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Okay, I know you may say, there are infinite numbers of possible state of an electron between its ground state and ionic state, since energy levels get continuous as electron gets farther away from the nuclear. But that's not what I mean.

Here is my concern: Any linear combination of eigenstates is also a state for an electron in an atom. For example, an electron in 3d orbital (n=3, l=2) is thought to have only five possible states: dxy, dxz, dyz, dx2-y2, dz2, with literally no m eigenvalue. Yet, what about there linear combinations? Aren't they also possible states of an electron in 3d?

Thanks for any help!

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    $\begingroup$ Well, if you have two or more degenerate eigenstates, then any their linear combination is also an eigenstate, and you have a continuum of those. $\endgroup$ – Ivan Neretin Aug 9 '18 at 5:26
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    $\begingroup$ You can have a superposition of any number eigenstates, not just for same n and l. It just won't be an eigenstate anymore (except for a few special cases). What do you mean by "no m eigenvalue".? $\endgroup$ – Feodoran Aug 9 '18 at 8:05
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For degenerate eigenstates any (normalized) linear combination of them will be again an eigenstate of same energy. This in fact means that there are infinitely many of them.

So where does the 5 (in case of $d$-orbitals) come from, when we actually have infinitely many? It means we can select at most 5 of them, which are orthogonal (or more generally speaking: linear independent). We are free to choose which 5, and thus there are (infinitely) many set of 5 $d$-orbitals.

Here are 2 possible sets (again for the example of $d$-orbitals) that are typically used:

  1. The general solutions are $Y_l^m(\theta,\phi)=f_l^m(\theta) g_m(\phi)$ for $l=2$ and $m=\{-2,-1,0,1,2\}$. Where we have \begin{align} f_l^m(\theta) &= (-1)^m \sqrt{\frac{(2l+1)(l-m)!}{4\pi (l+m)!}} P_l^m(\cos \theta) \\ g_m(\phi) &= e^{i m \phi} \end{align} and $P_l^m(\cos \theta)$ are the associated Legendre polynomials. As those are complex functions due to $g_m(\phi) = e^{i m \phi}$ (except for $m=0$), those are hard to plot or imagine.
  2. The real linear combinations from the above: \begin{align} d_{z^2} &= Y_2^0(\theta,\phi) \\ d_{x^2-y^2} &= \frac{1}{\sqrt2}\left[Y_2^{+2}(\theta,\phi) + Y_2^{-2}(\theta,\phi)\right] \\ d_{xy} &= \frac{1}{\sqrt2 i}\left[Y_2^{+2}(\theta,\phi) - Y_2^{-2}(\theta,\phi)\right] \\ d_{xz} &= \frac{1}{\sqrt2}\left[Y_2^{+1}(\theta,\phi) + Y_2^{-1}(\theta,\phi)\right]\\ d_{yz} &= \frac{1}{\sqrt2 i}\left[Y_2^{+1}(\theta,\phi) - Y_2^{-1}(\theta,\phi)\right]\\ \end{align}
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  • $\begingroup$ would the different linear combinations have different shape? $\endgroup$ – santimirandarp Sep 12 '18 at 3:45
  • $\begingroup$ @santimirandarp Yes, they do. $\endgroup$ – Feodoran Sep 12 '18 at 6:10
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The linear combination of eigen states of 3d will give you one of the 5 eigen states you mentioned. So there are only 5 of them possible for 3d.

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