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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?2012-03-29
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    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?2012-03-29
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    @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?2012-03-29
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    OK, I tidied up the $\TeX$. Some hints for the future: a) Function names shouldn't be italicized. Arturo had already corrected the ones in the wave functions, but you introduced another italicized sine function in the volume factor. If you just write `sin`, that gets interpreted as a juxtaposition of three variables, whose names get italicized. There are predefined commands like `\sin` for widely used functions; if you need a function name for which there isn't a predefined command, you can always use `\operatorname{name}`. (continued...)2012-03-29
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    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.2012-03-29
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    Now back to the content: I had originally pointed out that you haven't told us which part of the integral you're having trouble with, and I don't think you've responded to that.2012-03-29
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    @joriki thanks for all those changes. I see now that I was missing many things I didn't know about. I'll be more crareful in the future. About the content I have most problems with the angular parts, but I didn't post anything, because I'm not so sure about the radial either. I'll try posting a bit of my reasoning to see if it is easier to point out where I'm wrong.2012-03-29
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    added up until where I'm stuck2012-03-29
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    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.2012-03-29
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    I've corrected my attempt as best as I could based on your comments and I understand what you say, unfortunately I can't give up now. Thanks for your help; you did try.2012-03-29
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    *unfortunately I can't give up now*... Nobody is suggesting *to give up* here, on the contrary, but rather to go back to the basics that must be mastered before the symbols you fiddle with begin to even make sense for you.2012-03-29
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    I see, thank you Didier. Where do you suggest I start?2012-03-29
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    @joriki and others: I am cleaning up the comment a bit since some are obsoleted by the edits.2012-03-29

1 Answers 1

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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... ;-)2012-03-29
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    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$2012-03-29
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    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 ?2012-03-29
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    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.2012-03-29
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    @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.2012-03-29
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    @joriki: I just started with what margaritam gave us. `:p` Fixed by a glob of $2$ to $3$.2012-03-29
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    @margaritam: so for step (2) you should be computing instead $\int_0^\pi \sin^3\theta \mathrm{d}\theta = \int_0^\pi \sin\theta - \cos^2\theta\sin\theta \mathrm{d}\theta$2012-03-29
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    @joriki I did read your comment, but I couldn't follow the global process and decide if it was you or Willie Wong who was right, so I decided to answer Willie Wong's questions and that I could always come back later and do it again if something was indeed missing.2012-03-29
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    Thanks to both! The new answer then: 2) $\int_o^\pi \sin^3\theta d\theta = 0$. Isn't it? If all those things are right, how does it end up giving $1 \over \sqrt 2 \pi$?2012-03-29
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    @margaritam: No, the integrand is non-negative throught the integration region, how could the integral be $0$? Regarding the comments: OK, but you do have quite a non-interactive style. You could have just said something like "I don't know who's right so I'll answer Willie's questions first" instead of seemingly ignoring my comment. Also, all you need to do to find out who's right is count the factors of $\sin\theta$ -- one from each wave function and one from the Jacobian makes three.2012-03-29
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    @joriki I understand your comment, I will try to be more interactive. I have to leave now, but I'll try to give it a fresh look later today and see if I can compute it to the end. I'll post what I come up with. Thanks!2012-03-29