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In the paper Fundamental monopoles and multimonopole solutions for arbitrary simple gauge groups.- E weinberg

I am not being able to see one of the calculation. The author states (eqn 3.26) $$\langle x\mid \frac{1}{[-\nabla^2 + v^2(\alpha \cdot h)^2+M^2]^2}\mid x\rangle=\frac{1}{8 \pi}\frac{1}{[v^2(\alpha.h)^2+M^2]^{\frac{1}{2}}}.$$

I have no idea how he gets that answer. How does the power change to half? In the denominator, everything except $\nabla^2$ are scalars independent of $x$. Here $v, \alpha, h, M$ are all scalars independent of the position.

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Here is another question by OP about the Erick Weinberg paper. –  Qmechanic Dec 15 '12 at 17:05

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A suggestion:

Try inserting a Fourier basis $1 = \int dp |p\rangle\langle p|$. That will turn the differential operator in the denominator into a $p^2$. Then the integral will look like the integral of the product of two propagators, and you can go after it using the standard set of tricks for computing momentum space integrals.

This may or may not work. You haven't told us what $v$, $\alpha$, and $M$ are, and I'm much too lazy to click through to read the paper. But I'll bet it works, since there's a $\pi$ in the denominator on the right hand side.

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sorry, I forgot to put in what $v, \alpha, h, M$ are. $v, \alpha, h, M$ are all scalars independent of the position. Please could you tell me how to 'work out the propagators'? I am very new to QM, and QFT, and need help. Thanks in advance! –  user7757 Dec 15 '12 at 16:04
    
There's a number of little integral substitution and rearrangement tricks. In this case, I was thinking of the en.wikipedia.org/wiki/Schwinger_parametrization . The integral you need is also mentioned in the appendix of Chapter 11 of Volume I of Weinberg. –  user1504 Dec 15 '12 at 16:19

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