My wave function is $$ \Psi = A e^\left({-\frac{\left|x\right|}{2a}- \frac{\left|y\right|}{2b} -\frac{\left|z\right|}{2c}}\right)dx $$ and I need to normalize it. I tried to take an integral of it and I know that it's supposed to be equal to $1$ $$ \int_{-\infty}^{\infty}\left|\Psi\right|^2=1 $$ but now I'm stuck at that integral.

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    $\begingroup$ i think your integral is missing the differential $\endgroup$ – Kyle Kanos Nov 27 '19 at 18:47
  • $\begingroup$ For, say, the $x$ integration, break the range of integration into two: first integrate from $-\infty$ to $0$, then from $0$ to $+\infty$.This allows you to change $|x|$ to $-x$ in the first integral and to $x$ in the second. $\endgroup$ – G. Smith Nov 27 '19 at 18:50
  • $\begingroup$ I know that, but how do I deal with $y$ and $z$? $\endgroup$ – Viivi Aaltonen Nov 27 '19 at 18:51
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    $\begingroup$ Well if $\Psi$ is separable, then it might be pretty easy, right? $\endgroup$ – Kyle Kanos Nov 27 '19 at 18:53
  • $\begingroup$ The two key points are that i) the wavefunction doesn't contains a $dx$ factor and ii) that the normalization integral for a wavefunction depending on three space coordinates is supposed to be taken over a volume. Therefore your volume element should be $dxdydz$. $\endgroup$ – GiorgioP Nov 27 '19 at 19:13

This wavefunction is one over 3D space, given its dependence on $x,\,y$ and $z$. Hence, your normalization should be done over all space: $$ 1=\iiint_D\Psi^*\Psi\,\mathrm{d}V=\iiint_D\Psi^*\Psi\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z $$ where $D\subset\mathbb{R}^3$.

Additionally, if your function is separable, then $\Psi(\mathbf{x})=\psi(x)\phi(y)\xi(z)$ and the above can be written as, $$ \iiint_D\Psi^*\Psi\,\mathrm{d}V=\int\psi^*\psi\,\mathrm{d}x\cdot\int\phi^*\phi\,\mathrm{d}y\cdot\int\xi^*\xi\,\mathrm{d}z $$ And if it is not separable, then you might have a bit of work cut out for you in solving it, depending on the existence of cross terms, for instance.

  • $\begingroup$ @kylekeanos Maybe the purpose of this exercise is that the OP figures it out by himself with a few hints. $\endgroup$ – my2cts Nov 27 '19 at 19:56
  • $\begingroup$ I'll continue with this and ask if I get stuck again. Thank you. $\endgroup$ – Viivi Aaltonen Nov 27 '19 at 19:57
  • $\begingroup$ @my2cts I tried being vague with the response intentionally, so as to not provide too much. Do you think it gives away too much? $\endgroup$ – Kyle Kanos Nov 27 '19 at 20:09
  • $\begingroup$ @KyleKanos Maybe too critical on my side... $\endgroup$ – my2cts Nov 27 '19 at 21:22

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