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I have an expression that involves the Wigner 3j coefficient:

$$\left(\matrix{a&b&0\\0&0&0}\right)^{-1}$$

This simplifies to:

$$\left[\frac{\left(-1\right)^{a}\delta_{ab}}{\sqrt{2a+1}}\right]^{-1}$$

Which, in turn would be:

$$\frac{\sqrt{2a+1}}{\left(-1\right)^{a}\delta_{ab}}$$

What I'm unsure about is how the Kronecker Delta behaves when it's in the denominator. This would seem to me to indicate that the expression is finite when $a=b$ and infinite otherwise, is that correct?

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1 Answer 1

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Whether it makes sense to regard the value of expression as infinite will depend on the context; in some contexts "undefined" might be a better description. It usually makes sense to regard something as infinite when it approaches infinity, e.g. $1/r$ as $r$ approaches $0$ from above; but in this case there's no approach, and in particular there's no reason to prefer calling this $+\infty$ over calling it $-\infty$. This has nothing specific to do with the Kronecker symbol, which is just a convenient notation that captures the fact that this Wigner $3$-$j$ symbol vanishes unless $a=b$.

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