# How to normalize this wave function? [closed]

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.

• i think your integral is missing the differential – Kyle Kanos Nov 27 '19 at 18:47
• 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. – G. Smith Nov 27 '19 at 18:50
• I know that, but how do I deal with $y$ and $z$? – Viivi Aaltonen Nov 27 '19 at 18:51
• Well if $\Psi$ is separable, then it might be pretty easy, right? – Kyle Kanos Nov 27 '19 at 18:53
• 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 $dxdydz$. – 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.

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