I'm having to perform an integral of the following form, $$\int\frac{d^3\mathbf{p}}{(2\pi)^3}f(|\mathbf{p}|)\mathbf{\hat{p}}\cdot\mathbf{A}\exp\left(i\mathbf{p}\cdot\mathbf{B}\right)$$ Here, $\mathbf{A}$ is a $\mathbf{p}$ independent vector quantity. Except for the $\mathbf{\hat{p}}\cdot\mathbf{A}$ part this is a common type of integral in Field Theory calculations.

Generally one chooses the angle between $\mathbf{p}$ and $\mathbf{B}$ to be $\theta$. With that one writes $\exp\left(i\mathbf{p}\cdot\mathbf{B}\right)=\exp\left(i|\mathbf{p}||\mathbf{B}|\cos\theta\right)$. And we have a $d\theta$ integration coming from $d^3\mathbf{p}$ because $\hat{p},\hat{\theta}$ and $\hat{\phi}$ are being varied. Now my question is what happens to $\mathbf{\hat{p}}\cdot\mathbf{A}$?

My first guess was to write $\mathbf{\hat{p}}\cdot\mathbf{A}=\sin\theta\cos\phi A_x+\sin\theta\sin\phi A_y+\cos\theta A_z$. But what's bugging me is that the angle between $\mathbf{p}$ and $\mathbf{B}$ is $\theta$ why should that be same for the case of $\mathbf{\hat{p}}\cdot\mathbf{A}$. Moreover, while doing the angular integration the angle $\theta$ is measured with respect to the $z$-axis. Which direction should be fixed as $z$? Along $\mathbf{B}$ or along $\mathbf{A}$? Can someone help me reduce the the the angular part of the integration?

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    $\begingroup$ Without loss of generality, you can pick your coordinate axes in such a way that $\mathbf{B}$ lays on the $z$ axis while $A$ lays in the $xz$ plane. This is $\mathbf{B}=B\, \mathbf{\hat{k}}$ , $\mathbf{A}=A\, \left[ \sin(\theta_A) \mathbf{\hat{i}} + \cos(\theta_A) \mathbf{\hat{k}} \right]$ and $\mathbf{p} = p \, \left[ \sin(\theta) \cos(\phi) \mathbf{\hat{i}} + \sin(\theta) \sin(\phi) \mathbf{\hat{j}} + \cos(\theta) \mathbf{\hat{k}} \right]$. Integration over $\phi$ and $\theta$ gives you something proportional to $Apf(p)J_1(Bp)\cos(\theta_A)$, with $J_n$ the Bessel function of the first kind. $\endgroup$
    – secavara
    Commented Apr 8, 2020 at 8:42
  • $\begingroup$ @secavara Thanks! I see the key insight is that $\mathbf{A}$ can be taken to lie in the $xz$ plane. $\endgroup$ Commented Apr 8, 2020 at 8:48
  • $\begingroup$ No worries. Actually I just noticed that I didn't integrate properly. Taking into account the $p^2\sin(\theta)$ form the $d \mathbf{p}^3$, you get something proportional to $\frac{Af(p)\cos(\theta_A)\left(Bp\cos(Bp)-\sin(Bp)\right)}{B^2}$. $\endgroup$
    – secavara
    Commented Apr 8, 2020 at 8:58

1 Answer 1


The comment by @secavara is indeed correct, but there is a nice and simple trick as well to do this kind of integration in which you don't have to worry of the direction of $\mathbf{A}$, just follow, $$ \begin{align} &\int\frac{d^3\mathbf{p}}{(2\pi)^3}f(|\mathbf{p}|)\mathbf{\hat{p}}\cdot\mathbf{A}\exp(i\mathbf{p}\cdot\mathbf{B})\\ &=-i\mathbf{A}\cdot\boldsymbol{\nabla}_{\mathbf{B}}\int\frac{d^3\mathbf{p}}{(2\pi)^3}f(|\mathbf{p}|)\frac{1}{|\mathbf{p}|}\exp(i\mathbf{p}\cdot\mathbf{B}) \end{align} $$ Now one can consider $\mathbf{B}$ to be along $z$ direction and continue to do the integration as usual. The integral would be a function of $|\mathbf{B}|$ and in the final step you must calculate the gradient of $g(|\mathbf{B}|)$ with respect to $\boldsymbol{\nabla}_\mathbf{B}$ which, I presume, wouldn't be that difficult to calculate.


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