} \mathrm{d}\rho$$ This you can solve simply by repeated integration by parts, or by appealing to the Gamma function (whose values at positive integers are explicitly known).

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I hate to say it, but you're also missing a factor $\sin\theta$. Back to basics... ;-) – joriki Mar 29 '12 at 9:27
I love it. Thank you so much for your step by step answer. I'm gonna try to go through it answering your test questions. 1. $\sin\phi\cos\phi =0$ because $\sin4\pi = 0$. Also $\int \cos^2\phi d\phi = x/2 + (1/4)\sin2x + \kappa$. Since $\sin4\pi = 0$ >> $\int \cos^2\phi d\phi = x/2 = \pi$ – margaritam Mar 29 '12 at 10:14
2) $\int_0^\pi \sin^2\theta d\theta = \pi/2 -1/4$ ? || 3) $\int_0^\infty \rho^4 e^{-\rho}d\rho = 4!(1/\rho)^5 ? – margaritam Mar 29 '12 at 10:22 3)$\int_0^\infty \rho^4 e^{-\rho}d\rho = 4!(1/\rho)^5$? >> I don't really understand this last part. Basically I tried to apply the gamma function by trying to follow my text book's explanation. – margaritam Mar 29 '12 at 10:28 @margaritam: It really would be in your own interest to read comments; you're making things so unnecessarily inefficient and frustrating. As I wrote above, there's a factor$\sin\theta\$ missing in the answer. – joriki Mar 29 '12 at 11:12