- #1

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I was going through section 4.2 of Halzen and Martin, when I came across the following step

[tex]\Box^2 A^{\mu} = j^{\mu}_{(2)}[/tex]

where

[tex]j^{\mu}_{2} = -eN_{B}N_{D}(p_{D}+p_{B})^{\mu}e^{i(p_D-p_B)\cdot x}[/tex]

Now, according to the authors,

Since [itex]\Box^2 e^{iq\cdot x} = -q^2 e^{iq\cdot x}[/tex], the solution of the above equation is

[tex]A^{\mu} = -\frac{1}{q^2}j^{\mu}_{2}[/tex]

where [itex]q = p_D-p_B[/itex].

Question 1:

**Shouldn't this be the solution in Fourier space?**

My doubt stems from the fact that the expression for the transition amplitude [itex]T_{fi}[/itex] is

[tex]T_{fi} = -i\int d^{4}x j_{\mu}^{fi}A^{\mu}[/tex]

with [itex]A^{\mu}[/itex] being the vector potential of the muon, as obtained above, and [itex]j_{\mu}^{fi}[/itex] being the 4-current of the electron, given by

[tex]j_{\mu}^{fi} = -eN_{A}N_{C}(p_A+p_C)_{\mu}e^{i(p_C-p_A)\cdot x}[/tex]

Let [itex]j_{\mu}^{fi}[/itex] be denoted by [itex]j_{\mu}^{(1)}[/itex].

Then, according to me, the expression of the transition amplitude should be

[tex]T_{fi} = -i\int d^{4}x\,j_{\mu}^{(1)}\left(\frac{1}{(2\pi)^{4}}\int d^{4}q e^{-iq\cdot x}\frac{-j^{\mu}_{(2)}(q)}{q^2}\right)[/tex]

whereas the expression given by Halzen and Martin is

[tex]T_{fi} = -i\int d^{4}x\,j_{\mu}^{(1)}\left(\frac{-j^{\mu}_{(2)}(x)}{q^2}\right)[/tex]

Note that the [itex]j^{\mu}_{(2)}[/itex] which appears in their expression is not the Fourier Transformed version. While the final expression might just seem to turn out right, what I do not understand is how they could write the solution to [itex]A^{\mu}[/itex] quite so simply in

*position space*.

Isn't there a misprint somewhere?