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I am trying to work through Michael Tinkham's "Group Theory and Quantum Mechanics". In discussing crystal field theory he uses the following example:

We start with an atom at the origin. We want to see how the symmetry of the potential term of the Hamiltonian, and hence the eigenfunctions, are affected by placing ions around the atom.

To start, let's place an ion (which we will represent with a point charge) at (-a,0,0). This point charge will contribute a potential of:

$$ V = \frac{e}{[(x+a)^2 + y^2 + z^2]^{1/2}} $$

He then says that this can be expressed via a Taylor series expansion as:

$$ V = \frac{e}{a}\left(1 - \frac{x}{a} - \frac{r^2}{2a^2} + \frac{3}{2}\frac{x^2}{a^2} + \cdots\right) $$

I am trying to actually work out this Taylor expansion, so that I can see the terms and how they cancel as more ions are added to the crystal field.

I have brute forced my way through mixed partials and stuff for the second order terms of the expansion, but I was hoping for a summation-form expression of the expansion. I found the following thread and have been able to follow the rearrangement used: Electric dipole potential (Taylor expansion)

but I am struggling from there on how to do the actual expansion.

Following the rearrangement above the potential can be written as

$$ V = \frac{e}{r}\left( \frac{1}{\left(1+\frac{2ax}{r^2} + \frac{a^2}{r^2}\right)^{1/2}}\right) = \frac{e}{r}\left( \frac{1}{\left(1+v\right)^{1/2}}\right)$$

where $ r = (x^2 + y^2 + z^2)^{1/2} $.

When using the brute force method, I was able to expand around the origin because evaluating at (0,0,0) gave well behaved results; however, after rearranging, I need to know what value of v corresponds to (0,0,0), right? If you evaluate $ v = \frac{2ax}{r^2} + \frac{a^2}{r^2} $ at(0,0,0) you get undefined results. In multipole expansions I have found on different physics sites, they evaluate the expansion at v = 0. How is this mathematically justified? Am I misunderstanding something about the way these substitutions and rearrangements work?

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Do it like this: $$ \begin{aligned} V &= \frac{e}{\sqrt{a^2+2ax+r^2}} = \frac{e}{a} \left( 1 + \frac{2x}{a} + \frac{r^2}{a^2} \right)^{-1/2} \\ &= \frac{e}{a} \sum_{k=0}^{\infty} \binom{-1/2}{k} \left(\frac{2x}{a} + \frac{r^2}{a^2} \right)^k \\ &= \frac{e}{a} \left( 1 + (-1/2) \left(\frac{2x}{a} + \frac{r^2}{a^2} \right) + \frac{(-1/2)(-3/2)}{2!} \left(\frac{2x}{a} + \frac{r^2}{a^2} \right)^2 + \dots \right) \end{aligned} $$

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  • $\begingroup$ Thanks @hans. That removes the issue of denominators evaluating to zero. As a more broad question, am I right in my understanding of how to evaluate a substitution used in an expansion? By that, I mean do you evaulate $v = \frac{2ax}{r^2} + \frac{a^2}{r^2}$ (or whatever v is equal to in a given problem) at the values of (x,y,z) you want to expand around and use the value of v determined in evaluating the expansion? $\endgroup$
    – scmartin
    Jul 14 '16 at 13:37
  • $\begingroup$ @scmartin: I'm not quite sure I understand what you mean. The way I think about it is that I have an expression of the form $(1+v)^{-1/2}$, where $v$ stands for an expression such that $v \to 0$ when $(x,y,z) \to(0,0,0)$. This means that I can use the standard Maclaurin series for $(1+v)^{-1/2}$, and substitute $v=\dots$ into that, to obtain the expansion for my original expression in terms of $x$, $y$ and $z$ (around the origin). $\endgroup$ Jul 14 '16 at 15:29
  • $\begingroup$ That was what I was trying express, if inelegantly. I wanted to ensure I had a firm grasp on how to use v as a substitute in this case, so I could use the technique in other situations. Am I correct that I cannot expand around the origin with my original rearrangement, where r^2 shows up in denominators? $\endgroup$
    – scmartin
    Jul 14 '16 at 19:41
  • $\begingroup$ Yes, that's correct. $\endgroup$ Jul 14 '16 at 20:07

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