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I have this wave function of a system on a central potential: $V(r)$: $$\Phi(x,y,z)=C(x+y+z)e^{-\alpha r^2}.$$

And I'm asked a few things about probabilities. I don't have problems with that, because I understand how it's done. The main part is decomposing as a sum of eigenfunctions of the form. Changing the variables $x,y,z$ by their spherical forms, we have:

$$\Phi = Cr(\sin\theta\cos\phi+\sin\theta\sin\phi+\cos\theta)e^{-\alpha r^2}.$$

The part that depends on $\theta,\phi$ can be expressed as:

$$\sin\theta\cos\phi+\sin\theta\sin\phi+\cos\theta=\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell} a_{\ell m}Y_\ell^m(\theta,\phi).$$

Being $a_{\ell m}$ the scalar product of our function with the eigenfunctions. What I don't understand is that when we did it in class the teacher said that it was necessary to do it only for $\ell=0,1$, and that we didn't need to go any further. I don't remember why this was possible to say before making any calculations. Why is it?

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up vote 1 down vote accepted

In your case, the function on the LHS is very simple so you can more or less read off the coefficients (by inspection). The LHS is a linear combination of

$$ e^{i \phi} \sin \theta, \quad e^{-i \phi} \sin \theta \quad \mathrm{and} \quad \cos \theta.$$

Now the spherical harmonics are of the form $e^{i m \phi} P_{\ell}{}^m(\cos \theta)$, where the $P_\ell {}^m$ are associated Legendre polynomials. For even $m$, they are of degree $\ell$ in $\cos \theta$, otherwise they are of degree $\ell-1$ but with an extra factor $\propto \sqrt{1-\cos^2 \theta} = \sin \theta.$ So in this case, it's clear that you don't need to go beyond $\ell=1.$

To finish the calculation, look up the lowest spherical harmonics in your textbook and figure out the coefficients $a_{\ell m}.$

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Thank you very much. I give you the correct answer for completeness. –  MyUserIsThis Jun 16 '13 at 20:17
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What your teacher said seems correct. However, your statement "this was possible to say before making any calculations" is not obvious and does not follow directly from your teacher's statement. To prove that your teacher was right you can calculate the terms with l=0,1 and check that the sum of these terms equals the left-hand side of your last equation precisely. As the eigenfunctions form an orthogonal basis, the expansion is unique, so that would prove the teacher's statement.

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Thank you for your help. That's what I initially thought, but he made some reasoning before calculating, and doing that in the case when the highest value for $\ell$ is, let's say, $10$, doesn't seem very efficient. –  MyUserIsThis Jun 16 '13 at 20:19
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